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10554321 No.10554321 [Reply] [Original] [archived.moe]

Stupid questions thread.
Previously >>10527130
>>10554314 edition.

>> No.10554331

>>10554321
Thanks I feel anxious when there's no /sqt/

>> No.10554335

If I printed enough counterfeit money could I hypothetically devalue the supply enough to basically wipe away my countrie's debt?

>> No.10554337

>>10554335
no because the debt in measured in other countries currencies.
Double inflation = double debt

>> No.10554367
File: 77 KB, 911x662, FY2018-preliminary-to-whom-does-the-US-government-owe-money.png [View same] [iqdb] [saucenao] [google] [report]
10554367

>>10554337
Depends on the country and the debt - sometimes countries will issue debt that is to be repayed in their own currency, sometimes (especially if they need the debt to buy foreign goods) they will issue debt that is to be payed back in a foreign currency. The US debt is all owed in US dollars, for example, while Argentina has a bunch of bonds that have to be payed back in USD, since foreign lenders didn't trust them not to inflate their local currency.

So in theory, with enough counterfeiting you could inflate away a countries debt, or counterfeit the foreign currency you need to pay the money back in. However, actually faking enough money to be noticeable at the scale of even a small nation would be extremely hard to do without being caught.

>> No.10554405

Why is Tenshi so hot?

>> No.10554517

If you were to iterate a powerset operation on a finite set to it's "limit" at infinity, would you end up with a set of cardinality equal to the natural numbers, or the highest ordinal?

>> No.10554571

>>10554517
Which one is the highest ordinal?

>> No.10554578

>>10554517
Pick up a category theory text lad.

>> No.10554581

>>10554571
Sorry, what I mean is aleph-omega. As in, would the iterative application yield aleph-naught, or would it "reach" aleph-naught and then continue to increase in cardinality?

>> No.10554597

>>10554578
Do you have any recommendations, or would really any general text help for this type of problem? The wiki doesn't seem to have anything on the topic.

>> No.10554605

>>10554597
Handbook of categorical algebra.
You won't use much, just until the definition of categorical limits and colimits.

>> No.10554612

>>10554605
Thanks fren.

>> No.10554729

Is this the place to ask for help with my Biomechanics homeworkM

>> No.10554849

What is the good way to prove statements involving factorials. For example, I have to prove
[eqn]\binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}[/eqn]
but I don't know how.
Pls help I'm new to this stuff.

>> No.10554870

>>10554729
yes

>> No.10554892

>>10554849
There are n red balls and 1 blue ball. The number of ways to choose k of them is n+1 choose k.

The number of ways to choose k of them if black must be chosen is n choose k-1.

The number of ways to choose k of them if black is not chosen is n choose k.

Either black is chosen or not, so n+1 choose k equals n choose k-1 plus n choose k.

>> No.10554899

>>10554849
for these types of statements you're usually required to give either a technical "algebraic" proof or a more verbose "combinatorial" proof. because i dont fucks with the algebra, i'll just say that you have to come up with a story for both sides of the equation are counting and show that they describe the same amount.
in your example, the left hand side is counting the number of ways to choose k students in a class of n+1 people. on the right hand side, you're getting a similar story, take student #n+1: either he was chosen, in which case you need to pick k-1 more students from the remaining n, or he wasn't and you have to choose k out of the remaining n. go it mate?

>> No.10554936

>>10554849
Induction

>> No.10554943
File: 20 KB, 460x100, Annotation 2019-04-13 190905.png [View same] [iqdb] [saucenao] [google] [report]
10554943

can I learn all this and get ready for the final exam in 14 days?

>> No.10554960

>>10554892
>>10554899
Well I don't really understand meaning of any of that.
I'm not doing combinatorics, I'm learning calculus and none of that combinatorics stuff is even mentioned. Actually, this is an exercise after chapter on mathematical induction, where factorial function is an example of inductively defined function. Again, none of that stuff you guys are talking about is even mentioned so I'm probably expected to give algebraic proof (since you don't need to know anything about combinatorics to do that).
But it just seems so hard to do it algebraically. If you can, please give me a hint on how to do this albebraically. If I can't figure it out in either case I'll just read the solution in the book itself.
>>10554936
Still not sure how to do this inductively either, but anyway it is explicitly stated that no induction is required for that proof.

>> No.10554963

>>10554943
Stop shitposting and study.

>> No.10554964

>>10554943
How about you start studying and find out?

>> No.10554971

>>10554960
Option one: write out the extensive expression for both sides, and apply algebraic manipulation until you get 1=1. Then pay attention to what exactly it is that you did and apply it to the original formula to get the equality you want.
Option two: attempt induction on n, k and n+k.

>> No.10554974

>>10554960
Write out the definition of the right-hand side using factorials. Find a nice common denominator to add them, and see what happens

>> No.10554980
File: 13 KB, 657x527, R14kkDj.png [View same] [iqdb] [saucenao] [google] [report]
10554980

>>10554963
>>10554964
you're right, thank you friends

>> No.10555032 [DELETED] 

Can anyone give hints on this? topology
[math]Let $Y$ be connected, and assume a homeomorphism $B : Y \rightarrow Y$ exists such that $B \circ B(y) = y$. for all $y \in Y$ Prove that, for every continuous function $g :Y \rightarrow \mathbb{R}$ there exists $y \in Y$ such that $g(y) = g(B(y))$[/math].

This chapter had to do with generalization of intermediate value theorem in topology and I'm stuck on how that relates to this.

>> No.10555055

>>10554974
Hmm I did as you have said but I got
[eqn]\frac{n!(k(n-k)+1)}{k!(n-k)!}[/eqn]
I can't continue... I wonder if I made a mistake or I simply don't see what to do next. If you want I'll write out all the steps I applied.
>>10554971
>Option two: attempt induction on n, k and n+k.
How would I actually do that? Until now I used induction on expressions with only one variable i.e. [math]1+\dots+n=\frac{n(n+1)}{2}[/math].

>> No.10555061

>>10555055
I don't know why that didn't format correctly.

>> No.10555066

>>10555055
>i.e.
It always happens...

>> No.10555067

>>10555055
You can still use induction on one variable while treating the rest as parameters.

So here you can use induction over n and treat k as just a number (or 0,<=k<=n but it will work either way)

>> No.10555075

>>10555055
No, I meant attempt on n, and if it doesn't work try k, and if that doesn't work either try n+k.

>> No.10555133

>>10555067
>>10555075
Ok, I'll try induction.
But also please someone point out where is a mistake (I guess there is one) in my derivation.
[eqn]\binom{n}{k-1}+\binom{n}{k}=\frac{n!}{(k-1)!(n-k-1)!}+\frac{n!}{k!(n-k)!}[/eqn]
[eqn]\frac{n!k!(n-k)!}{(k-1)!(n-k-1)!k!(n-k)!}+\frac{n!(k-1)!(n-k-1)!}{k!(n-k)!(k-1)!(n-k-1)!}[/eqn]
[eqn]\frac{n!k!(n-k)!+n!(k-1)!(n-k-1)!}{k!(n-k)!(k-1)!(n-k-1)!}[/eqn]
[eqn]\frac{n!(k!(n-k)!+(k-1)!(n-k-1)!)}{k!(n-k)!(k-1)!(n-k-1)!}[/eqn]
[eqn]\frac{n!(k(n-k)!+(n-k-1)!)(k-1)!}{k!(n-k)!(k-1)!(n-k-1)!}[/eqn]
[eqn]\frac{n!(k(n-k)!+(n-k-1)!)}{k!(n-k)!(n-k-1)!}[/eqn]
[eqn]\frac{n!(k(n-k)+1)(n-k-1)!}{k!(n-k)!(n-k-1)!}[/eqn]
[eqn]\frac{n!(k(n-k)+1)}{k!(n-k)!}[/eqn]
There you have it.

>> No.10555139

>>10554870
Sweet.

On a gravity of 1.62m/s^2, I'm throwing a ball with the v0 of 25m/s, at an angle of 42degrees from the X axis.

-How do I calculate what's the highest it's gonna get?
-How do I calculate how long will it take it to fall back to the ground?

>> No.10555148

How would punching someone on the Moon or another low but non-zero gravity environment be different than on Earth?

>> No.10555169 [DELETED] 

>>10554849
Just put in the definition and put everything on the same denominator. It's a two minute exercise.

>> No.10555173

>>10555148
Do you mean in terms of damage inflicted, or are you wondering if you can punch somebody into orbit?
In terms of the former, there is no difference, because the energy you transfer via the mass of your arm will be the same, provided your stance allows for a stable platform. Concerning the latter, if you vector the punch correctly and the target is sufficiently low mass, and the planet is like the little prince's planet, it is possible.

>> No.10555186

>>10555173
Would you be able to plant your feet well enough to throw a punch? Lower normal force means lower friction, so I was guessing you'd probably slip or at least not be able to throw your weight behind the punch as easily.

>> No.10555195

>>10554849
[math] (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k [/math]

so

[math] \sum_{k=0}^{n+1} \binom{n+1}{k} x^k = (1+x)^{n+1} = (x+1) \cdot (1+x)^n = \sum_{k=0}^n \binom{n}{k} x\cdot x^k + \sum_{k=0}^n \binom{n}{k} x^k = \sum_{k=0}^n \left( \binom{n}{k-1} + \binom{n}{k} \right) x^k + x^{n+1} [/math]

>> No.10555197

[math]\sum_{k=0}^{n}(-1)^k \binom{n}{k} (n^2-kn)![/math]
How do I simplify this?

>> No.10555211

>>10555133
[math]\binom{n}{k-1} = \frac{n!}{(k-1)!(n-k+1)!}[/math]

>> No.10555217

>>10555197
Unless that's already the expansion of a known function, I don't think you'll have much luck reducing it.

But it leads me to another questions - are there some general results on how

[math] \dfrac { (n^a)! } { (n!)^a }, [/math]

say for a=2.

behaves. From experimenting with Mathematica, it appears that even the log of that still grows at least polynomialy.

