Question

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

Answer 1

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)`.