/sqt/ stupid questions thread

[Anonymous No.9670851]

This thread is for questions that don't deserve their own thread.
>give context
>describe your thought process if you're stuck
>try wolframalpha.com and stackexchange.com
>How To Ask Questions The Smart Way http://www.catb.org/~esr/faqs/smart-questions.html

prev >>9656198

[Anonymous No.9670865]

>>9670851
Anyone know about turbulent conditions ran thru the Navier Stokes equations which terms cancel if any? How would one design the control systems of a plane or rocket to account for local pressure?

[Anonymous No.9670935]

>>9670851
How do you justify [math]\frac{dv}{dt} dt = dv[/math]
Is there a proof or something that says that I can just simplify the dt? I don't understand why is possible to treat the differential as a simple number

[Anonymous No.9670942]

>>9670851
Can you ionize materials with electricity?

[Anonymous No.9670988]

>>9670935
The chain rule?

[Anonymous No.9670991]

>>9670942
You can ionize materials with a strong electric field, pushing the electrons off the nucleus.

[Anonymous No.9670998]

Do embers burn the material or are they just really hot? What is the difference between embers and fire? is it just a lack of burning gas? is there a mathematical model?

I read in my textbook that LEDs are sensitive to voltage. That they will either be on or off depending on if they have a very narrow range of voltage or not. In my experience, it doesn't seem to matter that much. What is the mathematical model?

Why do some explosives shatter material and why do others take paths of least resistance? Why can't you use high-explosives as a propellant for a gun? Is there a mathematical model?

[Anonymous No.9670999]

What creature has the sharpest fangs? Do they really win all that often?

[Anonymous No.9671000]

>>9670991
How about the current itself? If it's "flow" then there is exchange of electrons right?

[Anonymous No.9671001]

Does anyone remember where in electrodynamics this equation is used?

I think it's for solid non-conducting spheres, but I really can't remember.

[Anonymous No.9671004]

>>9671000
You can't flow a current through a conductor without putting electrons into the material at the same rate electrons leave the material. So you can't strip away all the electrons using only current. Maybe there's some way but it would be higher level.

[Anonymous No.9671006]

>>9670999
I think tiny spiders and the suchlike have the sharpest fangs. Like a 'daddy longlegs' has the sharpest. When I got a broom, they don't win at all and it doesn't matter how sharp their fangs are.

[Anonymous No.9671010]

>>9671004
Thank you anon

[Anonymous No.9671033]

Are there any ways to permanently fix bad brain chemistry like OCD?
>inb4 suicide

[Anonymous No.9671034]

A friend said that at cern they have created micro black holes that last for tenths of a second.

I told him that's a load of rubbish and that he should provide a citation.

He is wrong right?

[Anonymous No.9671035]

>>9671001
Never mind, Hyperphysics saved me once again.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html

The equation is actually in terms of r^3 not r, which makes a lot more sense since charge is not linearly dependent on radius (for spheres).

[Anonymous No.9671044]

>>9671033
How old are you? You can just grow out of it by putting yourself under extremely stressful conditions like 5 or 6 hard math/science courses in one semester which permanent fucks your GPA but changes you forever because you learn how to differentiate which things/people to give a fuck about while eliminating everything/everyone else from your life.

[Anonymous No.9671045]

I get why I should substitute y' here, but what's the point of substituting y? Seems pointless.

>>9671034
Yes, he's full of shit. Probably got redpilled on science by some /pol/tard.

[Anonymous No.9671046]

>>9671034
Yes, he's wrong, if black holes were created they would also be detected, which hasn't happened yet

[Anonymous No.9671050]

>>9671044
DO NOT DO THIS

[Anonymous No.9671051]

>>9671034
Your friend is supposed to keep quite about such topics. The black holes are actually portals through which we can communicate with super technologically advanced extra-terrestrials, which is how we learned about the Higgs Boson being real. Translation is slow, which is why we haven't learned everything all at once just yet.

[Anonymous No.9671055]

>>9671050
D's get degrees if you have research credits to boost your GPA back up above a 2.0

[Anonymous No.9671061]

When will science prove that anyone below 6'5 is not human?

[Anonymous No.9671062]

>>9671055
Maybe, but causing intense stress is not going to help with his anxiety disorder. It either won't do anything or it'll make it worse. He needs therapy.

[Anonymous No.9671071]

Assuming an oxygen free environment and high pressures, could wood be molten? Like... you have molten wood, it's wood, but its melted into a hot liquid.

[Anonymous No.9671076]

I'm trying to solve this integral equation
[eqn]\phi (x) = f(x) +\lambda\int_0^\infty\cos(xz) \phi(z) \, dz[/eqn]
where [math]f(x) = f(-x)[/math]
supposedly it can be done via fourier transforms. I know that since the right hand side has symmetry under the [math]x\to -x[/math], so [math]\phi (x)[/math] takes on that symmetry, which means this can be written
[eqn]\phi (x) = f(x) +\frac{\lambda}{2}\int_{-\infty}^\infty e^(-ixz) \phi(z) \, dz[/eqn]
since the evenness of the function means the imaginary part of the integral vanishes. Taking the fourier transform yields
[eqn]\tilde{\phi} (k) = \tilde{f}(k) + \frac{\lambda}{2} f(k)[/eqn]
the trouble I'm having is that from here I can't just solve to [math]\tilde{f}(k)[/math] and use the inverse transform since the equation contains [math]f(k))[/math] as well. Any ideas how to proceed from here?

[Anonymous No.9671077]

Why are nintendo fans autistic?

[Anonymous No.9671081]

I'm trying to solve this integral equation
[eqn]\phi (x) = f(x) +\lambda\int_0^\infty\cos(xz) \phi(z) \, dz[/eqn]
where [math]f(x) = f(-x)[/math]
supposedly it can be done via fourier transforms. I know that since the right hand side has symmetry under the [math]x\to -x[/math], so [math]\phi (x)[/math] takes on that symmetry, which means this can be written
[eqn]\phi (x) = f(x) +\frac{\lambda}{2}\int_{-\infty}^\infty e^{-ixz} \phi(z) \, dz = f(x) + \frac{\lambda}{2} \tilde{f}(x) [/eqn]
since the evenness of the function means the imaginary part of the integral vanishes. Taking the fourier transform yields
[eqn]\tilde{\phi} (k) = \tilde{f}(k) + \frac{\lambda}{2} f(k)[/eqn]
the trouble I'm having is that from here I can't just solve to [math]\tilde{f}(k)[/math] and use the inverse transform since the equation contains [math]f(k))[/math] as well. Any ideas how to proceed from here?

[Anonymous No.9671083]

>>9671071
You'd need to heat it to at least 300 celcius to bring it to melting point, but I don't see why not.

[Anonymous No.9671091]

>>9671071
>Classification: Carbon is a nonmetal
>Melting point: 3550 oC, 3823 K
>Note: At normal atmospheric pressure, carbon does not melt when heated, it sublimes. i.e. it undergoes a phase change directly from solid to gas. If the pressure is increased to 10 atmospheres carbon (graphite) is observed to melt at 3550 °C.
Wood is almost entirely carbon, or carbon and water if it's fresh. You'd have to heat it to the point that it was blasted apart into its constituent elements, but "wood" can be melted, insofar as the carbon and gases in its makeup will at some point become liquid given the right pressure. Of course, it would have ceased being what we consider wood long before that.

[Anonymous No.9671099]

>>9671081
oops I meant:

[eqn]\tilde{\phi} (k) = \tilde{f}(k) + \frac{\lambda}{2} \phi(k)[/eqn]
the trouble I'm having is that from here I can't just solve to [math]\tilde{\phi}(k)[/math] and use the inverse transform since the equation contains [math]\phi(k)[/math] as well. Any ideas how to proceed from here?

on the last couple of lines

[Anonymous No.9671145]

>>9670935
that expression doesnt make sense

[Anonymous No.9671146]

>>9671081
>>9671099
Define [math]\mathcal{F}[/math] the Fourier transform operator, then your equation (1) can be written as
[eqn]\left(I - \frac{\lambda}{2}\mathcal{F}\right)\phi = f\neq 0.[/eqn]
First we check Fredholm alternative: By Parseval's theorem [math]|\phi|_{L^2} = |\mathcal{F}\phi|_{L^2}[/math], hence if [math]\phi[/math] is a solution to the equation (2) [math](I - \lambda/2 \mathcal{F})\phi = 0[/math], then [eqn]\left(1 - \frac{\lambda^2}{4}\right)|\phi|^2_{L^2} = 0,[/eqn] which means that [math]\phi = 0[/math] whenever [math]\lambda \neq 2[/math] (if [math]\lambda = 2[/math] then there exists nontrivial solutions to (2) - such as the Gaussian - and (1) has no solutions), so we get solutions to (1) for every [math]\lambda \neq 2[/math].
Next we examine the invertibility of [math]I-\frac{\lambda}{2} \mathcal{F}[/math]. We are able to solve this equation of the Neumann series [math]R = \sum_{n = 0}^\infty \left(\frac{\lambda}{2}\right)^n\mathcal{F}^n[/math] is well defined. Now by Parseval again, the operator norm of [math]R[/math] is finite iff [math]|\lambda| < 2[/math], so the solution [math]\phi[/math] is given by [eqn]\phi = \sum_{n=0}^\infty \left(\frac{\lambda}{2}\right)^n\mathcal{F}^nf[/eqn] as long as [math]|\lambda| < 2[/math].

[Anonymous No.9671147]

>>9670999
The Ratel.
Wins every time, because it has the sharpest fangs.

[Anonymous No.9671169]

Why was the word "torsion" chosen to describe elements of a module which can be annilihated by the ring?
I don't see the connection with the physical definition of the word.

[Anonymous No.9671174]

>>9671145
Doesn't it mean the same thing as dv/dt = dv/dt? I know what he's talking about, and it is a consequence of the chain rule, though I'm not entirely sure how or why.

Strangely I got tripped up by this exact thing last night. Apparently a derivative function y'(t) conveyed in terms of functions of the form x(t) can just have every x(t), entire *functions* of t, substituted directly by t and still mean the same thing??

[Anonymous No.9671183]

>>9671145
A friend of mine was trying to learn the work-energy theorem and said that such thing pop up in the proof. Now I said to him that with ghetto math you can do that, but he told that still it doesn't make sense. I tried to make some sense of it but I can't.