>> No.10555223

>>10554321
If a 70kg body loses around ~1°c of temperature every hour after it dies and 1 calorie is needed to raise 1 gram of water 1°c then I figure:
70,000 *24h: 1680kc a day
Am I accounting for everything here? Is this right?
Do we really use most of our energy just keeping us warm?

>> No.10555229

>>10555197
My gut instinct says stuff cancels out. Try opening it up.

>> No.10555242

>>10555186
Well I wasn't exactly picturing a paved planetoid. If there is gravel, dirt or any terrain you can push against, that would be enough.

>> No.10555300

>>10555211
Yeah... fuck.

>> No.10555345

>>10555139
Okay, I managed to do the first one (28.5m), but still struggling with the second.

What do?

>> No.10555350

>>10555223
the energy isnt being used to keep us warm. The energy being used is what produces the warmth. there is a very distinct difference.

>> No.10555351

>>10554321
Sort of stumbled across an interesting idea. We have the placebo effect, cognitive bias and eye witness testimony all pointed towards a single potential effect. Belief has a profound effect on the way we interpret information. Could this be an in-built system rooted within the brain? Could hallucinations and other things entirely of the imagination also be effected to a large degree? Is this already covered in psychology to a the degree it should be? It's all very interesting and I want to know your opinion.

>> No.10555352

why do schools / unis arrange curriculum so that 3+ subjects or so are taken concurrently? are there enough cognitive benefits to merit this being the de facto standard?

for example, see:
>https://csd.cs.cmu.edu/sample-undergraduate-course-sequence

the a converse would be arranging curriculum as a series of individual classes, like:
>intro programming in haskell -> applications in haskell -> calculus I - III w/ applications in haskell -> linear algebra w/ applications haskell

each course of course would likely be about 4x shorter, since it's the only course you';d be taking at that time. could someone sight the cognitive benefits of the standard approach, where all the abvoe classes would basically be taken at once instead?

>> No.10555356

>>10555351
my opinion is that you shouldn't have a trip, especially so if you're an idiot with nothing of substance to say

>> No.10555358

>>10555351
My pet hypothesis is that prior beliefs/biases activate some kind of active gain mechanism in the brain to provide positive feedback to stimuli that are in-line with the prior.

t. EE with no background in cognitive psychology, so this is very unlikely to be a whole truth.

>> No.10555361

>>10555345
If y(t) is the altitude, solve y(t)=0

>> No.10555365

>>10555350
No way, you are saying it's waste heat? From what?
I thought for sure we only kept warm to keep our muscles ready at any time and situation unlike reptiles

>> No.10555366

>>10555356
I'm asking at least one important question we don't have an answer to. Do beliefs effect hallucinations? The rest is less important and well documented.

>> No.10555378

>>10555366
>Do beliefs effect hallucinations?
fucking DUH, retard. you think everyone envisioning jesus is pure chance? kill urself

>> No.10555386

>>10555378
>Being this butthurt over some letters that appear on your screen

It's like telling your enemies exactly what you don't like. You fucking sensitive little faggot.

>> No.10555411 [DELETED] 
File: 6 KB, 205x246, brainlet.png [View same] [iqdb] [saucenao] [google] [report]
10555411

How do I solve cos(10°)/sin(80°) without a calculator? According to my book the answer is 1 but I don't even know how to get there.

>> No.10555421
File: 517 KB, 650x390, Circle_cos_sin.gif [View same] [iqdb] [saucenao] [google] [report]
10555421

>>10555411
not very rigorous but:
since the cosine measures the X axis and the sine the Y axis you can tell that they are the same here since the sine is 10° away from being at full height and the cosine is 10° away from being fully to the right.

>> No.10555422

>>10555411
cos(10) = cos(90 - 10) = use the angle addition property

>> No.10555425

>>10555411
cos(10°)=cos(90°-80°)=cos(90°)cos(80°)+sin(90°)sin(80°)=sin(80°)

>> No.10555436 [DELETED] 

>>10555421
Not quite what I was looking for, but pretty interesting all the same.

>>10555422
>>10555425
That's what I was thinking initially, but I don't know how to find cos(80°) without a calculator. If it were something like cos(15°) then I could just use the unit circle for cos(45°- 30°).

>> No.10555441

>>10555436
cos(90°)=0 so it doesn't matter, that leaves you with cos(10°)=sin(90°)sin(80°) hence the result.

>> No.10555445

>>10555386
>calling me sensitive while clearly getting butthurt
lmao, you absolute idiot

>> No.10555449 [DELETED] 

>>10555441
Oh, okay. I realize I'm going full brainlet here, but then how do you know what sin(80°) is?

>> No.10555454

>>10555449
We don't, cos 10=1 sin 80 is just the equality you wanted to prove.

>> No.10555458 [DELETED] 

>>10555454
Oh jesus christ now I see, thanks anon.

>> No.10555470
File: 196 KB, 1478x796, file.png [View same] [iqdb] [saucenao] [google] [report]
10555470

does the author here just switch from the L^2 norm to the square of the L^2 norm without telling us? he did something similar earlier in the section. not sure if i'm just failing to understand the simplification properly

>> No.10555476
File: 45 KB, 1524x238, file.png [View same] [iqdb] [saucenao] [google] [report]
10555476

>>10555470
sorry, here's the referenced 2.67

>> No.10555477

>>10555470
You're referring to the sub and superscript of 2 in eq. 2.69 right?

>> No.10555480

>>10555477
yes. the subscript is denoting L^2 norm. the sqrt in the upper equation is expressing the Frobenius norm
d is a 1 dimensional matrix (big D isnt)

>> No.10555742

I need to find the degree of polarization of light that is reflected off a surface given the indices of refraction and incident angle. I've got the Fresnel coefficients and read into the Stokes vector but I'm not sure how to get from one to the other for an unpolarized source.

>> No.10555785

>>10555742
Pretty sure you can just assume the incident unpolarized wave is an equal proportion of p and s polarisation and then use the Fresnel coefficients to determine the resulting proportion

>> No.10555830
File: 1.18 MB, 2452x3000, 1533497190923.jpg [View same] [iqdb] [saucenao] [google] [report]
10555830

Please recommend books about the history of science. Wasn't there one by a nobel laureate?

>> No.10555839

>>10555785
What proportion exactly?

>> No.10556035

is inertial mass the same as gravitational mass?
specifically, does an object with high relative momentum "pull" more strongly gravitationally on an object in relative rest?

>> No.10556102

How do audio spectrograms work?
fourier transforms don't work instantaneously

>> No.10556170

>>10556102
not if you're using the fastest Fourier transform in the west

>> No.10556213

>>10555830
Modern Physics by Tipler and Llewellyn

>> No.10556237

>>10555470
Minimizing the L2 norm is the same as minimizing the square of the L2 norm. Usually, dealing with squares is simpler because they're easier to differentiate.

>> No.10556346

how were the summation properties/formulas discovered, just trial and error? or is there some method

like [math] \sum_1^n = n(n+1)/2 [/math]

>> No.10556351

>>10556346
the formula is guessed and then proved using induction

>> No.10556353
File: 128 KB, 548x575, are_you_fucking_serious (2).jpg [View same] [iqdb] [saucenao] [google] [report]
10556353

>>10556102
>something that plots a function of time
>Fourier transform

>> No.10556420

Can someone post that one meme of a trench that bases the depth of a math branch by difficulty/complexity?

>> No.10556446

>>10556420
God I hate that meme. Just google mathematics trench

>> No.10556468

>>10556446
kek, can't believe that works

>> No.10556524
File: 44 KB, 800x450, brainlettttt.jpg [View same] [iqdb] [saucenao] [google] [report]
10556524

If I know the current in a loop, and there's only 1 battery in the loop, how do I know how much of that current comes from the battery, and how much of it comes from the other loop(s)?

>> No.10556526

What transport equations did the scientists at the Manhattan project use to determine the critical mass of the Fat Man core?

>> No.10556527

Does a light path get manipulated by gravity itself (the influence of a mass) or by the curvature of space-time caused by gravity?

>> No.10556691

I have a graph where I plot 3 lines with some new data for two of the lines each day. The remaining line will be the absolute sum of the two other lines. Does anyone know what area of math I could look at if I wanted to find the optimal combinations of data for the highest absolute sum at, for example, 30 days from now?

>> No.10556860

>>10556346
> like [math] \sum_1^n = n(n+1)/2 [/math]
That one is quite trivial:
S(n)=1+2+...+(n-1)+n
S(n)+S(n) =
1+2+...+(n-1)+n +
n+(n-1)+...+2+1
= (n+1)+(n+1)+...+(n+1)+(n+1)
= n(n+1)
=> S(n)=n(n+1)/2
This can be extended to polynomials of any degree. If f(n) is a degree n polynomial, the difference f(n+1)-f(n) is a degree n-1 polynomial. So for the sum of terms given by a degree n polynomial, calculate the difference for a generic degree n+1 polynomial, equate coefficients, and solve the resulting system of linear equations.

A common approach is to find some transformation such that "shifts" the series, mapping each transformed term to the preceding or successive term. E.g. for the geometric series
S(n)=a+ar+ar^2+...ar^(n-1)
rS(n)=ar+ar^2+ar^3+...ar^n
rS(n)=S(n)-a+ar^n
=> (1-r)S(n)=(1-r^n)a
=> S(n)=a(1-r^n)/(1-r)

Also: difference equations are usually strongly analogous to differential equations. So if you can solve the differential equation, the solution of the difference equation usually has a similar form. E.g. dy/dx=k^x => y=A.k^x+B.

Telescoping is a common technique. If you can write f(n) as g(n)-(g-1) for some g(), then the sum is (g(1)-g(0))+(g(2)-g(1))+(g(3)-g(2))+...+(g(n)-(n-1)) = g(n)-g(0) as all of the intermediate terms cancel out (each positive matches an identical negative in the successive term, except for the first and last terms).

