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Deduce taylor series

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Deduce that the Taylor series about 0 of 1/sqrt(1-4x) is the series summation (2n choose n) x^n.
From this conclude that summation (2n choose n) x^n converges to 1/sqrt(1-4x) for x in (-1/4,1/4).
Then show that summation (2n choose n) (-1/4)^n = 1/sqrt(1-4(-1/4)) = 1/sqrt(2)

What I know:
Taylor series about 0 of f(x) = (1+x)^r, r is a real number given by summation (r choose n) x^n.
I know that (r choose n) can be rewritten as r(r-1)(r-2)..(r-n+1)/n!
and I know from a previous question that the Taylor series converges to f(x) for all x in (-1,1), and summation (2n choose n) x^n converges conditionally at x=-1/4.

How can I do this question with all this information? I am not sure how to piece it all together I am having a lot of trouble with this course.
 

Answers and Replies

Well, you know how to Taylor-expand the function [itex]\frac{1}{\sqrt{1+x}}=(1+x)^{-\frac{1}{2}}[/itex], it just your formula with r = -1/2.

So then just write -4x instead of x and you have the Taylor expansion for the given function:smile:

You might want to try and play around with the binomial coefficients ("r chose n") to bring it into the desired form.
 
Last edited:
162
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I have tried to do this, but I am stuck and cannot get them to equal each other.
For f(x)=(1+(-4x))^(-1/2)
The taylor series about 0 is:
sum (-1/2 choose n) (-4x)^n
expanding binomial coefficients:
sum -1/2(-1/2-1)(-1/2-2)...(-1/2-n+1) / n! x (-4)^n (x)^n
sum -1/2(-3/2)(-5/2)...(1/2-n)/n! x (-1)^n (4)^n (x)^n

While the series I am trying to get is:
sum (2n choose n) x^n
expanding binomial coefficients:
sum 2n(2n-1)(2n-2)..(2n-n+1)/n! (x)^n
sum 2n(n-1/2)(n-1)(n-3/2)...(n+1)/n! (x)^n

I am stuck now
 
162
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can anyone help me out, i am still stuck in same place
 

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