[Anonymous No.9671205]

>>9671169
Think about the cyclotomic field [math]\mathbb{Q}(a_m)[/math] acting on itself where [math]a_m = \exp(i2 \pi \frac{1}{m})[/math]. [math]a^{m} = e[/math] implies [math]a^{m+1} = a[/math], which looks like [math]a[/math] "twists back onto itself". A module is torsion then means that there are non-zero elements that "twists back onto itself", i.e. has torsion.
>>9671183
By Radon-Nykodym theorem, if [math]\nu \in \mathcal{M}[/math] is a measure that's absolutely continuous with respect to another measure [math]\mu[/math] on a Borel set [math]B[/math], then there exists a measurable function [math]f[/math] such that [eqn]\nu(A) = \int_B fd\mu[/eqn]. Since continuous functions are measures, absolutely continuous functions [math]r(t)[/math] such that [math]r(t) - r(t') = \int_{t}^{t'}d\tau r'(\tau)[/math] with a weak derivative [math]r'[/math] are absolutely continuous with respect to the standard measure [math]d\tau[/math], and hence automatically satisfies the Radon-Nikodym theorem with [math]f = r'[/math]. In addition, curves [math]\gamma \subset \mathbb{R}^n[/math] are Borel, hence
[eqn]E = \int_\gamma {\bf F}\cdot d{\bf r} = \int_\gamma {\bf F}\cdot {\bf r}' dt = \int_\gamma {\bf F}\cdot {\bf v} dt.[/eqn]

[Anonymous No.9671207]

>>9671205
It should be [math]\nu(B)[/math].

[Anonymous No.9671321]

Does anyone have a babel.hathitrust.org partner account?
I'm trying to download a book PDF but they only allow that for partner accounts.

Link: https://babel.hathitrust.org/cgi/pt?id=mdp.39015033508683

[Anonymous No.9671329]

>>9671045
perhaps its written that way to avoid confusing people who might try to reintroduce the original differential equation in y, but yes it is pretty pointless and you don't have to do that.

[Anonymous No.9671332]

>>9671044
>You can just grow out of mental illness by intentionally adding more stress to your life
hmmmmmmmmmm

[Anonymous No.9671367]

I'm doing a linear stability analysis for a two dimensional flow around a disk using both finite difference and psuedo-spectral method, before discretizing the infinite domain, I apply the inversion map z=r^(-1). For fd I use equidistant points while for ps I used Chebyshev's function and cosines.
Now the problem is that the eigenvalues (I'll talk only on the smallest magnitude one) are not converged with respect to the radial resolution, as the resolution is doubled, the eigenvalue is approximately halved. I've found that when I change from infinite domain to a finite domain, the eigenvalues converge and both methods give the same eigenvalues. The fastest convergence is for a domain of twice the disk radius and it slows down (until no convergence appears (for the time limit and computer's power)) as the domain's size increases.
Anyone have a take on this? I couldn't find papers on this kind of behavior.
What I did find is the possibility that the eigenspectrum is continuous and not discrete (in that case there is not much I can do) and that perhaps the discretization next to "infinity" (z=0) is not dense enough to describe the eigenfunction's behavior there and a more sophisticated map is needed.
Thank

[Anonymous No.9671460]

How do I solve this?

[Anonymous No.9671491]

guys I'm trying to verify equations in trigonometry and the one I'm working on is:

>secB + tanB = CosB/1 - SinB

I ended up turning the left side into 1 + SinB/CosB, but I don't know what the fuck I'm supposed to do now pls help.

[Anonymous No.9671492]

>>9671146
Thanks anon, that helped. However, I don't understand how you got to
[eqn]\left(1 - \frac{\lambda^2}{4}\right)|\phi|^2_{L^2} = 0.[/eqn]

[Anonymous No.9671496]

>>9671491
>CosB/1
What's with the 1?

[Anonymous No.9671504]

>>9671496
I should've written it as "CosB/(1 - SinB)", my bad.

[Anonymous No.9671506]

>>9671491
>verify
do you mean you're trying to solve the equation for a particular value, or verify a trigonometric identity, i.e. one that is true for any x?

[Anonymous No.9671508]

>>9671506
>verify a trigonometric identity
That one.

[Anonymous No.9671522]

>>9671508
right, I was asking because what you wrote doesn't hold for any B but I see you made a typographical error
1/cos x + sin x / cos x = cos x / (1 - sin x)
1 + sin x = cos^2 x / (1 - sin x)
(1 - sin x) (1 + sin x ) = 1 - sin^2 x = cos^2
1 = sin^x + cos^x
q e d

[Anonymous No.9671536]

>>9671508
When in doubt, multiply by the conjugate!

[Anonymous No.9671547]

>>9671492
[math]|\phi|_{L^2} = |\mathcal{F}\phi|_{L^2}[/math] and [math]\phi = \frac{\lambda}{2}\mathcal{F}\phi[/math] implies [math]|\phi|_{L^2} = \frac{\lambda}{2}|\mathcal{F}\phi|_{L^2} = \frac{\lambda}{2}|\phi|_{L^2}[/math].
Also to go forward with the Neumann series, if you assuming [math]\phi[/math] is absolutely continuous then [math]\mathcal{F}T = T\mathcal{F} = \mathcal{F}^{-1}[/math] where [math]Tf(x) = f(-x)[/math] is the inversion satisfying [math]T^2 = I[/math], so [math]\mathcal{F}^{2} = \mathcal{F} T^2 \mathcal{F} = \mathcal{F}^{-2}[/math], which implies that [math]\mathcal{F}^m = I[/math] for any [math]m \equiv 0 \mod 4[/math].
At the end you obtain
[eqn]
\phi = \sum_n\left(\frac{\lambda}{2}\right)^2(I
+ \mathcal{F}+ \mathcal{F}^{-1} + \mathcal{F}^2)f.[/eqn]

[Anonymous No.9671554]

>>9671522
>>9671536
Fug, thanks a bunch anons.

[Anonymous No.9671555]

>>9671547
fix your laytech mathfag

[Anonymous No.9671559]

>>9671555
[math]\sum_n \left(\frac{\lambda}{2}\right)^n[/math]

[Anonymous No.9671560]

>>9671547
Ah yes it does. I feel pretty dumb for not spotting that, but thanks a lot!
based avatarfriend

[Anonymous No.9671595]

Excel Solver sensitivity report. How do I answer this question?

[Anonymous No.9671620]

I'm not trying to figure out another 400 level math class to put in my schedule which is currently

Linear Algebra
Mathematical Statistics
Game Theory
Economics of Labor Markets
4xx

I'm between Abstract Algebra, Algebraic Geometry, Topology, and then a 400 level course in either ODEs or PDEs.

My algebra isn't very rich which is why I'm hesitant to take abstract algebra as well as algebraic geometry. I have to take topology but I can really take it whenever so that isn't much of an issue.

If it matters I'm doubling in math and econ and hoping to go to grad school for economics.

[Anonymous No.9671621]

>>9670851
I need an honest answer lads. Realistically how long would it take me to learn Math (not CS) Linear Algebra? Is it possible to learn it in around 2 weeks? I just want to get it done.

[Anonymous No.9671651]

>>9671621
Too vague to answer. Linear algebra is a fucking huge subject.
If you're reasonably smart it is probably possible to cram a linear algebra 1 course into two weeks if you have zero other obligations.

[Anonymous No.9671658]

>>9671651
>If you're reasonably smart it is probably possible to cram a linear algebra 1 course into two weeks if you have zero other obligations.
No fucking way. Maybe if you could study at peak performance on less than 4 hours of sleep per day.

[Anonymous No.9671668]

>>9671620
You should try to take both ODEs and PDEs at some point in your education, but do ODEs first. They're significantly easier to handle.

Also if your school even allows you to take algebraic geometry before you take algebra, your algebraic geometry course blows ass and is not worth your time.

[Anonymous No.9671672]

can someone please show me the resonance structures for this bastard (6-Bromo-4-chloroquinoline)? I'm trying to find the pKa for it using a set of data designating the logKaX values for Substituents X, but I would just like to see what the resonance structures are for this as from my understanding this helps with the solving the problem.

[Anonymous No.9671683]

Is the Linear Algebra course on MIT OCW good enough for the first contact with the subject? I want to study with Axler's book after finishing the course

[Anonymous No.9671696]

i'm trying to do a monte carlo tolerance analysis for an analog circuit, so i'm starting with a min and max value, and an assumed probability that a manufactured part will be in those bounds before its batch is tested and bad parts are tossed. so i want to generate random numbers based on a distribution like that (pic). i'm statistically illiterate though. can someone tell me how to calculate a standard deviation from those variables so i can feed it into a gaussian rng?

[Anonymous No.9671734]

What is a good assumption for the velocity, and pressure within the flue here?

burning LPG, temp at start of the flue is ~500'C, dimensions of the flue are
Heght: 40mm; Length: 321mm; Width: 56mm

would 2m/s and slightly under 1 bar sound about right?

[Anonymous No.9671739]

>>9671734

[Anonymous No.9671812]

what does it mean for the limit of a function to "exist"?
I always presumed it meant that we can assign one of [math] R=\ell\in\mathbb{R},``\infty",``-\infty" [/math] to the limit in a meaningful way so we may write [math] \lim_{x\to x_0}f(x)=R [/math], though i think im wrong

[Anonymous No.9671846]

>>9671812
that its finite. Because infinity is not a number.

[Anonymous No.9671883]

>>9671812
for a series [math]f_1, f_2, f_3, ...,f_n,...[/math] where
[eqn]\lim_{n\to\infty}f_n=f[/eqn] then for any [math]\epsilon\in \mathbb{R} [/math] such that [math]\epsilon>0[/math] there exists some [math]n[/math] for which [math]f - f_m<|\epsilon|[/math] for all [math]m>n[/math]

in other words, if a series tends to a limit then you can get arbitrarily close to the limit by computing enough terms.

[Anonymous No.9671886]

>>9671883
fug I meant [math]|f - f_m|<\epsilon[/math]

[Anonymous No.9671895]

when a bird is sitting on a power line, it doesn't get shocked because there's no path to ground.

however, why doesn't the bird's body act as a "ground" since its potential is 0? can't current travel around the bird's body to complete a circuit?

[Anonymous No.9671902]

>>9671895
its potential isn't 0 relative to ground, it's equal to whatever it's sitting on. think of the bird as a large resistor. a resistor across two points of equal voltage won't have any current flowing through it. if the bird were to manage to put one foot on a line and one foot on the ground then it would be shocked because its internal "resistor" suddenly has a voltage across it.

[Anonymous No.9671903]

>>9671895
>its potential is 0
potential is only really meaningful when defined as the difference in potential between two points.
The potential difference between the power line and the ground is very large, the potential difference between the power line and the bird is 0.

[Anonymous No.9671913]

>>9671846
>>9671883
thanks. whenever i saw that phrase that always seemed implied, i had just never seen it written down

[Anonymous No.9671922]

>>9670935
The two dt’s cancel eachother out

(X/X) = 1
(XY/X)= Y

[Anonymous No.9671923]

>>9671902
>>9671903

that clarifies that a lot. thanks guys

[Anonymous No.9671929]

>>9671595
Looks like cost accounting

[Anonymous No.9671937]

>>9671621
Depends on your mathematical maturity.

[Anonymous No.9671959]

why the fuck is lagrange error bound so retarded

[Anonymous No.9671968]

>>9671672
There are no resonance structures, but you can redraw it to show aromaticity like this.

