How do you find a power series representation for #x/(1-x^2)# and what is the radius of convergence?

1 Answer
George C.
Oct 24, 2015

Use the Maclaurin series for #1/(1-t)# and substitution to find:

#x/(1-x^2) = sum_(n=0)^oo x^(2n+1)#

with radius of convergence #1#.

Explanation:

The Maclaurin series for #1/(1-t)# is #sum_(n=0)^oo t^n#

since #(1-t) sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - t sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - sum_(n=1)^oo t^n = t^0 = 1#

Substitute #t = x^2# to get:

#1/(1-x^2) = sum_(n=0)^oo (x^2)^n = sum_(n=0)^oo x^(2n)#

Multiply by #x# to get:

#x/(1-x^2) = x sum_(n=0)^oo x^(2n) = sum_(n=0)^oo x^(2n+1)#

This is a geometric series with common ratio #x^2#, so it will converge if #abs(x^2) < 1#, which is when #abs(x) < 1#. That is, the radius of convergence is #1#.

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