Stupid question general

[Anonymous No.7570071]

I got this exercise from an analysis 2 book i use for self study.
Am i allowed to swap the derivatives like this? I marked that point with a question mark. Thanks
Also stupid question general I

[Anonymous No.7570124]

I don't quite understand what you are doing there.

Let <span class="math">a,b \in \mathbb{R}^2[/spoiler] then there exists and open sets <span class="math">U,V \subset \mathbb{R}[/spoiler] with <span class="math">(a,b) \in U \times V[/spoiler] and functions <span class="math">f: U \to \mathbb{R}[/spoiler] and <span class="math">g: V \to \mathbb{R}[/spoiler] such that for all <span class="math">x \in U[/spoiler] and all <span class="math">y \in V[/spoiler] you have
<span class="math">F(x,f(x)) = F(g(y),y) = 0[/spoiler].

Now differentiate those equations to get:

<span class="math">F_x(x,f(x)) + F_y (x,f(x)) f'(x) = 0[/spoiler]
<span class="math">F_x(g(y),y) g'(y) + F_y(g(y),y) = 0[/spoiler]

Plug in <span class="math">x=a[/spoiler] and <span class="math">y=b[/spoiler] then multiply the first equation with <span class="math">g'(b)[/spoiler] then subtract the equations and divide by <span class="math">F_y(a,b)[/spoiler] and you're left with

<span class="math">g'(b) f'(a) = 1[/spoiler]

[Anonymous No.7570459]

>>7570071
Yes, you are. Sometimes.

[Anonymous No.7570585]

Having trouble proving that <span class="math">A_5[/spoiler] has no subgroup of order 30. The hint says to prove by contradiction, assuming it has some subgroup <span class="math">H[/spoiler] with order 30. Then since <span class="math">A_5[/spoiler] has order 60, obviously <span class="math">A_5-H[/spoiler] is a left and right coset of <span class="math">H[/spoiler] in <span class="math">A_5[/spoiler]. I get everything up to that, but then it says to argue <span class="math">H[/spoiler] has an element of the form <span class="math">(i_1i_2)(j_1j_2)[/spoiler], and I can't figure out how to do that. Can't use the ideas of normal groups, factor groups, Cauchy's theorem, simple groups as those haven't been covered yet, basically I can only use the chapters up to cosets and Lagrange's theorem in Gallian.

[Anonymous No.7570740]

>>7570071
Can someone explain the process of retrieving this simplified form?

[Anonymous No.7572103]

What am I doing wrong in pic related? I need to prove the equality.

[Anonymous No.7572123]

>>7572103

<span class="math">(AB)^T = \left( \sum_j A_{ij} B_{jk} \right)_{i,k}^T = \left( \sum_j A_{kj} B_{ji} \right)_{i,k}[/spoiler]

<span class="math">B^T A^T = \left(B_{ki}\right)_{i,k} \left(A_{ki}\right)_{i,k} = \left( \sum_j B_{kj} A_{ji} \right)_{i,k}[/spoiler]

[Anonymous No.7572131]

>>7572103
You're messing yourself up by skipping steps. Work through it more carefully.

[Anonymous No.7572136]

>>7572103
<span class="math">(AB)^T = \left( \sum_j A_{ij} B_{jk} \right)_{i,k}^T = \left( \sum_j A_{kj} B_{ji} \right)_{i,k}[/spoiler]
<span class="math">B^T A^T = \left(B_{ki}\right)_{i,k} \left(A_{ki}\right)_{i,k} = \left( \sum_j B_{ji} A_{kj} \right)_{i,k}[/spoiler]

You always have to sum over the column index of the first matrix and the row index of the second matrix.

[Anonymous No.7572139]

0=/=0

first mistake

second mistake bad handwriting

third mistake dropping the '(x)' out of f(x)

fourth mistake is you cannot do algerbra with dx and dy unless you really know what you are doing, and you don't need it here (differential forms, advanced calculus)

[Anonymous No.7572180]

>>7572136
I don't follow the second part, are you sure about those indeces?

>>7572131
where did I skip steps?

[Anonymous No.7572217]

does the vibration of an atom consume energy? I don't understand how constant movement is possible without losing energy

[Anonymous No.7572230]

>>7572217
the total energy of the universe is conserved

[Anonymous No.7572248]

>>7572230
Wouldn't it need some kind of energy input to move an atom a tiny bit and then move it another tiny bit in another direction? How does it acclerate, declerate and change its path without using energy?

[Anonymous No.7572273]

>>7572248
don't think of atoms as isolated in a vacuum, in reality they interact with a fuckton of other atoms for every split of a second via countless collisions, therefore exchanging energy

[Anonymous No.7572284]

How do i solve without lhopitals?

[Anonymous No.7572286]

>>7572284
Learn to write from left to right first.

[Anonymous No.7572291]

>>7572284
multiply by the conjugate of the denominator

[Anonymous No.7572293]

>>7570071
Hello mathematicians. In about a year I will be taking Linear algebra (pure math undergrad) so like usual I decided to get ahead so that I can have an easy ride when the time comes and downloaded some linear algebra books.

The one I chose for now is: http://joshua.smcvt.edu/linearalgebra/book.pdf

And I'm now solving the first group of exercises. Exercises that tell "Solve for these set of equations" are pretty easy and straight forward but my monkey brain cannot understand what is the way of doing this:

Show that if ad − bc /= 0 then
ax + by = j
cx + dy = k
has a unique solution.

I wrote on paper my answer but I have to question that I can't answer for myself. Is my answer correct? And if so, why? So if someone who sees this as beyond trivial could show some procedure and explanation for this proof I would be more than happy.

tl;dr - help a fellow pure math master race that is just starting.

[Anonymous No.7572296]

>>7572293
The trick is noticing that

<div class="math">\det \left( \displaystyle{\vcenter{\rlap{\strut
\ulap{\qquad \atop \smash{a }}
\ulap{\qquad \atop \smash{ b }}
}\lower{1em}{\strut
\ulap{\qquad \atop \smash{ c }}
\ulap{\qquad \atop \smash{ d }}
}}} \right) = ad - bc</div>

[Anonymous No.7572307]

>>7572296
Okay, thanks for that.
So a little researching with your information made me find out that if ad-bc= 0 then
"The system of homogenous linear equations represented by the matrix has a non-trivial solution."
And, if this means, by the book's explanation, that it has infinite solutions.

Okay, I actually didn't know at all about the determinants of a matrix. I suppose the book expects you to know that. I also suppose that is the reason why we first do a "fundamentals" class before we even get to linear algebra.

Thanks for that, I actually tried to do row calculations with that to solve it.

[Anonymous No.7572312]

>>7572307
You can explicitly calculate the solution with GauB-elimination if you had that already.

[Anonymous No.7572430]

Can someone explain how to do partial fraction decomposition on this?:

(x^4 + 1) / (x^5 + 2x^3)

[Anonymous No.7573088]

I was walking towards my class at a community college when I went to glance in a classroom then after two seconds I walked away. The female professor that was teaching this class stopped what she was doing, opened the door, then said, "Can I help you?". I quickly said no and she said okay as she closed the door. How come the professor did that?

[Anonymous No.7573121]

>>7572430
first thing you do is factor as much as possible

[Anonymous No.7573225]

>>7572296
>The trick is noticing that
No it isn't, you just use Gauss elimination and in the process you will have to divide by ad-bc.

[Anonymous No.7573241]

>>7573088
because pic related

[Anonymous No.7573261]

NEW QUESTION

I did the conjugate shit and the denominator is still 0, does this limit not exist?

[Anonymous No.7573269]

>>7573261
solved it nvrmnd

[Anonymous No.7573270]

>>7573261
When you multiple by the conjugate sqrt(1+h)+1, you are left with just h in the numerator. Don't multiply out the denominator because the two h's will cancel and you are just left with 1/(sqrt(1+h)+1).

[Anonymous No.7573297]

>>7572430

Suppose we want

<div class="math"> \frac{A + Bx + Cx^2}{x^3} + \frac{Dx + C}{x^2 + 2} = \frac{x^4 + 1}{x^5 + 2x^3} <div class="math">

Now cross-multiply and equate coefficients of like-degree terms. I get these equations

<span class="math"> 2A=1,\ 2B=0,\ A+2C=0,\ C+D=1,\ B+E=0 [/spoiler]

Solving these gives coefficients

<span class="math"> A=1/2,\ B=0,\ C=-1/4,\ D=5/4,\ E=0 [/spoiler]

Then we have

\frac{x^4 + 1}{x^5 + 2x^3} = \frac{1/2 - x^2/4}{x^3} + \frac{5x/4}{x^2 + 2} = \frac{1}{2x^3} - \frac{1}{4x} + \frac{5}{4} \frac{x}{x^2 + 2}

where the RHS is ready to integrate.</div></div>

[Anonymous No.7573303]

>>7572430

Let's take

<div class="math"> \frac{A + Bx + Cx^2}{x^3} + \frac{Dx + C}{x^2 + 2} = \frac{x^4 + 1}{x^5 + 2x^3} </div>

Now cross-multiply and equate coefficients of like-degree terms. I get these equations

<span class="math"> 2A=1,\ 2B=0,\ A+2C=0,\ C+D=1,\ B+E=0 [/spoiler]

Solving these gives the coefficients

<span class="math"> A=1/2,\ B=0,\ C=-1/4,\ D=5/4,\ E=0 [/spoiler]

Then we have

<div class="math"> \frac{x^4 + 1}{x^5 + 2x^3} = \frac{1/2 - x^2/4}{x^3} + \frac{5x/4}{x^2 + 2} = \frac{1}{2x^3} - \frac{1}{4x} + \frac{5}{4} \frac{x}{x^2 + 2} </div>

where the RHS is now ready to integrate.

[Anonymous No.7573306]

>>7573303
Sorry, fucked up in the first equation, the Dx+C should be Dx+E obviously.

[Anonymous No.7573340]

>>7570740
Already did for you in the old thread.

>>7570825

[Anonymous No.7573357]

>>7572284

Use both conjugates:

<span class="math"> \frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \sqrt{5x-1}} \cdot 1 \cdot 1 = \bigg( \frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \sqrt{5x-1}}\bigg) \cdot \bigg( \frac{\sqrt{x+2} + \sqrt{3x-2}}{\sqrt{x+2} + \sqrt{3x-2}} \bigg) \cdot \bigg( \frac{\sqrt{4x+1} + \sqrt{5x-1}}{\sqrt{4x+1} + \sqrt{5x-1}} \bigg) [/spoiler]

<span class="math"> = \frac{(\sqrt{x+2} - \sqrt{3x-2})(\sqrt{x+2} + \sqrt{3x-2})}{(\sqrt{4x+1} - \sqrt{5x-1})(\sqrt{4x+1} + \sqrt{5x-1})} \cdot \bigg( \frac{\sqrt{4x+1} + \sqrt{5x-1}}{\sqrt{x+2} + \sqrt{3x-2}} \bigg) [/spoiler]

<span class="math"> = \frac{(x+2) - (3x-2)}{(4x+1) - (5x-1)} \cdot \bigg( \frac{\sqrt{4x+1} + \sqrt{5x-1}}{\sqrt{x+2} + \sqrt{3x-2}} \bigg) [/spoiler]

<span class="math"> = \frac{4-2x}{2-x} \cdot \bigg( \frac{\sqrt{4x+1} + \sqrt{5x-1}}{\sqrt{x+2} + \sqrt{3x-2}} \bigg) [/spoiler]

<span class="math"> = 2 \cdot \bigg( \frac{\sqrt{4x+1} + \sqrt{5x-1}}{\sqrt{x+2} + \sqrt{3x-2}} \bigg) [/spoiler]

Now the limit is easy:

<span class="math"> \lim_{x \to 2}\ \frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \sqrt{5x-1}} = \lim_{x \to 2}\ 2 \cdot \bigg( \frac{\sqrt{4x+1} + \sqrt{5x-1}}{\sqrt{x+2} + \sqrt{3x-2}} \bigg) = 2 \cdot \Big( \frac{3 + 3}{2 + 2} \Big) = 3. [/spoiler]

[Anonymous No.7573368]

How would you solve this?
I could solve all, but this actually confused me.

I need more precalc exercise...

[Anonymous No.7573398]

>>7573368
Here you are laddie

[Anonymous No.7573400]

>>7573398
Just to add to this full answer of mine, you don't actually have to do the math to figure this out. You're only able to know a point on g(x) which at least an x or y value with f(x). The only answer out of that which fits this is (12,1) which shares the y-value of one. Therefore it is the only possible answer anyway.
Just a quick tip to save calculation if there's something like this on an actual test/exam.

[Anonymous No.7573405]

>>7572217
Consider yourself in good company, anon. This bothered a lot of people before the development of quantum mechanics. They thought that an electron orbiting an atom ought to radiate electromagnetic waves -- as charges undergoing acceleration generally do.

The idea was that, as an electron orbits the nucleus it is under constant acceleration, directed inward, due to the coulomb force. Thus energy should be continually radiated away, in the amout determined by the Larmor formula. As a result of the loss of energy the electron would eventually spiral all the way down into the nucleus. Many calculations of the classical "lifetime" of the hydrogen atom were performed at the time. Classical electromagnetism generally predicts that all atoms should collapse in upon themselves within about 2 x 10^-11 seconds or so. Obviously this would be very bad, and it does not happen.

It was part of Bohr's great triumph to postulate that energy wouldn't be radiated unless the atom underwent a "quantum" transition. His full model predicted and explained the actual Rydberg radiation spectrum from Hydrogen-like ions, and He and Sommerfeld developed a fairly elaborate (but incomplete and inconsistent) theory that is now called "Old Quantum Theory" about what precise atomic transitions/radiations were allowed. (You may want to check it out.) But it didn't really explain WHY atoms didn't radiate.

Anyway, nowadays we know the Schrödinger equation in quantum mechanics predicts that all atoms must have a ground state from which they can't lose energy via radiation. But you're idea is basically right: atoms in excited states above the ground state will generally radiate their energy away -- and in a time scale comparable to the classical lifetime of the hydrogen atom.

[Anonymous No.7574313]

do iterated integrals always return volume beneath a surface over a region? what is the difference between double and iterated integrals?

[Anonymous No.7574318]

>>7574313
>iterated integrals

Sorry /sci/ can't help with such advanced topics. Better ask on stackexchange.

[Anonymous No.7574347]

these are all carbanions, right? what factors do i look at to determine stability? just the degree of substitution?

[Anonymous No.7575010]

Please excuse the trivial nature of this question
To 'remove' the log from the RHS, am I correct in doing this?
Trying to isolate L.

[Anonymous No.7575028]

Here's one for you guys:

If I have 20 presents and 12 kids, how many different ways can I distribute the presents? Not every kid has to have a present and, for example, one kid could have 20 presents.

Would this be the correct way of solving this:

20!
―
(20-12)!

I feel like I'm looking for a permutation, although I sometimes get confused with them. Is this the correct method to use for this question? If not, could someone steer me in the right direction?

[Anonymous No.7575031]

>>7575028

On further thinking, I think I might be looking for a combination rather than a permutation.

Is this right?

20!
―
12!(20-12)!

[Anonymous No.7575047]

>>7575028
>>7575031
I already answered this question for you last week in the old thread. Answer hasn't changed. It's neither a combination nor permutation question.

>>7561685

[Anonymous No.7575049]

>>7575010
Yes that's correct anon.

[Anonymous No.7575050]

>>7575047
Oh shit I never got around to seeing it and didn't expect the thread to still be there. Thanks!

[Anonymous No.7575127]

Been trying for the last hour. How to do trig sub? I'm on number 15

[Anonymous No.7575133]

>>7575127
I'm guessing they want you to use 1=csc^2 - cot^2

But if i were you i'd use a hyperbolic relationship instead, like:

1 = cosh^2 - sinh^2

As its faster to take the derivatives and cancel shit out etc.

[Anonymous No.7575146]

>>7575133
>>7575127
This is wat I think is right so far idk how to continue tho tbh I don't know wat a hyperbolic relationship is :/

[Anonymous No.7575209]

>>7575127
>>7575146
You're almost there anon. So far you used the trig substitution x=3sin(θ) to get

<span class="math"> \int{\frac{x^2}{\sqrt{9-x^2}}} dx = \frac{9}{2}\Big(2 \theta - sin(2 \theta) \Big) + C [/spoiler]

Now you just have to put the answer back in terms of x. Use these formulas

<span class="math"> sin(2\theta) = 2 sin(\theta)cos(\theta)\ \ \ \ \ sin(\theta) = \frac{x}{3}\ \ \ \ \theta = sin^{-1}(\frac{x}{3})\ \ \ \ cos(\theta) = \sqrt{1-(\frac{x}{3})^2} [/spoiler]

In the end I got

<span class="math"> \int{\frac{x^2}{\sqrt{9-x^2}}} dx = \frac{9}{2}sin^{-1}(\frac{x}{3}) - \frac{1}{2}\sqrt{9-x^2} + C [/spoiler]

BTW hyperbolic trig functions would be useful if you had

<span class="math"> \int{\frac{x^2}{\sqrt{9+x^2}}} dx [/spoiler]

but you don't.

