Question

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

`1+z^2+1/(1+z^2)`

Answer 1

`f(z)=2+sum_(n=2)^oo(-1)^nz^(2n)`

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 `1/(1+z^2)`.

Recall the following Taylor series expansion:

`1/(1-z)=sum_(n=0)^ooz^n`

We're going to want to get `1/(1+z^2)` in a similar form:

`1/(1+z^2)=1/(1-(-z^2))`

Thus,

`1/(1-(-z^2))=sum_(n=0)^oo(-z^2)^n`

`=sum_(n=0)^oo(-1)^nz^(2n)`

So,

`f(z)=1+z^2+sum_(n=0)^oo(-1)^nz^(2n)`

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

Write out the first two terms of `sum_(n=0)^oo(-1)^nz^(2n):`

`sum_(n=0)^oo(-1)^nz^(2n)=(-1)^0z^(2*0)+(-1)^1z^2+sum_(n=2)^oo(-1)^nz^(2n)`

`=1-z^2+sum_(n=2)^oo(-1)^nz^(2n)`

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

`f(z)=1+cancel(z^2)+(1cancel(-z^2)+sum_(n=2)^oo(-1)^nz^(2n))`

So, we got the `z^2` to cancel.

`f(z)=2+sum_(n=2)^oo(-1)^nz^(2n)`

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