# What is the Taylor series of f(z) = 1+z^2+1/(1+z^2) ? Please save me! Thank you :)

## $1 + {z}^{2} + \frac{1}{1 + {z}^{2}}$

Apr 25, 2018

$f \left(z\right) = 2 + {\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} {z}^{2 n}$

#### Explanation:

Since you did not mention the center of the Taylor series, I'm assuming it will be centered at $a = 0.$

For now, ignore the presence of the terms $1 + {z}^{2}$ and focus on finding a Taylor series expansion for $\frac{1}{1 + {z}^{2}}$.

Recall the following Taylor series expansion:

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

We're going to want to get $\frac{1}{1 + {z}^{2}}$ in a similar form:

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

Thus,

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

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

So,

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

Let's see if we can make some of the $1 + {z}^{2}$ vanish.

Write out the first two terms of ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {z}^{2 n} :$

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

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

Then, writing our function with the first two terms of the summation stripped out, we get

$f \left(z\right) = 1 + \cancel{{z}^{2}} + \left(1 \cancel{- {z}^{2}} + {\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} {z}^{2 n}\right)$

So, we got the ${z}^{2}$ to cancel.

$f \left(z\right) = 2 + {\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} {z}^{2 n}$

There isn't really a way to get the $2$ into the summation. That's fine, this happens with many series representations.