[Anonymous No.9672085]

>>9671547
Doesn’t this simplify more? F^2 = I when applied to even functions

[Anonymous No.9672202]

Do black holes evaporate due to annihilation?

[Anonymous No.9672225]

If the moon and sun cause the tides via gravitation, why don't lakes and puddles and other bodies of water not affected?

[Anonymous No.9672319]

>>9672225
They are, but it scales based on how large and open the body of water is, and the ocean is fucking huge dude.

[Anonymous No.9672343]

>>9672319
Why should scale make any difference?

[Anonymous No.9672414]

Alright, algebrafags. I got a ring theory question for you. I'm supposed to prove the following statement:

Suppose [math]f:A\rightarrow{B}[/math] is a homomorphism from [math]A[/math] onto [math]B[/math]. Suppose further that [math]J[/math] is an ideal of [math]A[/math] and [math]K_{f}\subset{J}[/math], where [math]K_{f}[/math] is the kernel of [math]f[/math]. Then [math]f(J)[/math] is an ideal of [math]B[/math].

Here's my proof so far:
Let [math]j\in{J}[/math]. Because [math]f[/math] is onto, we let [math]b\in{B}[/math] and [math]f(a)=b[/math], where [math]a\in{A}[/math].

Then [math]f(j)b = f(j)f(a) = f(ja)[/math]. And [math]f(ja)\in{f(J)}[/math] because [math]J[/math] is an ideal in [math]A[/math]. The [math]bf(j)[/math] case is proved similarly. Q.E.D.

My only problem is that I never used any information regarding the kernel. Is there a flaw with my proof? Thanks a ton in advance.

[Anonymous No.9672424]

Is it possible to escape the event horizon of a black hole if a larger black hole were to come by and suck you out of it?

[Anonymous No.9672444]

>>9672414
> Is there a flaw with my proof?
You didn't show f(J) is an ideal.

[Anonymous No.9672452]

>>9672414
You don't need any knowledge about the kernel to prove your statement. It holds for arbitrary ideals if f is surjective.

My only guess for why information about the kernel was given to you is that your prof/book wanted you to say that A/kerf = B and use the correspondence theorem instead of checking manually.

[Anonymous No.9672461]

>>9672452
That's a possibility, thanks for the info.

>>9672444
I thought that by showing that [math]f(j)b=f(ja)\in{f(J)}[/math] I showed that it absorbed products and is therefore ideal. Can you elaborate? I guess I forgot to literally say in my proof that 'this means [math]f(J)[/math] is ideal.'

[Anonymous No.9672468]

>>9672461
>Can you elaborate?
Reread the definition of ideal.

[Anonymous No.9672471]

>>9672414

what's the name of this course?

[Anonymous No.9672472]

>>9672424
Afaik, you'd be turned into black hole goo and important bits of information you carried in your constituent particles would disappear from reality. So whatever comes back out of the black hole is just going to be dark matter or Hawking radiation, and there would be nothing to indicate that you were ever even there.

[Anonymous No.9672473]

>>9672461
Ideals have more properties than just absorbing multiplication. They're subrings, and you didn't show that.
Normally this is a bit pedantic of him to nitpick but if this were on an abstract algebra exam you'd lose points for not showing that

[Anonymous No.9672474]

>>9672444
>>9672468
... Ahhh... it must also be closed to with addition and subtraction. Definitely a stupid quesiton. Thanks.

[Anonymous No.9672479]

>>9672472
Uhh no. Haven't you watched Interstellar? They fly into a black hole's event horizon and escape just fine by entering the 4th dimension. They don't turn into spaghetti powder lmao you dumb idiot.

[Anonymous No.9672572]

>>9672471
Looks like an introductory abstract algebra course.

[Anonymous No.9672598]

Does anyone know a good neuroscience book that goes over some of the structures of neurons in the brain? I've gone through a chapter of pic related and it's been great so far, but it seems to focus on building a good model of an individual neuron (correct me if I'm wrong here).
Context: I'm an electrical engineer / computer architect. I'd like to learn about what kinds of computational patterns neurons group into, so that I can turn that into circuits. I don't even know what field that is though, seems to be something between computational neuroscience and neuroanatomy.
I'm still going to finish this one regardless, but I'd like to know where to go when I'm done with it.

[Anonymous No.9672610]

>>9671953
hat's the same thing I did anon...

[Anonymous No.9672619]

>>9672598
neuroscience just backports ML memes. They don't know how the brain works.

[Anonymous No.9672642]

>>9672619
I do that too though

[Anonymous No.9672773]

>>9672085
[math]f[/math] is even iff [math]f\in\operatorname{Ker}(T -I)[/math]. If [math]f\in\operatorname{Ker}(T -I)\cap C^\text{abs}[/math] then [math]\mathcal{F}f = \mathcal{F}{-1}f[/math], so [math]\mathcal{F}^2f = \mathcal{F}\mathcal{F}^{-1}f = f[/math]. How it is not specified if [math]f[/math] is even or not from your question.

[Anonymous No.9672778]

>>9672773
Man it's 4:30am here.
[math]\mathcal{F}f = \mathcal{F}^{-1}f[/math].
>How
However*

[Anonymous No.9672815]

Pretty sure this is a stupid one. How do I find the intersections?
I've pulled an all nighter and that's the only problem I'm missing.

[Anonymous No.9672819]

>>9672815
I presume you know the center position and radius?

[Anonymous No.9672841]

>>9672819
Nope, it's just a Venn Diagram I made to represent the data.

In text form: U= 110; A= 63; B= 30; C= 50; A∩B∩C= 7; A-(B∪C)= 30; B-(A∪C)= 13; C-(A∪B)= 25.
I need to find A∩B, A∩C and B∩C. Can't figure it out for the life of me.

[Anonymous No.9672844]

>>9672841
ah not familiar with Venn's yet

[Anonymous No.9672849]

So permanent magnets aligned on the same pole will repel each other.

Why can't a motor be used that utilizes permanent magnets, repelling each other, with no other inputs, to do useful work? It would seem to a perpetual energy machine so there must be some reason why it can't work. Do they simply demagnetize themselves if they are manipulated to do work? Lazy fucking physics, I swear.

[Anonymous No.9672850]

>>9672844
No, missed a month straight of classes and I gotta get through 8 papers on my own.
Google got me through it so far but this one is killing me. Any idea how it's done?

[Anonymous No.9672851]

>>9672850
I mean I'm not familiar with Venn's yet, didnt mean to address you
we are in the same boat

[Anonymous No.9672855]

>>9672851
Oh, got it now senpai. Thanks anyways for trying though.

[Anonymous No.9672871]

>>9672815
This is solved very easily with the use of an augmented matrix
[eqn] \begin{bmatrix}
1 & 1 & 0\\
1 & 0 & 1\\
0 & 1 & 1
\end{bmatrix} \times \begin{bmatrix}
A\cap C\\
A\cap B\\
B\cap C
\end{bmatrix}
=\begin{bmatrix}
63-30-7\\
50-25-7\\
30-13-7
\end{bmatrix} [/eqn]
and row reduce to get 17,9,and 1
you can also use it to show that a total of 110 is impossible since the augmented matrix will reduce to an inconsistent matrix

[Anonymous No.9672920]

Should I take a course in complex analysis or abstract algebra?

Which one ramps up and gets good in the first week or so?

[Anonymous No.9672921]

How do I go about evaluating this summation if N is a random variable? It means the summation is always changing in length depending on the Poisson random variable.

Any resources or suggestions for attempting to evaluate this?

[Anonymous No.9672957]

>>9672921
In these types of questions, you need to use conditional probabilities

For instance, I guess that [math]b_i[/math] or such that you have no problem computing the sum is [math]N[/math] is fixed

I think you can calculate the sum as a function of N, something like [math] \sum_i^N b_i = f(N)[/math]

And then use conditional probabilities : the probability that sum is [math]f(N)[/math] is something like [math]f(N) \cdot p(Poisson = N)[/math]

I think, I'm a bit rusty in probability

[Anonymous No.9673028]

How do I calculate tg(a) manually?

[Anonymous No.9673043]

>>9672815
A=63, A-(B∪C)=30 => A∩(B∪C)=63-30=33
B=30, B-(A∪C)=13 => B∩(A∪C)=30-13=17
C=50, C-(A∪B)=25 => C∩(A∪B)=50-25=25

A∩(B∪C)=33, A∩B∩C=7 => A∩(B∪C-B∩C)=33-7=26
B∩(A∪C)=17, A∩B∩C=7 => B∩(A∪C-A∩C)=17-7=10
C∩(A∪B)=25, A∩B∩C=7 => C∩(A∪B-A∩B)=25-7=18

A∩(B∪C-B∩C)=26=(A∩B-C)∪(A∩C-B)=(A∩B-C)+(A∩C-B)
B∩(A∪C-A∩C)=10=(A∩B-C)∪(B∩C-A)=(A∩B-C)+(B∩C-A)
C∩(A∪B-A∩B)=18=(A∩C-B)∪(B∩C-A)=(A∩C-B)+(B∩C-A)

3 equations in 3 unknowns.

b+c=26 (1)
a+c=10 (2)
a+b=18 (3)

(1)+(2)-(3) => b+c+a+c-a-b = 26+10-18
=> 2c = 18
=> c = 9 (4)
(1)-(4) => b=26-9=17
(2)-(4) => a=10-9=1

(B∩C-A)=1
(A∩C-B)=17
(A∩B-C)=9
A∩B = (A∩B-C)+A∩B∩C = 9+7 = 16
B∩C = (B∩C-A)+A∩B∩C = 1+7 = 8
A∩C = (A∩C-B)+A∩B∩C = 17+7 = 24

Check:
A = (A-(B∪C))+(A∩C-B)+(A∩B-C)+A∩B∩C = 30+17+9+7 = 63
B = (B-(A∪C))+(B∩C-A)+(A∩B-C)+A∩B∩C = 13+1+9+7 = 30
C = (C-(A∪B))+(A∩C-B)+(B∩C-A)+A∩B∩C = 25+17+1+7 = 50
A∪B∪C = (A-(B∪C))+(B-(A∪C))+(C-(A∪B))+(B∩C-A)+(A∩C-B)+(A∩B-C)+A∩B∩C = 30+13+25+1+17+9+7 = 68+27+7 = 112

Dunno where you got 110 from.

