How do you find the maclaurin series expansion of #x^3/(1+x^2)#?

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Answer 1 · 2015-10-25T20:51:32

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

#x^3/(1+x^2) = sum_(n=0)^oo (-1)^n x^(2n+3)#

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 find:

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

Then multiply by #x^3# to get:

#x^3/(1+x^2) = x^3 sum_(n=0)^oo (-1)^n x^(2n) = sum_(n=0)^oo (-1)^n x^(2n+3)#

This is a geometric series with common ratio #-x^2#, hence the radius of convergence is #1#.