Math Checking/Explanations Thread

I like this new thread idea. More actual discussion on /sci/ and less shit posting I will start us off with one.

What is, If at all, the relationship between the centroid of a region and the equilibrium points in Lotka-volterra predator and pray systems?

Let [math]F:M \to N[/math] be a continuous map between smooth manifolds. I want to show that if smooth real valued functions on all of [math]N[/math] pullback to smooth real valued functions on [math]M[/math], then [math]F[/math] is smooth. I get the idea -- given a smooth chart [math](V,\psi)[/math] in [math]N[/math], we want to say that [math]\pi_1 \circ \psi \circ F[/math] is smooth ([math]\pi_1[/math] is a coordinate projection), but we can only say that smooth functions on *all* of [math]N[/math] pullback to smooth functions, right? What's the way around this?

I know it's something really simple, but it's bothering me. Will answer any equally stupid questions in return.

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It's been awhile since I've taken any kind of calculus, so I'd appreciate it if someone could check this for me, since I'm working on my own at the moment to refresh what I've lost over the years:
If I wanted to find the surface area of a rocket nozzle with a curve modeled by the equation [math] 3 + \left( \frac{\ln{(\frac{x}{2})}}{2} \right) [/math] with a domain 0.005 ≤ x ≤ 9 rotated about the x axis, would the equation be the following?
[eqn] \int_{0.005}^{9} 2 \pi \left( 3 + \left( \frac{\ln{\frac{x}{2}}}{2} \right) \right) \cdot (9-0.005) dx [/eqn]
Thanks for any help. Pic related is the equation and a sketch of the model.

>>7849203
I honestly can't help you here, the only thing I recognize there is centroid.

>>7849246
Fuck me I'm so out of my league on /sci/ math wise.
My gut says that substitution plays a role in solving your problem but beyond that no clue

>>7849253
Where are you getting this 9-0.005 term?

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>>7849253
>replying to a post just to say you don't know anything about how to solve it
>being THAT GUY

>>7849253
I'm on my ipad so I can't read the latex but I think you've got it right.

Surface area = 2π∫y√(1+y'^2)dx

>>7849276
The formula for surface area of a cylinder is
[math] 2 \pi \cdot R \cdot H + 2 \pi R^2 [/math]
where R is the radius (given by the function) and H is the height (given by the cutoff point 9 minus the intersection point 0.005). I left out adding the areas of the circles on either end, since obviously there's not a solid circle on the "go out shoot hot fire" point of a rocket nozzle.
Although, I'm not sure about adding that term it since I'm taking the integral from the same domain, 0.005 to 9.

>>7849280
Just didn't want to be a dick and ignore someone asking about it

>>7849246
Probably a bump function thingy:
For any point p in the domain U of your chart, you can define a smooth function [math]\rho_p[/math] that is 1 on a neighbourhood [math]W_p[/math] of p and 0 outside of U. Then you can extend [math]\rho_p \psi[/math] to 0 outside of U and define a smooth function.
Now, your argument (pulling back each projection) tells you that the pullback of this function is smooth. In particular, F is smooth on [math]F^{-1}(W_p)[/math]. Since we can do this at any point p in N (we take a chart at p, find a bump function and do what we just did), we find that F is smooth on an open cover of M

>>7849314
Ah, I get it now. Thanks man. I wonder how many more times I'm going to see arguments like this before my mind starts to go straight to bump functions.

>>7849300
>Although, I'm not sure about adding that term it since I'm taking the integral from the same domain, 0.005 to 9.
I don't think it should be there. That H in your formula is for a regular cylinder, and that comes as the result of integrating. To find to area of your shape, you want to add up the perimeters of a bunch of circles, so [math]\int_{0.005}^9 2\pi R(x) \mathrm{d}x[/math].