[Anonymous No.7575213]

>>7575209
sorry should be 9/4 in first eqn

[Anonymous No.7575220]

>>7575127
>>7575146
You're almost there anon. So far you used the trig substitution x=3sin(θ) to get

<span class="math"> \int{\frac{x^2}{\sqrt{9-x^2}}} dx = \frac{9}{4}\Big(2 \theta - sin(2 \theta) \Big) + C [/spoiler]

Now you just have to put the answer back in terms of x. Use these formulas

<span class="math"> sin(2\theta) = 2 sin(\theta)cos(\theta)\ \ \ \ \ sin(\theta) = \frac{x}{3}\ \ \ \ \theta = sin^{-1}(\frac{x}{3})\ \ \ \ cos(\theta) = \sqrt{1-(\frac{x}{3})^2} [/spoiler]

In the end I got

<span class="math"> \int{\frac{x^2}{\sqrt{9-x^2}}} dx = \frac{9}{2}sin^{-1}(\frac{x}{3}) - \frac{x}{2}\sqrt{9-x^2} + C [/spoiler]

BTW hyperbolic trig functions would be useful if you had

<span class="math"> \int{\frac{x^2}{\sqrt{9+x^2}}} dx [/spoiler]

but you don't + you have -.

[Anonymous No.7575650]

>>7575049
Thank you

[Anonymous No.7576078]

How to prove a bijection? I mean it's so trivial to write f(x)=y for all x I'm unsure if there's any more proof necessary

[Anonymous No.7576102]

What am I doing wrong? I figured the net torque and net force should both be equal to zero. Rearranged the net torque equation for the force exerted by support one and then substituted it into my net force equation. Sorry for poor form in my work.

[Anonymous No.7576385]

Am I suppose to wait until i'm accepted to a university to apply for financial aid, or am I suppose to apply now?

[Anonymous No.7576409]

>>7576385
You can start applying for scholarships before you're accepted.

[Anonymous No.7576635]

>>7576102
Your first equation is correct

<span class="math"> \sum F_{y} = F_1 + F_2 - m_p g - M_b g = 0 [/spoiler]

But your second equation has two small mistakes and one poor choice. You're finding the torques about the 2ND support, but why do that when a distance d=2.35m from support #1 is already given to you? It makes extra work to calculate distance from support #2. Although it's not a big deal in this problem, it could be in the next one. The two mistakes are that the distance from support #2 to itself is 0, not <span class="math">l[/spoiler], and you made a sign error there (which ends up not mattering because of the 0). Anyway, a correct 2nd equation would be

<span class="math"> \sum \tau_{y} = -F_2 0 + m_p g (l - d) + M_b g (\frac{l}{2}) - F_1 l = 0 [/spoiler]

Solving the corrected equations, you'll now get

<span class="math"> F_1 = m_p g (1 - d/l) + \frac{1}{2}M_b g [/spoiler]
<span class="math"> F_2 = m_p g (d/l) + \frac{1}{2}M_b g [/spoiler]

Some unsolicited advice from an anon who has solved a lot of physics problems: (1) always choose your coordinates very carefully because in some problems it helps big time, (2) keep everything in variables and wait until the very end to start putting in the numbers. It takes less work, is cleaner, and helps you to figure out algebra mistakes.

[Anonymous No.7576640]

Can the single split experiment on the uncertainty principle be explained using classical electrodynamics? Perhaps the light gets reflected by the edges of the slit?

[Anonymous No.7576704]

>>7576640
For coherent light of high enough intensity you can get the right answers for both the single and double slit experiments using classical electromagnetic waves. It is fundamentally a wave interference phenomenon though, not a reflection phenomenon.

However this doesn't explain the experimental results for very low intensity light, a regime where individual photonic behavior is important. Such experiments can be done. Similar experiments can also be done with electrons and even large molecules. The results are essentially the same and classical physics doesn't explain it. It requires quantum mechanics.

If you REALLY want to understand this stuff, I highly recommend watching this series from the master:

https://www.youtube.com/watch?v=eLQ2atfqk2c&list=PL8590A6E18255B3F4

Yes it's long (4 hours), but it will explain so much that you will figuratively thank me later. It's also quite entertaining.

[Anonymous No.7576726]

>>7576704
what difference does changing the light's intensity for the experiments make

[Anonymous No.7576764]

>>7576726
At low intensity individual photons are passing through the slit, one-by-one, and their particle (as opposed to wave) behavior becomes evident. Individual flashes of light will appear on a screen instead of a nice diffraction pattern. Classical electromagnetism doesn't explain this.

The question of where the photon will hit the screen becomes important. And classical physics also doesn't explain why the final destination should vary probabilistically for identically prepared photons passing through the slit.

It turns out that the different possible paths the photon can take (ie places along the slit where the photon may pass through) give rise to interference terms that predict a diffraction-like probability distribution for the photons' destination.

Strangely, quantum electrodynamics says the photon actually takes *all* possible paths through the slit, each path with a slightly different phase, and when this "sum over histories" is calculated (using Feynman's path integral technique) it gives the actual probability distribution that is observed for the photon. Same story for electrons and other particles.

[Anonymous No.7576798]

so i have this problem here

find the number of solutions of
cos(2x)-2sin(x) = k, -pi<=x<=pi

so for this, i have to use intermediate value theorem

my question is concerning the method to use to solve this

what i do is this: first i show this function is continuous on the interval
here i don't know if i have to use formal definitions and proofs, or just write "f(x) is continuous because x, sin(x), cos(x) are continuous"

then i find maximum and minimum values, so i derivate f(x)

once i get those, i have to use the ivt corollary to find the number of solution, and it says to cut the interval into parts

how do i proceed to do that? do i cut parts into this
[a,b],[b,c],[c,d] and see what solutions are in it? are the maxima and minima the solutions? i'm quite lost here

[Anonymous No.7576861]

>>7576764
>And classical physics also doesn't explain why the final destination should vary probabilistically for identically prepared photons passing through the slit.
Perhaps some quality of the slit changes in a probabilistic fashion.

[Anonymous No.7576987]

>>7576078
Either simply exhibit the inverse if you can find it or prove that it is both injective and surjective (if you are working with a "concrete" function, showing that it is surjective will usually give you the inverse but for more difficult problems it might be necessary to do both separately)

[Anonymous No.7576989]

>>7576635
Thank you very much.

[Anonymous No.7577002]

>>7576635
Although, I don't really understand why the torque on the second support should be equal to zero. Would it not pivot about its own fulcrum, rather than about support 1?

[Anonymous No.7577005]

>>7576861
Perhaps, but what quality would that be exactly? Particles that interact with the barrier generally don't make it to the screen. The very few that are reflected don't have the diffraction pattern, but only add background noise/experimental error. And generally only the shape of the slit matters to the diffraction pattern, not the material composition of the barrier, vibrations, temperature, etc (within reason, of course).

Now, if you could propose a way to explain the diffraction/interference pattern for PARTICLES without without invoking the principles of quantum mechanics I would really love to hear about it. Many people dislike QM philosophically, and others are always looking for alternative explanations. The problem is that QM has worked so well for so long in such a wide variety of situations that it would be very hard to explain it away with an ensemble of ad hoc loopholes. The burden is on the skeptics at this point, but I do believe in science we should always keep an open mind about alternatives. After all, Newtonian gravity survived for 300 years until Einstein supplanted it with general relativity. Who's to say you won't do the same to quantum mechanics? I do look forward to reading your future papers in the Physical Review.

[Anonymous No.7577024]

>>7577002
It pivots about the fulcrum, yes. But the idealization here is that the force from support #2 acts at one point, and that point is the *same* place about which you're calculating the torques. The definition of torque is distance times the perpendicular component of force: (<span class="math">\vec{\tau} = \vec{r} \times \vec{F}[/spoiler]) and that distance is *zero*.

So forces that act exactly at the location around which you are calculating the torques do not contribute to the sum. Those forces would contribute if you chose to compute the torques about some other point though. For example, if you chose to compute torques about the support #1, like I would prefer, then F2 would now contribute to the sum, but F1 no longer would do so.

[Anonymous No.7577034]

>>7577024
Right, that makes sense. Thank you again.

[Anonymous No.7577119]

>>7576764
how do they know when only each photon passes the slits one at a time?

[Anonymous No.7577139]

>>7577119
You can count the flashes.

These guys were the first to make a true, isolated single-photon source in 1974:
http://journals.aps.org/prd/abstract/10.1103/PhysRevD.9.853

Watch this video. These guys were able to do it with electrons:
https://www.youtube.com/watch?v=ToRdROokUhs
http://iopscience.iop.org/1367-2630/15/3/033018/article

[Anonymous No.7577198]

>>7577139
In which field of engineering covers stuff like these?

[Anonymous No.7577205]

HOW TO PROVE THIS BY INDUCTION ?!

n! > ((n+1)/2)^n for n>1

[Anonymous No.7577206]

>>7570071
>Am i allowed to swap the derivatives like this?
No.

[Anonymous No.7577210]

Given two velocoties dx, dy which are two different directions. How can I calculate acceleration?

[Anonymous No.7577219]

>>7577139
>of all the photons passing through the crystal, only one gets separated into a pair of lower energy photons
how?

[Anonymous No.7577240]

I have a function of two variables and I want to see how its second derivative looks like with respect to a single variable r=sqrt(x^2+y^2). How to do? I get the wrong answer everytime

[H No.7577246]

>>7577205
That isn't true. Have you got the greater than the wrong way round?

[H No.7577250]

>>7577210
>Given two velocoties dx, dy
What? dx and dy are differentials (or differential forms) - velocity is just a function. Do you mean dx/dt and dy/dt?

[Anonymous No.7577257]

>>7577246
Oh yeah, sorry. It's the other way around..

[Anonymous No.7577260]

>>7577246
So,

n! < ((n+1)/2)^n for n > 1;

Prove by induction.

[H No.7577265]

>>7577198
Optics is part of physics, not engineering.

[H No.7577269]

>>7577219
Where is that said? I don't have access to the article.
I assume it's the opposite of second harmonic generation.

[Anonymous No.7577273]

>>7577269
it wasn't from one of your links
https://youtu.be/1MaOqvnkBxk?t=2m39s

[Anonymous No.7577280]

>>7570071
I have a question on what the fuck was my prof talking about;

Z isn't a subset of Q because Q has equivalence classes instead of numbers.

I thought this could be bypassed by finding an isomorphic set to Q with numbers instead of equivalence classes. But do we call that Q? How do we differentiate between the two?

[Anonymous No.7577309]

Could someone help with pic related? An answer and an explanation would be lovely. I know its simple but im stumped

thanks

[H No.7577320]

>>7577260
So, start by proving it for n = 2.
Then you assume that it is true for some general k>1, and show that this implies that it is also true for k+1. So just write down the equality for general k:
> k! < ((k+1))/2)^k
That you assume to be true, and from that you need to derive the result for k+1, which is:
> (k+1)! < ((k+2)/2)^k+1

[Anonymous No.7577358]

>>7577205
>>7577260
First we need a little lemma. From the first two terms of the Binomial Theorem (also usually proved by induction) we have for any n>0
<div class="math"> (n+2)^{n+1} = \Big((n+1) + 1\Big)^{n+1} = (n+1)^{n+1} + (n+1)\cdot(n+1)^n +\ ...\mathbb{positive\ terms}...\ > 2 (n+1)^{n+1} </div>
Therefore
<div class="math"> 2\Big(\frac{n+1}{n+2}\Big)^{n+1} < 1. </div>
Now the main proof...

In the base case (n=2) we just have <span class="math"> 2! = 2 < 2.25 = (3/2)^2 [/spoiler]

For the induction, we show that if the proposition is true for n=m it must be true for n=m+1 as well:
<div class="math"> (m+1)! = (m+1) \cdot m! < (m+1) \cdot \Big(\frac{m+1}{2}\Big)^m = 2 \Big(\frac{m+1}{m+2}\Big)^{m+1} \Big(\frac{m+2}{2}\Big)^{m+1} < \Big(\frac{m+2}{2}\Big)^{m+1}. </div>
We used the lemma in the last step.

[Anonymous No.7577394]

>>7577309
Suppose it were cyclic. Let A be it's generator, so that for any x∈R we have x=nA for some n∈Z. Clearly A cannot be zero, for then A*n = 0 for all n. So A ≠ 0. How do you generate A/2? A/2 ∈ R, so A/2=nA for some integer n. But this implies n=1/2. Contradiction.

[Anonymous No.7577406]

>>7577309
If it were cyclic, then it would be at most countable

[Anonymous No.7577423]

Well this has blown my mind, I've been challenged to find a function with a maximum point at x=2 and a minimum at x=-6.

Naturally I did the following:
f'(2)=0
f'(-6)=0
So, taking an obvious choice, (x-2)(x+6) would do it.

Expand, x^2 + 4x - 12 =0
Integrate, f(x)=x^3 /3 +2x^2 -12x + c [I have no idea how to find this c]
However simple evaluation, or checking f'' of any of the values I've mentioned reveals me to be laughably wrong.

Is there a general approach to problems like this? It's certainly got my interest, although I can see some easy ways to cheat (e.g. let f(-6)=-200 and f(2)=5 and let f(x)=0 everywhere else).

[Anonymous No.7577453]

>>7577423
Just doodle what you want then try to see if you can do it with a piecewise affine function (if you don't require that the extrema be strict then you can just take any constant function, otherwise a well chosen piecewise affine function should do the trick)

[Anonymous No.7577471]

>>7577423
f'(2) = 0
f''(2) = -1
f'(-6) = 0
f''(-6) = 1

[Anonymous No.7577493]

>>7577423
Your idea is basically right, anon. But take df/dx= -3(x-2)(x+6) to get the max/min properly swapped and make the integration a little nicer. You can drop any constant of integration, because it only shifts the whole graph up or down and has no effect on the max/min.

Plot y = f(x) = 36x - 6x^2 - x^3 and you will see it has the desired features with zero derivatives at x=2, -6.

http://www.wolframalpha.com/input/?i=Plot+y%3D36x-6x^2-x^3++from+x%3D-11+to+x%3D6

[Anonymous No.7577569]

>>7577358
I still don't understand the process at all. Can you try again and formulate each step you take? This question in one of the first exercises for mathematical induction in my analysis book and so far proving inequalities by induction does not make any sense to me at all.

Can you offer some guidance?

[Anonymous No.7577576]

I'm at the midpoint in the semester, feeling kind of depressed. Grades are good, classes aren't really a problem...having trouble getting through the days. Lost pretty much all my motivation to do anything that isn't class related. Any advice?

[Anonymous No.7577586]

>>7577576
Welcome to math dpt. Enjoy your stay, cope however you can. I smoke a lot and drink scotch nightly.

[Anonymous No.7577660]

Are all protons in an atomic nuclrus located on its surface (do atomic nuclei follow the principle of charges being located on the surface of an object)?

Does a particular nucleon actually continue to exist as a discrete entity in a nucleus, or is it more accurate to describe a nucleus as a soup o' quarks?

Is it realistically possible to heat something enough that relativistic effects measurably increase its mass? If not, what temperature would be required for such a mass increase to be detectable, assuming unlimited strength of materials etc., but still using real-world detection sensitivity?

[Anonymous No.7577683]

>>7577394

Thank you based sci anon.

[Anonymous No.7577924]

Doing some preparation tasks for a control lab and this is the last question. Have spent a couple of hours trying to solve for the roots but can't for the life of me get it right. Anyone here that knows how i should solve this problem ? (It's the last one i need done). Would really appreciate some help.

[Anonymous No.7577930]

>>7572248
think of a ball in a frictionless divot in the ground rolling back and forth

[Anonymous No.7578053]

>>7577423
Do you mean global or local maximum ?

[Anonymous No.7578054]

>>7577569

OK anon, lets take something much simpler involving inequalities. I will go suuuuupppeeerrr slow, so don't take it as an insult. This will be long. I want to see if I could explain this to my 8yo neice. I would appreciate some constructive criticism if you find any points unclear, etc.


Sometimes you have a proposition P(n) that depends on a parameter n, and you want to prove it is true for all n=0,1,2,3,... The proposition could be anything that makes a definite statement, it doesn't even have to be an equality or an inequality. I will provide such an example later.

But for starters, let's say P(n) is the statement that "n < 2^n". How do we prove P(n) for all n=0,1,2,3,... ?

Is P(0) true? Well what is the statement of P(0)? It is that "0 < 2^0". Well 2^0=1 and 1>0, so yes, it's definitely true.
Is P(1) true? P(1) says that "1 < 2^1". 2^1=2 and 2>1, so yes, it's true too.
Is P(2) true? P(2) says that "2 < 2^2". 2^2=4 and 4>2, so yes, it's true too.
Is P(3) true? P(3) says that "3 < 2^3". 2^3=8 and 8>3, So yes, it's true too.
Is P(4) true? P(4) says that "4 < 2^4". 2^4=16 and 16>4, So yes, it's true too.

Ok you can see this is gonna get very tedious, it involves a lot of checking. And we can never really finish this process -- it will go on forever -- so we need a better way anon.

The Induction Principle solves this problem. One step toward the idea of induction is this:

Suppose that some genius came to you and said he could do some clever tricks to show that, for some definite natural number, say n=17362, that P(17362) implies P(17363). That would be cool. Then, if we could somehow establish P(17362) is true, we would be able to use the genius's tricks to conclude that P(17363) is also true, for free. We wouldn't have to check P(17363) directly ourselves.

This would be useful, but of limited usefulness. After all it can only potentially get us one step further.

[Anonymous No.7578058]

>>7578054
>>7577569

But now suppose the genius came back and said, you know what anon, i discovered that my clever trick is more general that I initially thought. I can extend my argument to show very generally that if P(k) is true for some arbitrary natural number k, then P(k+1) would also have to be true. That is, the truth of P(k) implies the truth of P(k+1).

OK now we are in a different universe, anon. Here's why. If we just go back to our last check -- that P(4) is true -- and use it in conjuction with the generalized trick of the genius dude -- but specialized to the case that k=4 -- we can conclude that P(5) is also true.

