How do you find the power series representation for the function f(x)=1(1+x)2 ?

2 Answers
Wataru
Sep 7, 2014

By Binomial Series,
1(1+x)2=(1+x)−2=∞∑n=0(−1)n(n+1)xn

Let us review the binomial series.
(1+x)α=∞∑n=0C(α,n)xn,
where C(α,n) is a binomial coefficient defined by
C(α,n)=α(α−1)(α−2)⋅⋯⋅(α−n+1)n!

Let us first the binomial coefficients for f(x)=(1+x)−2.
Since α=−2, its binomial coefficient looks like
C(−2,n)=−2(−2−1)(−2−2)⋅⋯⋅[−2−(n−1)]n!
=−2(−3)(−4)⋯[−(n+1)]n!
by factoring out all −'s in the numerator,
=(−1)n[2⋅3⋅4⋅⋯⋅(n+1)]n!=(−1)n(n+1)!n!
by dividing the numerator and the denominator by n!,
=(−1)n(n+1)

Hence, we have the binomial series
1(1+x)2=∞∑n=0(−1)n(n+1)xn.

mason m
Mar 11, 2017

1(1+x)2=∞∑n=0(−1)n+1nxn−1 for |x|<1

Explanation:

We have the standard power series:

11−x=∞∑n=0xn

From this we write the power series for 11+x:

11+x=11−(−x)=∞∑n=0(−x)n=∞∑n=0(−1)nxn

Note that 1x+1=(x+1)−1, so ddx(11+x)=−(x+1)−2:

ddx(11+x)=−1(x+1)2=ddx∞∑n=0(−1)nxn

Which can be rewritten as:

−1(1+x)2=∞∑n=0(−1)nddxxn=∞∑n=0(−1)n(nxn−1)

Reversing the signs by multiplying both sides by −1:

1(1+x)2=−∞∑n=0(−1)n(nxn−1)=∞∑n=0(−1)n+1nxn−1

Which is convergent for |x|<1.

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