Binomial expansion of a function with x raised to a power

[Dixanadu]

Hey guys.

So I need to know how to Binomial expand the following function
$\frac{1}{(1-x^{2})}$.

I need this because I have to work out $\prod^{∞}_{i=1}$$\frac{1}{(1-x^{i})}$ for i up to 6. But I have to do it with Binomial expansion. If i can learn how to do $\frac{1}{(1-x^{2})}$ then the rest of the powers should be the same.

I was under the impression that $\frac{1}{(1-x^{2})}$ can be binomial expanded as

$1+(-1)(-x^{2})+(-1)(-2)\frac{(-x^{2})^{2}}{2!}+(-1)(-2)(-3)\frac{(-x^{2})^{3}}{3!}+...$

Is that correct?

Thanks guys!

 

[RobinSky]

Maybe this will help:

http://en.wikipedia.org/wiki/Binomial_series#Special_cases

Good luck!

 

[Office_Shredder]

This is correct, and you probably want to observe that
(-1)/1! = -1
(-1)(-2)/2! = 1
(-1)(-2)(-3)/3! = -1

and you can probably guess the pattern as you continue.

 

[HallsofIvy]

Wait, wait- don't tell me. I'm still working on it!

 

[CellCoree]

Hello!

Yes, your understanding of binomial expansion for \frac{1}{(1-x^{2})} is correct. In general, the binomial expansion of a function with x raised to a power can be written as:

(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...

In your case, n is equal to -2, so the expansion becomes:

(1+x)^{-2} = 1 - 2x + \frac{(-2)(-3)}{2!}x^2 + \frac{(-2)(-3)(-4)}{3!}x^3 + ...

Simplifying this, we get:

\frac{1}{(1-x^2)} = 1 + x^2 + \frac{3}{2}x^4 + \frac{5}{3}x^6 + ...

So, to find \prod^{∞}_{i=1}\frac{1}{(1-x^{i})} for i up to 6, we can use this expansion and substitute the values of i for x. For example, for i=1, we get:

\frac{1}{(1-x)} = 1 + x + x^2 + x^3 + ...

For i=2, we get:

\frac{1}{(1-x^2)} = 1 + x^2 + x^4 + x^6 + ...

And so on. We can then multiply these expansions together to get the final result. I hope this helps! Let me know if you have any further questions.