Huh? Succinctly: we know P(4) is true, and (thanks to the genius) we know P(4) implies P(5), so we can conclude P(5) must in fact be true too.

We can continue this line of reasoning by specializing the genius's trick to the case k=5. Thanks to the genius dude's triumph, P(5) automatically imples P(6). By knowing P(5) is true from three sentences ago, we can conclude P(6) is true as well.

OK, so the idea of induction is that this chain of simple logic can effectively be continued forever. There is a rigorous proof of this, based on set theory, but I won't go into it now. Just accept for now that we live in a universe where the simple logical chain continues forever.

So the procedure for doing induction proofs involves two steps. One is to get a genius to prove that generally P(k) implies P(k+1). This has to work for the entire range of k values that we're interested in, so it has got to be very *general*. The second step is something simple we can do ourselves. Just check some usually trivial "base case" like P(0) is true to get the induction chain going.

Without the base case the induction can't get started. Without all the implications the induction can't continue. So we must have both.

[Anonymous No.7578061]

How do I prove that the determinant of the transpose of a matrix is the same as the determinant of the original matrix WITHOUT assuming you can do Laplace with both rows and columns?

[Anonymous No.7578064]

>>7578058
>>7577569

The bad news is that, unfortunately, there is no genius to help us out. We have to be the genius. So let's go back to our problem and see what we can say about it.

P(k) says that "k < 2^k". Oh, OK that is what we are trying to prove, for ALL k values. So how do we prove what we're already trying to prove? Well, we're not gonna focus on ALL k values right now. We're only going to focus on some PARTICULAR value of k. Maybe k is 17362, maybe something else. It is an arbitrary, unspecified value of k, but we'll think about it as a fixed specific value. We are not even going to prove P(k). We don't even CARE about the truth of P(k) right now. All we care about is this: does P(k) imply P(k+1)? That is, if we just outright blatantly ASSUMED P(k) was true, could we use that to prove that P(k+1) was also true? Let's try.

So ASSUME k < 2^k. YAY! Are we done? Have we concluded that P(n) for all n? Can we go home now? No no, remember this is all just playing a game of impersonating the genius for a moment. What ARE we trying to prove? We need to show that P(k+1) must also be true, granting our assumption of P(k). Well, what does P(k+1) say? It says, "(k+1) < 2^(k+1)." We have no idea if it's true, but it doesn't look so hard if we can tentatively use that k < 2^k.

[Anonymous No.7578068]

>>7578064
>>7577569

Let's multiply k < 2^k through by 2. We get 2k < 2^(k+1). That's pretty close to the form of P(k+1), but we're not quite there yet on the LHS. What else do we know? Well we haven't said what k is, on purpose, but we do know that whatever it may be, it is certainly an natural number (0,1,2,3...). So we can definitely say that 0 <= k. Let's add k to both sides of this. We get k <= 2k. But wait! we also have that 2k < 2^(k+1). If we combine these, we get that k <= 2k < 2^(k+1). YAY! ..... Oh but wait .... Damn! Looks like we got k < 2^(k+1), but that is not what P(k+1) says. It says "(k+1) < 2^(k+1)". Awww F_@#%!! we can't just go add 1 to only the LHS of k < 2^(k+1) to get P(k), because that might violate the inequality. Are we doomed?

Look a little closer at what went wrong. The problem seems to be with low values of k. IF we had started with 1 <= k, then we could add k to both sides to get k+1 <= 2k. Combining THIS with 2k < 2^(k+1) would give k+1 <= 2k < 2^(k+1). Then we could conclude P(k+1).

OK that's awesome, but it's not legit, because 1 <= k is false. The case k=0 is possible. Remember, we have to be able to regard k as COMPLETELY ARBITRARY for the chain of implications to work. Our implication chain couldn't get going if we didn't establish that P(0) implies P(1).

But wait... couldn't we... just... establish this last implication separately? I mean, if we have proved that P(k) implies P(k+1) with the additional requirement that 1 <= k, (ie, k=1,2,3,...), we're only missing one of the possible values for k -- namely k=0.

If P(0) is true, then 0 < 2^0 = 1. If we add 1 to both sides of that, we get 1 < 2 = 2^1. So P(1) is in fact implied by P(0). All is not lost! There really is no problem! We used different arguments for different k's, but we do now know that P(k) implies P(k+1) for ALL k=0,1,2,3,....

[Anonymous No.7578078]

>>7578068
>>7577569

OK, but do you see the potential pitfall with the induction implication step? If our argument fails for even ONE of the implications, the whole thing can blow up. Also we can legitimately break up the implication that we are trying to prove into different cases for different k's. We just must make sure that we cover ALL the cases in the end, because we have to be able to regard k as completely arbitrary in order to link together the infinite logical chain. Sorry for being so repetitive about this crucial point.

So to summarize, we've proved P(k) implies P(k+1) for any natural number k. GENIUS! Earlier, we showed that P(0) was true. So we can declare by the power of the Induction Principle that P(n) is true for all n=0,1,2,3....


---

Now here is an exercise. I will give another induction proof, and you tell me if it is acceptable or, if something is wrong, what.

Theorem: All of the people in a room must have the same birthday.

Proof.
We will establish the theorem by induction. Let P(n) be (something like) the statement that "If exactly n people are in a room they all have the same same birthday."

Base case. If n=1 then P(1) is obviously true. If only one person is in a room, then everyone in the room has the same birthday.

Induction step. Assume P(n) is true for some n=k, where k=1,2,3,... is some arbitrary number. We want to prove that P(n) is also true for n=k+1. That is, we must prove P(k) implies P(k+1).

[Anonymous No.7578081]

>>7578078
>>7577569

Our induction assumption is that in any room with any k people, they must all have the same birthday. Suppose there is an (arbitary) room with exactly k+1 (arbitrary) people in it. Pick a person, call that person A. Remove A temporarily from the room. The room now has k people in it, so by the induction hypothesis they all must have the same birthday. Bring A back into the room. Pick a different person B =/= A. Bring B out of the room. The room again has k people in it, so by the induction hypothesis they all must have the same birthday. Bring B back into the room. Both A and B share the same birthday with everyone besides them in the room. Therefore they have the same birthday as each other as well. So we can conclude that all k+1 people must share the same birthday. This is P(k+1). So P(k) implies P(k+1).

By the Induction Principle, the fact that P(1) is true, and P(k) implies P(k+1), we finally conclude that P(n) is true for all n. Therefore everyone in a room must have the same birthaday.

Q.E.D.

[Anonymous No.7578099]

Hello scientific people,

I need some geometry help here.

I have been trying my best to solve one problem involving intersection curves, but my solution doesn't seem to quite work. The solution I've found doesn't verify the sphere's equation.

I am asked to find the equation of the intersection curve between a sphere and an parabolic cylinder.

I have used a graphing program to plot out my solution, and it seem to work as the curve appears to be the intersection curve graphically.

If my solution works, shouldn't it verify the sphere's equation?

Normally, I'd ask a classmate for advice, but literally nobody has been able to solve it. Any idea?

I determined the resulting shape should be an ellipse, calculated its dimensions and then used the y=x^2 relationship from the parabolic cylinder to get the solution.

The sphere has been parametrized to :
x=2*sin(t)cos(u)
y=2*sin(t)sin(u)
z=2*cos(t)

and the parabolic cylinder to:
x=t
y=t^2
z=u

My solution being
x=1.249*cos(t)
y=1.5625*cos(t)^2
z=2*sin(t)

[Anonymous No.7578260]

>>7578081
>>7577569

Hopefully you did your birthday problem homework anon, so now lets explain your problem.

What is the P(n) that you're interested in?
Your P(n) is the statement that <span class="math"> ``\ n! < \Big(\frac{n+1}{2}\Big)^n \ " [/spoiler], which needed proof for n=2,3,4...

Lets go back and look at the proof I offered. >>7577358
Hopefully the base case is trivially clear.

The lemma from the Binomial Theorem is also pretty trivial. It just says

<span class="math"> 2\Big(\frac{n+1}{n+2}\Big)^{n+1} < 1,\ \ \ \ \ if\ \ n>0. [/spoiler]

I wanted to clearly separate it out from the main part of the induction proof in order to keep it logically independent. It's just a simple mini-"theorem" we are going to make use of; Like you might use the fact that a Euclidean triangle has 180 degrees in a more complicated geometry proof.

Now let's go step-by-step through the induction part. We want to prove P(m) implies P(m+1) for m=2,3,4,... We need the implication to hold for ALL those values, for reasons beaten to death in my epic posts above. So we let m be some arbitrary but fixed value in our range, and we tentatively assume P(m) is true in order to try to prove P(m+1).

What does P(m+1) actually say? It says

<span class="math"> ``\ (m+1)! < \Big(\frac{m+2}{2}\Big)^{n+1} \ " [/spoiler]

Proving this is our goal. Let's begin. We start with the LHS of P(m+1) and will try somehow to work our way to the RHS.

[Anonymous No.7578274]

>>7578260
>>7577569

<span class="math"> (m+1)! = (m+1) \cdot m! [/spoiler]

That much is clear about factorials. Next I am going to use the induction hypothesis, P(n), as I am free to do. I will multiply both sides of P(m) by (m+1) to get

<span class="math"> (m+1) \cdot m! < (m+1) \cdot \Big(\frac{m+1}{2}\Big)^m [/spoiler]

Next just rearrange the RHS to be more P(m+1)-like. This is just plain algebraic manipulation.

<span class="math"> (m+1) \cdot \Big(\frac{m+1}{2}\Big)^m = 2 \cdot \Big(\frac{m+1}{2}\Big)^{m+1} = 2 \Big(\frac{m+1}{m+2}\Big)^{m+1} \Big(\frac{m+2}{2}\Big)^{m+1} [/spoiler]

Finally I will use the lemma on part of the RHS. I am basically multiplying both sides of my lemma by the quantity

<span class="math"> \Big(\frac{m+2}{2}\Big)^{m+1} [/spoiler]

to get

<span class="math"> 2 \Big(\frac{m+1}{m+2}\Big)^{m+1} \Big(\frac{m+2}{2}\Big)^{m+1} < \Big(\frac{m+2}{2}\Big)^{m+1}. [/spoiler]

Now each of these steps involved an equality or an inequality. I can chain them all together to get

<span class="math"> (m+1)! < \Big(\frac{m+2}{2}\Big)^{n+1} [/spoiler]

This is our sought-for P(m+1). Assuming P(m) we have proved P(m+1). Therefore P(m) implies P(m+1). The only extra conditions I used were that m is an integer and m>0. These are general enough for ALL the implications we need.

So having done a base case P(2), and having shown P(m) implies P(m+1), all that remains is to call upon the mighty power of the Induction Principle in order to conclude that P(m) is true for m=2,3,4,...

[Anonymous No.7578380]

Wuzzup /sci/ I need some help in my physics class. Pic is wat I'm doin (it's an online thing I printed out to show work required) but it says the distance I got is wrong. What am I doin wrong?

[Anonymous No.7578889]

Can you overdose on melatonin? I've taken 20mg so far and I'm not tired enough.

[Anonymous No.7578930]

I wonder if anyone actually answers these, but I got a simple one.

arc length | (x y z) = (5 t ,2 t^2, 4/3sqrt(10)t^(3/2)) | t = 0 to 4 (parametric curve) anyway it was given as from point (0,0,0) to (20, 32, (32/3)sqrt(10)) instead of t from 0 to 4,

How to get from "(0,0,0) to (20, 32, (32/3)sqrt(10))" to "t from 0 to 4"

[Anonymous No.7578968]

>>7578930
at point B (20, 32, etc) x = 5t = 20
solve for t.

you'll see that it works for all of them. i.e.
At B: y = 2t^2 = 32. should come out to 4.

[Anonymous No.7579474]

>>7577493
Thanks anon, worked perfectly.

[Anonymous No.7579542]

There's an idiot in my CS classes that always shitposts and gives bad advice on our class forums. How do I get him to quit?

He's failed calc1 5 times and intro to programming 4 times at the community college, yet he stills believes hes better than everyone, and constantly complains about how hard and unrealistic classes are.

[Anonymous No.7580469]

>>7579542
Really? That's awesome anon. Why would you ever want him to stop? His posts must be comedy gold! Please post some examples?

[Anonymous No.7580621]

>>7570071
wow you got fancy hand writing, r u a grill?

[Anonymous No.7580650]

>>7579542
Maybe he is. He just finds pleasure in finding out how to annoy people in the most efficient of ways. Clearly you are one of the "I can annoy this guy by pretending to be stupid"-kind of a guy.

[Anonymous No.7580715]

>>7574347
Degree of substitution and resonance of the negative charge.

But don't mind me. I've only taken high school chemistry.

[Anonymous No.7580722]

So for my problem, the truck accelerates and given the mass of the box as well as the acceleration of the truck, I need to find the minimum coefficient of static friction that will keep the box from moving. Now, I found the coefficient using Newton's second law, setting the sum of forces to zero, and using the acceleration of the truck to find the force the static friction must be and then solving for the coefficient, but I don't see how it's justifiable considering it's just the box's inertia that causes it to slide back. Am I missing something here?

[Anonymous No.7580734]

>>7580722
You need to specify the forces acting on which object and try not to use same symbols for forces and accelerations as that can cause confusion. We need the force on the box forward to be as large as ma, since F = ma.

so µ mg = ma, gives µ = a/g, i.e. coefficient of friction must be at least a/g.

[Anonymous No.7580753]

>>7580734
Right, I got the answer, but what I don't understand is the justification for using the acceleration of the truck. If I were looking at a free body diagram of the box, the forces acting upon it would be its weight, the normal force, and the force of friction?? Wouldn't there have to be a fourth force in order for there to be a force of friction to begin with? And how would I represent that looking at just the box?

[Anonymous No.7580765]

>>7580753
The box needs to accelerate as fast as the truck does or it would start slipping or "falling behind" relative to the truck. The problem was calculating what would be necessary to avoid that slipping.

The box would remain still in left-right direction if no other forces acted on it horizontally. But it would remain still not in relation to the truck but in relation to the "reference frame" or "inertial" frame. In this case the inertial frame is the earth. A coordinate system moving along with the earths motion and rotation so that it is aligned with the earth gravity field.

[Anonymous No.7580794]

>>7578099
>If my solution works, shouldn't it verify the sphere's equation?

Yes, if your solution really sat on the intersection, it would.

We have the conditions <span class="math">y = x^2[/spoiler] and <span class="math">x^2 + y^2 + z^2 = 4[/spoiler]. We can mostly keep your parameterization for x(t) and y(t), but your z(t) is wrong.

First, what is the max value of x? On the sphere it is x=2, but <span class="math">2^2 + 2^4 > 4[/spoiler] so this can't be the case for the intersection. So what is the right max value? Substitute <span class="math">y=x^2[/spoiler] in the sphere equation to get <span class="math">x^2 + x^4 + z^2 =4.[/spoiler]

Clearly x=m will be a max when z=0, so to determine the max we solve the equation
<div class="math"> m^2 + m^4 = 4 </div>
Use the quadratic formula to get
<div class="math"> m = \sqrt{\frac{\sqrt{17} - 1}{2}} \approx 1.249621...</div>
My solution (using the same value of m) is

<span class="math"> x(t) = m \cos(t) [/spoiler]
<span class="math"> y(t) = m^2 \cos^2(t) [/spoiler]
<span class="math"> z(t) = m \sin(t) \sqrt{1 + m^2 + m^2 \cos^2(t)} [/spoiler]

It's obvious that <span class="math"> y(t) = (x(t))^2 [/spoiler] satisfies the parabolic cylinder condidion for all t, but the sphere condidion is harder to see. Lets prove that my solution really lies on the sphere:

<span class="math"> \Big(x(t)\Big)^2 + \Big(y(t)\Big)^2 + \Big(z(t)\Big)^2 =
m^2 \cos^2 t + m^4 \cos^4 t + \Big( m^2 \sin^2 t \Big)\Big( 1 + m^2 + m^2 \cos^2 t\Big) [/spoiler]

<span class="math"> = m^2 \cos^2 t + m^4 \cos^4 t + m^2 \sin^2 t + m^4 \sin^2 t + m^4 \sin^2 t \cos^2 t [/spoiler]

<span class="math"> = m^2 (\cos^2 t + \sin^2 t) + m^4 \sin^2 t + m^4 (\sin^2 t + \cos^2 t) \cos^2 t [/spoiler]

<span class="math"> = m^2 + m^4 \sin^2 t + m^4 \cos^2 t [/spoiler]

<span class="math"> = m^2 + m^4 [/spoiler]

<span class="math"> = 4 [/spoiler]

This shows the solution above is correct. Your answer is not because it is missing the last term (involving cos) under the square root in my z(t). So you had <span class="math"> m \sqrt(1+m^2) = 2. [/spoiler]

[Anonymous No.7580829]

>>7578930
Not sure if you needed the arc length anon, but I got

<span class="math"> s = \int_0^4 \frac{ds}{dt}\ dt = \int_0^4 \sqrt{\dot{x}^2 + \dot{y}^2 + \dot{z}^2\ }\ dt [/spoiler]
<span class="math"> = \int_0^4 \sqrt{(5)^2 + (4 t)^2 + (2 \sqrt{10 t})^2\ }\ dt [/spoiler]
<span class="math"> = \int_0^4 \sqrt{5^2 + 16 t^2 + 40 t^2\ }\ dt [/spoiler]
<span class="math"> = \int_0^4 \sqrt{(5 + 4t)^2}\ dt [/spoiler]
<span class="math"> = \int_0^4 (5 + 4t)\ dt [/spoiler]
<span class="math"> = 5 t + 2 t^2\ \vert_0^4 [/spoiler]
<span class="math"> = 5 \cdot 4 + 2 \cdot 4^2 [/spoiler]
<span class="math"> = 52 [/spoiler]

[Anonymous No.7580840]

If mass, momentum and energy are the same and gravity is "emitted" by matter, do really fast or really hot objects also put out a stronger gravitational pull?

