How do you find the power series for $f(x)=1/(1-x^2)$ and determine its radius of convergence?

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Answer 1 · 2017-06-25T03:05:33

$1/(1-x^2) = sum_(n=0)^oo (x^2)^n = 1 + x^2 + x^4 + x^6 + x^8 + cdots$

$-1 < x < 1$

We know that the power series for $1/(1-x)$ is:

$1/(1-x) = sum_(n=0)^oo x^n = 1 + x + x^2 + x^3 + x^4 +cdots$

So, we can just plug in $x^2$ for $x$ to get our new power series:

$1/(1-x^2) = sum_(n=0)^oo (x^2)^n = 1 + x^2 + x^4 + x^6 + x^8 + cdots$

We can find our radius of convergence with the geometric series test. Since this sum is in the form $sumar^n$ , we can use the rule that $|r| < 1$ .

$|x^2| < 1$

$x^2 < 1$

$-1 < x < 1$

This is our interval. Now to check the endpoints, and then we'll be finished.

$sum_(n=0)^oo((-1)^2)^n = sum_(n=0)^oo1^n = 1+1+1+1+cdots$

$sum_(n=0)^oo(1^2)^n = sum_(n=0)^oo1^n = 1+1+1+1+...$

Both of these sums diverge, so our interval is:

$-1 < x < 1$

Final Answer

How do you find the power series for f(x)=1/(1-x^2)f(x)=1/(1-x^2) and determine its radius of convergence?