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)

1 Answer
Apr 25, 2018

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.