I am struggling with this calculation, any idea:

Details here:

Sorry for asking this, I know most of Sup Forums is just full retard but maybe

Hudson Cooper

I am struggling with this calculation, any idea:

Details here:

Sorry for asking this, I know most of Sup Forums is just full retard but maybe

Dylan Hill

It's 3.14

David Wood

What steps have you tried?

Or are you struggling because wolfram isn't giving you what you want?

What do you expect the answer to be?

What do you think this problem is trying to show?

Henry Lewis

42

Jason Walker

You have to be a bit more precise here OP. Is C considered small or large? Is T arbitrarily in] 0,1[ or is it meant to approach one of the boundaries?

Aaron Rivera

If course you can't prove it. Its a theorical question. Why do you need it?

Ian Carter

C is generally large, but firs we should do it without this approximation.

I have to prove that in the result there is an 1/C^(d-1) term

It is connected to a much more complex problem in network science.

Ian Nelson

It makes me want to do a change a variables.

Take C theta/2 = phi

It will put the C in the sinus

Then if C is large you will be able to say that sin(2 phi/C) ~ 2 phi / C

And there is your factor.

Thomas Gomez

Hum wait I think I got bamboozled, it's not gonna work that easy but I think it's still something to try.

Anthony Wright

You say you have a great that the result is proportional to C^(1-d). Any reason why? You're asking for an approximation but aren't telling us how good do you want it to be. Maybe you could give us some context or some clue about how good do you want the approximation to be?

This looks good to me tbh

Jacob Martinez

A guess*, not a great

Ethan Gonzalez

Nah it was stupid of me, the boundaries of the integral will tends to 0 and plus infinity, where the approximation I wanted to used become ineffective.

Ryder Sanders

But maybe my idea can be applied if we can the integral in half before doing the approximation. Showing that the infinity part is small, because of the sin bounded term, and that the other is 1/C^(d-1).

Jacob Gonzalez

*if we cut the integral in half

Logan Miller

it is connected to a complex hyperbolic embeddig problem, I can not explain now, but the numerical result shows that 1/C^d-1 should appear

Jace Thomas

Try my method, I believe it will work.

Julian Brown

Grayson King

yep, trying

Noah Rivera

You can cut the integral like this [0,1] and [1,2/C]

And the second part will be inferior to a constant times 1/C due to the fact that 1/T>1, so the integral will converge.

Jonathan Sullivan

did you do the usual, like wolfram alpha, matlab syms, maple (if you have it)?

John Hughes

Hum wait...

Ah crap it's to late in the night for this, I believe it won't work.

I'm dumb tonight.

Isaiah Perez

Yeah nah

Caleb Nguyen

I have tried wolfram mathematica 9 with no results

but this is something! It leads toward some expansion type solution.

Matthew Gutierrez

this looks like rubbish. there is probably a trick to this that is not very hard, or this is in a table somewhere. good luck