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

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:

$1 = \left(2 + {x}^{2}\right) \left(\frac{1}{2} - \frac{1}{4} {x}^{2} + \frac{1}{8} {x}^{4} - \frac{1}{16} {x}^{6} + \ldots\right)$

$= \left(2 + {x}^{2}\right) {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left({2}^{n + 1}\right) {x}^{2 n}$

So dividing both sides by $\left(2 + {x}^{2}\right)$ we get:

$\frac{1}{2 + {x}^{2}} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left({2}^{n + 1}\right) {x}^{2 n}$

... provided the sum converges.

This power series is a geometric series with common ratio $- {x}^{2} / 2$, which will converge when $\left\mid - {x}^{2} / 2 \right\mid < 1$, that is when $\left\mid x \right\mid < \sqrt{2}$.

So the radius of convergence is $\sqrt{2}$.