# How do you do the taylor series for expanding 1/(2+x^2), about a = 0 up to order n = 2?

May 24, 2015

Firstly we would find the Taylor series for the function centered at a=0, thus moreso finding is Maclaurin series.

we can make use of the common Maclaurin series of:

$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$

Now lets try and write our function to look similar to the common Maclaurin Series:

$\frac{1}{2 + {x}^{2}} = \frac{1}{1 + \left({x}^{2} / 2\right)} = \frac{1}{1 - \left(- {x}^{2} / 2\right)}$

now we can just substitute $\left(- {x}^{2} / 2\right)$ in place of $x$ in our common Maclaurin series, thus we get:

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

which can simplify a bit:

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

Now to get the expansion up to order n=2, we just follow the sum and get:

${\left(- 1\right)}^{0} {\left({x}^{2} / 2\right)}^{0} + {\left(- 1\right)}^{1} {\left({x}^{2} / 2\right)}^{1} + {\left(- 1\right)}^{2} {\left({x}^{2} / 2\right)}^{2}$

which can simplify to:

$1 - {x}^{2} / 2 + {x}^{4} / 4$

note that this series is only convergent at $\left\mid {x}^{2} / 2 \right\mid < 1$

thus:

$- 1 < {x}^{2} / 2 < 1$

$- 2 < {x}^{2} < 2$

$- \sqrt{2} < x < \sqrt{2}$