Find \frac{1}{i+z} as a Power Series in z | Learn to Use Taylor Series

[aks_sky]

I need to find \frac{1}{i+z} as a power series in z.

I want to know if am doing this right.

If i use the taylor series here by doing
<br /> f(z) = z^i<br />

<br /> f&#039;(z) = i z^{-1} z^i<br />

<br /> f&#039;&#039;(z) = i (i-1) z^{-2} z^i<br />

This taylor series is just for z= i +1, but i tried using it for my problem but i don't seem to get the right answer.
this is the taylor series that i should be using but how do i find f(i) here?
<br /> <br /> f(z) = f(i) + f&#039;(i) (z-i) + f&#039;&#039;(i) (z-i)^2 + \cdots<br /> <br />cheers

 

[Mark44]

If you needed to find the power series for 1/(1 - x), you could do it by getting the various derivatives, but an easier way would be just to do the long division. I would give that a shot first with the function you have. The thing about a power series is that it is unique, so however you get it, that's the one.

 

[aks_sky]

oh yup that's cool.. i just had another question.. will i be finding the derivative of f(z) or f(i) for the taylor polynomial?

 

[Dick]

I'm not sure you are getting it. f(i) is a constant. And you don't have to take any derivatives. As Mark44 said, you can expand 1/(i+z) in a power series without taking any derivatives. Write it as (1/i)*(1/(1-iz)). Now expand 1/(1-iz) as a geometric series. 1/(1-x)=1+x+x^2+x^3+... Does that ring a bell?

 

[Chrisas]

power series in z.

You can do any of the short cuts given already (long division / geometric series), but if you want to grind the Taylor expansion machinery, then you are doing two things wrong.

I read the quotation above to mean an expansion about z=0. What you are trying to do is an expansion about z=i. That is mistake number 1. (As an aside, I don't think you can do that because z=i is a singularity and I don't think your power series can be done around a singularity. However, I may not be 100% correct in this)

Your second mistake is that the function f(z) = 1 / (i+z). Why did you use z^i? Take the derivatives of the full f(z) and plug in z=0 the get the series coefficients.

If you use the function as given, and expand around z=0, then you should get the same thing as the short cut methods.

 

[HallsofIvy]

You could also use the fact that the sum of the geometric series
\sum_{n= 0}^\infty r^n= \frac{1}{1- r}
That's similar to Mark44's suggestion.

Oh, and
\frac{1}{i+ x}= \frac{1}{i(1+ \frac{x}{i})}= -i\frac{1}{1- (-\frac{x}{i})}

 

[aks_sky]

sweet as.. got it.. cheers