>> No.10557093

>>10556102
Short-time Fourier transform (STFT).
Only a small "window" of the signal is transformed (fraction of a second), and the window moves over the signal (usually by a distance that is shorter than the window length, so the windows overlap).
The frequency spectrum for that small snippet is manipulated and transformed back - the signal gets synthesized by adding the snippets back together continuously (overlap-add-method).

>> No.10557100

why is this?

[math]\|A\|_F = \sqrt{ \Tr{AA^T} }[/math]

>> No.10557104

>>10557100
that should be the trace operator... guess we don't support that

>> No.10557128

>>10556446
>he works on a field near the top
Heh, nothing personal kid.

>> No.10557153

>>10556526
they have a grad student stand in the desert and clap two plutonium half-spheres of successively larger sizes together til something interesting happened

>> No.10557164

>>10557104
>>10557100
oh nm i get it now

>> No.10557300

>>10556353
if you don't know what i'm talking about, take small brain somewhere else

>> No.10557318
File: 5 KB, 204x247, download.jpg [View same] [iqdb] [saucenao] [google] [report]
10557318

So I'm kinda lost.
Covering magnetic fields, and one of the first formulas I get is for flux.
[math]flux = \int_c BdS,[/math]
(B is magnetic induction, S is the area of the contour, might be translating these somewhat wrong)
Okay, seems reasonable enough, but next chapter it says the law of conservation is:
[math]\int_c BdS = 0, [/math]
Does this mean that the flux is always equal to zero? I mean, I know it doesn't, but what am I missing here?

>> No.10557331

>>10557318
>hey guys, the book is saying that the amount of energy in the system is constant, but also that objects can have different amounts of energy through time wtf?

>> No.10557343

>>10557331
Oh, the second fomula just implies that it's constant?
How come it isn't just::
[math]flux = \int_c BdS = 0[/math]
My college taught us literally 0 things about integrals before shoving them onto us.

>> No.10557441

How do I show the identity
[math]
exp[\hat{A}+\hat{B}] = exp[\hat{A}] + \int_0^1 dz exp[z(\hat{A}+\hat{B})] \hat{B} exp[-z\hat{A}] exp[\hat{A}]
[/math]
?

>>10557318
I think the key is what you are integrating over
Flux is stuff moving through an area (dS)
>One way to better understand the concept of flux in electromagnetism is by comparing it to a butterfly net. The amount of air moving through the net at any given instant in time is the flux. If the wind speed is high, then the flux through the net is large. If the net is made bigger, then the flux is larger even though the wind speed is the same. For the most air to move through the net, the opening of the net must be facing the direction the wind is blowing. If the net is parallel to the wind, then no wind will be moving through the net. The simplest way to think of flux is "how much air goes through the net", where the air is a velocity field and the net is the boundary of an imaginary surface.

I don't get what you mean with the law of conservation sentences

>>10557343
Well look at it this way
We know that energy is conserved always
Now consider you take a big rock in your arms and walk up a stair and place down the rock and walk down again
You obviously expended energy, we could write down integrals for that, dW = F.ds
What matters is that your energy is lowered
But yet, the total energy is conserved! How does that work?!
It works because we look only at the system that is you, in which energy is not conserved
But in the system that is the universe, energy is conserved
We don't even have to go that big, we can just look at the you-rock system
You lost energy but the rock gained energy, your energy
Assuming perfect everything the rock gained all of the energy that you expended
So energy is not conserved in the you system, nor in the rock system, but it is in the you-rock system
I don't know what you mean with your flux = 0 but I hope this makes it a bit clearer

>> No.10557452
File: 1.23 MB, 495x341, green lantern is fucking stupid.gif [View same] [iqdb] [saucenao] [google] [report]
10557452

Did someone ever figure out the nerve circuitry responsible for feeling pleasure during orgasm? When will be finally able to simulate and stimulate without burning the skin on our dicks?

>> No.10557480

>>10557452
Yes, it's called dopamine.
We also have substances that give us that, they're called drugs.

>> No.10557489

>>10557318
>>10557343
For an open surface:
[math]\Phi_B = \iint_S {\mathbf B} \cdot d{\mathbf S}[/math]
For a closed surface:
[math]\oint_S {\mathbf B} \cdot d{\mathbf S} = 0[/math]
or (can't remember if oiint works here)
[math]\oiint_S {\mathbf B} \cdot d{\mathbf S} = 0[/math]
IOW, for a closed surface, the net flux is zero; any flux which enters the region bounded by the surface must leave it. There's no such thing as a magnetic monopole; flux goes around in loops with no start or end points.

>> No.10557490

>>10557480
I was thinking more of electrodes in your shaft.

>> No.10557493

>>10557490
Literally no reason to go put shit near your shaft when you can pop a pill and get the same result, no?

>> No.10557591

>>10557343
No, the second formula implies that you should read the actual text and see what it is you're integrating over instead of just noting down the formulas.

>> No.10557641

>>10557591
Yeah, I understood the concepts, it was just the formulas messing with me.
Mostly it was just me noticing the difference between [math]\oint_s[/math] and [math]\int_s[/math].

>>10557489
>>10557441
I think I get it now, thanks!

>> No.10558016

>>10557480
Or you could go next level and take a bunch and drugs *and then* fuck for exponential returns.

>> No.10558082 [DELETED] 

I'm stumped lads, how do I get this into rectangular or polar form for the purpose of graphing? I typed it into cymath and it just told me to fuck off. I'm fully aware I'm retarded btw.

r = 1 + cos θ

>> No.10558090

>>10558082
[math]( cos \theta + cos^2 \theta , ~ sin \theta + sin \theta cos \theta)[/math]

>> No.10558268

How the fuck do I calculate charge-to-mass ratio given initial velocity, displacement, time, and E? I don't think it's possible to use kinematics to get mass, so that makes me think I need to use E (units are N/C), but I can't figure out how. Any help would be appreciated.

>> No.10558531

>>10558268
Calculate acceleration from displacement, velocity and time (a=v^2/2s). F=ma, F=qE => ma=qE => q/m = a/E.

>> No.10558719
File: 13 KB, 692x92, file.png [View same] [iqdb] [saucenao] [google] [report]
10558719

how do i take the gradient of:

[math]
f(\boldsymbol x) = \frac{1}{2} \| \boldsymbol A \boldsymbol x - \boldsymbol b \|_2^2
[/math]

?

i know it's supposed to look like pic related but i'm unclear on how to get to it. i know that we can rewrite as

[math]
f(\boldsymbol x) = \frac{1}{2} (\boldsymbol A \boldsymbol x - \boldsymbol b)^T \cdot (\boldsymbol A \boldsymbol x - \boldsymbol b)
[/math]

not sure how to get any closer though... when i work the multiplication out it doesn't seem like i'm on the right track.

>> No.10558732 [DELETED] 

>>10558719
the \cdot shouldn't be there; omitting it, it's equivalent to dot product

>> No.10558737

Idiot here. I need to calculate some probabilites that are adding onto each other like this
>1 in 6 chance to get the item you want
>1 in 13 for it to be the color you want
>1 in 20 for it to have something else you want
>1 in 50 and so on on.
Each of these are sequential. If step 1 didn't happen, step 2 is irrelevant because step 1 didn't occur. How do I calculate what the likely hood is to complete each of these steps? Obviously, the first is 1 in 6, but how do I find out the second or third and so on?

>> No.10558738

>>10558719
Expand [math]||Ax-b||=\sqrt{tr(Ax-b)^T(Ax-b)}[/math]

>> No.10558739

>>10558719
It's just the chain rule, you should look up Matrix Calculus for how to extend a lot of your single variable understanding of calculus to derivatives-by-vectors, which the gradient is (derivative of a scalar function by a vector x)

>> No.10558748

>>10558739
i know that.. i'm just not sure clear on the algebraic simplification to get to the point prior the derivative actually being taken

>> No.10558758

>>10558748
But that's the thing, it's already as simple as it needs to be for application of a vector derivative. Expanding it out is more of a hindrance than a help. A straightforward application of the derivative gives you (1/2) * 2 * A'(Ax - b) immediately

>> No.10558764

>>10558758
ah - ok thanks for pointing that out

>> No.10558770

>>10558764
np, the only real trick that Matrix calculus adds on top of vector or single variable calc is keeping the transposes and ordering correct, which can change whether you're considering your vectors to be rows or columns. Otherwise most of the rules work with the same intuition as single variable

>> No.10558811

What's the point of adding so many different collection types when they all serve the same purpose. For instance, queues, stacks, and lists all store multiple objects. Most are just implementations of lists anyways so why add these extra collections.

>> No.10558849

>>10558811
they have different interfaces and so internally can be implemented differently. therefore certain operations will be more performant with one type than another

>> No.10559789
File: 69 KB, 1260x444, Screen Shot 2019-04-15 at 3.53.34 PM.png [View same] [iqdb] [saucenao] [google] [report]
10559789

herps me

>> No.10559981 [DELETED] 

>>10559789
I'm pretty sure there's a section in Concrete Mathematics that goes into detail on how to construct these sequences using that example.

>> No.10560071

>>10559789
This is quite straightforward once you dispel any assumptions you may have about what constitutes a "polygon". Given any ordered set of points, a polygon can be formed by joining adjacent points. The resulting polygon won't necessarily be convex and may self-intersect.

Also, the question should probably have stated that no pair of lines are parallel, i.e. each pair of lines has a point of intersection.

There n! ways to choose an ordering of n lines, each of which forms a polygon. But a cyclic permutation (of which there are n) yields the same polygon, as does reversing the order. So there are n!/n/2 = (n-1)!/2 distinct polygons.

>> No.10560098

>>10554321
Are there any notations for "complex numbers" that avoid [math]i[/math]?

>> No.10560103

>>10560098
[math] \sqrt{-1} [/math]

>> No.10560118

>>10560103
I mean along the lines of tuples or vector spaces.

