Laurent series (Complex)

Jul 2006
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44
Using term-by-term differentiation, expand \(\displaystyle f(z)=\frac{1}{(z+3)^2}\) in a Laurent series valid for \(\displaystyle |z| > 3\).

Ok, so, suppose I get a few derivatives:

\(\displaystyle f'(z)=-2(z+3)^{-3}\)

\(\displaystyle f''(z)=6(z+3)^{-4}\)

\(\displaystyle f'''(z)=-24(z+3)^{-5}\)
...

There's infinitely many. What next?
 

Drexel28

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Nov 2009
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Using term-by-term differentiation, expand \(\displaystyle f(z)=\frac{1}{(z+3)^2}\) in a Laurent series valid for \(\displaystyle |z| > 3\).

Ok, so, suppose I get a few derivatives:

\(\displaystyle f'(z)=-2(z+3)^{-3}\)

\(\displaystyle f''(z)=6(z+3)^{-4}\)

\(\displaystyle f'''(z)=-24(z+3)^{-5}\)
...

There's infinitely many. What next?
Pattern.

Do you have to do it this way? Note that \(\displaystyle f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2}\)
 
Jul 2006
364
44
I guess I have to, since that's what the question asks.

I know the pattern is:

\(\displaystyle f^{(n)}(z)=(n+1)!(-1)^n(z+3)^{-(n+2)}\)

but I don't know what to do with it.
 
Jul 2006
364
44
Doing it the other way, I suppose what happens is:

\(\displaystyle
f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2} = \frac{1}{z^2}\left(\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\right)^2
\) for \(\displaystyle |z|>3\).

But the whole series is squared. How am I suppose to write out the first few terms?
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
Doing it the other way, I suppose what happens is:

\(\displaystyle
f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2} = \frac{1}{z^2}\left(\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\right)^2
\) for \(\displaystyle |z|>3\).

But the whole series is squared. How am I suppose to write out the first few terms?
Hint: \(\displaystyle \frac{1}{(1+z)^2}=-\left(\frac{1}{1+z}\right)'\)...
 
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Jul 2006
364
44
Ok, so do I subs that in one of the factors and write it in terms of d/dz ?

I'm not sure what do with that...
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
Ok, so do I subs that in one of the factors and write it in terms of d/dz ?

I'm not sure what do with that...
\(\displaystyle \left(\frac{1}{1-x}\right)'=\left(\sum_{n\in\mathbb{N}}x^n\right)'=\sum_{n\in\mathbb{N}}nx^{n-1}\)

Adapt that.
 
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Jul 2006
364
44
I guess you want me to take the deriv of both sides?

\(\displaystyle f'(z) =\frac{2}{z^2}\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\cdot \sum_{n=0}^\infty n \left(-\frac{3}{z}\right)^{n-1}\cdot \frac{3}{z^2}\)

is that right? I could simplify this and all, but I don't quite see where this is going...