[Anonymous No.7580849]

>>7580840
but deeds in males

[Anonymous No.7580897]

If gravity bends light, does it also bend electrostatic fields?

[Anonymous No.7580948]

if im trying to argue that X belongs in NP by use of a certificate to verify the yes instance. how would i go about doing that with a problem that requires me to partition the set S with subsets C into the sets (A, B) s.t. no entire set C is contained in either A or B

could i just have c1 be {1,2} and c2 be {3, 4} then let A = {1, 3} and B = {2, 4}? i dont really understand how i get this verifier

[Anonymous No.7581006]

>>7578054
>>7578058
>>7578064
>>7578068
>>7578078
>>7578081
>>7578260
>>7578274

Hello, Anon.

I just wanted to thank you for all of your posts. They definitely helped me forge those concepts into my mind. The first few posts were unnecessarily stretched out in explaining the concepts. They could probably be briefer, but maybe you could use them in such form for your 8-year old niece. As for the result of your posts on my understanding, even though I understood the principle of induction in the first place, they helped me solidify that understanding, albeit I am still a bit puzzled with the application. I'm sure that's a matter of exercise though. If you're willing, I could use some more exercise examples. Since you also intuitivelly reached for the binomial theorem lemma, I am wondering if there are any similar fundamental mini-"theorems", as you called it, which I could use in proving more inequalities like this one.

Once again, I give you my deepest thanks for your help.

[Anonymous No.7581024]

>>7580840
>>7580897
>If mass, momentum and energy are the same and gravity is "emitted" by matter, do really fast or really hot objects also put out a stronger gravitational pull?

You've got these concepts a bit fuxx0rezed up, anon. Mass, momentum, and energy are not the same. However, they all enter into the stress-energy tensor <span class="math"> T_{\mu \nu} [/spoiler] of general relativity. Einstein's Field equation (without a cosmological constant) is
<div class="math"> R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = \frac{8 \pi G}{c^4} T_{\mu \nu} </div>
The LHS describes the curvature of spacetime (where <span class="math">g_{\mu \nu}\ [/spoiler] is the metric on the spacetime and <span class="math">R_{\mu \nu}\ [/spoiler] is the Ricci curvature tensor). This is the side of the equation that you will spend loads of time solving for, if you take a general relativity (GR) course.

The RHS is the input side: the stress-energy tensor contains all the matter, energy, and momentum of the physics you're interested in.

The point of GR is that the force of "gravitational pull," as it is often imagined and discussed, doesn't exist. There is matter, momentum, and energy, and they curve spacetime according to the field equation. Objects moving in the warped spacetime take the straightest possible path they can (the geodesic), but those still end up being curved paths.

Hotter or faster moving objects will increase the stress-energy tensor, and once you solve for the spacetime, you'll find it will be slightly *more* curved as a result.

(Also, if you take a watch and wind it up or heat it up, you will have very slightly increased its inertia. This just comes from special relativity.)

[Anonymous No.7581027]

>>7581024

>>7580840
>>7580897 (You)
are not the same person

[Anonymous No.7581030]

>>7581024
>>7580840
>>7580897
>If gravity bends light, does it also bend electrostatic fields?

You got this a bit messed up too. Electromagnetic fields (propagating waves or static field) extend out into spacetime -- curved spacetime. Don't think of it as gravity bending light. Instead, think of it as a large mass like a star bends spacetime according to the field equation above. Light (or fields) take the straightest possible path through the warped spacetime.

Also, you can think of it the other way around as well. Light and static EM fields can bend space! For electromagnetic fields,
<div class="math"> T_{\mu \nu} = \frac{1}{4 \pi} \Big( F_{\mu}^{\ \ \lambda} F_{\lambda \nu} - g_{\mu \nu} F_{\alpha \beta} F^{\alpha \beta} \Big) , </div>
where the electromagnetic field tensor is <span class="math"> F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} [/spoiler], and <span class="math"> A_{\mu} [/spoiler] is the EM 4-vector potential.

So fields can be inputs that cause spacetime to curve -- like matter, energy, and momentum do. This means that the spacetime and the fields propagating in them become highly coupled. (Propagating fields curve the spacetime which changes the fields which curves the spacetime differently which...) So now you see why doing GR can be a real fookering pain in the ass!

[Anonymous No.7581036]

>>7581027

>>7580840
>>7580897
>are not the same person

Yes but I posted a two-post answer (hit post length limit) and each post is relevant to the other.

>>7581030
>>7581024

[Anonymous No.7581077]

Any ideas how to prove that? I've no idea what to do with that arctan.

[Anonymous No.7581103]

>>7581077

In pic related <span class="math"> \tan(\theta) = \frac{x}{1} , [/spoiler] so <span class="math"> \theta = \arctan(x) . [/spoiler]

Also <span class="math"> \cos(\theta) = \frac{1}{\sqrt{1+x^2}} . [/spoiler] Therefore

<span class="math"> \cos(\arctan (x))) = \frac{1}{\sqrt{1+x^2}} [/spoiler]

[Anonymous No.7581358]

Perhaps this is a more general question.
What is there left to do research about?
Haven't we pretty much figured out the major topics of all sciences by now? I feel like there are only niche research topics left.
I want to become a scientist, but I don't want to spend an entire life doing research about something that might not even affect anything at all.
tl;dr: Is there truth in the statement "Born too late to explore earth and too early to explore space"?

[Anonymous No.7581367]

>>7581358
The truth is (imho)
>there is A LOT to be done, but nothing that will make you famous
At most moments in time, shit seems to be solved. Newton did his thing 16xx and while Lagrange, Poisson, Bernoulli etc. became famouse also for solving mechanics problems, they were basically extending Newtons framework and the math of the time. There are 10000 scientists from Lagranges time not being mentioned today and most people now also don't get a mention.
Then again, Planck famously got told science is solved and he shouldn't become a scientist, and Kelvin said it too - then some new ideas came up in the last 20 years before 1900 and the following decades were great.

So tl;dr it's solved in that for becoming famous, you have to accidentally get into the right time frame. If you just like science, maybe go for it anyway.

[Anonymous No.7581431]

Help? I don't know where to begin.

[Anonymous No.7581467]

think it's not right

[Anonymous No.7581512]

I have stupid question. Program says answer doesn't have correct dimensions?

[Anonymous No.7581520]

>>7581512

Dimensions of momentum would be Kg m/s not s^2.

[Anonymous No.7581531]

>>7581520
right ya I just got that as well, thanks

[Anonymous No.7581549]

>>7581006

I'm glad the posts were helpful. And after typing them out, I'm really glad you did make it back to read them.

OK, I came up with some more induction exercises -- actually problems for you to think about. These are things off the top of my head that I remember solving at some point. Pick some and give them a try. They're all quick and brief, if you have the right idea. They start out easy and get a little harder (by my opinion/recollection).

Some require lemmas, that you will have to provide. I'll give hints or sketches if you ask. If you post sketches of the answers, I can "grade" them for you. (Don't bother with a long detailed solution, just see if you can get all the critical ideas down succinctly.)

1) Did you find out what was up with that birthday theorem? Do you buy the argument that everyone must have the same birthday? If not, where *precisely* does the proof break down?

2) Show that the number of subsets of a set of n elements is <span class="math"> 2^n [/spoiler].

3) Prove that every positive integer can be expressed as the product of an odd
number and a power of two (possibly <span class="math"> 2^0 [/spoiler]).

4) Show that if n straight lines are drawn on a plane it is possible to color all the
regions formed with only two colors so that no two bordering regions share the same color. (Bordering means share a segment of one of the lines.)

(split due to post length limit)

[Anonymous No.7581552]

>>7581549
>>7581006

5) Suppose n>1 is an integer, show <span class="math"> 1 + 1/sqrt{2} + 1/sqrt{3} + 1/sqrt{4} + ... < 2 sqrt{n} [/spoiler].

6) Easy case of Fermat: This Pythagorean triplet is well-known: <span class="math"> 3^2 + 4^2 = 5^2. [/spoiler] But prove that <span class="math"> 3^n + 4^n = 5^n [/spoiler] has no other solutions for integers n > 2.

7) For all n=1,2,3... prove that <span class="math"> 5^n + 2 \cdot 11^n [/spoiler] is always divisible by 3, and that <span class="math"> 2^{6n-2} + 3^{6n-2} + 5^{6n-2} [/spoiler] is always divisible by 38.

8) Show that <span class="math"> 3^{n+1} [/spoiler] divides evenly into <span class="math"> 2^{3^n} + 1 [/spoiler] for all n=0,1,2,...

9) Prove for all integers n > 1: <span class="math"> \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \ ... \ + \frac{1}{n^2} < 1 . [/spoiler]

10) Suppose a teacher gives a quiz and wants the students to grade each others papers. Teacher collects the papers and distributes them back so that each student receive a paper back to grade, but no student receives his own paper back. If there are n students, in how many ways N(n) can such a redistribution be done? (Probably needed hint: show by induction that the answer is <span class="math"> N(n) = n! \cdot \Sum_{k=0}^{n} \frac{(-1)^k}{k!} . [/spoiler]

Hopefully the level is about right. I have easier ones and much harder ones.

[Anonymous No.7581560]

>>7581549
>>7581006

5) Suppose n>1 is an integer, show <span class="math"> 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ... \frac{1}{\sqrt{n}} < 2 \sqrt{n} [/spoiler].

6) Easy case of Fermat: This Pythagorean triplet is well-known: <span class="math"> 3^2 + 4^2 = 5^2. [/spoiler] But prove that <span class="math"> 3^n + 4^n = 5^n [/spoiler] has no other solutions for integers n > 2.

7) For all n=1,2,3... prove that <span class="math"> 5^n + 2 \cdot 11^n [/spoiler] is always divisible by 3, and that <span class="math"> 2^{6n-2} + 3^{6n-2} + 5^{6n-2} [/spoiler] is always divisible by 38.

8) Show that <span class="math"> 3^{n+1} [/spoiler] divides evenly into <span class="math"> 2^{3^n} + 1 [/spoiler] for all n=0,1,2,...

9) Prove for all integers n > 1: <span class="math"> \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \ ... \ + \frac{1}{n^2} < 1 . [/spoiler]

10) Suppose a teacher gives a quiz and wants the students to grade each others papers. Teacher collects the papers and distributes them back so that each student receive a paper back to grade, but no student receives his own paper back. If there are n students, in how many ways N(n) can such a redistribution be done? (Probably needed hint: show by induction that the answer is <span class="math"> N(n) = n! \cdot \sum_{k=0}^{n} \frac{(-1)^k}{k!} . [/spoiler]

Hopefully the level is about right. I have easier ones and much harder ones.

[Anonymous No.7581622]

Quick question. Does anyone know what is the best statistic to include on a scatterplot? Since it's a linear regression, would you want to include R^2, R, or p-value or something else?

[Anonymous No.7581665]

First comment in 4chan

[Anonymous No.7581668]

>>7581665
welcome!

[Anonymous No.7581762]

Trying to find a probability for this situation, I'm getting it wrong somewhere but I don't know where.
I set B = defective item and A = item breaks when dropped from 1m height
I need to know what is the probability of B assuming A never occurs in 4 trials.

B = 0.285
A for B = 0.285*0.795
A for B' = (1-0.285)*0.095

From that I got A' = 0.7055

Assuming a binomial distribution I keep getting a result of 0.000047 for 4 trials, but that seems wrong. I think it is because I am counting the probability of B in each trial even though it is the same item being dropped each time, but I'm not sure how to calculate it properly.

[Anonymous No.7581798]

>>7581549
>>7581560

I'll go through them properly as soon as I can (next few days), but I'll try to see if I can think about some of them right now.

1) I don't believe the proof generally, since it relies on the induction assumption that any k people had the same birthday in the first place. We can't really analytically resolve the problem with the given information.

2) Is n an element of natural numbers?

3) z = (2z+1)*2^n where n is an element of natural numbers? If so, I would test the base case and then assume induction assumption for k = n and then somehow try to use that assumption somewhere in the case of k + 1

4) No immediate clue.

5) Use sum of arithmetic series formula (prove it too, if necessary), evaluate for base case and try and prove k + 1 through algebraic manipulation somehow.

I would use similar methods on 9). No immediate clue with the rest.

[Anonymous No.7581835]

>>7581762
Ok, I figured it out, I was looking at it from the wrong angle. I just calculated the probability of the item not breaking if it was defective (a), then if it was not defective (b), and then did \frac{0.285*a}{0.285*a + (1-0.285)*b}.

[Anonymous No.7581948]

Stupid General Question here, I'm thinking A but I don't want to lose marks where I don't have to

[Anonymous No.7582105]

>>7581622
Typically r^2 Is the value that should be posted on the graph with the equation of the line
However if you're doing a proper linear regression analysis you need to test for the statistical significance of r, followed by a lack of fit F-test to tell whether there is a statistically significant linear relation

[Anonymous No.7582133]

Okay so I'm probably going to seem retarded but can someone explain the sum_AVERAGE_DA column in this picture? I'm working on a mapping program and this category is under the attribute table for neighborhood traffic. So I'm being asked what that indicates and what the numbers listed for each neighborhood are measuring.

[Anonymous No.7582632]

>>7581668
Thanks :-)

[Anonymous No.7582804]

I'm doing a problem where you are supposed to use eulers formula to solve a second order differential in matlab. The inital condition was given in degrees, thus I used sind in the script and that gave a curve visually identical to a sine/cosine curve. However switching to sin and then converting initial condition to radians resulted in an oscillating curve, completely different. Am I just stupid for not understanding what happened?

[Anonymous No.7582919]

Ok I am too stupid to google that shit.

The Earth is moving around the sun. Our solar system is rotating around the galaxy center. Towards what constelation(s) do the galaxies gravitate to? Any keywords I can use to search for literature on this topic?

[Anonymous No.7583135]

Need binary help. How do I go about making the largest number?

[Anonymous No.7583229]

>>7582919
Constellations are collections of stars, so galaxies do not move towards these.

Galaxies can form clusters and then there is further structure on a larger scale than this. Keywords might be "large scale structure", "galactic clusters".

[Anonymous No.7583566]

>>7583135
All ones except the sign.

[Anonymous No.7583799]

testing

[Anonymous No.7583806]

i'm trying to evaluate

<span class="math"> \int \frac{x^3}{e^{x}-1} dx [/spoiler]

Doing it by hand i managed to get

<span class="math"> -x^{3} Li_{1}(e^{-x}) -3x^{2} Li_{2}(e^{-x}) -6x Li_{3}(e^{-x}) -6Li_{4}(e^{-x}) [/spoiler]

Where

<span class="math"> Li_{s} (z) [/spoiler]

are the polylog functions.

But wolfram gives a different answer, like

<span class="math"> 3x^2 Li_{2}(e^{x}) -6x Li_{3}(e^{x}) +6 Li_{4}(e^{x}) -\frac{x^{4}}{4} +x^3 Li_{1}(e^{x}) [/spoiler]

I can't check my result, but why does the minus signs alternate when the argument is the inverse of mine, and there is a suspicious x^4 term that i don't have?

Anyone?

[Anonymous No.7584243]

>>7583806
>I can't check my result

Don't forget the basics, differentiation is the inverse of integration. This is the textbook way though: https://math.stackexchange.com/questions/99843/contour-integral-for-x3-ex-1

[Anonymous No.7584263]

>>7584243
That won't do. He seems to want the *indefinite* integral, so differentiating the numerical residue constant from your link reference is useless.


>>7583806
Both results are correct, anon.

I took derivatives of both versions of the indefinite integral, and they both worked out to be the same integrand, EXCEPT for one small problem. It looks like you must have made a sign error when transcribing the WA result to /sci/, because the term <span class="math"> + \ x^3 Li_1(e^x) [/spoiler] ought to have a minus sign.

This issue comes up in black-body radiation problems. It turns out that

<span class="math"> \frac{x^3 e^{-x}}{1-e^{-x}} = \frac{x^3 }{e^x - 1} = - \frac{x^3 e^x}{1-e^x} - x^3 [/spoiler]

You expanded and integrated the LHS, like any reasonable human being would who is used to seeking out a convergent Taylor series to work with, whereas WolframAlpha must have internally integrated the RHS like a stupid machine.

Anyway, that's where the <span class="math"> - \frac{x^4}{4} [/spoiler] came from.

[Anonymous No.7584291]

>>7584243
Nevermind, sorry, on first glance I assumed your two separate thoughts were connected.

>>7583806
BTW, the signs don't need to alternate in your version anon because the half of the terms that involve differentiating the polylogs get an automatic minus sign via the chain rule on <span class="math">e^{-x}[/spoiler]. The WA result with <span class="math">e^x[/spoiler]has to provide for those itself in order to get the proper cancellation of terms.

[Anonymous No.7584303]

I might be a idiot but I don't see how
2sinx cosx-sinx=0
becomes
sinx (2cosx - 1)=0

[Anonymous No.7584313]

>>7584303
Factor out sin
2sinxcosx - sinx = 0
Pull out sin
Sinx(2cosx - 1) = 0

[Anonymous No.7584404]

>>7584303

What >>7584313 said. Use a(b+c) = ab + ac. Try it with numbers until you believe it.