>> No.10560124

>>10560118
(a, b) for the number a + ib is sometimes used but the multiplication isn't really obvious when you write it like that.

>> No.10560127

>>10560118
>I mean along the lines of tuples or vector spaces.
We give [math]\mathbb{R}^2[/math] the operations [math](a, ~ b)+(c, ~ d)=(a+c, ~ b+d)[/math] and [math](a, ~ b)*(c, ~ d)=(ac-bd, ~ bc+ad)[/math].

>> No.10560128

Does anything actually ever fall into a black hole from the perspective of an outside observer?

When a star collapses into a black hole, what happens to the mass that was at the center of the star when the black hole formed? Does if just get pushed out by the event horizon? It can't pass the event horizon because nothing can do that from the perspective of an outside observer right?

>> No.10560137

>>10558811
Because each one describes certain problems better. You can use a list for almost anything, but if it makes sense to use a queue, then using one will make your code easier to read an more maintainable.

>> No.10560229

>>10558737
Please respond, I'm too dumb for this.

>> No.10560274

>>10560229
Ignore the order. Effectively, all conditions have to be met, so the required probability operation AND, which means that the required arithmetic operation is multiplication.

Think of the fractions as a bunch of unlikely outcomes. If you multiply them, then the result is something more unlikely. For example if you multiply a coinflip by another coinflip and want two heads in a row, then the probability of that happening is 1/2 * 1/2 = 1/4. The reason for this is that the number of possible outcomes increases to 4:
head, head
head, tails
tails, head
tails, tails

>> No.10560278

>>10558531
I tried that, but either it's wrong or I fucked up somewhere. I have these givens:
v0=2.32m/s, d=.0698m, t=0.2s
object is travelling upwards, so:
g=9.8m/s^2
and
E=3.6*10^3 N/C
I can find Vf, and then acceleration, but even doing that I still get the wrong answer.

>> No.10560345

>>10560274
So 1/6 and 1/13 would be 1/78, right?
for all three it would be 1/6 x 1/13 x 1/20 = 1/1560?
Alright, thank you very much. That helped.

>> No.10560413

>>10560345
I believe so, unless the probabilities are somehow dependent ( https://www.khanacademy.org/math/ap-statistics/probability-ap/probability-multiplication-rule/a/general-multiplication-rule )

That doesn't seem to be the case for the first two probabilities in your question. They're independent because choosing an item doesn't remove a color from existence.

>> No.10560427

>>10560413
No I don't think so. The only thing linking is that the first chance must have occurred before moving on to the second, but the first thing doesn't change or influence anything in the second run. Again. thank you very much.

>> No.10560559

>>10560278
> I have these givens:
> v0=2.32m/s, d=.0698m, t=0.2s
d and t are the distance and time until the object comes to rest? If that's the case, d=v0*t+(1/2)*a*t^2 => a = 2*(d-t*v0)/t^2 = -19.71 m/s^2
Otherwise, state the exact problem.
For constant acceleration, you have distance, initial velocity, final velocity, time and acceleration. Given any four you can find the fifth.
> object is travelling upwards, so:
> g=9.8m/s^2
If the object is charged and in an electric field, then gravity won't be the only force.
q/m = a/E assumes that the acceleration is due to the electric field alone. If you have gravity in the mix, then the actual acceleration experienced by the object will be the sum (or difference) of acceleration due to gravity and acceleration due to the field.

>> No.10560774

>>10560559
The problem reads:
>In a region where there is a uniform electric field that is upward and has magnitude 3.60×10^4 N/C , a small object is projected upward with an initial speed of 2.32 m/s . The object travels upward a distance of 6.98 cm in 0.200 s. What is the object's charge-to-mass ratio q/m (magnitude and sign)? Assume g = 9.80 m/s2, and ignore air resistance.
I found that a=-19.71m/s^2 using that kinematics equation, but after that I don't know what to do. a/E doesn't give me the right answer using 19.71 (positive or negative) for a and 3.6*10^4 for E.

>> No.10561043

>>10560774
> The problem reads:
Okay, so -19.71 m/s^2 is the total acceleration. -9.81 m/s^2 of that is gravitational, leaving -19.71-(-9.81)=-9.9 m/s^2 for the electric field, which gives q/m=a/E=-9.9/3.6e4=-2.75e-4 C/kg.

PS:
> Given any four you can find the fifth
Correction:
Given any three you can find the other two. The mean speed is 0.0698m/0.2s = 0.349 m/s = (v0+v1)/2 => v1=2*0.349-2.32 = -1.622. v1-v0=-1.622-2.32=-3.942, a=(v1-v0)/t=-3.942/0.2=-19.71.

>> No.10561332

>physics professor's teaching style never clicked with me
>textbook sucks
>only kind of half understanding what I'm doing at any given time
>would you look at that, finals are in two weeks

are there any resources anyone can recommend so that I can fill in the gaps in my Calc-Physics I course knowledge?
I get lost in the equations not knowing how the fuck they figured how to use what and where they're pulling what they're using out of their ass, and something just isn't clicking

>> No.10561372

>>10561043
Wow, I'm stupid. Thanks for your help. I didn't think you were supposed to combine the acceleration values.

>> No.10561385

>>10561332
For physics, if you're having trouble with figuring out which equation to use, write your givens out and then find the equation that has the givens you have, and what you're trying to find. In Phys 1, you're probably doing mostly kinematics, force (including gravity and friction), centripetal force, and maybe torque.
If you can give an example of a problem, I might be able to help more. Khan Academy is a great source though if you want get a better understanding.

>> No.10561397
File: 74 KB, 567x368, 2019-04-15-191446_567x368_scrot.png [View same] [iqdb] [saucenao] [google] [report]
10561397

>>10561385
I don't have any problems I need to solve right now, but here's one we were assigned recently to give you an idea.

I can write out all of the givens but no equations ever just pop out at me, and then the way the textbook manipulates the equations is extremely foreign to me and the straightforward/elegant ways my calculus textbook uses.

>> No.10561402
File: 7 KB, 696x88, 4help_15.png [View same] [iqdb] [saucenao] [google] [report]
10561402

Little stumped on this one.

>> No.10561427

>>10561402
course?

>> No.10561435

>>10561402
I dunno lad, what's the definition of average again?

>> No.10561436

>>10561435
1/b-a integral from a to b of f(x) dx

>> No.10561440

>>10561427
calc 1

>> No.10561442

>>10561436
Yeah, and the integral of [math]4x^{-2}[/math] is [math]-8/x[/math]. Evaluated at 1 we have -8, at c it's -8/c. So it's [-8/c-8]/(1-c)=1.

>> No.10561445

>>10561442
>that entire formula
I meant [-8/c+8]/(c-1)=1.

>> No.10561461

>>10561442
>>10561445
the answer isn't 1, I already tried that

>> No.10561467
File: 18 KB, 617x168, 2019-04-15-195328_617x168_scrot.png [View same] [iqdb] [saucenao] [google] [report]
10561467

>>10561461
are they looking for something like this?
I wouldn't think so since that's Calc II for me, but maybe your course introduced this concept in Calc I

>> No.10561481

>>10561397
If you're given a formula sheet (which I'm guessing you are), don't worry too much about being able to just know what equation to use. As long as you can piece together what you have, and figure out what you need, you should be able to figure out what equation to use. If you can categorize the problem, it will make it easier to do this though. "Angular speed" should be a pretty big hint, and a "uniform rectangular rod" might also be a hint at moment of inertia.
In my high school physics classes, my teacher sometimes had us solve problems by simplifying equations until we got down to what we needed to find rather than giving us values for variables and having us find an actual value. I found that to help a lot because you get a better understanding for the relationship between some variables.

>> No.10561484

>>10561467
okay I randomly guessed and it said the answer 4 is correct, could someone explain this to me?

>> No.10561490

>>10561484
the average of a function over an interval (a,b) is 1/(b-a) times the integral over (a,b)
so set that equal to one and solve for c
[math]\displaystyle 1= \frac{1}{c-1} \int_{1}^{c} 4 / x^2 dx= \frac{4}{c-1} (-1/c + 1) = \frac{4}{c-1} \frac{c-1}{c} \\
1 = 4/c\\
c = 4
[/math]

>> No.10561530

>>10561490
it all makes sense now

>> No.10561556

>>10561481
We're not given a formula sheet which is a big part of the problem
Mr professor pretty much just writes down solutions from his lecture notes on the board without explaining them (and gets them wrong anyway half the time) so I'm having a real difficult time getting any intuition like you're describing in your second paragraph

>> No.10562243

-1=i*i=sqrt(-1)*sqrt(-1)=sqrt(-1*-1)=sqrt(1)=1
why not

>> No.10562591

How do I go about calculating the following: "A graph has 30 vertices, 63 edges and all vertices have the degree 3, 4, or 5. It is known that there is one vertex with the degree 3. How many vertices have degree 4 and how many have degree 5?"

>> No.10562787

>>10562591
How about you write down the equations for

X = summation (number of vertices of degree D) times D
N = number of vertices

X will be twice the amount of edges in the graph because it counts every edge twice. Then you'll have a system of equations with two variables, because X is 1*3 + a*4 + b*5, N is 30 is 1+a+b

>> No.10562847

What to do if I read the solution but still don't get it?

>> No.10562858

>>10562243
>sqrt(-1)*sqrt(-1)=sqrt(-1*-1)
Incorrect

>> No.10562870

>>10562847
check what specific step in the solution you do not understand and start from there.

>> No.10562910
File: 57 KB, 645x729, pqafkb6d9ba01.jpg [View same] [iqdb] [saucenao] [google] [report]
10562910

>>10562870
Heh I still don't get it...

>> No.10562913

>>10562910
Reread the definitions and try to picture them in your head.