[Anonymous No.9673055]

I'm trying to evaluate the following integral, but I'm not getting the correct answer.
[eqn]I = \frac{1}{2 \pi} \int^\infty_\infty i e^{-i k x} \left( \frac{1}{x + i} - \frac{1}{x-i} \right) dx. [/eqn]
Mathematica says the answer is [math]e^{-|k|}[/math].
Now this can obviously be done via contour integration. By the normal ML estimates, if [math] k > 0 [/math] the contour should be the semicircle closed in the lower half plane, if [math]k < 0[/math] it should be closed in the upper half plane. Hence this should evaluate to

[eqn] I = \left\lbrace
\begin{matrix}
-\mathrm{res} \left( -\frac{e^{-i k x} }{x - i}, \, i \right), & k<0 \\
-\mathrm{res} \left( \frac{e^{-i k x} }{x +i}, \, -i \right), & k > 0 \end{matrix} \right. [/eqn]

Now those residues are just simple poles and evaluate to
[eqn] \mathrm{res} \left( \frac{e^{-i k x} }{x - i}, \, i \right) = e^{ k } \qquad \mathrm{and} \qquad -\mathrm{res} \left( \frac{e^{-i k x} }{x + i}, \, -i \right) = -e^{- k } [/eqn]
so that
[eqn] I = \left\lbrace
\begin{matrix}
e^{k}, & k<0 \\
-e^{-k}, & k > 0 \end{matrix} \right. = - \mathrm{sign} \, ( k ) \, e^{-|k|} [/eqn]
and not the correct answer [math]e^{-|k|} [/math]. The thing is I can't find a flaw in what I've done. The residues seem to be correct and Mathematica agrees with them, and the contours seem to be correct.
Any ideas where I might have fucked up?

[Anonymous No.9673060]

I'm trying to evaluate the following integral, but I'm not getting the correct answer.
[eqn]I = \frac{1}{2 \pi} \int^\infty_{-\infty} e^{-i k x} \left( \frac{1}{x + i} - \frac{1}{x-i} \right) dx. [/eqn]
Mathematica says the answer is [math]- i e^{-|k|}[/math].
Now this can obviously be done via contour integration. By the normal ML estimates, if [math] k > 0 [/math] the contour should be the semicircle closed in the lower half plane, if [math]k < 0[/math] it should be closed in the upper half plane. Hence this should evaluate to

[eqn] I = \left\lbrace
\begin{matrix}
i \, \mathrm{res} \left( -\frac{e^{-i k x} }{x - i}, \, i \right), & k<0 \\
i \, \mathrm{res} \left( \frac{e^{-i k x} }{x +i}, \, -i \right), & k > 0 \end{matrix} \right. [/eqn]

Now those residues are just simple poles and evaluate to
[eqn] \mathrm{res} \left( -\frac{e^{-i k x} }{x - i}, \, i \right) = -e^{ k } \qquad \mathrm{and} \qquad -\mathrm{res} \left( \frac{e^{-i k x} }{x + i}, \, -i \right) = e^{- k } [/eqn]
so that
[eqn] I = \left\lbrace
\begin{matrix}
-i e^{k}, & k<0 \\
i e^{-k}, & k > 0 \end{matrix} \right. = i \, \mathrm{sign} \, ( k ) \, e^{-|k|} [/eqn]
and not the correct answer [math]-i e^{-|k|} [/math]. The thing is I can't find a flaw in what I've done. The residues seem to be correct and Mathematica agrees with them, and the contours seem to be correct.

[Anonymous No.9673064]

>>9672921
The result will be another random variable. It's unlikely that this variable would follow any common distribution, or have a tractable mass function.

[Anonymous No.9673123]

>>9673064
>>9672957
Are you sure? I only doubt because I've simulated the problem using Python which generates roughly 10 000 $b_i$ and then sums them up over their randomly generated N. Then I filter out the sequences in which the sum equalled 0, and find their proportion over the total sequences.

After doing that whole process 10 000 times my computer seems pretty confident that it's 0.1355xxxxxx, but every time I run it those first 4 numbers never change.

I formally put all the information for the question here, which might provide a little more context to what I need to work towards.
[statsDOTstackexchangeDOTcumSLASHquestionsSLASH340632]

If you need any extra information just ask.

[Anonymous No.9673124]

>>9673060
That's not how contour integrals work.

[Anonymous No.9673137]

>>9673060
You forgot a sign when closing in the lower-half plane. The integral along the complex contour has to be counter-clockwise, so the limits swap and that's the sign that you're missing.

[Anonymous No.9673142]

>>9673060
check this
http://wwwf.imperial.ac.uk/~jdg/eejordan.pdf

[Anonymous No.9673151]

>>9673137
cheers anon! I knew it would be something simple like that.

[Anonymous No.9673168]

Given the simple condition triangle 1 similar to triangle 2 in an unknown trapezium (could be right angle one, could be equal-sided could be a wild fucking thing) how do you assert proportional side according to similarity rules. Is there some kind of technique/rule of thumb? Right now I just use my gut because "obviously" I can put AC:AC since other sides arent equal by definition (trapezium).

I understand it is
[math]
\frac{AB}{CD} = \frac{BC}{AC} = \frac{AC}{AD}
[/math]

but cant always rely on intuition, in some cases triangles can look VERY different because drawing always approximate.
I feel like I'm missing something?

[Anonymous No.9673175]

>>9673168
there's probably something I'm missing from your post, but if the triangles are similar then by definition they are only differ only by an overall scale, a rotation, or a reflection. Since in a trapezium triangles share one side, it follows that their overall scale is equal.

[Anonymous No.9673190]

>>9673175
No, it is not the fact they are similar I have issues with, they are similar by definition of the exercise.

I'm having issue with sides cross-referencing, like, some exercises can give very vague description so your drawing could be WAY off (like, in the end it may turn out triangle is right-angled but you dont know it at first and draw random one - just for example). In that particular case what is the rule of thumb which sides are corresponding? I mean, I know whcih ones, I already said that I need the explanation if there is a rule that makes it easier to see

[Anonymous No.9673263]

>>9672871
>>9673043

That's the part that I'm mostly struggling with. The assignment says: There's a total of 110 people that are divided between those groups. How many of those 110 won't take part in any group?
But I did something similar to the second anon and also got 112. According to the assignment there should and I don't know why I'm two elements over the limit when it says there should even be a leftover.
Maybe there's an error in the exercise?

Thanks a lot for taking the time to answer guys.

[Anonymous No.9673270]

is there some simple method to extract square/cubic roots out of big numbers you dont know by memory?

[Anonymous No.9673366]

>>9673270
other than just using a calculator you can use the Newton's method:
[eqn]x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}[/eqn]
to find the square root of S, you need to solve x^2 - S= 0, thus you have
[eqn]x_{n+1} = x_n - \frac{{x_n}^2 - S}{2x_n} = \frac{1}{2} \left( x_n + \frac{S}{x_n} \right)[/eqn]
start with some estimate, for say 5000 we might make an initial guess of 50,
[eqn] x_1 = 50
x_2 = (50 + 5000/50)/2 = 75
x_3 = (75 + 5000/75)/2 \approx 71
[/eqn]
you can check 71^2 = 5041 which is a pretty good estimate for just the 3rd iteration.

[Anonymous No.9673372]

>>9673366
formatting
[eqn] x_1 = 50 \\ x_2 = (50 + 5000/50)/2 = 75 \\x_3 = (75 + 5000/75)/2 \approx 71
[/eqn]

[Anonymous No.9673394]

>>9673366
>>9673372
I see, thanks.

[Anonymous No.9673421]

>>9671668
If I could fit 2 or 3 grad courses in my schedule which would you recommend I take? Right now I think I'm doing

Semester n

Topology (Under)
Linear Algebra (Under)
Mathematical Statistics (Under)

Semester n+1

Analysis I (Under)
Abstract Algebra (Under)

Semester n+2

Analysis II (Under)
Complex Analysis (Under)
PDEs (Under)

Semester n+3

ODEs (Under)
Insert Graduate Course

Semester n+4

Insert Graduate Course
Possibly 2nd Graduate Course

[Anonymous No.9673430]

I have a question about biology or maybe chemistry.

Say you have two identical people consuming identical diets and exercising regularly in an identical manner. The only difference between these two people is that one is testosterone. Obviously the one injecting testosterone will put on more muscle mass. However, does this mean the person injecting testosterone would be defecating smaller amounts of stool? The mass has to be coming from somewhere (the food). So if the mass is coming from the food then more of it is coming from the food for the one taking testosterone. So he must be taking smaller shits, right?

[Anonymous No.9673534]

>>9671367
Dude just use multigrid lmao

[Anonymous No.9673592]

goys how do I verify is an identity? Everything I try turns it into a mess, I don't know what I'm doing.

[Anonymous No.9673756]

>>9673592
what does "identity" means?

[Anonymous No.9673768]

>>9673592
I've got:

sec0^2-1^2 = tan0^2

or
(sec0-1)(sec0+1) = tan0^2

I have no idea what sec0 means so can't simplify further but you should know what?
Can you express sec0 via tan or sin/cos?

[Anonymous No.9673946]

Is there a formula for base conversion without converting the numbers to base 10?

[Anonymous No.9673978]

>>9673592
Let [math] \displaystyle \Phi(\theta) =\frac{\sec\theta-1}{\tan\theta}\,,\quad \theta\in\operatorname{dom}(\Phi) [/math] and note that [math] \operatorname{dom}(\Phi)=\{\operatorname{dom}(\tan):\tan(x)\neq 0\}\cap\operatorname{dom}(\sec)=\{x\in\mathbb{R}:x\neq 0,\pm\pi/2, \pm\pi,...\} [/math].
It suffices to show that [math] \displaystyle\Phi(\theta)=\frac{\tan\theta}{\sec\theta+1} [/math] for all [math] \theta\in\operatorname{dom}(\Phi) [/math]. To this end, multiplying through by [math] \tan\theta(\sec\theta+1) [/math] yields [eqn] \begin{align}\tan\theta\cdot\Phi(\theta)\cdot \sec\theta+1&=\sec(\theta)^2-1\\ &=\tan(\theta)^2\,;\\ \Phi(\theta)&=\frac{\tan(\theta)^2}{\tan\theta}\frac{1}{\sec\theta+1}\\&=\frac{\tan\theta}{\sec\theta+1}\,,\quad \theta\in\operatorname{dom}(\Phi)\,.\end{align} [/eqn]
Dividing through by sec theta +1 is valid since it is never zero in the domain of Phi

[Anonymous No.9673987]

>>9673592
multiply both sides by the denominator
[eqn](\sec \theta - 1 )(\sec \theta + 1 ) = \sec^2 \theta - 1 = \tan^2 \theta
[/eqn]

[Anonymous No.9674026]

>>9673978
Imagine being this rigorous in a trig course.

[Anonymous No.9674079]

How to solve for the constants when there are two non elementary integrals????

[Anonymous No.9674094]

>>9673043
>>9673263
>68+27+7 = 112
should be 102

[Anonymous No.9674109]

>>9674079
i've not done it, but im guessing with the initial values you'll end up with integrals of the form [math] \displaystyle \int_0^0 f \,dt [/math] for y(0) and then for y'(0) you can use the fundamental theorem of calculus

[Anonymous No.9674217]

>>9674109
Thanks. Solved it in the end

[Anonymous No.9674284]

Are there any non-academia fields of research that someone in the applied track can get into? I really want to do research but I can not become a professor I would be doing my students a disservice.

Also I know it's not really research but Operational Research in the military sounds really cool but I have a feeling that it might not end up being what I expected.

