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