>> No.10562920

>>10562910
Then ask about that specific step

>> No.10562963

>>10562913
>>10562920
Ok, I have to prove
[eqn]\sum^l_{k=0}\binom{n}{k}\binom{m}{l-k}=\binom{n+m}{l}[/eqn]
I'm also given a hint to apply binomial theorem to [math](1+x)^n(1+x)^m[/math].
So I did. I more or less figured out by myself that
[eqn]\sum^n_{k=0}\binom{n}{k}x^k\times\sum^m_{j=0}\binom{m}{j}x^j=\sum^{n+m}_{l=0}\binom{n+m}{l}x^l[/eqn]
But then comes the part that I don't understand. The author just says that from this formula it is obvious that coefficent of [math]x^l[/math] on the left-hand side is [math]\sum^l_{k=0}\binom{n}{k}\binom{m}{l-k}[/math] and then it completes the proof.

>> No.10562971

>>10562963
>then it completes the proof
No, it's better to say "with that it completes the proof".

>> No.10563033

Am I fucked for getting into grad school without doing research in undergrad?

>> No.10563311

>>10562963
It is obvious. Try it with n = 3 and m = 4 for example.

You can also prove it by counting number of ways for choosing [math]l[/math] elements from two disjunct sets [math] A \text{ and } B[/math] of sizes [math]n \text{ and } m [/math] respectively, in two different ways. One is obvious (RHS), and the other one is obtained by finding number of ways to choose [math]l[/math] elements by taking [math]k[/math] elements from [math]A[/math] and [math]l-k[/math] elements from [math]B[/math] and then summing from [math] k = 0 \text{ to} k = n[/math] (LHS).

>> No.10563333
File: 7 KB, 284x168, intersect.png [View same] [iqdb] [saucenao] [google] [report]
10563333

I want to be able to calculate where lines(vectors?), spheres, cubes etc intersect in 3D space but dont even know where to start. what do? anyone have some good learning material for that stuff?

>> No.10563341

>>10563333
Do you know analytic geometry in plane?

>> No.10563357

>>10563341
my knowledge of maths doesnt go much further than the basic trig I learned in school unfortunately

>> No.10563366

>>10563341
ok i shouldve looked that up before replying, while i've never heard of it with that name, quite a lot of that stuff does look familiar. so yea, i somewhat do know analytic geometry i guess

>> No.10563399

For ξ and η elements of a Hilbert space, what does ξ|η stand for?

>> No.10563425

>>10563366
Okay. Even if you don't know calculus, any beginner text on vector calculus typically covers how to describe line and shapes and planes in the 3D space in the first chapters. Check that out. Basic lin alg covers that too, but it's more abstract.

>> No.10563466

>>10563311
Desu it is more clear when you phrase it like that.
But even though I see why it is true when I look at the original equation I still don't understand what I need to do to the second one to obtain the first one. In other words, how do I algebraically manipulate the second equality to obtain the first one?
>summing from k=0 tok=n (LHS)
Did you mean "to [math]l[/math]"?

>> No.10563478

>>10563333
nice quads bro

>> No.10563508

>>10563425
>>10563478
thanks, shouldnt be too hard to learn that stuff eh

>> No.10563511

I don't understand this question, any help would be appreciated:
>For the system equations in problem 1, determine what the value of y(t) will be after a long time (i.e as t approaches infinity) if the initial conditions are i. y(0)=0, and ii y.(0)=0.1
Problem 1 simply lists equations, here is an example
>1. a. y'=1-y^2
Am I trying to find c1 and c2 after taking the anti-derivative, or should I be finding a limit or some combination of both? Perhaps I'm completely off base. This is for a Linear Engineering Systems class, so it's Linear Algebra and differential equations.

>> No.10563520

>>10563466
I meant from [math]k = 0 \text{ to } k = n[/math], but since [math]a \choose b = 0 \text{ for } b > a [/math] they are the same.

>> No.10563526

>>10563520

[math]{{a} \choose {b} } = 0 \text{ for } b > a[/math]

>> No.10563536

>>10561556
Oh, that is strange. I've never not been given a formula sheet, even on the AP test and in college I get one. My best advice in that case is to try working through a few problems (even ones you've already done) and doing them how I previously described; don't use values, just variables. When I was in physics in high school, we usually had 10-20 problems due online biweekly through WebAssign, and when you're doing that many problems, you become so used to using the equations you start to memorize them without really trying.
I suppose if you're really feeling SOL you could put the equations on a note card and try to look at them while your teacher isn't looking, or if you have a calculator you can take notes in, store them in that.

>> No.10563591

A solution to Snake is to come up with a route that passes through every square on the map exactly once and ends where it started. It's easy to create such a route for any grid with at least one even side, but it seems impossible for any grid with two odd sides (except 1x1). How do you prove this?

>> No.10563606

>>10563511
You're not supposed to use integration. The point is to describe the behaviour of the system, not solve it explicitly, since a given system of differential equations will often have a solution that can't be expressed in terms of standard functions. Post all the equations.

>> No.10563630

So I just learned about the equivalence of norms in euclidean spaces, and I know the formal definition + the gist of the proof. But I feel like I don't quite grasp the usefulness of that property yet. The name "equivalence" makes it seem like a very strong relation, as if norms inherited several properties because of this relation. But from the definition of equivalence I can't see why.

So how powerful is the equivalence of norms? What properties are shared between equivalent norms, that I should know of?

>> No.10563793

>>10563606
So, I should be looking at the set of equations, rather than each one? If I was to find the equilibrium points, would I do it individually, or for the entire set?
Here is the list of equations:
a. y' = 1-y^2
b. y' = y^2-1
c. y' = y(y-1)(y+1)
d. y' = (y-1)^2
e. y' = (y+1)(y-1)^2
f. y' = (y^2+1)(y-1)
g. y' = cos(y)

>> No.10563824

>>10563511
The formula for y'(t) tells you how the value of y changes over time.

When t=0, the rate of change of y is 1-0^2 = 1, so y will get bigger. Once y has increased to, say, 0.5, the rate of change is 1-0.5^2 = 0.75, so it's still increasing but more slowly. When y=1 the rate of change is 1-1^2 = 0, so the value of y remains stationary at 1.

You want to plot a vector diagram. As there's only one variable, this will be a one-dimensional plot, i.e. a straight line. Find all the stationary points (for the equation you gave, the points where 1-y^2 = y) and mark them. Then work out how the value of y behaves inbetween the stationary points (you'll find it points towards one of the stationary points), and mark this with arrows. By looking at your diagram you'll be able to see where y will end up if it starts at 0 and 0.1.

You should get stationary points at y=1 and y=-1. 1 will be a stable point and -1 will be an unstable point, so:

any initial value greater than -1 will converge to 1
if the initial value is -1 exactly it will stay at -1
any initial value less than -1 will tend towards negative infinity

>> No.10563841

>>10563591
https://en.m.wikipedia.org/wiki/Eulerian_path

>> No.10563871

>>10563793
Nah, do each one individually. Sometimes you might get multiple variables per set, example:

x' = y-x
y' = 2x-y

In which case you'd need a 2D plot and probably graphing software to make it easier. But all yours are single-variable, they're separate questions. Find the equilibrium points individually.

>> No.10563872

>>10563824
> (for the equation you gave, the points where 1-y^2 = y)

This is wrong, it should be the points where 1-y^2=0.

>> No.10563910
File: 304 KB, 1488x1737, 20190416_210400.jpg [View same] [iqdb] [saucenao] [google] [report]
10563910

>>10563793
Seems like it doesn't matter if you start at 0 or 0.1, you always end up at the same point regardless. I would've expected at least one difference. Probably a good idea to do them yourself to make sure.

>> No.10564120

Okay so I feel retarded but I'm having trouble with my calc III homework. basically I have to find the line integral of the gradient of [math]f(x,y,z) = xy^3z^2[/math] with the curve [math]r(t) = < e^{tcos\left(t^2+1\right)},ln(t^2+1), \frac{1}{\sqrt{t^2+1}}>[/math]. I found the gradient of f to be [math]<y^3z^2, 3xy^2z^2, 2xy^3z>[/math] and then plugged in r(t) to give me [eqn]\left(f\:\circ \:r\right)(t) = <\frac{ln^3\left(t^2+1\right)}{t^2+1},\:\frac{3e^{tcos\left(t^2+1\right)}ln^2\left(t^2+1\right)}{t^2+1},\:\frac{2e^{tcos\left(t^2+1\right)}ln^3\left(t^2+1\right)}{\sqrt{t^2+1}}>[/eqn]. I then had to take the derivative of r(t) with respect to t which gave me: [eqn]<\mathrm{e}^{t\cos \left(t^2+1\right)}\left(\cos \left(t^2+1\right)-2t^2\sin \left(t^2+1\right)\right),\:\frac{2t}{t^2+1},-\frac{t}{\left(t^2+1\right)^{\frac{3}{2}}}>[/eqn]. I use this to find the dot product and take the integral on the bounds provided to me and get: [eqn]\int _0^1\:\frac{ln^3\left(t^2+1\right)\mathrm{e}^{t\cos \:\:\left(t^2+1\right)}\left(\cos \:\:\left(t^2+1\right)-2t^2\sin \:\:\left(t^2+1\right)\right)}{t^2+1}+\:\frac{6te^{tcos\left(t^2+1\right)}ln^2\left(t^2+1\right)}{\left(t^2+1\right)^2}+\:\frac{-2te^{tcos\left(t^2+1\right)}ln^3\left(t^2+1\right)}{\left(t^2+1\right)^2}dt[/eqn] and there is no fucking way that shit is even remotely possible. Should I just try to integrate this or am I doing something horrendously wrong? sorry if the formatting is fucked btw, this is my first time using it.

>> No.10564138

>>10564120
>line integral of the gradient

>> No.10564158

>>10564138
>just looked back at my textbook
>fundemental theorum of line integrals
[eqn]\int\limits_{C}{{\nabla f\centerdot \,d\,\vec r}} = f\left( {\vec r\left( b \right)} \right) - f\left( {\vec r\left( a \right)} \right)[/eqn]
>this could have been so much easier for me
time to an hero

>> No.10564159
File: 76 KB, 500x672, 65763543354354.jpg [View same] [iqdb] [saucenao] [google] [report]
10564159

If I have two objects and I reduce the distance between them by half each second, how come there is a moment when they touch? Does that means the second before they touch they are at a distance that can not be divided by 2? And if so, what is that distance?