[Anonymous No.9674296]

>>9674284
Roche

[Anonymous No.9674304]

>>9670851
Let's say I have "34.66 years", and I want to express it in years and months. 1 month is 1/12th of a year. So I divide .66/(1/12) = 7.92, which means 34.66 years roughly equals 34yrs and 8 months. Is my reasoning correct?

[Anonymous No.9674312]

I'm trying to work through a homework proof that basically comes down to "Riesz Representation Theorem does not hold on infinite-dimensional vector spaces without further constraints"
Can somebody maybe provide me with a quick explanation of why that is? I'm swinging way out of my league with this class and I just need to scrape by and get out of everybodys way

[Anonymous No.9674314]

>>9674296
>Roche
Muh fucking dick. I've always wanted to work in the Military or Pharmaceuticals. I am a burger, the American pharmaceutical companies should have math jobs as well right?

[Anonymous No.9674328]

>>9670935
dv is not really defined for you, so you could think of it as a generalised function, where it only makes sense when you apply an integration

[Anonymous No.9674337]

>>9671169
algebraic topology. The homology groups having torsion subgroups (for example a Z/2Z group for RP^2) usually correspond to non-orientability.

[Anonymous No.9674349]

>>9671812
You are right, we say the limit is in [math]\mathbb R\cup \{\pm\infty\}[/math]. Intuitively, limits don't exist when they never converge. For example, the sequence (-1)^n is bounded but never converges, or the sequence n for n odd and 0 for n even also tends to infinity and 0 at the same time. In essence, it boils down to what the limsup and liminf are.

[Anonymous No.9674361]

>>9672920
complex is goat but abstract algebra is essential and probably harder

[Anonymous No.9674448]

can someone explain what im misunderstanding about what's in the box
i understand the proof afterwards, but i dont understand why [math] \deg g(X), \deg h(X)>0 [/math] should follow from f(X) being reducible in Z[X]. for example, if say [math] f(X)=4 [/math], then f(X) is reducible in Z[X] since 4 is not a unit, but deg 4=0

[Anonymous No.9674456]

>>9674448
Reread your definition of "reducible".

[Anonymous No.9674480]

>>9674448
I BELIEVE that the lemma should also assume that degf>0. Then primitivity eliminates your ability to factor out a constant.

[Anonymous No.9674488]

>>9674349
i feel like im missing something really obvious, but i can't see how what i've wrote contradicts the definition

[Anonymous No.9674489]

>>9674480
>>9674488
>>9674488

[Anonymous No.9674493]

How do I know when to drop out?

[Anonymous No.9674499]

>>9674493
when your other option is suicide

[Anonymous No.9674522]

>>9674499
Well without school I'm a 19 year old virgin withing with my parents who hate me and no job qualifications; is that close enough?

[Anonymous No.9674535]

>>9674522
why would you want to drop out then?

I lost my virginity at 20 after I decided to get fit after being fat for a long time. It wasn't as hard as the autists at R9K might make you believe, especially if you're in college, and Im an autist too. If you feel you're too brainlet, you should try going to a trade school, that is if, I'm assuming, you don't want a shitty office job. You can make a good salary after not too long and if you're mobile, you can really start raking in the money if you're good.

[Anonymous No.9674610]

If I neglect the definitions of the electric field and magnetic field in terms of q and r and focus on solving all electrodynamic and electrostatic problems by solving for the potentials first, will I be fucking myself over? Or is it a completely fair strategy to become good at finding the potentials in order to find the fields?

[Anonymous No.9674619]

>>9674522
Just get your degree, even if it's not with a good GPA. You're still in a far better position than if you had done nothing at all. Realistically, you don't need a huge salary to be happy, even with a family. If you make only 30k a year, the government will help you out by not asking for taxes and if you're not a retard with money then you will actually be comfortable. I'm getting a double degree in chem and phys and I don't care for a high salary. I know I'm frugal enough to be good with 30k a year if things come to that. The reason poor people struggle so much isn't just because they make very low salaries, it's also because they spend recklessly.

[Anonymous No.9674808]

How do I stop my self-made anatomy anki deck from going so fucking quickly? Like I'll say I'm good and then the card won't show up for a whole week, and then if I say I'm good again it's gone for the next month.

[Anonymous No.9674850]

>>9674610
Gauge potentials are more fundamental in the sense that they are the ones that arise naturally as connection 1-forms on a [math]G[/math]-principal bundle in gauge theories, not the physical fields themselves. However, the kind of equations they satisfy depends on which gauge you chose for them. The Coulumb gauge leads to the scalar potential satisfying the Poisson equation in electrostatics, for instance.

[Anonymous No.9674957]

>>9674619
Why bother with a double degree if you don't really care about a high salary?

[Anonymous No.9674971]

>>9671547
>>9671559
Use the TeX previewer in the "Reply to Thread No.xxxx" box. It's in the top left corner.

[Anonymous No.9674977]

>>9674971
Refresh your page.

[Anonymous No.9675049]

This has been on my mind a lot recently.

How do you accurately rank based on scores? For example, in the picture (which is a screenshot of the website MyAnimeList), they have a way of ranking things that makes a lot more sense than in IMDb.
For example, if you look at "Top Animations" in IMDb, you consistently see shows/movies that have very few votes out-ranking more popular films.

With the MAL list, though, the #4 show has a slightly higher score than the #5 show, despite the 5th show having many more members. This system is apparently slightly biased towards less people voting (but I may be wrong here)

I guess my question is, what would a good formula be for ranking shows?
Something I came up with quickly, and works decently for sufficiently large numbers of votes, is:
[math]
Score = (Avg) \cdot [ log_{10}(V) + 1]
[/math]
Where V is the number of people that have rated a show, and "Avg" is the average of the scores.

One small problem is that, with this, it's possible for a show with an 8.5 average to be ranked higher than a show with a 9.0 (although in this case, if the show with an 8.5 had [math] b [/math] people voting, and the show with a 9.0 had [math] a [/math] people, then [math] b \geq a^{1.05} [/math] for this to happen)

Is this a good scheme? How can I improve it?

[Anonymous No.9675075]

>>9674957
I don't care about a high salary because I don't need much money to live comfortably. If I have a family then I actually would need a high salary. However, because I know I most likely never will be in a relationship (despite the fact that I seriously want to have children), I don't really have to worry about money. What I'm going after instead is a job/profession where I do something that I find really interesting. Anything less than that and I won't know what to do with myself considering I don't believe I have it in me to commit suicide.

Anyway, a double major doesn't guarantee a higher salary, and I've even heard that it can lower your salary because employers will assume that you know only 75% of each major as opposed to 95% of one major for someone who didn't double major (this negative effect shouldn't apply to me because I intend on going to graduate school). Essentially, I'm double majoring solely because I find two subjects interesting enough that I want to know a lot about both. What I plan on doing in the future is to focus on something that the two topics have in common. In my case I'll be working on something related to materials science or physical chemistry.

[Anonymous No.9675081]

I have very minimal understanding of genes and how they work so hopefully someone can answer my question.
I have a girlfriend that I deeply wanted to marry and start a family with. She is 50% asian and 50% white, while I am basically 100% white.
I worry that if we were to have kids, the kids would be "off" in a sense. I.E. they will have a below average IQ, will have trouble socializing or fitting in to groups, etc.
If I had a child with this woman, would it be that bad? I'm aware that badly race mixed kids have trouble fitting into groups and have trouble learning, but if she is already 50% white does that mean the effect wouldn't be as bad?

[Anonymous No.9675143]

>>9675081

>>>/pol/168315412

[Anonymous No.9675414]

How long does it take to write a 5-6000 word BEng Mech Eng dissertation? I've got a final exam on the 23rd but the dissertation is due the 27th, should I start doing a bit of it now?

[Anonymous No.9675486]

What are some cool current math theories or issues that I can look up on youtube for fun?

[Anonymous No.9675635]

>>9670295
i dont know if i put the arrows on the right way, but i essentially follows from pic related

[Anonymous No.9675663]

>>9675635
not him but are you just rolling with the fact blue angle is 90 deg? because it wont be if F1^2 != f^2 + other-side^2
and from the looks of just via drawing their modul length does not work that way.

Unless I'm reading it wrong and you just need to prove statement F1 = F*cos(theta) then I guess it's a given

[Anonymous No.9675666]

>>9675486
none. anything new can only be found in research papers, potentially a high level lecture. you won't find the quantum woo bullshit on any self-respecting math channels

[Anonymous No.9675743]

Idiot here. Is there a different between proportional and relative proportional relationships or is it just wording?

[Anonymous No.9676133]

>>9675143
i would trust /sci/ more than pol, thats why im asking here

[Anonymous No.9676142]

>>9675743
Relative proportional isn't a standard term. First time I'm seeing it.

[Anonymous No.9676147]

>>9675743
define "proportional relationship" and "relative proportional relationship"

[Anonymous No.9676150]

If energy in the universe is constant, wouldn't a universe need to exist before ours to contain the energy present in our universe (pre big bang)

[Anonymous No.9676154]

>>9676150
define what you mean by "energy in the universe"

[Anonymous No.9676155]

>>9676154
the total sum of energy in the universe. Kinetic, chemical etc

[Anonymous No.9676227]

How do I prove that a random geometric distribution Z of parameter p is (1-p)^k for all Z<=k?

[Anonymous No.9676233]

>>9676155
It's theorized that the total sum of energy in our universe is zero.

[Anonymous No.9676240]

>>9676150
Is the total energy in the universe really constant though? Because the universe is expanding, the potential energy (gravitational, electromagnetic, etc) increases naturally.

[Anonymous No.9676243]

>>9675081
>>9676133
What is your source to begin with on race-mixed offspring having lower intelligence? What the fuck?

[Anonymous No.9676248]

>>9676243
there is a noticeable drop in IQ in blacks when compared to say, Asians.
Am i saying blacks are not as intelligent? Well, thats where an argument is formed. I would argue that intelligence is based off of how the child is raised, how good the education is, etc.

I mentioned it because I was hoping to hear multiple views on the subject, maybe I didn't word it as best as I could. At the end of the day, I just want the best for a child, you know? If you have any input on it I would appreciate hearing it.

[Anonymous No.9676257]

Do you mind looking at my stack question? It's about latin squares, I've come up with a conjecture that is hard for me to prove or disprove.

https://math.stackexchange.com/questions/2740177/orthogonal-reduced-latin-squares

[Anonymous No.9676262]

>>9676248
>She is 50% asian and 50% white, while I am basically 100% white.
> I worry that if we were to have kids, the kids would be "off" in a sense. I.E. they will have a below average IQ,

These are your words. What do blacks have to do with it? Why would the kids have a below average IQ if the parents themselves aren't below average? I don't understand where you got the information to begin with here.

[Anonymous No.9676263]

>>9676243
an article i have read in the past http://thealternativehypothesis.org/index.php/2016/04/15/human-race-are-real-race-is-a-valid-scientific-category/

some things in here i disagree with, im just linking what I have read when i did personal research on the topic.