[Anonymous No.7584570]

>>7581798

>I'll go through them properly as soon as I can (next few days)
No hurry. If this thread gets pruned or archived, I'll keep an eye out for you on another "Stupid Questions" thread. Just mention "mathematical induction" (spelled correctly so my control-f can find it).

>1) I don't believe the proof generally, since it relies on the induction assumption that any k people had the same birthday in the first place. We can't really analytically resolve the problem with the given information.
OK, this is why its a great problem for you anon, because it illustrates a "hole" in your understanding of induction. Either the proof is correct or it's flawed. Ponder and understand why your basic answer can *never* be a valid criticism of a mathematical induction proof. (No exaggeration!) There is a definitive, starkly specific answer that is well hidden by design, and you should work find it -- even if you have to formalize the entire argument. This is probably the best exercise of the lot for you, and really worth thinking hard about.

>2) Is n an element of natural numbers?
The n is whatever can count the number of subsets, so yes n=0,1,2,...

>3) z = (2z+1)*2^n where n is an element of natural numbers? If so, I would test the base case and then assume induction assumption for k = n and then somehow try to use that assumption somewhere in the case of k + 1
Right, but this one is included because it requires a slight modification of that idea that is sometimes required.

>4) No immediate clue.
Keep thinking anon, you'll get it.

(hit post length limit)

[Anonymous No.7584572]

>>7581798
>>7584570

>5) Use sum of arithmetic series formula (prove it too, if necessary), evaluate for base case and try and prove k + 1 through algebraic manipulation somehow.
Unfortunately this is neither an arithmetic nor geometric series, and can't be simply summed in closed form. (It's an incomplete instance of something called the Riemann Zeta function, which has bedeviled mathematicians for centuries...) But there is a much easier way, anon.

>I would use similar methods on 9). No immediate clue with the rest.
9 requires a certain "lemma" which I'll supply if you need it. Keep on thinking...

All in all I think these are great questions for you, targeted at about the right level. They are the kinds of things that are short and easy to prove, one you have the right idea. Finding that right idea might take a lot of thought. In any case they will stretch your mind. However, try to think about them for just the right amount of time. If you dwell too long, you won't really be learning, you'll waste your time. If you think too little, you won't really appreciate or grok the answers or hints given. Don't worry about not getting everything on your own, ask for hints or just ask to discuss if your approach is on track.

One last piece of unsolicited advice, it may seem too "schoolboi," but if you actually write out in quoted words what your P(n) statement is, it can really help to clarify your thinking.

[Anonymous No.7584690]

>>7581762

Seems to me that this problem has nothing to do with the binomial distribution. Instead, I think it is asking about how a weakly discriminating drop test can be made more strongly discriminating by independent repetition.

Assuming the drop tests are actually independent (which in real life I would probably disagree with because a drop could indeed weaken the product -- but whatever, this is a typically contrived textbook problem without the subtleties that would be present in actual test data) then you can just use Bayes' Theorem directly to get
<div class="math"> P(B| 4 \times A') = \frac{P(B)\Big[P(A'|B)\Big]^4}{P(B)\Big[P(A'|B)\Big]^4 + P(B')\Big[P(A'|B')\Big]^4} </div>
<div class="math"> = \frac{P(B)\Big[1-P(A|B)\Big]^4}{P(B)\Big[1-P(A|B)\Big]^4 + \Big[1-P(B)\Big]\Big[1-P(A|B')\Big]^4} </div>
Guessing that your undescribed numbers mean P(A|B) = 0.795 and P(A|B') = 0.095, with P(B) = 0.285, then I get
<div class="math"> P(B| 4 \times A') = \frac{(0.285)\Big[1-0.795\Big]^4}{(0.285)\Big[1-0.795\Big]^4 + \Big[1-0.285\Big]\Big[1-0.095\Big]^4} \approx 0.001048 </div>
P(B'| 4 \times A') = 1 - P(B| 4 \times A') \approx 0.998952

So if the product passes the drop-test 4 times it is highly likely (999 times vs 1 in 1000) that it is not defective.

protip: These numbers or everything here could be wrong, but if you don't include the actual text of the problem, you make it pretty hard for anons to give a good reply or even be bothered to respond at all based only on your not-working-out interpretation of the problem.

[Anonymous No.7584693]

>>7581762

Seems to me that this problem has nothing to do with the binomial distribution. Instead, I think it is asking about how a weakly discriminating drop test can be made more strongly discriminating by independent repetition.

Assuming the drop tests are actually independent (which in real life I would probably disagree with because a drop could indeed weaken the product -- but whatever, this is a typically contrived textbook problem without the subtleties that would be present in actual test data) then you can just use Bayes' Theorem directly to get
<div class="math"> P(B| 4 \times A') = \frac{P(B)\Big[P(A'|B)\Big]^4}{P(B)\Big[P(A'|B)\Big]^4 + P(B')\Big[P(A'|B')\Big]^4} </div>
<div class="math"> = \frac{P(B)\Big[1-P(A|B)\Big]^4}{P(B)\Big[1-P(A|B)\Big]^4 + \Big[1-P(B)\Big]\Big[1-P(A|B')\Big]^4} </div>
Guessing that your undescribed numbers mean P(A|B) = 0.795 and P(A|B') = 0.095, with P(B) = 0.285, then I get
<div class="math"> P(B| 4 \times A') = \frac{(0.285)\Big[1-0.795\Big]^4}{(0.285)\Big[1-0.795\Big]^4 + \Big[1-0.285\Big]\Big[1-0.095\Big]^4} \approx 0.001048 </div>
<span class="math"> P(B'| 4 \times A') = 1 - P(B| 4 \times A') \approx 0.998952 [/spoiler]

So if the product passes the drop-test 4 times it is highly likely (999 times vs 1 in 1000) that it is not defective.

protip: These numbers or everything here could be wrong, but if you don't include the actual text of the problem, you make it pretty hard for anons to give a good reply or even be bothered to respond at all based only on your not-working-out interpretation of the problem.

[Anonymous No.7584751]

Find the tangent line of 1/sqrtX at (1,1)

[Anonymous No.7584756]

>>7584751
<span class="math"> y = - \frac{1}{2} x + \frac{3}{2} . [/spoiler]

[Anonymous No.7584758]

>>7584756
This took me way longer than it should of. Thanks.

[Anonymous No.7585103]

HELP PLEASE!!!

my real analysis textbook has "left as an exercise for the reader" the proof of the following:

for all functions f,g,h continuous on [0,1], <span class="math"> sqrt{int_{0}^{1} (f(x)-g(x))^2 dx} [/spoiler] =< <span class="math"> sqrt{int_{0}^{1} (f(x)-h(x))^2 dx} + sqrt{int_{0}^{1} (g(x)-h(x))^2 dx} [/spoiler]

[Anonymous No.7585199]

>>7584693
I am making a sample actuary exam for teh lulz and I was wondering what the little accent next to your A means. Like right now I have something like

<span class="math">
\P(A \cup B')
\<span class="math">

What does the accent mean?[/spoiler][/spoiler]

[Anonymous No.7585208]

>>7585199
Let me try that again:

<span class="math">P(A \cup B')[/spoiler]

[Anonymous No.7585296]

>>7585103

that's just the triangle inequality for the L2 norm, look it up,

I can't be asked to prove it right now but I might come back to this thread later

[Anonymous No.7585306]

>>7585208
Set complement?

[Anonymous No.7585313]

>>7585306
Isn't <span class="math">A^{C}[/spoiler] or <span class="math">U \setminus A [/spoiler] the official notation for that?

[Anonymous No.7585316]

>>7585313
I don't think there's an "official" notation but it is normally one of those two. Is there any more context you can give?

[Anonymous No.7585323]

>>7585316
Question was like this:

You have <span class="math">P(A \cup B) = \cdots[/spoiler] and <span class="math">P(A \cup B') = \cdots[/spoiler]. Calculate <span class="math">P(A)[/spoiler].

Note that I don't need help with the question itself, only the notation of <span class="math">B'[/spoiler] is a bit strange and I don't know exactly what it means.

[Anonymous No.7585327]

>>7585199
>>7585208
>what the little accent next to your A means
>P(A∪B')

The accent is the quick way to write the "complement" of an event, as in the set complement. Sometimes it is written as <span class="math">A^C[/spoiler].

P(A∪B') means "the probability event A occurs but event B does not."
P(A'|B) means "the probability event A doesn't occur, given that event B does."
etc...

Any probability book will use and explain the symbols.

[Anonymous No.7585330]

>>7585327
Thank you so much!

[Anonymous No.7585646]

>>7585103

What this guy said >>7585296

Define the following

<span class="math">A_2=\int_0^1 \Big(f(y)-g(y)\Big)^2 dy \ge 0,\ \ \ \ B_2=\int_0^1 \Big(f(y)-h(y)\Big)^2 dy \ge 0,\ \ \ \ C_2=\int_0^1 \Big(h(y)-g(y)\Big)^2 dy \ge 0.[/spoiler]

These are all nonnegative, because the integrands are nonnegative (being squares of real-valued quantities). We want to prove that <span class="math">\sqrt{A_2} \le \sqrt{B_2} + \sqrt{C_2}.[/spoiler]

Let D be dot product of the differences

<span class="math">D=\int_0^1 \Big(f(y) - h(y)\Big)\Big(h(y)-g(y)\Big) dy[/spoiler]

For the Cauchy-Schwarz part, we start with another squared integrand,

<span class="math">0 \le \int_0^1 \Big[\Big(f(x)-h(x)\Big) C_2 - \Big( h(x)-g(x)\Big) D \Big]^2\ dx[/spoiler]

<span class="math">=\int_0^1 \Big[\Big(f(x)-h(x)\Big)^2 C_2^2 - 2 \Big(f(x)-h(x)\Big) \Big( h(x)-g(x)\Big) C_2\ D + \Big( h(x)-g(x)\Big)^2 D^2 \Big] \ dx[/spoiler]

<span class="math">= B_2 \, C_2^2 - 2 D\, C_2 D + C_2 D^2[/spoiler]

<span class="math">0 \le C_2 \Big[B_2 C_2 - D^2 \Big] [/spoiler]

Now suppose <span class="math">C_2[/spoiler] vanished. Since our functions f,g,h are continuous, this would imply that h=g identically on [0,1]. In this case the triangle inequality degenerates into a trivially true equality as <span class="math">A_2=B_2[/spoiler]. Therefore assume <span class="math">C_2 > 0[/spoiler] and drop it from the last inequality to get

<span class="math">D \le \sqrt{ B_2 C_2 }.[/spoiler]

It doesn't matter if D is negative. We may now write

<span class="math"> \Big(\sqrt{A_2}\Big)^2 = A_2 = \int_0^1 \Big(f(x)-g(x)\Big)^2 dx = \int_0^1 \Big[\Big(f(x)-h(x)\Big) + \Big(h(x)-g(x)\Big)\Big]^2 dx = B_2 + 2 D + C_2 \le B_2 + 2 \sqrt{ B_2 C_2 } + C_2 = \Big( \sqrt{B_2} + \sqrt{C_2} \Big)^2[/spoiler]

Finally take square roots to arrive at the desired triangle inequality.

[Anonymous No.7585826]

>>7585646
you beautiful bastard

[Anonymous No.7585940]

>>7585826
Thanks anon (I'll take that as a compliment). It's good to know there are individuals here on /sci/ who can truly appreciate a cleanly and tightly constructed argument.

[Anonymous No.7585996]

>>7570071

hi, I'm stuck on #53

any help would be awesome, not quite sure i know how to simplify one variable raised to a negative power by a different variable raised to another negative power

[Anonymous No.7586007]

>>7585996
y^4 / x^2, remember that a negative exponent is just the inverse of the number raised to the positive exponent

[Anonymous No.7586013]

>>7586007
thank you very much!!

[Anonymous No.7586017]

>>7586007

so #54 is going to be x^9/y^6 ?

[Anonymous No.7586029]

Is there a way to calculate the thermal energy of an object?

for instance if my object ways one gram and has a temp of 5 kelvin, is there a formula to figure out its thermal energy?

[Anonymous No.7586044]

>>7586029
look up heat capacity anon

[Anonymous No.7586052]

>>7585646
when i see things like this i'm intimidated by how smart someone has to be to deal with this

holeee shit

[Anonymous No.7586070]

What is the smallest value above 0 I can make with 4 bits? I get that a decimal goes somewhere, but I am unsure if I need a bit in the whole numbers. Is .0001 legit or does it have to be 0.001?

[Anonymous No.7586093]

How would I go about finding out if functions are dependent or independent, like 1/1+x^2?

[Anonymous No.7586104]

>>7586044
forgive me if I am wrong, but that gives how much energy I need to heat something up rather than how much thermal energy it contains

right?

[Anonymous No.7586201]

Why is clock arithmetic not reversible?

[Anonymous No.7586209]

>>7586104

Yes, that is astute of you, anon. But imagine heating up your object from absolute zero to it's current temperature. The the total heat energy needed is

<span class="math"> \Delta Q = \int_{T=0}^{T_{current}} C(T)\ dT = \int_{T=0}^{T_{current}} m \cdot c(T)\ dT, [/spoiler]

where m is the mass and c(T) is the specific heat capacity temperature at T.

(Note this doesn't include any latent heats that also must be accounted for if you pass through any phase transitions during the warming.)

Assuming c(T) is fairly constant (I don't know if this is a good assumption for the situation you have in mind, anon) up to <span class="math">T_{current} = 5 K[/spoiler], for example, then

<span class="math"> \Delta Q = \int_{T=0}^{T_{current}} m \cdot c(T)\ dT \ \approx \ m \cdot c \cdot T_{current} [/spoiler]

[Anonymous No.7586214]

>>7586209
Sorry, meant to say c(T) is the "specific heat capacity at temperature T"

[Anonymous No.7586222]

how would you prove f(n) = a[sub]n using induction?
for a problem like f(n) = 2^n -1, and a[n] = 2*a[n-1]+1, a[1]=1?
i proved a 2 base cases, n=1, n=2, to both be true. i don't get how you prove f(k+1) = a[k+1] though..

[Anonymous No.7586242]

>>7586209
You'll have to forgive me since its been several years since I've done this type of math, but basically I take the mass times the desired temperature times the current temperature?

[Anonymous No.7586262]

>>7586070
>Is .0001 legit or does it have to be 0.001?
.0001 is legit, the leading zero in 0.0001 can be assumed.

>>7586201
>Why is clock arithmetic not reversible?

Most all arithmetic is not reversible, anon. If I summed two whole numbers and gave you the result, say 5, could you tell if my sum was 2+3 or 1+4? No you couldn't.

>>7586222
>Prove <span class="math">a_n = f(n) = 2^n - 1[/spoiler] if <span class="math">a_n = 2 a_{n-1} + 1[/spoiler] and <span class="math">a_1 = 1[/spoiler].
You say you got the base case already taken care of, anon. So let's proceed to the induction step.
Assume for an arbitrary k=1,2,3... that <span class="math">a_k = f(k) = 2^k - 1[/spoiler]. Then start from the given recursion relation to get

<span class="math">a_{k+1} = 2 a_{k} + 1[/spoiler]
<span class="math">a_{k+1} = 2 f(k) + 1[/spoiler]
<span class="math">a_{k+1} = 2 (2^k - 1) + 1[/spoiler]
<span class="math">a_{k+1} = 2 \cdot 2^k - 2 + 1[/spoiler]
<span class="math">a_{k+1} = 2^{k+1} - 1[/spoiler]
<span class="math">a_{k+1} = f(k+1)[/spoiler]


>>7586242
Well it depends, anon. What is the specific situation you have in mind?

[Anonymous No.7586294]

>>7586017
>so #54 is going to be x^9/y^6 ?
No anon. I'm not that other anon, but I get

<span class="math"> \Big( \frac{x^{-3}}{y^2} \Big)^{-3} = \frac{x^{(-3)(-3)}}{y^{2(-3)}} = \frac{x^9}{y^{-6}} = x^9 y^6 [/spoiler]

[Anonymous No.7586309]

Real quick

Why are galaxies flat? Like they're just shit coming together, and if shit cam together and made a round sun and stuff, why can't the milky way be round too?

Inb4 Google you dumbfucking ass I did and can't get shit

[Anonymous No.7586365]

>>7586309
>>7586309
OK anon, real quick...

The reason for the disk shape in a galaxy or proto-star has to do with the competition between gravity and the conservation of angular momentum.

Gravity tries to draw matter together, but all the matter has to orbit its center of mass in order to conserve angular momentum. It can't just all fall immediately into the center.

As tidal forces and collisions occur, the disk gradually becomes more defined, and gravity shapes the disk as matter above and below the galactic plane are drawn down toward the plane of the disk.

As tidal forces and collisions continue, the matter slowly spirals in to the center of mass, picking up rotational speed as it does so.

Eventually so much matter is near the core that it becomes densely packed enough for the mean free path of the particles to become short enough that collisions dominate the dynamics over gravity. Pressure increases and if nuclear fusion starts at the core, there is enough thermal energy and pressure to resist further collapse (until much later when the nuclear fuel runs out). The matter is in an oblate spheroid shape, and radiation from the star pushes away some of the matter that had yet to fall into the star. Planets also congeal via the same process and you may end up with a solar system.

The mean free path of atoms in the sun is much smaller than the mean free path between star collisions in a galaxy. Most of the stars in a galaxy haven't had nearly enough time to spiral all the way down into the center yet. But those that have are so many in number and total mass that the gravity at the center of the galaxy becomes overwhelming, and the center collapses into a black hole.

[Anonymous No.7586380]

What's the difference between Newtonian, Hamiltonian, and Lagrangian mechanics? How does one go about studying these areas?