>> No.10564186

>>10564138
thank you anon for pointing out how retarded I was being, it was actually really easy when I stopped small-braining.

>> No.10564212

>>10564159
>If I have two objects and I reduce the distance between them by half each second, how come there is a moment when they touch?
If we are talking about classic rigid bodies, this isn't true. There is no "smallest distance"

>> No.10564242

>>10563871
>>10563910
Ok, so that all makes sense, but I'm still wondering if I should be considering all the equations at once, or individually like
>>10563606 said.
I would think individually, since it seems to make more sense with the work we've done so far in class. However, this is my first experience with linear algebra/dif eq, so I'm not sure how to interpret "system equations".
>>10563824
Ok, so I'm assuming I'm looking at the equations individually then? I found equilibrium points for each equation, and now I'm working on making sense of plugging in t=0 and getting either y=0 or y=.1. When I first read it, I assumed it to be an initial value problem, but seeing the responses I got, it seems I misunderstood the problem. If it's not IVP, then it would appear to be a limit, but that doesn't seem right. I understand what you mean in your last 3 statements, I'm just confused as to why the problems is asking about plugging in t=0 and getting either y=0 or y=.1.
I really appreciate everyone's help, it's definitely giving me a better understanding of this.

>> No.10564506
File: 78 KB, 1270x402, file.png [View same] [iqdb] [saucenao] [google] [report]
10564506

why are the equal signs far apart on the left and close together on the right?
i'd like the right hand side to look like the left
how do?

>> No.10564517

>>10564492
That's because of the chain rule.

>> No.10564526 [DELETED] 

>>10564506
Attempt.
P(y = y \mid x = x) = \frac{ P(y = y, x = x)}{P(x = x)}
[latex]$P(y = y \mid x = x) = \frac{ P(y = y, x = x)}{P(x = x)}$[/latex]

>> No.10564536

>>10564506
Attempt.
P(y = y \mid x = x) = \frac{ P(y = y, x = x)}{P(x = x)}
[eqn]P(y = y \mid x = x) = \frac{ P(y = y, x = x)}{P(x = x)}[/eqn]

>> No.10564538
File: 86 KB, 1226x424, file.png [View same] [iqdb] [saucenao] [google] [report]
10564538

>>10564536
renders properly on 4chins but not on my local

>> No.10564540
File: 57 KB, 482x549, 1493405349916.jpg [View same] [iqdb] [saucenao] [google] [report]
10564540

How the fuck do I generate a square wave with a period of 1s in Matlab?

>> No.10564548

>>10564538
I don't know man. Does /g/ have a qtddtot?

>> No.10564751
File: 8 KB, 884x469, vecTriangle.png [View same] [iqdb] [saucenao] [google] [report]
10564751

Is it possible to make a triangle with the values on the right?
How can it be explained with math?

>> No.10564781

>>10564751
Sure.
[math]y=90[/math]
[math]a=\sqrt{c^2-b^2}[/math]
[math]b=\sqrt{c^2-a^2}[/math]
[math]c=\sqrt{a^2+b^2}[/math]
[math]x=h[/math]

>> No.10564786

>>10564781
>x=h
Forgot to erase that.

>> No.10564800

>>10564751
No, since the largest side is always opposite to the largest angle, and b is opposite to y.
>>10564781
Setting up values doesn't show it actually exists.

>> No.10564826

>>10564786
And your angles
[eqn]\sin(x) = \frac{a}{c}[/eqn]
[math]h = 90-x[/math]
This would've been in the first post but I had to look up laTEX for sine (and now I feel retarded).

>> No.10564848

>>10564800
>the largest side is always opposite to the largest angle
How can this be explained with math?

>> No.10564870

>>10564800
>Setting up values doesn't show it actually exists.
But in this case it does. All I described was a right triangle where one side is inbetween the other side and the hypotenuse in length. The sum of all angles in a Euclidean triangle is 180, so y has to be 90, or the triangle non-Euclid. FUCK!
I just realized that doesn't work because y can't be a right angle if c is the hypotenuse. I mean, assuming we're only using positive values.

>> No.10564871

>>10564848
Apply the sine law to the biggest>90 and biggest<90 cases.

>> No.10564918

>>10564871
Thank you

>> No.10565495

>>10564212
Please explain it to me as if I was dumb, because I'm afraid that's the case and I really want to get this.

>> No.10565511

>>10560098
Complex numbers can be represented by particular real 2-by-2 matrices

https://en.wikipedia.org/wiki/Matrix_(mathematics)#Applications

>> No.10565515

>>10565495
https://en.wikipedia.org/wiki/Geometric_series

>> No.10565518

>>10565495
https://en.wikipedia.org/wiki/Zeno%27s_paradoxes#Achilles_and_the_tortoise

>> No.10565821
File: 437 KB, 1920x1080, 1554456584839.jpg [View same] [iqdb] [saucenao] [google] [report]
10565821

>>10565515
>>10565518
I'm afraid these do not explain this event at all.
Both tell me that theoretically the distance can always be smaller, ok, makes sense, but in practice there is a moment in time when both objects touch, and a time when they do not. A time when the distance between is 0 and a time when it is not. Maybe at a near atomic level there is a moment when 2 objects are so close that if let go they slam together and end up touching, idk, I'm pulling this theory totally off my ass. But could that be an explanation?

>> No.10565878

>>10564540
you uninstall and run python instead, like a sane person.

>> No.10565955
File: 350 KB, 1200x1000, 1550617206182.jpg [View same] [iqdb] [saucenao] [google] [report]
10565955

>>10554321
How do I learn enough to pass a calc 3 exam in 1 day?

>> No.10565972

>>10565821
https://en.wikipedia.org/wiki/Half-life
Not directly related to distance, but still.

>> No.10566010

>>10556035
yes they are same and it has been experimentally proven momentum is not equal to mass but there is another thing called relativistic mass,though i am not sure if it affects gravitational mass.

>> No.10566163

>>10564242
>Ok, so that all makes sense, but I'm still wondering if I should be considering all the equations at once, or individually

Each equation is a separate question. Solve each one individually. Period.

>so I'm not sure how to interpret "system equations"

In your case, each 'system' only has one equation. If this doesn't make sense to you, just don't think about it. 100% you need to solve each equation individually.

>I found equilibrium points for each equation, and now I'm working on making sense of plugging in t=0 and getting either y=0 or y=.1

What? No, you're not supposed to try and "get" 0 or 0.1, those are the values you start with. Look at the graphs on >>10563910, that's the approach you need to take. Imagine you dropped a ball at 0 and it moved according to the arrows (which represent whether y'(t) is positive or negative for that value of y). If the arrow points left, the ball rolls left. If the ball reaches a stationary point, it stops there. And so on. Wherever the ball ends up (either at a stationary point or rolling off into infinity) is your answer. Do this for each equation, finding out where the balls ends up if it starts at 0, then if it starts at 0.1.

For systems with multiple equations, you get more complicated answers, like the ball might be caught in a loop. But this can't happen with one-equation systems, which is all you're dealing with right now.

t is just a time variable, it parametrises the equation. y(t) means "the value of variable y at time t". If y(0)=0.1, that means y's initial value is 0.1. What you want to find is where y ends up.

>> No.10566172

[eqn]
e(x)=\sum_{j=0}^{\infty}\frac{x^j}{j!}
[/eqn]

Starting from this definition only, how do I prove that e(0)=1?

>> No.10566175

>>10566172
Just substitute x = 0 lol.

>> No.10566200

>>10565821
>I'm afraid
go see your shrink then

>> No.10566201

>>10565955
I'm struggling to reach that .75 boys. Wish me luck.

>> No.10566202

>>10565821
>I'm pulling this theory totally off my ass.
yes, that's the explanation?

>> No.10566222

>>10566172
lmao just substitute x by 0 and remember 0^0=1

>> No.10566232

>>10565821
.5+.25+.125+.0625+...=1 Right?
Provided the sum converges, it turns out that summing an infinite number of terms takes a finite amount of time and results in a finite distance. That's why the objects touch. Zeno didn't get that, hence the "paradox" (it isnt a paradox).

>> No.10566362

>>10566175
>notices your undefined term
>0^0 what's this?

>> No.10566471
File: 73 KB, 1200x1200, 814266.0.jpg [View same] [iqdb] [saucenao] [google] [report]
10566471

I have a knife that I use for woodcarving and cutting modeling clay- when it gets cold large chunks of the stuff are hard to work. Will dragging a knife through a soft but dense material dull it?
Surprisingly hard to google

>> No.10566589 [DELETED] 
File: 4 KB, 153x123, 1489686137131.jpg [View same] [iqdb] [saucenao] [google] [report]
10566589

amirite?
[math]$\forall a\in A$ on a topology $T$ if $a \in B \subseteq$ A, $B\cap C=\{a\}$ and $B,C$ are open, then $a$ is not limit point of $B$.\\

Proof. Assume for contradiction that $a$ is a limit point of $B$. Then every open set containing $a$ must contain some point $z\in B$, but the only points that $B$ and $C$ share is a.$\bot$\\
$\square$[/math]

>> No.10566596
File: 55 KB, 581x525, fx7un7vpmgh21.png [View same] [iqdb] [saucenao] [google] [report]
10566596

[math]\forall a\in A[/math] on a topology [math]T[/math] if [math]a \in B \subseteq A, B\cap C=\{a\}[/math] and [math]B,C[/math] are open, then [math]a[/math] is not limit point of [math]B[/math].

