How should I expand 1/(1+x)^n around x=0?

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SUMMARY

The discussion focuses on expanding the function 1/(1+x)^n around x=0 using Taylor series. Participants emphasize the importance of the Taylor expansion, specifically how to derive the polynomial form of (1+x)^-n. The conversation highlights the application of the Taylor series formula, f(x)=f(0)+f'(0)(x)+f''(0)/2!(x)^2+f^{(3)}(0)/3!(x)^3+..., to achieve the desired expansion. Key insights include the necessity of calculating derivatives at x=0 to construct the series accurately.

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Homework Statement



How should I expand 1/(1+x)^n around x=0?

Homework Equations


The Attempt at a Solution

 
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Did you try a Taylor expansion?
What did you get? Where did you run into problems?
 
mfb said:
Did you try a Taylor expansion?
What did you get? Where did you run into problems?

I know that (1+x)^n could be expanded easily by binomial theorem, but what I need here is to expand (1+x)^-n into polynomial form, not the reciprocal of a polynomial
 
I don't see how your answer is related to my post.
You can just calculate the Taylor expansion.
 
liyz06 said:

Homework Statement



How should I expand 1/(1+x)^n around x=0?

Homework Equations



The Attempt at a Solution

Are you familiar with Taylor Series ?

The Taylor expansion for a function, f(x), expanded about x = a is:

## \displaystyle f(x)=f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots \ . ##

So, expanding about x = 0 gives:

## \displaystyle f(x)=f(0)+\frac {f'(0)}{1!} (x)+ \frac{f''(0)}{2!} (x)^2+\frac{f^{(3)}(0)}{3!}(x)^3+ \cdots \ . ##
 

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