[Anonymous No.7586538]

>>7586380

OK, all three of these are huge subject areas, so I'm just going to give you a very tiny taste of what each one is like and what the main objects of study are in each. If you wanted more, you'd be reading a mechanics textbook like Goldstein.

Newtonian mechanics is founded on Newton's laws of motion. It's generally done in terms of finding all of the forces that act on some body, and then using the 2nd law of motion,

<span class="math"> \vec{F} = m \vec{a} [/spoiler]

to find some unknown, say, the trajectory of the body at every future point in time given some initial conditions of the motion. To do this you generally need to do some vector calculus, solving differential equations of the form

<span class="math"> \frac{d^2 \vec{r}}{dt^2} = \vec{a} = \frac{\sum \vec{F}(\vec{r},\dot{\vec{r}},t)}{m} [/spoiler]

Newtonian mechanics is the most "practical" for understanding how blobs of matter like vehicles, bridges, and buildings roughly behave under stresses and strains. It is generally considered to be the most elementary formulation of mechanics. It's so easy that video game makers, schoolboys, and even engineers can do it.

Modern physicists rarely touch this shit.

[Anonymous No.7586545]

>>7586380

OK, all three of these are huge subject areas, so I'm just going to give you a very tiny taste of what each one is like and what the main objects of study are in each. If you wanted more, you'd be reading a mechanics textbook like Goldstein.

Newtonian mechanics is founded on Newton's laws of motion. It's generally done in terms of finding all of the forces that act on some body, and then using the 2nd law of motion,

<span class="math"> \vec{F} = m \vec{a} [/spoiler]

to find some unknown, say, the trajectory of the body at every future point in time given some initial conditions of the motion. To do this you generally need to do some vector calculus, solving differential equations of the form
<div class="math"> \frac{d^2 \vec{r}}{dt^2} = \vec{a} = \frac{\sum \vec{F}(\vec{r},\dot{\vec{r}},t)}{m} </div>
Newtonian mechanics is the most "practical" for understanding how blobs of matter like vehicles, bridges, and buildings roughly behave under stresses and strains. It is generally considered to be the most elementary formulation of mechanics. It's so easy that video game makers, schoolboys, and even engineers can do it.

Modern physicists rarely touch this shit.

[Anonymous No.7586547]

Excuse me for being a basic bitch, but I saw something today that stated the following:
<span class="math">\frac{a-b}{b} = \frac{a}{b}-1[/spoiler]
I don't understand how that conclusion was reached. Would someone mind explaining the process?

[Anonymous No.7586549]

>>7586380

OK, all three of these are huge subject areas, so I'm just going to give you a very tiny taste of what each one is like and what the main objects of study are in each. If you wanted more, you'd be reading a mechanics textbook like Goldstein.

Newtonian mechanics is founded on Newton's laws of motion. It's generally done in terms of finding all of the forces that act on some body, and then using the 2nd law of motion,

<span class="math"> \vec{F} = m \vec{a} [/spoiler]

to find some unknown, say, the trajectory of the body at every future point in time given some initial conditions of the motion. To do this you generally need to do some vector calculus, solving differential equations of the form
<div class="math"> \frac{d^2 \vec{r}}{dt^2} = \vec{a} = \frac{\sum \vec{F}(\vec{r},\dot{\vec{r}},t)}{m} </div>
Newtonian mechanics is the most "practical" for understanding how blobs of matter like vehicles, bridges, and buildings roughly behave under stresses and strains. It is generally considered to be the most elementary formulation of mechanics. It's so easy that video game makers, schoolboys, and even engineers can do it.

Modern physicists rarely touch this shit.

[Anonymous No.7586551]

how does one solve this?
forces in point b and e are asked

[Anonymous No.7586554]

>>7586551
mass of the cinders is 10 kg each

[Anonymous No.7586556]

>>7586547
(a-b)/b = (a/b) - (b/b)

[Anonymous No.7586560]

>>7586380
>>7586549

The Lagrangian mechanics formalism is all about finding a Lagrangian function <span class="math"> L(q,\dot{q},t) [/spoiler] of some generalized coordinates q and velocities <span class="math"> \dot{q} [/spoiler] that describes the physical system you're interested in. The main idea considers the overall action S of your physical system, which is given by a time-integral over the Lagrangian:
<div class="math"> S = \int_{t_1}^{t_2} L(q(t),\dot{q}(t),t)\ dt . </div>
Generally, out of every potential path q(t) that *could* describe the time evolution of your system, you want to find that *particular* q(t) which minimizes the overall action S. It turns out that the actual path that Nature chooses for physical systems to follow is generally the one that minimizes this action. This is sometimes called the Principle of Least Action for unknown reasons. You use techniques from the calculus of variations to find the minimizing path q(t). Actually, you usually end up finding the Euler-Lagrange equations of motion, which frequently look like this
<div class="math"> \frac{\partial L}{\partial q} - \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q}}\Big) = 0 . </div>
So you're often back to solving differential equations, as in the Newtonian framework.

However, the Lagrangian viewpoint is extremely helpful and deep, especially if there are symmetries in your system. There is an important theorem by Emmy Noether that helps you to easily enumerate all of the conserved quantities from the symmetries in your Lagrangian. For example, conservation of energy, momentum, angular momentum, charge, etc are all explained in this formulation in terms of underlying symmetries in the physics of the system or the universe.

[Anonymous No.7586561]

>>7586556
Thanks. I feel so silly now.

[Anonymous No.7586562]

>>7586561
np lol.
you happen to be good at mechanics? :D
I feel silly for skipping too much classes
I'm stuck with >>7586551

[Anonymous No.7586566]

>>7586562
If I were good with with mechanics, I would probably have a better understanding of basic algebra, haha. I'm a software developer. Sorry I can't be of help.

[Anonymous No.7586569]

>>7586380
>>7586560

In quantum field theory, this is very important and powerful when used in conjunction with Lie group symmetries. Because of this, and the fact that Lagrangian mechanics turns out to be the nicest formulation for describing relativistic physics and the physics of fields, particle physicists mostly use the Lagrangian viewpoint when developing new models for particle interactions. Also, it is usually easiest to start with the action when trying to derive the Feynman rules for calculating Feynman diagrams for perturbative particle interaction processes using this formalism. Remember all that hype about the Higgs particle and CERN? That was basically all about testing the features/mechanism involving a few terms in a Lagrangian.

Anyway, if you can write down a Lagrangian for your system, you can generally say a hell of a lot about it even if you can't solve the whole theory (eg, equations of motion).

See previous pic related for the Lagrangian of the Standard Model of particle physics.

[Anonymous No.7586572]

>>7586380
>>7586569

Hamiltonian mechanics generally starts with a function called the Hamiltonian,

<span class="math"> H = H(p,q,t) . [/spoiler]

It is related to the Lagrangian via something called a Legendre transformation, and it also describes a great deal about the physical system of interest.

The Hamiltonian formulation elevates momentum p to an equal status to the coordinate q. (In advanced treatments of Hamiltonian physics, no real distinction is made between coordinates and momentum.) Appropriately then, Hamiltonian mechanics trades second order equations of motion for double the number of coupled first-order equations. This is what the Hamiltonian equations of motion look like:
<div class="math"> \dot{q} = \frac{\partial H}{\partial p} </div>
<div class="math"> \dot{p} = - \frac{\partial H}{\partial q} </div>
The different signs there give rise to an important type of differential geometry called symplectic geometry which has a very rich structure. The preferred viewpoint of Hamiltonian mechanics is that of phase space (p,q) (see pic related for a phase portrait), where the symplectic structure gives rise to many powerful theorems (eg, Liouville's theorem, KAM theorem) about the dynamics of classic physical systems. Hamiltonian mechanics is where research in modern classical mechanics resides. (Yes, classical mechanics is still being studied.)

The Hamiltonian viewpoint is also very powerful as a platform from which you can make departures into other areas of physics. Pertubation theory, nonlinear dynamics, chaos, many-body physics, statistical mechanics, quantum mechanics are all areas of physics which have Hamiltonian mechanics as the basic starting point.

[Anonymous No.7586578]

>>7586380
>>7586572

In particular, the objects and quantities that Hamiltonian mechanics (HM) defines are very well suited for promoting to analogous objects in quantum mechanics (QM). For example, the coordinates, momenta, energy all have natural analogues as operators in the operator formalism of QM; the canonical transformations of HM get promoted to unitary transformations in QM; the Poisson brackets of HM get promoted to commutators in QM, etc. Also, the Hamiltonian-Jacobi equation of HM is analogous to the Schrödinger equation of QM. HM is essentially the "geometrical optics" approximation to the full "wave mechanics" of QM; the two are very deeply connected.

Turns out that a lot of physicists live entirely in the Hamiltonian world.

OK, sorry for being so long winded anon. Half the stuff I said was probably incomprehensible to you, but I just love this stuff and couldn't help it.


TL;DR Read a book or take some physics classes.

[Anonymous No.7586580]

Didn't look here...

∫ √(x^3 + x^4) dx

please help me

[Anonymous No.7586585]

A basin of water can be filled using two different pipes. If only the first pipe is used, it takes 45 minutes until the basin is completely full. If only the second pipe is used, it takes 30 minutes to fill the basin. How long does it take to fill the basin if both pipes are used at the same time?

[Anonymous No.7586642]

>>7586585
18 minutes

[Anonymous No.7586651]

>>7586642
I would be interested to know how you calculated it, i couldn't find a way to get that result myself. Also thanks for the reply.

[Anonymous No.7586655]

>>7586578
>>7586572
>>7586569
>>7586560
>>7586549
Not the guy who asked the quuestion but this is awesome, thanks

[Anonymous No.7586657]

>>7586585
>>7586642
>>7586651
So the first pipe fills 1/45 of a tank per minute and the second fills at 1/30 per minute.
[math\frac{1}{45}+\frac{1}{30} = \frac{5}{90}][/math]
That means they can fill 5 basins in 90 minutes, but since we want one basin...
<span class="math">\frac{5}{90}=\frac{1}{x}[/spoiler]
<span class="math">x=\frac{90}{5}=18[/spoiler]
So yeah, it takes 18 minutes to fill the one basin using both the pipes.

[Anonymous No.7586659]

>>7586651
So the first pipe fills 1/45 of a tank per minute and the second fills at 1/30 per minute.
<span class="math">\frac{1}{45}+\frac{1}{30} = \frac{5}{90}][/spoiler]
That means they can fill 5 basins in 90 minutes, but since we want one basin...
<span class="math">\frac{5}{90}=\frac{1}{x}[/spoiler]
<span class="math">x=\frac{90}{5}=18[/spoiler]
So yeah, it takes 18 minutes to fill the one basin using both the pipes.

[Anonymous No.7586690]

>>7586659
Thanks a lot, i get it now.

[Anonymous No.7586713]

>>7586572
>>7586569
>>7586560
>>7586549
thanks a lot, fam

[Anonymous No.7586733]

>>7586651
>>7586659
>>7586690
I did it by
x(x/30 + x/45)
we know x, nor do we need it so you can fill in any value
which will yield 18

[Anonymous No.7586734]

>>7586733
we don't know x*

[Anonymous No.7586738]

>>7586580
You're not going to like the answer anon...

<span class="math"> \int \sqrt{x^3 + x^4}\ dx = \int x \sqrt{(x + \frac{1}{2})^2 - \frac{1}{4}}\ dx [/spoiler]

Use the substitution <span class="math">y = 2x + 1[/spoiler]

<span class="math"> = \int \Big(\frac{y}{2} - \frac{1}{2}\Big)\sqrt{\Big(\frac{y}{2}\Big)^2 - \frac{1}{4}\ }\ \Big(\frac{dy}{2}\Big)[/spoiler]

<span class="math"> = \frac{1}{8} \int y \sqrt{y^2 - 1}\ dy\ \ -\ \ \frac{1}{8} \int \sqrt{y^2 - 1}\ dy[/spoiler]

The first integral is easy, and the second is a standard form which you can integrate with cosh() or sec() substitution.

<span class="math"> = \frac{1}{24} \Big(y^2 - 1\Big)^{\frac{3}{2}}\ \ -\ \ \frac{1}{8} \Big[ \frac{1}{2} y \sqrt{y^2 - 1} - \frac{1}{2} \ln\Big(y + \sqrt{y^2 - 1}\Big) \Big] + C [/spoiler]

<span class="math"> = \frac{1}{24} \Big((2x+1)^2 - 1\Big)^{\frac{3}{2}}\ \ -\ \ \frac{1}{8} \Big[ \frac{1}{2} (2x+1) \sqrt{(2x+1)^2 - 1} - \frac{1}{2} \ln\Big((2x+1) + \sqrt{(2x+1)^2 - 1}\Big) \Big] + C [/spoiler]

<span class="math"> = \frac{1}{3} \Big(x^2 + x\Big)^{\frac{3}{2}}\ -\ \frac{1}{8} (2x+1) \sqrt{x^2+x} + \frac{1}{16} \ln\Big(2 x + 1 + 2 \sqrt{x^2+x}\Big) + C [/spoiler]

I took the derivative of this big ugly mess to check it, and I did in fact get <span class="math">x \sqrt{x^2 + x}[/spoiler] for the derivative, so I think it is correct.

[Anonymous No.7586782]

>>7586551
>>7586554

Is there any more info given in this problem? Seems to me that in addition to the masses of the cinders you need the mass of the rectangular support and the full length of the left part of it (that runs along b to a) and its thickness to solve this properly.

Anyway, I would start with the top cinder. There are only two contact forces on it, plus the weight from gravity. One you have those, you get the three contact forces on the lower cinder. Then proceed to the contact forces on the support.

Use the torques about b to get the perpendicular force applied at e. Finally then you can work out the support force at b.

[Anonymous No.7586788]

>>7586782
the L beam's mass can be neglected
also it's pure force question, no torque etc
I got the solution which is:
Fxb = 100 N, Fyb= 136.6 N, Fye = 39.9 N

but I don't get how they get this

[Anonymous No.7586833]

>>7586738
wow, I didn't think the solution would be this easy.

I thought I'd have to sue the other complicated binome integral transformation.

Thank you so much.

[Anonymous No.7586834]

>>7586833
>suing math
Oh you americans.

[Anonymous No.7586836]

>>7586788

OK, well I worked it out using the method that I outlined, and neglected the support like you said. But all my numbers are slightly off from what you reported. I got

Fxb = 98.1 N
Fyb = 133..6 N
Fye = 36.25 N

I don't know what could explain the slight discrepancies, except for the missing factors that I mentioned.

[Anonymous No.7586837]

>>7586836
we use g = 10 m/s^2
could you please post your calculations & drawings etc kind anon

[Anonymous No.7586841]

>>7586738
wait, shouldn't the "y" transformation be [ x+ 1/2 =y ]?

[Anonymous No.7586844]

Can someone give me a qualitative explanation of what 'action' is in physics and why it's the difference between an object's kinetic and potential energy?

[Anonymous No.7586861]

>>7586837
>we use g = 10 m/s^2
Ah, that would explain some of it

>could you please post your calculations & drawings etc kind anon
Sure. I didn't make a drawing, I used yours. But let me write up my equations... gimme a while


>>7586841
>wait, shouldn't the "y" transformation be [ x+ 1/2 =y ]?

No because then you'd have 1/4's everywhere in all your square roots.


>>7586844
>why it's the difference between an object's kinetic and potential energy?

It's not anon. It's the time integral of the Lagrangian (which is that difference).

See here >>7586560

[Anonymous No.7586910]

>>7586861
>But let me write up my equations... gimme a while
I'm waiting ... please don't leave me hanging anon, I need this quite badly

[Anonymous No.7586954]

>>7586837
>>7586910

Sorry anon, I got distracted by something important, but I wasn't gonna bail on you.

I will use these

(Radius) R = (180 mm)/2 = 90mm
(Length) L = 600 mm

Like I said, the method is like I outlined. These are the forces ON THE SUPPORT at the various points:

<span class="math"> F_x^{(a)} = - 2 M g \sin(\theta) [/spoiler]
<span class="math"> F_y^{(c)} = - M g \cos(\theta) [/spoiler]
<span class="math"> F_y^{(d)} = - M g \cos(\theta) [/spoiler]


Sum of torques about point (b) is zero:

<span class="math"> \sum \tau^{(b)} = - R \cdot F_x^{(a)} + R \cdot F_y^{(c)} + 3R \cdot F_y^{(d)} + L \cdot F_y^{(e)} = 0 [/spoiler]

This gives
<span class="math"> F_y^{(e)} = \frac{2 M g R}{L}\Big(2 \cos(\theta) - \sin(\theta) \Big) \approx 36.96 N [/spoiler]


Sum of forces ON THE SUPPORT in x-direction is zero:

<span class="math"> \sum F_x = F_x^{(a)} + F_x^{(b)} = 0 [/spoiler]

This gives
<span class="math"> F_x^{(b)} = 2 M g \sin(\theta) \approx 100 N [/spoiler]
Sum of forces ON THE SUPPORT in y-direction is zero:

<span class="math"> \sum F_y = F_y^{(b)} + F_y^{(c)} + F_y^{(d)} + F_y^{(e)} = 0 [/spoiler]

This gives
<span class="math"> F_y^{(b)} = 2 M g \Big(1-\frac{2R}{L}\Big) \cos(\theta) + \frac{2 M g R}{L} \sin(\theta) \approx 136.2 N [/spoiler]


Let me know if you have questions.

