How can you find the taylor expansion of #1/(2+x^2)# about x=0?

1 Answer
Oct 9, 2015

See explanation...

Explanation:

Without using differentiation, you can basically write out a power series which when multiplied by #2+x^2# gives #1#. Choose each successive term to cancel out the remainder left by the previous terms:

#1 = (2+x^2)(1/2 - 1/4 x^2 + 1/8 x^4 - 1/16 x^6 +...)#

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

So dividing both sides by #(2+x^2)# we get:

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

... provided the sum converges.

This power series is a geometric series with common ratio #-x^2/2#, which will converge when #abs(-x^2/2) < 1#, that is when #abs(x) < sqrt(2)#.

So the radius of convergence is #sqrt(2)#.