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

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.