Finding power series representation of the fucntion $f(x) = \frac{1+x^2}{1-x^2}$.

Question

So I want to find the power series representation of the function $f(x) = \frac{1+x^2}{1-x^2}. $ So how do I solve this? I can start by reducing the function to the known form of geometric series but I have a summation at the numerator. How do I overcome that?

$\frac{1+x^2}{1-x^2} = \frac{(1+x)^2 -2x}{1-x^2} = \frac{1+x}{1-x} - \frac{2x }{1-x^2}$, now second part I can solve, it would be $2x \sum_{n =0}^{\infty} x^{2n}.$ How do I go about the first part? Please give some hint. Thank You.

score 2 · Accepted Answer · answered Feb 03 '20 at 19:41

2

I would go straight using with $X=x^2$

$${1\over 1-X}=\sum_{n=0}^\infty X^n$$

This leads to

$${1+x^2\over 1-x^2}=\left(1+x^2\right)\sum_{n=0}^\infty x^{2n}$$

Now rearranging the sum

$${1+x^2\over 1-x^2}=1+2\sum_{n=1}^\infty x^{2n}$$

answered Feb 03 '20 at 19:41

marwalix

16,773

score 1 · Answer 2 · answered Feb 03 '20 at 19:40

1

Simpler:

Substitute $u=x^2$ in $$\frac{1+u}{1-u}=(1+u)\sum_{k=0}^\infty u^k=\sum_{k=0}^\infty u^k+\sum_{k=1}^\infty u^k.$$

answered Feb 03 '20 at 19:40

Bernard

175,478

But then it isn't in the form $\sum a_n x^n$, right? How do I compute its radius then? – Rick Feb 03 '20 at 19:44
I left to you to end the computation. – Bernard Feb 03 '20 at 19:46
Oh, I got it!! First-term is 1 – Rick Feb 03 '20 at 19:50
1

And the remaining coefficients are constant for even-degree terms. There reains to find the radius of convergence with Hadamard's formula. – Bernard Feb 03 '20 at 19:52
Actually the second answer just wrote that already. – Rick Feb 03 '20 at 19:54
I thought it was more instructive to let some computations to you… – Bernard Feb 03 '20 at 19:57

Finding power series representation of the fucntion $f(x) = \frac{1+x^2}{1-x^2}$.

2 Answers2