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

  • Thread starter Thread starter liyz06
  • Start date Start date
liyz06
Messages
6
Reaction score
0

Homework Statement



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

Homework Equations


The Attempt at a Solution

 
Last edited by a moderator:
Physics news on Phys.org
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 \ . ##
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top