My question: Why don't you guys use
https://en.wikipedia.org/wiki/Standard_gravity

[Anonymous No.7586998]

>>7586954
damn
more complex than I thought
thanks anon

[Anonymous No.7587122]

>>7586998
wonder why my post with the image of the problem got removed
>>7586551
was it copyrighted or something? lmao

[Anonymous No.7587132]

A bank can make one of two types of loans. It can lend money on home mortgages where it has a 75% probability of earning 110 million kronor and 25 % probability of earning 70 million kronor. Alternatively it can lend money to IT firms where it has 25 % probability of earning 400 million and a 75% probability of losing 160 million (due to defaults). Richard, the manager of the bank, who will make the lending decision, receives 1 % of the bank’s earnings as a bonus. He believes that if the bank loses money he can walk away from his job without repercussions but without bonus. What investment will Richard make if he only cares about maximizing his own private earnings? What investment would the owners of the bank prefer?

[Anonymous No.7587264]

Is the solution manual incorrect or am I just retarded. I am asked to find an equation of the plane such that the point (-5,1,0) lies in it, and that it is parallel to the plane 3x-2z=6. Sounds easy enough, althought the solution manual says "Since the plane is parallel to 3x − 2z = 6 the normal vector is given by <3,0,-2>". How can this be true if the two planes are parallel? Wouldn't that be the direction vector for the two planes?

[Anonymous No.7587420]

>>7587264
Parallel planes share normal vectors. If V is perpendicular to plane 1 and plane 1 is parallel to plane 2 V is perpendicular to plane 2

[Anonymous No.7587440]

>>7587264
Two planes are parallel if their normal vectors are parallel, thus you can use the normal vector for the given plane to construct yours

[Anonymous No.7587448]

>>7587420
>>7587440
Thanks guys. I just got through working with lines in space, whose equations use direction vectors parallel to them, and then absent-mindedly began working with planes. Just goes to show how important it is to truly understand and think about what's going on as opposed to applying route formulas.

[Anonymous No.7587486]

>>7586262
>Well it depends, anon. What is the specific situation you have in mind?

specifically I want to find out how much thermal energy (in joules) is in a ball of plasma at a certain temperature. I'm sure this is very complicated but an estimate should suffice

[Anonymous No.7587499]

>>7587132
mortgages:
0.75 * 110 + 0.25 * 70 = 100 mil / odds
IT firm:
0.25 * 400 - 0.75 * 160 = -20 mil / odds

bank will obviously chose for the mortgages

as for richards perspective:
mortgages:
0.75 * 1.1 + 0.25 * 0.70 = 1 mil / odds
IT firm:
0.25 * 4 = 1 mil / odds

should also pick the mortgages since it ensures a bonus of 700k

[Anonymous No.7587504]

>>7587499
for richard it doesn't matter since they both have the same ratio
but I don't see why he should pick the other since it's gambling for no reason

[Anonymous No.7587572]

>>7587486

Ah, should have said so, anon... That changes everything. I got thrown when you mentioned

>"for instance if my object ways one gram and has a temp of 5 kelvin"

I was thinking about a solid object at 5 Kelvin (5 degrees above absolute zero), like a little block of copper or something.

A plasma temp is probably over 5000 Kelvin! Anyway, yes, if you only want to estimate the *thermal* energy in a quasi-equilibrium (and not say electromagnetic energy if the plasma is highly charged or contained by a strong magnetic field) of a monatomic/fully dissociated gas then just use

<span class="math"> E = \frac{3}{2} N k_B T = \frac{3}{2} n R T , [/spoiler]

where N is the number of particles or <span class="math"> n = N/N_A [/spoiler] is the number of moles (these should count everything, even the ionized electrons). <span class="math"> k_B [/spoiler] is the Boltzmann constant or use <span class="math"> R = k_B \cdot N_A [/spoiler] is the ideal gas constant, <span class="math"> N_A [/spoiler] is Avogadro's Number. T is the temperature in Kelvin.

(The other way would work in theory, but you would have to account for a lot of phase changes, etc as you heated from 0 to high temp -- clearly not a good way at all!)

[Anonymous No.7587600]

>>7587122
Maybe the new janitors are exercising their enormous newfound powers over /sci/.

Uh oh, shouldn't have said anything... now waiting for my ban.

[Anonymous No.7587641]

>>7570071
what is this gibberish? where are the numbers? I only 2s and some 0s

[Anonymous No.7587706]

Anyone can help me with baby number theory

How do I find the smallest achievable value such that all greater values are achievable by adding A and/or B together when A and B are co-primes. i.e for 5 and 7 the answer is 24.

[Anonymous No.7587712]

I'm supposed to find the normal line of f(x) = 2.5 + 3.5(4^p) at the y intercept.
I know I need a line perpindicular to the slope, and I find that the slope is 3*ln(4)*4^p, but I don't know where to go from there. Any ideas?

I figured that at the y-intercept p is just 0, and the slope of the normal line would just be -1/(3*ln(4)), but apparently that's wrong.

[Anonymous No.7587749]

So I have this analysis problem that I had to translate because germanfag

I have a bunch of rules how to get coefficients from others, for example a_k = c_n + c_-n
But how do I start?

I noticed that the original function looks suspiciously like the real standard form pictured at the bottom, but thats a sum so I can't just get a_k and b_k from that right?

[Anonymous No.7587757]

>>7587706
I think the answer is (A-1)(B-2) btw, but not sure how to prove it.

[Anonymous No.7587768]

>>7587757
*(B-1)

[Anonymous No.7587771]

can you multiply an integral by x/x to help solve for a u substitution? rather then 7/7

[Anonymous No.7587786]

>>7587771
bump

[Anonymous No.7587852]

>>7587749
You can define an inner product:
<span class="math"> \langle f, g \rangle:=\int_0^{2\pi} f(x)\overline{g(x)}dx [/spoiler]
and prove that the e^{ikx} actually form an orthonormal basis, and hence by the projection theorem, each c_k is given by <span class="math"> \langle f, e^{ikx} \rangle [/spoiler]

[Anonymous No.7587873]

> how the fuk do I find this shit

https://gyazo.com/ae246557363cbc83f224c176c589b214

[Anonymous No.7587892]

>>7587749
learn 2 latex brah

[Anonymous No.7588046]

What do I get if I derive sine? Would also like to know what cosine is.

[Anonymous No.7588049]

>>7587749

In general anon, for a more arbitrary function f(x), you can do real Fourier analysis like this guy described >>7587852 .

But for babby problems (trigonometric polynomial) that's overkill. Instead just use these

<span class="math">\cos(x) = \frac{e^{i x} + e^{-i x}}{2}\ \ \ \ \ \ \ \sin(x) = \frac{e^{i x} - e^{-i x}}{2i}[/spoiler]

to get

<span class="math">f(x) = 4 + 2 \cos(3 x) - 2i \sin(3 x)[/spoiler]
<span class="math">= 4 + 2 \Big( \frac{e^{3 x} + e^{-3 x}}{2}\Big) - 2 i \Big( \frac{e^{3 x} - e^{-3 x}}{2i} \Big)[/spoiler]
<span class="math">= 4 + (e^{3 x} + e^{-3 x}) - (e^{3 x} - e^{-3 x})[/spoiler]
<span class="math">= 4 + 2 e^{-3 x}[/spoiler]

<span class="math">f(x) = \sum_{k=-3}^{3} c_k e^{i k x} = 2 e^{-3 x} + 0 e^{-2 x} + 0 e^{- x} + 4 e^{0} + 0 e^{x} + 0 e^{2 x} + 0 e^{3 x}[/spoiler]

So your Fourier coefficients are <span class="math">c_k = 2,0,0,4,0,0,0[/spoiler].

[Anonymous No.7588050]

>>7588049
wew I was actually trying to wrap my inferior mind around fourier analysis, you might have just saved my life and I am forever in your debt

[Anonymous No.7588084]

>>7588049

what if the function is like this?
<span class="math">f(x) = (e^{ix}-e^{-ix})(e^{ix}+e^{-ix})[/spoiler]

[Anonymous No.7588128]

>>7588084
>>7588050

Fourier analysis is good to know anon, don't let me dissuade you from learning it. But it is a high-class tool, and you don't really need it for these problems.

For <span class="math">f(x)=(e^{ix}−e^{−ix})(e^{ix}+e^{−ix}), [/spoiler] you just have to multiply it out:

<span class="math">f(x)=e^{ix}e^{ix} - e^{-ix}e^{ix} + e^{ix}e^{-ix} - e^{-ix}e^{-ix}[/spoiler]

<span class="math">f(x)=e^{2ix} - e^{0} + e^{0} - e^{-2ix}[/spoiler]

<span class="math"> f(x)=e^{2ix} - e^{-2ix} [/spoiler]

<span class="math"> f(x) = \sum_{k=-3}^{3} c_k e^{i k x} = 0 e^{-3 i x} - 1 e^{-2 i x} + 0 e^{- i x} + 0 e^{0} + 0 e^{i x} + 1 e^{2 i x} + 0 e^{3 i x} [/spoiler]

[Anonymous No.7588141]

If Dave is standing next to a silo of cross-sectional radius
r = 9
feet at the indicated position, his vision will be partially obstructed. Find the portion of the y-axis that Dave cannot see. Enter your answer using interval notation.

I'm having trouble finding the tangent point.

[Anonymous No.7588144]

>>7588128
oh i got kinda hung up on the i but we are getting the complex coefficients after all.
would <span class="math">a_2[/spoiler] in that example be 4 and <span class="math">b_2[/spoiler] be <span class="math">0i[/spoiler]?

[Anonymous No.7588163]

>>7570071
Look up the implicit function theorem, it solves the exercise pretty quickly

[Anonymous No.7588170]

>>7588144

If you mean the coefficients from this Fourier expansion

<span class="math">f(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty} \Big( a_k \cos(k x) + b_k \sin(k x) \Big)[/spoiler]

Then for the first problem you posted

<span class="math">f(x) = 4 + 2 \cos(3 x) - 2i \sin(3 x) = 4 + 2 e^{-3 x}[/spoiler]

All <span class="math">a_k = 0[/spoiler] except <span class="math">a_0 = 8[/spoiler] and <span class="math">a_3 = 2[/spoiler]; and all <span class="math">b_k = 0[/spoiler] except <span class="math">b_3 = - 2 i[/spoiler] .

And for the second one

<span class="math">f(x)=(e^{ix}−e^{−ix})(e^{ix}+e^{−ix}) = e^{2ix} - e^{-2ix} = 2 i \sin(2 x)[/spoiler]

All <span class="math">a_k = 0[/spoiler]; and all <span class="math">b_k = 0[/spoiler] except <span class="math">b_2 = 2 i[/spoiler] .

[Anonymous No.7588174]

>>7588141
Draw the radius from the origin to the point of tangency, and a line parallel to the y axis that passes through that point, then just trig your way through.

[Anonymous No.7588187]

>>7588170
huh, for the second one i thought that <span class="math">a_k = c_k + c_{-k}[/spoiler] (disregard my answer earlier)

I also don't really understand how you determined b_2

[Anonymous No.7588190]

>>7588046
Let me reword this: can someone explain the proof that d/dx [sine(x)] = cos(x)?

[Anonymous No.7588214]

I made a lens that focuses light coming from any direction into a thin beam going in the same direction, by having a glass sphere with a shell made of some sort of material that has a <1 index of refraction.
I've been testing it with large beams of light, with the intent of concentrating the beams with the lens into small beams.
However, I honestly have no idea what I'm doing.
What I've found is that thinning the shell will, to some extent, help focus the beam. Dunno if it causes more light to scatter out of the lens, probably does.
Bringing the IoR down close to zero has a much more profound effect in focusing the beam, but definitely causes more light to scatter out.
Anyone know how I can optimise this to have an extremely narrow beam without losing 99% of the light going into the lens?

[Anonymous No.7588233]

>>7588187

Yes, it is easy to prove these rules are always true: <span class="math"> a_k = c_k + c_{-k} , [/spoiler] and <span class="math"> b_k = i (c_k - c_{-k}) [/spoiler] for k=1,2,3,.... Also <span class="math"> \frac{a_0}{2} = c_0 [/spoiler] .

Let's check them for the two problems.

For the first problem the only nonzero ones are <span class="math">a_0 = 8,\ \ a_3 = 2,\ \ b_3 = -2i,\ \ c_0=4,\ \ c_{-3} = 2,\ \ c_3 = 0.[/spoiler]

<span class="math">c_0 = 4 = \frac{8}{2} = \frac{a_0}{2} [/spoiler] (works)
<span class="math">c_3 + c_{-3} = 0 + 2 = 2 = a_3[/spoiler] (works)
<span class="math">i (c_3 - c_{-3}) = i(0 - 2) = -2i = b_3 [/spoiler] (works)

So the relations between a's, b's, and c's all work for your first problem.

Now in the second problem, the only nonzero ones are <span class="math">b_2 = 2i,\ \ c_{-2}=-1,\ \ c_2 = 1.[/spoiler]

<span class="math">c_0 = 0 = \frac{0}{2} = \frac{a_0}{2} [/spoiler] (works)
<span class="math">c_2 + c_{-2} = 1 + (-1) = 0 = a_2[/spoiler] (works)
<span class="math">i (c_2 - c_{-2}) = i(1 - (-1)) = 2i = b_2 [/spoiler] (works)

So the relations between a's, b's, and c's all work for your second problem too.

Why in the second problem is <span class="math"> b_2 = 2i [/spoiler] ? It's easy anon, use the first identity I showed you:

<span class="math">\sin(\theta) = \frac{e^{i \theta} - e^{-i \theta}}{2i} \ \ \ \ \ \theta = 2x[/spoiler]

<span class="math">f(x)=(e^{ix}-e^{-ix})(e^{ix}+e^{-ix}) = e^{2ix} - e^{-2ix} = 2 i \sin(2 x)[/spoiler]

[Anonymous No.7588272]

>>7588046
>>7588190

sin and cos are standard trigonometric functions. Read the Wikipedia page for the definitions.

What part of the proof don't you understand, anon? This an abbreviated standard proof:
<div class="math">\frac{d}{dx} \sin(x) = \lim_{h \rightarrow 0} \frac{\sin(x+h) - \sin(x)}{h}</div>
<div class="math">= \lim_{h \rightarrow 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}</div>
<div class="math">= \sin(x) \, \Big( \lim_{h \rightarrow 0} \frac{\cos(h) - 1}{h} \Big) + \cos(x) \, \Big(\lim_{h \rightarrow 0} \frac{\sin(h)}{h} \Big)</div>
<span class="math">= \sin(x)\,(0) + \cos(x)\,(1) [/spoiler]

<span class="math">= \cos(x) [/spoiler]

Or, you can do it with Taylor series,
<div class="math"> \sin(x) = \sum_{k=0}^{\infty} \frac{(-1)^k }{(2k+1)!} x^{2k+1} \ ,\ \ \ \ \ \cos(x) = \sum_{k=0}^{\infty} \frac{(-1)^k }{(2k)!} x^{2k} </div>
So then
<div class="math"> \frac{d}{dx} \sin(x) = \frac{d}{dx} \sum_{k=0}^{\infty} \frac{(-1)^k }{(2k+1)!} x^{2k+1} = \sum_{k=0}^{\infty} \frac{(-1)^k }{(2k+1)!} \frac{d}{dx} x^{2k+1} = \sum_{k=0}^{\infty} \frac{(-1)^k }{(2k+1)!} (2k+1) x^{2k} = \sum_{k=0}^{\infty} \frac{(-1)^k }{(2k)!} x^{2k} = \cos(x) </div>

[Anonymous No.7588282]

>>7588272
I get that a trig identity is used to expand sin(x+h), but how does the sin(x) and cos(x) get out of the numerator?

[Anonymous No.7588294]

>>7588282

OK, I think your problem is with algebra, anon. Calculus will make little sense until you understand the basic algebraic manipulations like this one. Here is a true equation:
<div class="math"> \frac{a \cdot b + c \cdot d}{e} = a \frac{b}{e} + \frac{c}{e} d </div>
Do you understand how the a and the d got out of the numerator? The proof just does the same kind of thing.

[Anonymous No.7588307]

>>7588294
I see it know. Thanks anon. I've been having simple issues like this lately.

[Anonymous No.7588316]

>>7588307
No problem, anon. Any other steps you're not getting?

This stuff isn't hard, but there are a lot of little details to know. My unsolicited advice is to solve as many practice problems as you can. That's really the only way to grok it.

[Anonymous No.7588320]

>>7588316
*now
I'm good for now. From what I heard, it gets easier to deal with proofs the more you see them.

[Anonymous No.7588333]

>>7588320
>From what I heard, it gets easier to deal with proofs the more you see them.

Yeah, I definitely agree with that.

Always, read the proofs. If you get lost try to keep reading through them anyway. Sometimes a later point clarifies what you were stuck on. Don't dwell too much on any one thing. If the proofs aren't making sense, do some problems, then come back to it. Sometimes what seemed baffling before can appear trivial later. Something just gets sorted out in the subconscious.

If you go back to re-read older proofs that you didn't understand before, you'll be surprised how much more they make sense after time has passed.

Whatever you do, try not to stress out about it too much, it just takes time to learn this stuff. Stress inhibits learning anyway.

Best of luck, anon.

[Anonymous No.7588376]

Does anyone have a more formal/better proof of this? It just isn't sitting with me

[Anonymous No.7588450]

why can't anything go faster than the speed of light? what theoretically happens if you do?

[Anonymous No.7588459]

how hard is Calculus of Variations

[Anonymous No.7588496]

>>7588141

This is a nice doge problem, anon. I like it. You must have a cool teacher, whoever made this up.

Anyway, to solve it you don't need calculus or trigonometry or anything fancy. Just calculate the area of the blue triangle in pic related in two different ways and equate them.