[Anonymous No.9676276]

>>9676262
i was just giving an example to show there is difference in IQ and race. asians have higher iq scores than whites as well.

[Anonymous No.9676313]

>>9676276
Okay, but I don't understand why you're worried about statistics at all when it's an issue of two individuals. Say you have an IQ of 130 and your wife an IQ of 135. Your kids are receiving these genetics (simplified), not the average of whites, nor the average of Asians. Statistics aren't useful here.

[Anonymous No.9676375]

Let [math] n [/math] be and even, positive integer and let [math] \pi\colon \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} [/math] be a bijective map.
Show that there are [math] x,y \in \mathbb{Z}/n\mathbb{Z} [/math] with [math] x \neq y [/math] and [math] x - y = \pi(x) - \pi(y) [/math].

I saw it posted as a problem on /mg/ a bit back but didnt see anyone prove it.
the only approach i can think of is using the pigeonhole principle, but im lost on the details

[Anonymous No.9676376]

>>9676313
thats true, i didnt think about it that way. thank you

[Anonymous No.9676406]

>>9676276
>>9676248
Your "kids" can be the most brilliant geniuses because of your retarded gene magic attempt, but because of that they also can have super-high cancer chance so point is moot

dont try to fuck around herem 8

[Anonymous No.9676484]

>>9676376
It's very, very important to understand the limitations of statistics. Lying with statistics is probably the most prevalent form of lying on the internet. I suggest looking into it.

[Anonymous No.9676527]

Why without AC (but in ZF + some other axiom, like AD or countable choice) all subsets of reals are Lebesgue-measurable? Immeasurable sets, like Vitali's set don't become measurable without AC, so it must be the case that such immeasurable sets don't exist. But why should that be the case? We use AC to construct Vitali set, but even without that it exists, but that contradicts all subsets of R being measurable. Or maybe such sets don't don't exist, but that means power set of R in ZFC is different that power set of R in say [math]\textrm{ZFA}_\omega[/math] or ZF+AD, but that doesn't seem right.

[Anonymous No.9676620]

>>9676484
I wasnt trying to lie or anything, I appreciate you pointing it out though.

[Anonymous No.9676931]

>Doing field theory
>Talks about propagators
>[math]\frac{1}{\Box}[/math] shows up

Now, I already know what [math]\Box[/math] is, but what the hell do they mean by dividing by a differential operator? Is this abuse of notation?

[Anonymous No.9676965]

>>9676527
If you take the Solovay model of ZF+'stuff', the Vitali set doesn't exist. You need AC to put your hands on it so to speak. For your second question, the powerset is subject to interpretation by the model, it is not absolute, so it will be can be a different set depending on the model.

[Anonymous No.9676971]

>>9676931
Yes, it's to be understood on the Fourier side as [eqn]\frac{1}{p^2 + m^2}[/eqn]. But once you generalize to distributions on [math]H^2[/math] it's not an abuse of notation anymore, since the Fourier transform is an isometric automorphism.

[Anonymous No.9676974]

>>9676971
Nice, thanks anon.

[Anonymous No.9677212]

how do i think about

[math]x^{\frac{n}{c}}[/math]

where x,n, and c are all greater than 1?

im in cal 2 and this shit bothers me. how do i differentiate shit like that? for example, [math]x^{\frac{3}{5}}[/math]

[Anonymous No.9677352]

Can someone explain to me what tensors are?
scalars are 1 x 1
vectors are n x 1
matrices are n x p

tensors are n1 x n2 x n3 x ... x nn

Why am I mistaken?
For practical purposes, "tensors are to matrices what matrices are to vectors" did me well enough for the exams, but I'm curious.

[Anonymous No.9677421]

Can someone explain this to me? I'm doing one of the problems, and I found the Fourier series for f(x), but I don't know where to go from there.

[Anonymous No.9677536]

How do I calculate the g-force of an angular acceleration? How fast can humans spin before they die?

[Anonymous No.9677634]

>>9677352
>scalars are 1 x 1
>vectors are n x 1
These are false.

[Anonymous No.9677673]

>>9677634
explain

[Anonymous No.9677679]

>>9677673
In what sense do either of those statements carry truth? Calling scalars 1x1 doesn't make much sense in terms of matrix multiplication since the product of an axb matrix and a cxd matrix (in that order) only makes sense when b=c, but you can multiply any matrix by a scalar, there's no notion of 'dimension' to them. And vectors in infinite-dimensional vector spaces (i.e. the majority of vector spaces) can not be described as "nx1".

[Anonymous No.9677686]

>>9677679
hurr durr

[Anonymous No.9677687]

>>9677686
>hurr durr
Not an argument.

[Anonymous No.9677737]

okay I get it but why the fuck would anyone want to do this

if im writing a program, i would much fucking rather just have a transformation function that just maps [x,y] -> [x+h,y+k] directly.
something like r(x,y,h,k){ return [x+h,y+k]; } I mean.

why the hell would I want to have to compute a matrix multiplication? isn't that much more work than what's necessary? what the fuck is the point

[Anonymous No.9677745]

>>9677737
it's general. a lot of functions can be represented as matrices. your function works too

[Anonymous No.9677747]

>>9677421
just throw your bn into 2.9 bro
you said you found your bn through fourier decomposition of your f(x) right when y=0 right

[Anonymous No.9677750]

>>9677536
you divide your angular acceleration by [math] g=9.81 [/math] to find your g force from angular acceleration and then you look up when humans die from the blood not circulating from g force

[Anonymous No.9677756]

>>9677212
just use the power law [math] \frac{d}{dx}x^m = m \cdot x^{m-1} [/math]
you just set [math] m=\frac{n}{c} [/math] so [math] \frac{d}{dx}x^{\frac{3}{5}} = \frac{3}{5} \cdot x^{\frac{3}{5}-1} = \frac{3}{5} \cdot x^{-\frac{2}{5}} [/math]

>>9677352
the way tensors were explained to me is that they transform nicely under coordinate transforms (simple derivatives) while matrices do not transform this way

[Anonymous No.9677764]

>>9677745
when people talk about 3d graphics they say it's matrix multiplication. here it is in this book in fact. if you dont want to read all this it's just making a matrix for (x,y,z)->(x/(1-z/d),y/(1-z/d),0). if I were gonna do this on a computer, again I would just do it directly without fucking with matrix multiplication at all. but they extend it out into homogeneous coordinates, and then finally say "to obtain the (x',y') coordinates you actually needed corresponding to ((x/(1-z/d),y/(1-z/d)), after multiplying (x,y,z,1) by the homogeneous 4x4 matrix, just divide the x and y of the vector product by the last thing in that vector". WHY THE FUCK WOULD I WANT TO GO THROUGH ALL OF THAT? that's effectively just undoing the entire process of constructing this homogeneous matrix in the first place. why not just have a function (x,y,z)->(x/(1-z/d),y/(1-z/d),0) DIRECTLY? this section is literally called "computer graphics". why would i want to apply anything in here to computer graphics if it's just more steps?

[Anonymous No.9677779]

>>9674304
yup

>>9672849
it would just reach equilibrium really fast. magnets and their magnetic fields do not do work due to one of maxwell's equations [math] \div \cdot B = 0 [/math]. if you learn about relativity you'll see that magnetic fields are just a "shadow" of an electric field (if you look at the system the right way with the right velocity they turn into electric fields) so i believe that's why magnetic fields don't do work

>>9671959
>lagrange
>retarded
pick one

>>9677764
idk m8 but would you find it easier if you wanted to make a different transformation and had a base matrix like that and just slightly modify it. i'm sure if you can code your function then it doesn't matter. as well aren't the matrices you're transforming like the example in the textbook? i'm pretty sure the matrices you look at can get extremely long (for not tiny pictures). can your function handle a ton of a matrix and output a matrix as easily as in that example?

[Anonymous No.9677780]

>>9677779
[math] \nabla \cdot B = 0 [/math]*

[Anonymous No.9677801]

>>9677777
whomst'd've

[Anonymous No.9677808]

>>9677756
>they transform nicely under coordinate transforms (simple derivatives) while matrices do not transform this way
Can you illustrate with an example?

[Anonymous No.9677822]

>>9677808
no kek
here's another reason: you can have rank 2 tenors which take the covariant (lower indice) form [math] T_{\mu \nu} [/math] which can usually be represented as a matrix [math]A_{ij}[/math] but a rank 3 covariant tensor looks like [math] T_{\mu \nu \lambda} [/math]. if you can find a matrix representation for 3 indices then i suppose you could call matrices the same as tensors

[Anonymous No.9678086]

>>9677737
> okay I get it but why the fuck would anyone want to do this
Given a function, the only thing you can do is pass arguments and receive a result.

Given a matrix, you can find its inverse, or conclusively determine that it doesn't have an inverse; in the latter case, you can identify the set of arguments which map to the same value. You can multiply matrices; this is equivalent to function composition, but the cost of a matrix multiply remains constant while the cost of evaluating a chain of composed functions increases as you add more functions.

Linear algebra is widely used in anything relating to computational geometry (games, CAD).

[Anonymous No.9678100]

>>9677764
> if I were gonna do this on a computer, again I would just do it directly without fucking with matrix multiplication at all.
Then you'd be an idiot. Actual 3D APIs that people actually use (i.e. DirectX and OpenGL) actually use matrix multiplication (and they use homogeneous coordinates).

> why would i want to apply anything in here to computer graphics if it's just more steps?
Generality. Almost every modern graphics API has the concept of a transformation matrix; typically a 4x4 matrix for 3D APIs, 2x3 (i.e. a 3x3 with the bottom row assumed to be [0 0 1]) for 2D. That allows any combination of translation, rotation, scale and (for 3D) projection transformations to be specified as a single parameter which is handled by a single code path (or dedicated hardware).

[Anonymous No.9678132]

>>9677352
a multilinear map from the cartesian product of copies of a vector space and its dual to the field of the vector space.

[Anonymous No.9678144]

I don't want to waste 8-9 hours a night anymore sleeping when others are getting by on 6, but my half-awake self doesn't have the discipline to stop sleeping after 6 hours if I don't have to. I figure the next best option is to stay awake for 22-24 hours at a time and sleep 8 hours, this gives me an extra two hours of awake time per day. How harmful is this, and will it lower my mental ability? Is it worth the extra two hours I'd get? Ignore that this puts me on a 30-32 hour cycle, it's not a problem.

[Anonymous No.9678149]

How fucked am I for physics 2 if physics 1 took my ass for a sexy ride?

[Anonymous No.9678151]

(0,0,0) is an example of a conservative vector field, right?

[Anonymous No.9678153]

>>9677737
it all comes down to efficiency. computing the determinant or inverse of a small matrix is much less computationally intensive than finding the inverse of a transformation in general. The bonus of homogeneous coordinates is that you go up into a higher dimensional ambient space where matrix multiplication is justified, that is, you can represent linear transformations on the original coordinates, and then you project down to your original space. The thing with translation is that it is not technically a linear transformation since you need to be in an affine setting, while it is a linear transformation in the homogeneous setting.

