S scorpion007 Jul 2006 364 44 May 14, 2010 #1 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?

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 MHF Hall of Honor Nov 2009 4,563 1,566 Berkeley, California May 14, 2010 #2 scorpion007 said: 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? Click to expand... Pattern. Do you have to do it this way? Note that \(\displaystyle f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2}\)

scorpion007 said: 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? Click to expand... Pattern. Do you have to do it this way? Note that \(\displaystyle f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2}\)

S scorpion007 Jul 2006 364 44 May 14, 2010 #3 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.

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.

S scorpion007 Jul 2006 364 44 May 14, 2010 #4 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?

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 May 14, 2010 #5 scorpion007 said: 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? Click to expand... Hint: \(\displaystyle \frac{1}{(1+z)^2}=-\left(\frac{1}{1+z}\right)'\)... Reactions: scorpion007

scorpion007 said: 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? Click to expand... Hint: \(\displaystyle \frac{1}{(1+z)^2}=-\left(\frac{1}{1+z}\right)'\)...

S scorpion007 Jul 2006 364 44 May 14, 2010 #6 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...

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 May 14, 2010 #7 scorpion007 said: 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... Click to expand... \(\displaystyle \left(\frac{1}{1-x}\right)'=\left(\sum_{n\in\mathbb{N}}x^n\right)'=\sum_{n\in\mathbb{N}}nx^{n-1}\) Adapt that. Reactions: scorpion007

scorpion007 said: 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... Click to expand... \(\displaystyle \left(\frac{1}{1-x}\right)'=\left(\sum_{n\in\mathbb{N}}x^n\right)'=\sum_{n\in\mathbb{N}}nx^{n-1}\) Adapt that.

S scorpion007 Jul 2006 364 44 May 14, 2010 #8 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...

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...

O Opalg MHF Hall of Honor Aug 2007 4,039 2,789 Leeds, UK May 15, 2010 #9 Drexel28 said: Hint: \(\displaystyle \frac{1}{(1+z)^2}=-\left(\frac{1}{1+z}\right)'\)... Click to expand... Maybe it's easier to use Newton's binomial theorem to expand \(\displaystyle (1+z)^{-2}\).

Drexel28 said: Hint: \(\displaystyle \frac{1}{(1+z)^2}=-\left(\frac{1}{1+z}\right)'\)... Click to expand... Maybe it's easier to use Newton's binomial theorem to expand \(\displaystyle (1+z)^{-2}\).

S scorpion007 Jul 2006 364 44 May 15, 2010 #10 Ah, I figured it out guys. The example I needed was on this page: Pauls Online Notes : Calculus II - Power Series and Functions

Ah, I figured it out guys. The example I needed was on this page: Pauls Online Notes : Calculus II - Power Series and Functions