In the first way, consider the x,y-intercepts of the tangent line. The area of the triangle is one-half the base <span class="math">x_{int}[/spoiler] times height <span class="math">y_{int}.[/spoiler]

In the second way, since the radius in red is perpendicular to the tangent line, the area is one-half the base (hypotenuse of the blue triangle) <span class="math">\sqrt{x_{int}^2 + y_{int}^2}[/spoiler] times the height (the radius) r. So we get

<span class="math">Area = \frac{1}{2} x_{int}\ y_{int} = \frac{1}{2} r \sqrt{x_{int}^2 + y_{int}^2}[/spoiler]

Solve for <span class="math">y_{int}[/spoiler] in the equation

<span class="math">x_{int}\ y_{int} = r \sqrt{x_{int}^2 + y_{int}^2}[/spoiler]

to get

<span class="math">y_{int} = \frac{x_{int}\,r}{\sqrt{x_{int}^2 - r^2}}.[/spoiler]

The portion of the y-axis that Dave cannot see is twice this:

<span class="math">2 y_{int} = 2 \frac{12 \cdot 9}{\sqrt{12^2 - 9^2}} = \frac{72}{\sqrt{7}}.[/spoiler]

OK, now let's solve the problem the standard way by determining the tangent line using a derivative...

[Anonymous No.7588500]

Do electromagnetic charges also curve spacetime like gravity?

[Anonymous No.7588507]

>>7588496
>>7588141

The circle has equation <span class="math">x^2 + y^2 = r^2.[/spoiler] We'll focus on the upper semi-circle <span class="math">y = \sqrt{r^2 - x^2},[/spoiler] which has derivative

<span class="math">y' = - \frac{x}{\sqrt{r^2 - x^2}}.[/spoiler]

Let's call the point of tangency <span class="math">(x_t,y_t)[/spoiler] with tangent slope <span class="math">y_t'[/spoiler]. The equation of the tangent line is given by

<span class="math">y - y_t = y_t' (x - x_t)[/spoiler]

<span class="math">y - \sqrt{r^2 - x_t^2} = \Big( - \frac{x_t}{\sqrt{r^2 - x_t^2}} \Big) (x - x_t)[/spoiler]

The two points of this line that we care about are at <span class="math">(x_{int},0)[/spoiler] and <span class="math">(0,y_{int}).[/spoiler]

Putting these points into the equation of the tangent line gives two relations:

<span class="math">(x_{int},0):\ \ \ \ \ 0 - \sqrt{r^2 - x_t^2} = \Big( - \frac{x_t}{\sqrt{r^2 - x_t^2}} \Big) (x_{int} - x_t)[/spoiler]

<span class="math">(0,y_{int}):\ \ \ \ \ y_{int} - \sqrt{r^2 - x_t^2} = \Big( - \frac{x_t}{\sqrt{r^2 - x_t^2}} \Big) (0 - x_t)[/spoiler]

Use the first one to find <span class="math">x_t[/spoiler]:

<span class="math">x_t = \frac{r^2}{x_{int}}.[/spoiler]

Then use the result to eliminate <span class="math">x_t[/spoiler] in the second relation and solve for <span class="math">y_{int}[/spoiler]. We end up with

<span class="math">y_{int} = \frac{x_{int}\, r}{\sqrt{x_{int}^2 - r^2}}.[/spoiler]

Notice this is exactly the same formula that we found before, but with harder work doing it this way.

Long live doge.

[Anonymous No.7588547]

>>7588450
>why can't anything go faster than the speed of light? what theoretically happens if you do?

I wrote up an explanation of this a few threads back, but it got pruned. See here:

>>/sci/thread/S7542983#p7564823
>>/sci/thread/S7542983#p7564826


>>7588459
>how hard is Calculus of Variations

Compared to what? It's harder than calculus. It depends how deep you go into it, and if your study is proof-based or problem-based. If you apply it to fluids or fields, you need to know PDEs first. If you already know about L2 theory or Banach spaces, it's not that hard at all. If you only want to learn to do Langrangian mechanics >>7586560 , it's quite easy.


>>7588500
>Do electromagnetic charges also curve spacetime like gravity?

Dubs checked. I answered a similar question earlier in this thread:
>>7581024
>>7581030

[Anonymous No.7588577]

>>7587771
>>7587786

You can generally multiply something by 1 without changing it.

[Anonymous No.7588729]

How do you calculate the time it takes for a capacitor to reach full capacity given a certain voltage?

[Anonymous No.7588832]

Are graduate level classes, in general, a lot more lax?

I've only taken two, but the professors for both have been really cool. If you didn't get the homework done, they don't care, and just let you turn it in the next class. My current professor just said because we've all been doing homework consistently, and doing well on these weekly quizzes he gives, he doesn't see the point in a mid-term that's just going to stress us out.

Just my experiences with graduate level classes as an undergrad have been great.

[Anonymous No.7588939]

Help me out please

The shape of a hill is described by the height function
{math} h(x,y)=\frac{1}{({2+x^2+y^4})^{1/2}}{math}
Let {math}r_0=(1,1){math}
Find a vector of a path through {math}r_0{math} where the height remains the same

I know that {math}x^2+y^4=2{math} satisfies the solution but how do I find a vector of this?

[Anonymous No.7588946]

Help me out please

The shape of a hill is described by the height function
<span class="math"> h(x,y)=\frac{1}{({2+x^2+y^4})^{1/2}}<span class="math">
Let r_0=(1,1)
Find a vector of a path through r_0 where the height remains the same

I know that x^2+y^4=2 satisfies the solution but how do I find a vector of this?[/spoiler][/spoiler]

[Anonymous No.7588955]

Help me out please

The shape of a hill is described by the height function
{math} h(x,y)=\frac{1}{({2+x^2+y^4})^{1/2}}{\math}
Let
{math}r_0=(1,1){\math}
Find a vector of a path through {math}r_0{\math} where the height remains the same

I know that {math}x^2+y^4=2{\math} satisfies the solution but how do I find a vector of this?

[Anonymous No.7588970]

>>7588955
Tried 4 times to format it now, this'll have to do

[Anonymous No.7589187]

>>7588955
I'm not sure what "vector of a path" means.... you mean the tangent vector at r_0? Then just take the gradient of h at r_0 and rotate it 90 degrees.

[Anonymous No.7589196]

>>7588376
write (r,phi) = (i,j) * R where R is a rotation matrix. Then (i,j) = (r,phi)*R^t, where R^t is the transpose of R.

[Anonymous No.7589204]

>>7588049
you dropped your i's , but well done
.

[Anonymous No.7589277]

Imagine you have a rpg 1 where yo roll the speed stat of an human you roll a 6 sided dice 3 times and get the sum
Now imagine a rpg 2 where to get the speed stat of a human you roll a 100 sided dice once.


Is there some formula to convert the stat of a char in a rpg to the other rpg?

[Anonymous No.7589294]

>>7578081
>>7578078
I don't want to spoil this, but this is a really cute exercise and will help anyone trying to understand induction.

[Anonymous No.7589414]

I'm on third year of Med school, but I have to start thinking about my future, I don't want to become a specialist right after school and start babysitting people, I would like to study a little more and do Shit science related, what are my choices? Sorry for engrish

[Anonymous No.7589450]

What does L wza describe?

[Anonymous No.7589606]

I'm solving a ODE with runge-kutta method. My function has two inputs, phi which is a solution from another ODE, and u which the method should solve for. basically f(phi,u)=sin(phi)+u^2
With the runge-kutta method where k2 and k3 have the input changed to f(y+k1/2*h), do I add k1*h/2 to both phi and u or just the variable which corresponds to the solution?

[Anonymous No.7589751]

>>7588955
>>7588970

OK, anon. You're right that

<span class="math">h(x,y) = \frac{1}{\sqrt{2+x^2+y^4}}\ \ \ \ \ \vec{r_0}=(1,1)[/spoiler]

means that the level curve of h through the given point is described by the equation

<span class="math">x^2 + y^4 = 2.\ \ \ \ \ \ (*)[/spoiler]

On that entire curve, for every point (x,y) satisfying that equation, h has the fixed value

<span class="math">h(x,y)=\frac{1}{\sqrt{2+2}} = \frac{1}{2}[/spoiler]

By "a vector of a path" I think they're just asking for a parametrization of the curve:

<span class="math">\vec{r}(t) = \Big(x(t), y(t), z(t)\Big)[/spoiler]

There are an infinite number of choices for a valid parametrization. An unimaginative one would be something like

<span class="math">x(t) = t[/spoiler]
<span class="math">y(t) = \pm (2 - t^2)^{\frac{1}{4}}[/spoiler]
<span class="math">z(t) = h\Big(x(t),y(t)\Big) = \frac{1}{2}[/spoiler]

This is OK, but it has problems. One problem is that it suddenly ends at <span class="math">t = \pm\sqrt{2}[/spoiler]. Another problem is that there are TWO branches of the curve, due to the <span class="math">\pm[/spoiler] in y(t). So this is really TWO vector paths. You could drop the negative branch, since only the positive one goes through (1,1), but then you're throwing away half of the level curve.

(post continued due to length limit)

[Anonymous No.7589764]

>>7589751
>>7588955
>>7588970

One way to get a single vector path/branch that describes the whole level curve in an infinite loop would be to try polar-like coordinates:

<span class="math"> x = r\,\cos(t+\delta),\ \ \ y = \sqrt{r}\,\sin(t+\delta)[/spoiler]

(WARNING: these are not really polar coords!)

Substituting these into your equation (*) above, and solving for r(t) gives
<div class="math">r(t) = \sqrt{\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}}</div>
So we obtain the path components
<div class="math"> x(t) = \sqrt{\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}}\,\cos(t+\delta)</div>
<div class="math"> y(t) = \Big(\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}\Big)^{\frac{1}{4}}\,\sin (t+\delta)</div>
<div class="math"> z(t) = \frac{1}{2}</div>

(see pic related)

I chose to include the <span class="math">\delta[/spoiler] so that the path passes through <span class="math">\vec{r_0} = (1,1,1/2)[/spoiler] when t=0. (You could drop it for simplicity if you want, thereby shifting the time values.)

Personally, I prefer this vector path, but the cost is in the complication of
<div class="math">\delta = \arccos \Big( \frac{\sqrt{5} - 1}{2} \Big) = \arcsin \Big( \sqrt{\frac{\sqrt{5} - 1}{2}} \Big)</div>
I'll leave it for you to show that when

<span class="math"> t=0:\ \ \ \ x(0)=1,\ \ y(0)=1 [/spoiler]

I double-checked that it does, but the calculation is a bit ugly.

[Anonymous No.7589777]

>>7589751
>>7588955
>>7588970

One way to get a single vector path/branch that describes the whole level curve in an infinite loop would be to try polar-like coordinates:

<span class="math"> x = r\,\cos(t+\delta),\ \ \ y = \sqrt{r}\,\sin(t+\delta)[/spoiler]

(WARNING: these are not really polar coords!)

Substituting these into your equation (*) above, and solving for r(t) gives
<div class="math">r(t) = \sqrt{\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}}</div>
So we obtain the path components
<div class="math"> x(t) = \sqrt{\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}}\,\cos(t+\delta)</div>
<div class="math"> y(t) = \Big(\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}\Big)^{\frac{1}{4}} \, \sin(t+\delta)</div>
<div class="math"> z(t) = \frac{1}{2}</div>

(see pic related)

I chose to include the <span class="math">\delta[/spoiler] so that the path passes through <span class="math">\vec{r_0} = (1,1,1/2)[/spoiler] when t=0. (You could drop it for simplicity if you want, thereby shifting the time values.)

Personally, I prefer this vector path, but the cost is in the complication of
<div class="math">\delta = \arccos \Big( \frac{\sqrt{5} - 1}{2} \Big) = \arcsin \Big( \sqrt{\frac{\sqrt{5} - 1}{2}} \Big)</div>
I'll leave it for you to show that when

<span class="math"> t=0:\ \ \ \ x(0)=1,\ \ y(0)=1 [/spoiler]

I double-checked that it does, but the calculation is a bit ugly.

[Anonymous No.7589782]

>>7588970
the gradient of h is

(-x/(x^2+y^4+2)^(3/2), -(2 y^3)/(x^2+y^4+2)^(3/2))

at (x,y)=(1,1) the gradient is

(-1/8, -1/4). This is orthogonal to the level set at r. Thus (1/4,-1/8) is tangent to the level curve at r. A path along the level curve will have a tangent vector at r of the form k* (1/4,-1/8) for some k.

[Anonymous No.7589797]

>>7588729

You mean like in an DC RC circuit, anon? Mathematically (according to the usual model) it takes an infinite time. Physically there is a time when the last electron enters the capacitor, but not really because thermal noise will make the charge in the capacitor constantly fluctuate anyway.

That's why we don't talk about the capacitor reaching "full" capacity, but only the <span class="math">\tau = RC[/spoiler] time constant when it reaches a fraction 1/e of "full" charge.

See https://en.wikipedia.org/wiki/RC_time_constant

[Anonymous No.7590134]

>>7588214
Anyone?

[Anonymous No.7590204]

>>7589797
I meant when the current through the capacitor essentially reaches zero

[Anonymous No.7590224]

I'm trying to draw configurations using the Cahn-ingold-prelog rules. There's a chirality center on the carbon under the OH, but I was stuck on something.

There's going to be a hydrogen going into the page, but how do you know whether to place the wedge on the right or left of the hydrogen? The book doesn't mention this, and just seems to put the wedge on the left usually, but sometimes on the right.

[Anonymous No.7590278]

>>7588729
C= q/V

CV = q

differentiating both with respect to time

C(dV/dt) = dq/dt = I

using kerchoffs current law and modeling as a simple RC circuit

C(dV/dt) + V/R = 0

dV/V = -(dt/RC)

integrating both sides, and exponentiating gives

v(t) = ke^(-t/RC)

with k = V_source when you solve for t = 0

[Anonymous No.7590316]

>>7589277

OK, anon, the main issue is that when rolling three 6-sided dice, the summed values are not all equally likey. The possible values for the sum run from 3 to 18, but the middle values are much more likely to appear than the extremes. For example, there is only one way to get a sum of 3 (1+1+1), but there are 27 ways to get a sum of 10 (eg, 1+3+6, 4+2+4, 5+1+4, ...). The total number of outcomes for rolling 3 x 6-die is 6^3=216, which are distributed unevenly into the 16 possible values of the sum. When rolling a 100-die once, this issue doesn't arise; all values are equally likely. So you will have to make bins with groups of the 100-die outcomes to get approximately the same probabilities for the 3 x 6-dice rolls.

I'll put a table in the following post.

[Anonymous No.7590325]

>>7590316
>>7589277

3x6-roll=3 #ways= 1/216 0.463% 0.463% 100-roll=1 #ways= 1/100 1% 1%
3x6-roll=4 #ways= 3/216 1.389% 1.852% 100-roll=2 #ways= 1/100 1% 2%
3x6-roll=5 #ways= 6/216 2.778% 4.630% 100-roll=3-5 #ways= 3/100 3% 5%
3x6-roll=6 #ways= 10/216 4.630% 9.259% 100-roll=6-9 #ways= 4/100 4% 9%
3x6-roll=7 #ways= 15/216 6.944% 16.204% 100-roll=10-16 #ways= 7/100 7% 16%
3x6-roll=8 #ways= 21/216 9.722% 25.926% 100-roll=17-26 #ways= 10/100 10% 26%
3x6-roll=9 #ways= 25/216 11.574% 37.500% 100-roll=27-37 #ways= 11/100 11% 37%
3x6-roll=10 #ways= 27/216 12.500% 50.000% 100-roll=38-50 #ways= 13/100 13% 50%
3x6-roll=11 #ways= 27/216 12.500% 62.500% 100-roll=51-63 #ways= 13/100 13% 63%
3x6-roll=12 #ways= 25/216 11.574% 74.074% 100-roll=64-74 #ways= 11/100 11% 74%
3x6-roll=13 #ways= 21/216 9.722% 83.796% 100-roll=75-84 #ways= 10/100 10% 84&
3x6-roll=14 #ways= 15/216 6.944% 90.741% 100-roll=85-91 #ways= 7/100 7% 91%
3x6-roll=15 #ways= 10/216 4.630% 95.370% 100-roll=92-95 #ways= 4/100 4% 95%
3x6-roll=16 #ways= 6/216 2.778% 98.148% 100-roll=96-98 #ways= 3/100 3% 98%
3x6-roll=17 #ways= 3/216 1.389% 99.537% 100-roll=99 #ways= 1/100 1% 99%
3x6-roll=18 #ways= 1/216 0.463% 100.000% 100-roll=100 #ways= 1/100 1% 100%

Columns 3 and 4 are the individual and running probabilities for the 3x6-dice rolls.
Columns 7 and 8 are the same for the 1x100-die roll.

Some may quibble with my choice of bins, but I like these partly for stylistic reasons.

I pastebin-ed the table in case /sci/ messes up the formatting: http://pastebin.com/9NdSNy7Q

[Anonymous No.7590737]

>>7590325
Just give me fifty G's on eight the hard way fam

[Anonymous No.7591045]

New thread is here

>>7588958

[Anonymous No.7591097]

Am I doing it right?
Any rigorosen details I forgot?
I am going to work and buy myself a belt, sonce the other one ripped apart.

[Anonymous No.7591227]

>>7591097

<span class="math">\frac{x^2 + 7x + 10}{x+5} = \frac{(x+2)(x+5)}{x+5} = x+2 \rightarrow -3[/spoiler]

Good luck belt shopping anon.

[Anonymous No.7591700]

>>7589751
>>7589777
>>7589782
Thanks very much mate, a real help