Also, it allows you to do any type of linear transformation since it has a more general setting. While translations and rotations and scaling may be rather easy to represent, stretches and shears may not be so easy. Further, if you want to calculate a series of operations, with matrices it's as simple as matrix multiplication on your coordinates.

[Anonymous No.9678339]

>>9676375
Let d(x)=π(x)-x If d(x)≡d(y) (mod n) for distinct x,y we’re done. Otherwise d is a bijection so:
0≡Σπ(x) -Σx≡Σd(x)≡0+...+(n-1)≡n(n-1)/2
But for even n the right expression isn’t 0 mod n, hence d can’t be a bijection.
Where all sums are from x=0 to x=n-1.

[Anonymous No.9678490]

>>9678100
But I'm right in saying it's more computationally complex right
It takes more steps than if you just had a special function for whatever type of transformation you were using that you figure out on paper beforehand

As in like if you want to rotate points in R^3 about an axis, then scale them by n, then shift then up 5, then shear them, then perspective project them onto the xy plane, yeah I see how it would be very nice to just be able to specify that without thinking about it and have it work, but it would be less steps if you just multiplied all the matricies for those transformations yourself beforehand and then came up with a general form vector describing them, and then just had a simple function of a vector that returns a vector. I mean I would think that's a shit ton faster then multiplying matricies every time you want to apply this transformation to a point.

[Anonymous No.9678561]

>>9678149
General physics 2 is usually seen as more difficult, but some people will find it easier for no apparent reason, like myself.

[Anonymous No.9678568]

>Please disregard >>>/pol/ propaganda in bottom right corner
Why are AIs consistently racist? I'm curious as to why this occurs outside of nazi autism.

[Anonymous No.9678610]

Never done any vector calculus, but I'm supposed to apply it now.

For [math]\mathbf W \in \mathbb R^{m\times n}, w_{jk}[/math] being the element of W at the jth row and kth column, am I correct in understanding that [math]\frac{\partial \mathbf W}{\partial w_{jk}}[/math] is a n x m matrix, with a 1 at the jth row and kth column, and zeros everywhere else? If so, is there any nice way to write this matrix? What about [math]\frac{\partial \mathbf W \mathbf x}{\partial w_{jk}}[/math], where x is a vector of size m? Or would taking the derivative like this be easier: [math]\frac{\partial \mathbf W \mathbf x}{\partial \mathbf W}[/math]

[Anonymous No.9678855]

My left nostril is irritated/infected. Which is a more likely cause?
>the germs from the dollar i used to snort the heroin
>irritation from the heroin itself

[Anonymous No.9678896]

>>9678490
> multiplying matricies every time you want to apply this transformation to a point.
Except you don't do that. You typically transform large numbers of points with the same matrix. The overhead of constructing the matrix is negligible by comparison. Overall, multiplying many points by a matrix is computationally cheaper than applying a series of simplified transformations (assuming those transforms are actually simplified; e.g. rotation about an arbitrary axis can't be made any cheaper than transforming a point with a 3x3 matrix, so you may as well throw any scaling transformations into the same matrix rather than adding another step).

And if you have a perspective projection you're going to have to do projective division at some point (which is why "3D hardware" is a thing), so there's no penalty to just using homogeneous coordinates throughout (2D graphics doesn't normally use perspective projection, hence the use of 2x3 matrices with w assumed to be 1 rather than 3x3).

[Anonymous No.9678935]

>>9678568
> Why are AIs consistently racist?
Because when developers need to construct a training set containing pictures of people, they end up constructing a training set containing pictures of white people. Or more generally, training data tends to reflect the social environment of the developers.

[Anonymous No.9679012]

>>9678151
doesn't a conservative vector field have to satisfy [math] \nabla \times F = 0 [/math] because [math] \nabla \times (\nabla f) = 0 [/math] (you can prove this pretty easily)
so yeah that vector field is conservative

[Anonymous No.9679062]

>>9678610
Pls respond

[Anonymous No.9679089]

Help me I am mentally disabled.
How do I even prove that something is divisible. For example how would I prove that pic is divisible by 4? I've spent 3 hours thinking and googling.

[Anonymous No.9679094]

>>9679089
what is the exponent level for x with coefficient 4?

[Anonymous No.9679103]

>>9679094
If I have not lost all my senses, it would be two!
But I still don't really get how to prove it.

[Anonymous No.9679113]

>>9679089
The integral of the first function should be 4 times larger than the second function. Assuming calculus is already proved, this should prove that division works.

But then again, you need to perform division in order to integrate. So maybe just plot a function, and calculate the area (choose a function with an area easy to calculate considering you have to plot from negative infinity to positive infinity).

[Anonymous No.9679116]

>>9679089
Is x an integer? If so, n would be more common.
I think the simplest way to do it would be to find all the roots and write it as a product of (x-r_1)(x-r_2)(x-r_3), then prove that at least 2 of the terms are divisible by or one is divisible by 4. If that does not work you could try a proof by induction, and if that does not work then maybe complete induction

[Anonymous No.9679137]

>>9679089
you probably can try to use group method but I'm having issues

best 've got is

(x+1)(2+1)x(x+1)

[Anonymous No.9679150]

>>9679137
It's x(x+1)(3x+1)

>>9679089
It's not divisible by 4, for f(2)=42

[Anonymous No.9679152]

>>9679150
>It's x(x+1)(3x+1)
no, that just one of variations

[Anonymous No.9679164]

>>9679152
I have no idea what you're talking about

(x+1)(2+1)x(x+1) = 3x^3 +6x^2 +3x

[Anonymous No.9679167]

Can you experience time if you travel at 100% the speed of light? Ive heard, that if you start traveling at 100% c, you would travel across the universe in 0 seconds.
According to the formula for time dialation, if you travel at c, 1 second for you would be "infinity" seconds for the observer. Does this mean someone traveling at the speed of light wouldnt experience time?
Can something with mass travel at 100% c, why not?

[Anonymous No.9679168]

>>9679116
>>9679137
>>9679150
>>9679152
Yea sorry, it's an integer and I'm sorry that it doesn't work. I actually have a different exercise but I didn't want to be told the solution so I tried changing it up so I can apply the principle once it was explained to make sure I understood it.

Now I did what >>9679116
you said and formed it to (2n2+n)(n+1). I have to prove divisibility by three but I'm not sure how I go about it still.

[Anonymous No.9679171]

>>9679164
my bad you are right fucked up along the lines grouping

[Anonymous No.9679176]

>>9679168
>(2n2+n)(n+1)
You can still split up the left term further (n=0 is a root)
Then you have three terms, and for every n, one of the three will be divisible by 3 (you have to show this). And as I already said, if that is too hard, you can try induction

[Anonymous No.9679189]

>>9677747
Well, in my case, there are two different functions f(x). They do have the same coefficients, though. I'm just not sure what to do at this point. Do I plug the coefficients into T in 2.9? Or do they go somewhere else?

[Anonymous No.9679198]

>>9679176
I'm not allowed to use induction actually. Every solution to similar problems I found uses induction.
It's n (2n+1) (n+1) now and I actually have to prove divisibility by 6. Somehow this task seems needlessly frustrating. I can't show that one of those terms is divisible by 6 but they will always yield a result divisible by 6. Is there any way to prove this directly? I'm beating myself up over this.

[Anonymous No.9679204]

>>9679198
A hint: Prime factorization of 6

[Anonymous No.9679208]

>>9679204
Right, two and three, which is what you get if you say n=1 but how does that prove this is always the case?

[Anonymous No.9679211]

>>9678935
"no"

[Anonymous No.9679219]

>>9679208
How do you prove that a*b*c is divisible by 2 when a,b and c are integers?

[Anonymous No.9679221]

I want to get into astronomy, non academically. What are some great resources for this?

[Anonymous No.9679222]

>>9679211
>There's a logical explanation, but I refuse to believe for no good reason
You're a very smart one, aren't you?

[Anonymous No.9679224]

>>9679219
I don't?
I'd try getting a 2 isolated in front of the integers I think.

[Anonymous No.9679233]

>>9679224
You would try to prove that 2 divides one of the three (since then it also divides the product). You do the same here. And again for 3

[Anonymous No.9679234]

>>9679012
[math] \nabla\times F=0 [/math] does not imply [math] F [/math] is conservative.

[Anonymous No.9679240]

>>9678896
>transform large numbers of points with the same matrix
wat
so you're saying if the transformation matrix is R and the set of wire-frame points is P, then computing RP is cheaper than looping over P (a 2-dimensional array of points) and applying a simplified vector->vector function to each vector in this loop?

[Anonymous No.9679243]

>>9679233
NOT ONE OF THE THREE IS DIVISIBLE BY 2 OR 3 FOR ALL INTEGERS THOUGH.

[Anonymous No.9679245]

>>9679234
https://en.wikipedia.org/wiki/Vector_calculus_identities#Curl_of_the_gradient

[Anonymous No.9679253]

>>9679243
>n (2n+1) (n+1)
For any given n, either n or n+1 will be even since they are two consecutive integers. Therefore the product of the two will always be even. Etc

[Anonymous No.9679267]

>>9679253
Alright, I get how this works. But I'm still not sure how to prove it. Do I have to prove the prime factors keep jumping terms? This is so strange.

[Anonymous No.9679283]

>>9679245
https://en.wikipedia.org/wiki/Conservative_vector_field#Irrotational_vector_fields

[Anonymous No.9679287]

>>9679267
The proof that it will always be divisible by 2 is in my previous post. For the proof of the divisibility by 3, you have to consider 3 cases:
n mod 3 = 0
n mod 3 = 1
n mod 3 = 2
For each of those cases, you have to show that n, 2n+1 or n+1 is divisible by 3. The hardest part will be showing the second case

[Anonymous No.9679292]

>>9679287
Well thank you for all your help. I'll see where I'll get.

[Anonymous No.9679371]

>>9670851
Please help me. This apears in my physiscs problem. I prefer pure maths but cant find the solution even if translated

[Anonymous No.9679375]

>>9679371
I think i remember lagrange said something about this??

[Anonymous No.9679377]

>>9679234

[Anonymous No.9679393]

>>9679371
>>9679375
>>9679385
lmfao

[Anonymous No.9679396]

>>9679393
its me

[Anonymous No.9679407]

>>9679371
here you go anon
[math] y=57x\pm\frac{806}{5} [/math]

[Anonymous No.9679408]

>>9679407
meant [math] \frac{806}{5}\pm 57 x [/math] rather

[Anonymous No.9679409]

>>9679407
how did you do it?