Proof. Assume for contradiction that [math]a[/math]is a limit point of [math]B[/math]. Then every open set containing[math]a[/math]must contain some point [math]z\in B[/math], but the only points that [math]B[/math] and [math]C[/math]share is [math]a[/math].[math]\bot[/math]

[math]\square[/math]

>> No.10566626

>>10566596
YOU WHAT
TOPOLOGIES ARE CLOSED UNDER FINITE INTERSECTION LAD
{a} IS FUCKING OPEN, IT ISN'T A LIMIT POINT OF ANYTHING

>> No.10566643

>>10564242
If the system converges to a stable equilibrium, then y'=0 at that point. So substitute y'=0 and solve. If there are multiple solutions, the initial conditions will determine which one it converges to.

>> No.10566659

Is differential equations useful for higher math courses? Everything so far has been cookbook work that only works for specific problems, and apparently that's fine for sciences, but if you come across those "unsolvable" problems in other studies I don't feel like they would just drop the problem.

>> No.10566702

>>10566202
No Anon, that's the disclaimer.

>> No.10566754

>>10566626
Right, but is my proof good? It's just a stupid question. That's acknowledged. Why you are in a stupid questions thread acting surprised is a different question.

>> No.10566762

>>10566626
Wrong.
The topology T could contain nothing but {X,O,{a},{c,d},{a,c,d},{b,c,d,e}} and b would be a limit point of a subset {a,b,c)

>> No.10566869

>>10566762
>implying {b} is open in your example
Viper.
>>10566754
No, it's horrendous. Still correct, but for the wrong reasons.

>> No.10567011

Why does Europa have so much water?

If all the moons around Jupiter formed from leftovers of the same blob of matter, why wouldn't the water be more evenly distributed amongst them? Is it possible the more volcanically active ones DID have water originally, but outgassed and ejected them all due to tidal forces (because closer to Jupiter)?

Or is it just "one of the moons got a fuckload of water for some reason"? If so, what the hell is the reason?

>> No.10567911
File: 86 KB, 1540x370, file.png [View same] [iqdb] [saucenao] [google] [report]
10567911

did the author make an error transcribing this formula? seems incorrect to me

>> No.10567937

>>10567911
>turns a function of n variables into a function of one variable times a function of n variables
I hate stats so much.
But no, the formula isn't incorrect. It's just Bayes's theorem.

>> No.10567958
File: 42 KB, 1138x346, file.png [View same] [iqdb] [saucenao] [google] [report]
10567958

>>10567937
ok so please help me understand how the examples map to the definition. see pic related. let's take the first line.

ostensibly, the first part of the first line -- P(a | b,c) -- maps to the second part of the right hand side: a maps to x^i, so b and c must map to x^1 and x^2 (i.e., x^i-1). clear enough. but the first part of the right hand side is P(x^1)... the second part of the first line of pic related is P(b,c), so two values, rather than one. we have an impedance mismatch here afaict

what am i missing?

>> No.10567965

>>10567958
It's literally "the probability of a, b and c is the probability of a happening given that b and c happened, times the probability of b and c happening."
The second line is the usual Bayes, and the third line is just swapping the values in.

>> No.10567972

>>10567965
i hear you and i understand the intuition. i'm just struggling to understand the notation. what specifically does x^1 represent?

>> No.10567976

>>10567972
The event x^(1). Formally, x^(1) is a subset of the set of events that we consider as "x^(1) happened" and also call x^(1).

>> No.10567979

>>10567976
but in the example it's "a", right?
so why does nothing in the example reflect P(a)?

>> No.10567997

>>10567979
Same same. a, x^(1), heads, you can call the event whatever.
It doesn't reflect P(a) because it doesn't need to. There is no "immanent order of events", so we can consider it a and b happening simultaneously to be the same as a happening and then b happening or b happening and then a happening.
Consider something like a is "a dice falling on an even number" and b is "a dice falling on two".
Both events happen simultaneously, but we nonetheless know that if it falls on two, then it must fall on an even number, or that if it fell on an even number, it has a 1/3 chance of having fallen on 2.
Notation wise, this reads P(b)P(a|b)=P(a)P(b|a);

>> No.10568006

>>10567997
i apologize if i'm being dense so bear with me.

the whole point of the example is to show how a joint probability of many variables can be decomposed into conditional distributions over only 1 variable

P(x^1) is the one variable
P(a,b,c) = P(a|b,c)*P(b,c) ostensibly is an example of the decomposition being applied
where is the one variable in from the decomposition

>> No.10568034
File: 28 KB, 1044x124, file.png [View same] [iqdb] [saucenao] [google] [report]
10568034

>>10567997
>>10568006
here is how wikipedia defines the chain rule.
i hope you can see why to me it seems like the book didn't define it properly
afaict the author is being inconsistent with notation and should not have referred to P(x^1)

>> No.10568045

>>10568034
Anon, the comma and the cup stand for the same thing.
>>10568006
Yes.

>> No.10568052

>>10568045
>Anon, the comma and the cup stand for the same thing.
i know....?

>> No.10568062

am i being retarded about this? i feel like i havent gotten a straight answer to the question

>> No.10568099

fuck sorry

>> No.10568211
File: 6 KB, 676x139, nword.png [View same] [iqdb] [saucenao] [google] [report]
10568211

okay, a little confused here

>> No.10568238

>>10568211
literally just divide the interval into two simpler subintervals

>> No.10568264

How do we get people (especially young people) re-invested into society and healthy social structures?

>> No.10568292

>>10568211
int f(x) from 0 to 1 + int f(tx) from 1 to 2

>> No.10568295

>>10568264
by getting rid of weak fathers

>> No.10568320

>>10568295
Weak fathers is just another symptom i think. I grew up without a father (mine died when i was under 10 due to heroin overdose) and despite having only a mom I turned out better than plenty of others so that's not a magic bullet in either case

>> No.10568326

>>10568320
>Weak fathers is just another symptom i think
weak fathers are a symptom of previous weak fathers

>> No.10568333

>>10568326
Honestly, fair enough. It seems unfair to put the onus onto the guys but i suppose we are supposed to be the tough ones. We're the ones that are supposed to crack down on our children and make them be good and nice to the people around them, one way or another. The real question is how do we solve that.

>> No.10568439

Two things ive been thinking about. Harnessing energy from the earths magnetic field, and mass planting trees ti decrease carbon pollution

>> No.10568496

>>10568439
>Harnessing energy from the earths magnetic field
doable but ultimately not as worthwhile as solar or fusion (if doable).
>mass planting trees ti decrease carbon pollution
We can just suck out co2 of the air and sequester it into solid forms, its not even that costly anymore. The problem is methane is a more potent greenhouse gas. Artificial leaves will also probably start being using in architecture fairly soon when they are able to be readily produced as theyll be easier to maintain than natural plants

>> No.10568803

If you have the equation of a circle, how would you the find the values for c in y=c-x if you know that the straight line touches the circle at one point?

>> No.10568857

>>10568803
I assume you mean a circle around the origin.
If you want the line to be tangent to the circle and have slope -1, there's obviously two possibilities for that point: one is at [math]\vartheta = {\pi \over 4}[/math], the other at [math]\vartheta = {5\pi \over 4}[/math].
In cartesian coordinates, solve for [math] {r \over \sqrt{2}} = c - {r \over \sqrt{2}} [/math]

>> No.10569246

>>10568496
ok nice, i had no idea i was shooting in the dark. lol

and im not sure where your getting your source.
http://www.cleanerandgreener.org/resources/air-pollution.html

>> No.10569358

>>10568803
you can set the two equations equal to each other in terms of c

>> No.10569821

>>10554849
Write out the right side. Get it to a common denominator. Add up. Done

>> No.10569847 [DELETED] 
File: 3 KB, 436x91, 267.gif [View same] [iqdb] [saucenao] [google] [report]
10569847

How do you go from step 1 to step 2?

>> No.10569854

>>10567911
>>10567958
He just split up the probability from left to right on top and from right to left on the bottom. This is possible because P(A,B) = P(B,A) and P(C)*P(D) = P(D)*P(C)

>> No.10569863

>>10569358
I know that this is the answer somehow but I keep fucking it up. I just end up spending 20 lines of algebra to prove c=y+x or something dumb.

The circle is [math]x^2 +y^2 -4x-2y-3=0[/math] and through graphing the values for c are 7 and -1 but how I arrive at those values are missing me

>> No.10570061

>>10569863
Substituting y=c-x into the equation for the circle gives you a quadratic in x:
2x^2-2(c+1)x+c^2-2c-3=0
Applying the quadratic formula gives you
x = ((c+1) ± sqrt(-c^2+6c+7))/2
The discriminant is -c^2+6c+7. The equation has two real solutions if the discriminant is positive, one solution if it's zero, and no real solutions if it's negative. The line is tangent to the circle when the equation has a single solution, i.e. when the discriminant is zero:
-c^2+6c+7=0
=> -(c-7)(c+1)=0
=> c=7 or c=-1

>> No.10570080

>>10568264
Probably get rid of social media and celebrity culture and replace with STEM role models and people with moral fiber in the limelight.

>> No.10570149

>>10570061
Thank you very much. I tried a variation of this but I think I got confused when it came to the values for the discriminant.

>> No.10570236

If stuff falling towards an event horizon, in the reference frame of a remote observer, appears to slow down and stop, doesn't that mean the apparent matter flux across the EH is zero (in said reference frame)? Does that mean BHs cannot be observed to grow?

>> No.10570455

Am I supposed to write it as "10ml" or "10 ml"? I'm assuming the latter, but I see the former a lot and don't know if I'm actually the one in the wrong.

>> No.10570507

>>10570455
There should be a space between the number and the units.

>> No.10570822 [DELETED] 
File: 21 KB, 767x251, really now.png [View same] [iqdb] [saucenao] [google] [report]
10570822

Opine
This is wrong?

>> No.10571015

What notation do I use if I want to get min(two non-numerical values) by supplying a function to map the values to numbers? e.g.:

min(F(x), F(y)), if F(x) < F(y) then it would give x and not F(x)

Of course I could just say "x if F(x) < F(y) otherwise y" .