>>7849253
your equation will give you the volume of your figure times some unnecessary extra bit. sorry i cant do laTex, but the equation you are looking for is to integrate to find the length of that curve, then rotate that equation about the axis

>>7849363
That makes sense,
so this would be the correct equation? [eqn] \int_{0.005}^{9} 2 \pi \left( 3 + \left( \frac{\ln{\frac{x}{2}}}{2} \right) \right) dx [/eqn]

>>7849372
I thought I would have to square the function in order for it to give me the volume?
like this [eqn] \int_{0.005}^{9} 2 \pi \left( 3 + \left( \frac{\ln{\frac{x}{2}}}{2} \right) \right) ^2 \cdot (9-0.005) dx [/eqn]

>>7849392
>>7849281
Yo dis right here is the correct equation. Let ds=infitesimal length of the curve then

ds^2=dy^2+dx^2
ds^2/dx^2=dy^2/dx^2 +1
ds/dx=√(1+ dy^2/dx^2)
ds=√(1+dy^2/dx^2)dx
∫ds=∫√(1+(dy/dx)^2)dx = s

Thats it for the arc length. To get the surface area all you have to do is think about it intuitively and rotate that length around the axis. This gives us, where in this case y=the radius=whatever ln function for your rocket you got.

Surface area= ∫2πy√(1+(dy/dx)^2)dx

>>7849551
Okay back up

I can understand how you got from [math] \frac{ds^2}{dx^2} = \frac{dy^2}{dx^2} +1 [/math] to ∫ds=∫√(1+(dy/dx)^2)dx = s (i'm not gonna work that out in latex), but why is [math] ds^2 = dy^2+dx^2 [/math] ?

>>7849551
>>7849579
Oh, wait, I think I got it,
The length of the arc is like a curvy version of pythagoras' theorem,
Going up to the last step there is just algebraic, but then you have a Y in the final integral and I don't know where it comes from
[math] \int_{0.005}^{9} 2 \pi \; why \; \sqrt{1+ \left( \frac {dy}{dx} \right) ^2} dx [/math]
was that a typo or can you please explain why it's there?
Thanks

>>7849579
It is the definition of a riemannian metric for a flat 2-dimensional space. Comes from the general definition: [math]d{s^2} = {g_{\mu \nu }} \cdot d{x^\mu } \otimes d{x^\nu }[/math]

>>7849613
No, the y is very important. The is the radius at which you rotate the curve around whatever axis, in this case the x axis. Similarly, if you were rotating about the y axis you would use x for the radius. The the "curvy pythagorean theorem". Yes it is kind of like that, but not really, because you could think of them as infitesimals and if you zoom into a curve to infinite it will look straight. So it makes sense that it is just an infitesimal triangle. There is a bit of a better proof that is not nearly as unrigorous using limits.

>>7849670
This is beyond me, reminding me to get studying..

>>7850036
Lol didnt mean the sage it was from another retarded meme thread.

>>7850036
So do I put the function [math] f(x) = 3 + \left( \frac{ \ln{ \frac{x}{2}}}{2} \right) [/math] where the Y is, or where [math] \frac{dy}{dx} [/math] is?

>>7850069
Put the first function in where y goes, and put the derivative of that function in dy/dx

>all this boring calculus and analysis

i'm trying to compute the hilbert series [math] \sum \dim A[n] t^n [/math] for [math] A=k[x_1, ..., x_m] [/math]

it turns out that [math] \dim A[n] = \binom {m+n-1} {n} [/math], and supposedly this lets us write [math] \sum \binom {m+n-1} {n} t^n = (\sum t^k)^m = \frac{1}{(1-t)^m} [\math]

can anyone give a combinatorial argument for why that equality between the power series with binomial coefficients and the product of m geometric series should hold?

>>7850247
[math] \sum \binom {m+n-1} {n} t^n = (\sum t^k)^m = \frac {1} {(1-t)^m} [/math]

I'm doing a proof right now that states for any proposition, f, that uses only ^ or \neg there's an equivalent proposition, f', that has an equal amount of ^ and \neg in the proposition.

I have all the cases proven except for one. If there's more ^ than \neg I can't figure out how to prove it. I know in the case where there's 2n more of ^ than \neg you can just add that many \neg in front of the equation and it would be equivalent but I don't know how to do it for odd numbers.

How can I prove that any even number greater than 2 can be expressed as the sum of two primes?

>>7850264
using the mochizuki sieve

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>>7849280
GET OUT OF MY OFFICE