[Anonymous No.9679415]

>>9679409
lucky guess

[Anonymous No.9679424]

>>9679415
come on

[Anonymous No.9679446]

>>9679424
ok then
-you have two equations [math] \{y=mx+c, y=-5x^2\} [/math].
-they intersect when [math] 5x^2+mx+c=0 [/math].
-the roots to that are [math] x=\frac{-m\pm\sqrt{m^2-20c}}{10} [/math].
-you want the two x intersections to be a distance 1 from each other so:
[eqn] \left|\frac{-m+\sqrt{m^2-20c}}{10}-\frac{-m-\sqrt{m^2-20c}}{10}\right|=\left|\frac{2\sqrt{m^2-20c}}{10}\right|=1 [/eqn]
-then [math] m^2-20c=25 [/math]
-also [math] |y_1-y_0 |=57,\ |x_1-x_0|=1 [/math]
-so [math] m =\frac{y_1-y_0}{x_1-x_0} =\pm 57 [/math]
-finally [math] c=806/5 [/math]

[Anonymous No.9679460]

>>9679446
why did you replace 57 for m?

[Anonymous No.9679520]

what am i doing wrong for pic related?
i end up with
[eqn] \displaystyle I=\int_0^{2\pi}\int_0^{\pi/2}\int_0^{1/\sin\varphi}\rho\sin \varphi\,d\rho\,d\varphi\,d\theta\,, [/eqn]
which evaluates to
[eqn] I=\pi \int_0^{\pi/2}\frac{1}{\sin \varphi}\,d\rho\,d\varphi\,d\theta\,, [/eqn]
which does not converge.

>>9679460
what do you mean?

[Anonymous No.9679580]

>>9678610
why would it be? That is only the case if W=w_jk

Look up jacobian matrix

[Anonymous No.9679596]

Should I learn Golang, Fortran and Julia? What are those languages good for? Which one should I learn between Maple, Matlab and Mathematica?

I'm a mathematician btw.

[Anonymous No.9679603]

>>9679089

[Anonymous No.9679608]

>>9679520
>[math]\int_0^{1/\sin\varphi}[/math]
why?

[Anonymous No.9679614]

>>9679520
>what do you mean?
you already knew 57 was the coeficient why change it?

[Anonymous No.9679618]

>>9679520
why would they ask you to use spherical coordinates here? that's retarded. Use cylindrical (duh) coordinates instead

[Anonymous No.9679629]

>>9679608
through pic related. im guessing it's either here or in the substitution ive made the mistake

[Anonymous No.9679634]

>>9670851
How much would it cost out of pocket (with no health insurance) to determine if someone has suffered from a stroke in the past? A rough estimate will suffice.

[Anonymous No.9679653]

General strategy on how to prove this?

[Anonymous No.9679667]

>>9679653
[math]Aw_1[/math] is literally just a vector [math]v[/math]. If you take the transpose of this vector, then the coordinates of the vector dont change. However, note that the transpose switches the direction of the multiplication. So for the matrix multiplication that follows to make sense, you need to take the transpose of [math]v[/math].

[Anonymous No.9679681]

>everything is made by subatomic particles
>the universe was created with the big bang
>elements formed
>they are constituted basically by protons, electrons, neutrons
>protons define the element, while neutrons and electrons, their properties
>protons and neutrons attract each other
>masses have a similar attraction, called gravity
>interactions within atoms result in radioactivity and similar effects
>concepts like temperature, time, speed, etc allow us to describe reallity
>energy can propagate through materials or vaccum - as physical / eletromagnetical waves

I finished high school a few years ago, and that's all I remembered that could be used in a narrative about how the universe works, all the rest was memorized nomeclature, formulas, etc. that I don't really understand.

How I can go on from here to study physics and chemistry?
Preferably, I would like to learn these things in the chronological order they were discovered, so that I understand the reasoning behind models.

[Anonymous No.9679723]

>>9679681
You probably want to learn the things in the order they're presented in a class actually. It's really easy to say "boy oh boy I can't wait to learn about stuff" but it's another thing to actually sit down and study and work and comprehend stuff, live with structure and stick to a plan

[Anonymous No.9679747]

>>9679653
you can write any dot product as [math] v^T w [/math]

[Anonymous No.9679839]

>>9670998
First question - why would embers be hot? They're still burning dude, just not much. A mathematical model would be too complicated.

Your textbook is most likely correct. I bet you haven't tested voltage well enough.

Literally what do you think bullets are provided by? Anything too powerful could break your gun. Stop asking about a mathematical model and learn some more math so you can do it yourself

[Anonymous No.9679840]

>>9679240
> so you're saying if the transformation matrix is R and the set of wire-frame points is P, then computing RP is cheaper than looping over P (a 2-dimensional array of points) and applying a simplified vector->vector function to each vector in this loop
Both are linear in the size of P, but the matrix multiply has a fixed cost per vertex regardless of the number of transformations from which the matrix is composed. A large part of the reason that linear maps are important is that they're closed under composition: composing two linear maps produces another linear map (and the use of homogeneous coordinates extends this to projective maps).

The number of practical situations where the "simplified" vector->vector function is going to be noticeably cheaper than a matrix multiply is fairly small. That basically corresponds to optimising the cases where specific elements are known to be 0, 1 or -1. Anything with an arbitrary rotation (i.e. not about one of the coordinate axes and not a right angle) automatically means 9 multiplies.

And bear in mind that fixed-function (pre-GPU) 3D hardware would automatically perform the matrix multiply on every vertex sent to it; if you don't want any transformation, you'd use an identity matrix.

[Anonymous No.9679871]

>>9670851
what does it means when your pee has a strong ammonia odor?
also, what do you people think about starting a thread for asking medical questions? will there ever be on?

[Anonymous No.9679890]

>>9676240
The first law of thermodynamics suggests that it is.

[Anonymous No.9679957]

>>9677352
>tensors are to matrices what matrices are to vectors
I don't agree. Vectors are rank 1 tensors, matrices can represent rank 2 tensors. that's just going up in ranks. Tensors are a more general class of objects that include scalars (rank-0), vectors and rank-2 tensors.
an nth-rank tensor has [math]3^n[/math] components ([math]\mathrm{N}^n[/math] in N-dimensional space) which transform under rotations like
[eqn]
A_{\alpha \beta \cdots \omega}' = \underbrace{\sum_i \sum_j \cdots \sum_w}_{n \text{ summations}}
\left(
\frac{\partial x'_\alpha}{\partial x_i}
\frac{\partial x_\beta'}{\partial x_j} \cdots \frac{\partial x'_\omega}{\partial x_w} A_{i j \cdots w}
\right)
[/eqn]
e.g. for a vector
[eqn]
A_\alpha' = \sum_i
\frac{\partial x'_\alpha}{\partial x_i}
A_i
[/eqn]
another way to think about it is that tensors represent geometrical entities that don't change under rotations of the coordinate system (but the components do change). that said, tensors of higher rank than vectors are extremely hard if not impossible to visualise, so a geometric interpretation may not be helpful.

[Anonymous No.9679973]

>>9679596
>Which one should I learn between Maple, Matlab and Mathematica?
matlab is best for numeric computations, whereas maple and mathematica are symbolic languages best for computer algebra
numpy/scipy and octave (literally just an open source clone of matlab) are good free alternatives to matlab
i've not had much experience with maple but mathematica is the tits

[Anonymous No.9680008]

>>9679957
>under rotations
More than just rotations...

[Anonymous No.9680051]

if 0.999...=1 then does 0.98999...=0.999...? and so on?

[Anonymous No.9680054]

>>9680051
>if 0.999...=1 then does 0.98999...=0.999...?
no, but 0.98999...=0.99

[Anonymous No.9680485]

>>9670851
Hey anons please help out a brainlet

Is this function:

f(x)=(x*LARGE_PRIME)%2^64

Guaranteed to have no collisions for any input x in the range [0, 2^32] ? I assume it's correct that there is no chance of a collision since LARGE_PRIME is necessarily co-prime with the modulus, 2^64, right?

[Anonymous No.9680490]

>>9680485
IIRC past 4th billion there is still high chance of collision regardless so it is better to think in terms of more efficient and faster hash computation

[Anonymous No.9680495]

>>9680490
How is there a chance of collision, what is an example pair of inputs that collide?

[Anonymous No.9680746]

>>9680485
That depends, do you consider 2 to be big? Jokes aside, no.

[Anonymous No.9680829]

Anyone help me solve this? I simpliy it to log5[ (x+7/x-7)] = 2, but idk how to progress from here.

[Anonymous No.9680834]

>>9680829
Convert to exponentials.

[Anonymous No.9680835]

>>9680834
Ok, so then I have (x+7)/(x-7) = 5^2 = 25

so I go ahead and simplify and end up with 24x = -182
x = -7.58

The multiple choice has 7.58 listed as a solution but not negative 7.58

[Anonymous No.9680838]

That should be positive 182

[Anonymous No.9680840]

>>9680838
>>9680834
yeah, sorry I realized my mistake and deleted the post. Thanks anon.

[Anonymous No.9680855]

new thread

>>9680852
>>9680852
>>9680852

[Anonymous No.9680865]

>>9680485
It's guaranteed to have no collisions in the range [0,2^64). And LARGE_PRIME doesn't need to be either large or prime; it just needs to be coprime to 2^64 (i.e. odd).

If k is coprime to n, then for all a, b:
k*a ≡ k*b (mod n) <=> a ≡ b (mod n)

[Anonymous No.9681009]

>>9679603
But you didn't really prove that if
n mod 3 = 1
that (2n + 1) is divisible by three. You merely inserted 1 and showed that is is true for n = 1. That's not how you prove something right?

[Anonymous No.9681028]

>>9681009
do you know about modular arithmetic?

[Anonymous No.9681029]

>>9681028
>>9681009
if n mod 3 =1, then 2n+1 = 2*1+1 = 3 = 0 mod 3

[Anonymous No.9681030]

>>9681028
I don't think so.

[Anonymous No.9681033]

>>9681028
>>9681029
Isn't mod the rest after a division? modulo operation?

I'm guessing it's not.

[Anonymous No.9681035]

>>9681030
well, it is a fancy way of saying what the remainder after division by 3 is. So if you divide an integer n and get a remainder 1, then if you multiply (check) n by 2, then the remainder after division by 3 is 2. Similarly, if you add 1 to n, the remainder after division by 3 is 1+1=2. Putting these facts together, then if n gives remainder 1, then 2n+1 gives remainder 0, that is, 2n+1 is divisible by 3

[Anonymous No.9681048]

>>9681035
Well, right I knew that. I also get the logical steps to the conclusion now, can I just assert that the remainder after multiplying n by 2 is 2?
Thank you for your patience by the way, I feel like I understand everything when it's explained to me but reaching that point myself is really challenging.

[Anonymous No.9681055]

>>9681048
The following fact is true [math]x\equiv n\pmod m\implies 2x\equiv 2n\pmod m[/math], so yes, it is true, and i suggest you prove it if you can (read what the definition of modulo is!)

[Anonymous No.9681067]

>>9681055
I'll try, thank you very much!
I am honestly fighting with the fact with how to prove formally statements that are obviously true. It requires some mental gymnastics but I'll get over it, thank you again.

[Anonymous No.9681069]

>>9681067
first step is knowing the correct definitions, second is understanding the statement you're trying to prove