>> No.10571089

>>10555351
This is kinda covered. Humans have incredibly strong pattern recognition. This pattern recognition is a larger part of our conscious minds than most people suspect. As such, doing things that fit the pattern of something can produce an illusory effect since our own perceptions don't play as big a role in our thoughts.
Sorry if this isn't understandable I'm a tired drunk esl

>> No.10571098

>>10555366
>>10555378
>>10571089
Also yes culture and beliefs are more or less the source of where many hallucinatory images come from. Likewise, environment, culture, and expectations have a huge effect on drugs like LCD, alcohol, and others.

>> No.10571608

Submitting this one for stupidest question of the year please be gentle

When an equation has π in it how many decimals of π are you supposed to use.

>> No.10571641

>>10571608
Use the pi function on your calculate but by hand anyone using more than 3 is a try hard.

>> No.10571647

How much of basic lab work in any given field could be done with people who never studied the subjects?

Like can you ask someone who has never studied molecular biology to do Western blotting after giving him of her a training?

Just wondering how much "science" stuff is accessible to most yet locked behind a wall.

>> No.10571651

>>10571641
>using decimals while calculating by hand
Who does this? Just use 22/7 or 355/113

>> No.10571670

>>10571608
Depends, if the formula has Pi, you can use less than if the formula has Pi^10 in it.

>> No.10571952

Is just working through the recommended textbooks enough to be able to solve Putnam problems? Currently working though Spivak and a lot of Putnam problems still give me alot of trouble.

I know they’re not the point of math but it’d still be nice to be able to solve them.

>> No.10572023

>>10571608
It depends on the precison you want in your result.
If you're an engineer, 3.14 covers you. If you're a mathematician, you might as well use literally no digits. An astronomer would use a whole lot more.

>> No.10572251

Why does matter settle into a flat ring around a planet instead of a full sphere?

>> No.10572257

>>10572251
> Be earth
> Have a moon that's a full sphere

>> No.10572264

>>10572257
You're funny anon

>> No.10572266

>>10571608
As many as the most precise number is in that formua. A circle with radius 1.1 m doesn't have a circumference of 6.9115038378975451246178154432149063452337726786252328 m. 1.1 m implies a precision of 0.05 m, so stay at that precision and just write 6.9 m

>> No.10572425

Taylor Polynomials. Obviously they're very useful and I already know of their applications. Obviously, they work.

Yet, what I'm having trouble seeing/building intuition for is, if you necessarily have to have an explicit function to take its derivative (as far as my up-to-calc II knowledge goes, this is probably what's limiting me), how then do you use a taylor polynomial to model a function without an explicit definition?

>> No.10572496

>>10572425
What exactly do you mean? Implicit functions?

Remember that you don't need an algebraic expression inside the Taylor expansion. The nth derivatives _at a particular point_ are used.
For finding these, even numerical methods might suffice depending on the application.

>> No.10572606
File: 191 KB, 253x280, firefox_iEYypnb2Ru.png [View same] [iqdb] [saucenao] [google] [report]
10572606

I'm in a draw of 30 people. Two people will be picked.
So first it'd be 1/30, then 1/29.
How do I work out the overall probability of being picked in this draw?
It's not simply 1/15 either, is it?

>> No.10572639

>>10572606
[math] \left( \matrix{ 30 \\ 2 } \right) [/math] possible draws. 29 of them have you in it.

>> No.10572728

>>10572606
They pick one person among thirty, and then pick the second among 29, giving 29x30. Since we don't care about order, that's 29x15 possible choices. 29 include you, so that makes 1/15 chance of you being picked.

>> No.10572747

>>10572606
It actually is. Don't know how to tex it, but 29 / >>10572639 is 1/15. Alternative way to calculate:
1/30 + 29/30 * 1/29 = 1/15
[getting picked on the first pick] + [not gettin picked on the first pick and getting picked on the second one]

>> No.10572986
File: 1.13 MB, 535x301, tumblr.gif [View same] [iqdb] [saucenao] [google] [report]
10572986

Take a prime natural N
Subtract a prime natural X, where X < N
If N-X < X then add N to the difference and subtract X.
If you do this for N steps, you'll end up at 0.
I'm a brainlet and I don't understand this. I know it's because of modular arithmetic, but why does it always seem to come up to N steps with primes?

>> No.10573028

>>10572986
https://en.wikipedia.org/wiki/Field_(mathematics)
Read the introduction and then the classical definition.
https://en.wikipedia.org/wiki/Characteristic_(algebra)
Read the introduction and then the case for fields.

>> No.10573037

>>10573028
That was kinder than I was expecting, and very helpful.
Thanks much anon

>> No.10573201
File: 7 KB, 661x131, bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb.png [View same] [iqdb] [saucenao] [google] [report]
10573201

how do I evaluate this if there isn't an elementary way to integrate sqrt(sin(x))

>> No.10573282 [DELETED] 

>>10572425
If a function [math] f(x) [/math] exists, explicit or not. All the taylor series asks for are the functions derivatives, [math] f^{(n)} (x) [/math]. Some times, that's all you would know about a function: its value at a particular x, and the first and second derivative....sometimes more.

Let say you have a graph of a function in a spread sheet. You know the value of the function at particular points. You also know an approximation of the derivative of a function at particular points.....and so on. You can use this to reconstruct an approximate explicit function.

Taylor expansions are also useful to solve for difficult limits. Because a function behaves almost idiotically to it's partial taylor expansion in the neighborhood of it's limit point of interest.

>> No.10573295

>>10572425
If a function f(x) exists, explicit or not. All the taylor series asks for are the functions derivatives, f(n)(x). Some times, that's all you would know about a function: its value at a particular x, and the first and second derivative....sometimes more.

Let say you have a graph of a function in a spread sheet. You know the value of the function at particular points. You also know an approximation of the derivative of a function at particular points.....and so on. You can use this to reconstruct an approximate explicit function.

Taylor expansions are also useful to solve for difficult limits. Because a function behaves identically to it's taylor expansion in the neighborhood of it's limit point of interest.

>> No.10573298

>>10573201
You integrate the square. Every slice of your integral is actually a disk around the x-axis with radius y(x) and thickness dx. To get the volume, you calculate y^2 * dx. To sum all disks up, you integrate.

>> No.10573305

>>10573201
You don't have to integrate sqrt(sin(x)). The square root actually makes the problem easier.

>> No.10573314

>>10573295
Alright, now it's starting to make more sense. I'm sure once I get to diffeq it'll make a lot more sense.

>>10573298
>>10573305
Thank you, it's been 4 months since we've covered it, and I'd completely forgotten the methods to find solids of rotation.

>> No.10573318

>>10573314
You either remember the formula [math]\integral \left(y(x)\right)^2 \mathrm{d}x[/math] or you remember what you're trying to achieve, which is basically the thing with the disks from >>10573298

>> No.10573361

How do I calculate local sidereal time

>> No.10573364

>>10573314
>I'm sure once I get to diffeq it'll make a lot more sense.
You don't necessarily deal with the Taylor series in diffeq. You deal with the more general power series.

Search for "Method of Frobenius"
https://en.wikipedia.org/wiki/Frobenius_method

You can solve for some differential equations using this method.

>> No.10573471
File: 3 KB, 262x42, helpimdumb.png [View same] [iqdb] [saucenao] [google] [report]
10573471

So I need prove that pic related's a equivalence relation but I also need to list the equivalence classes the relation has, but if x^2=y^2 then x=y, right? so why do I need to list every equivalence class if that's the case?

>> No.10573526

>>10573471
>but if x^2=y^2 then x=y, right?
youre forgetting negatives

>> No.10573531

is campbell biology a good textbook to teach myself freshman biology?

>> No.10573534

>>10573471
Gee, I wonder what are the equivalence classes. Surely that must be a problem for someone on Euler's level.

>> No.10574882
File: 48 KB, 420x375, 1350587791929.jpg [View same] [iqdb] [saucenao] [google] [report]
10574882

Why does mathematics use 0 instead of +0 and -0?

We have no problem thinking of infinity as a number that's constantly expanding, so why isn't zero treated as a number that's constantly collapsing into itself? Why isn't it given a positive or negative sign like other values?

And I understand that the direct answer is "The math checks out", but just because an answer works in mathematics doesn't mean it has to be 100% correct right? Neutral zero being a substitute for positive zero and negative zero might work in certain cases, but in others, such as divided by zero, it does not. Zero might be such a small number that you can essentially say that it is worth nothing, but is that actually correct? Or just a hotfix solution?

>> No.10575013

>>10574882
Maybe you'd like to read this essay by GZA, the genius? There's a lot of sci in it:

>The importance of the “O” it’s shape and what it means to me and some of the things it reminds me of by GZA, “genius”
>first of all the “O” represents a cipher the numeral 0 of a thing or person of no importance, [...]
>The lens too a camera, binoculars, or telescope are all round they are like eyes which are also round enabling us to rotate or move it any direction. [...]
>When dealing with mathematics or numbers the “O” represents 0 or nothing but when you put it next to any number it multiplies that number by 10 but when you remove the “O” the number is devised by 10, it’s the only number that has no value unless it stands next too another number.
>Quote “together we stand, separated we fall”.[...]
> The “O” or circular shape plays an important role in our lives, imagine if tires weren’t round the bicycle, the car, roller skates, the skateboard, the baby stroller, or even horse carriage probably wouldn’t exist, if so then travelling would be extremely difficult and uncomfortable.
>The steering wheel in most motor vehicles has a circular pattern similar to the earth which is a clock in its self thats how we got 24 hours in a day. Even in the world of music the circular shape plays an important role from the vinyl records too the turn tables we play them on they are all round, from the compact disc’s that stores our music the headphones we use to listen to the songs, the drums that holds the beat too the woofers in the speakers are round, VHS, Cassettes tapes and film also play in circular motion.

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