How do I construct the following Laurent series (clipped off Wolfram Alpha)?
I know that the numerator can be written as $-1+\frac{\pi}2 z-...$
Alternatively (without the Laurent series), how can I see that the pole at $1$ is of order $2$?
Asked
Active
Viewed 965 times
1
-
1To investigate the singularity at $z=1$, determine that the order of the zero in the denominator is $4$ and the order of the zero in the numerator is ..., then compare these two. – GEdgar Mar 28 '13 at 21:09
-
Order 1 is incorrect. – GEdgar Mar 29 '13 at 00:20
-
@GEdgar Oh sorry, I made a mistake, you're right of course. But why does [denom's zero's order]-[numerator's zero's order] give the function's pole's order? Is this a theorem, or is it merely a good guess for further verification (using say DonAntonio's suggestion below)? Thanks! – ryang Mar 29 '13 at 16:12
-
Yes, it is a theorem. It's great for poles. When there are essential singularities in both numerator and denominator, this will not help, though. – GEdgar Mar 29 '13 at 16:12
-
@GEdgar Why does having essential singularities in both num and denom invalidate this theorem? – ryang Mar 29 '13 at 17:03
-
Because "order of zero" or "order of pole" is undefined at an essential singularity. So we cannot subtract to get information about the singularity of the quotient. – GEdgar Mar 29 '13 at 19:19
-
@GEdgar Thanks. Btw, do the order of a zero and the multiplicity of a zero refer to exactly the same thing? – ryang Mar 29 '13 at 19:22
2 Answers
3
Hint: Here is a start $$ \frac{\sin\left(\frac{(z-1)\pi}{2}+\frac{\pi}{2}\right)-1}{(z-1)^4}(1+(z-1))^{-5}=\frac{\cos\left(\frac{(z-1)\pi}{2}\right)-1}{(z-1)^4}(1+(z-1))^{-5} .$$
Now, just expand the series and collect terms of powers of $z-1$.
Mhenni Benghorbal
- 47,431
2
By l'Hospital, check that
$$\lim_{z\to 1}f(z)\,$$
isn't finite, so you have a pole. Now check that also
$$\lim_{z\to 1}(z-1)f(z)\,$$
isn't finite, but
$$\lim_{z\to 1}(z-1)^2f(z)\,$$
is finite , so the pole has order two.
Added on request: Let us write:
$$\sin\frac{\pi}{2}z=\sin\left(\frac{\pi}{2}(z-1)+\frac{\pi}{2}\right)=\cos\frac{\pi}{2}(z-1)=1-\frac{\pi^2}{8}(z-1)^2+\frac{\pi^4}{24\cdot 16}(z-1)^4-\ldots$$
$$\frac{1}{z^5}=\frac{1}{(1+(z-1))^5}=\left(1-(z-1)+(z-1)^2-(z-1)^3+\ldots\right)^5=1-5(z-1)+\ldots$$
The first expansion above is true for any $\,z\,$, the second though only for $\,|z-1|<1\,$ , which we can assume for $\,z\,$ pretty close to $\,1\,$ , thus
$$\frac{\sin\frac{\pi}{2}z-1}{z^5(z-1)^4}=\frac{1}{(z-1)^4}\left(-\frac{\pi^2}{8}(z-1)^2-\ldots\right)\left(1-5(z-1)+\ldots\right)=$$
$$=\frac{1}{(z-1)^4}\left(-\frac{\pi^2}{8}(z-1)^2+\frac{5\pi^2}{8}(z-1)^3+\ldots\right)=-\frac{\pi^2}{8(z-1)^2}+\frac{5\pi^2}{8(z-1)}+\ldots$$
The advantage in the above is that you see $\,z=1\,$ is a double pole and also you can see the residue!
DonAntonio
- 211,718
- 17
- 136
- 287
-
Hey thank you! I like this definitive answer, but differentiating the numerator and denom multiple times (l'Hopital's) did get a bit tedious and hardgoing. – ryang Mar 29 '13 at 15:58
-
Yes, I know but since you asked "alternatively, without Laurent Series", well... – DonAntonio Mar 29 '13 at 16:20
-
Oh, you mean it's easier to construct a Laurent series for this problem? Pray tell how! I understand Mhenni's working, but I don't know what to do from there; also it seems clever and I prefer a more general(enlightening) technique if possible. – ryang Mar 29 '13 at 16:58
-
See the stuff I just added to my answer. When one's interested only in the residule or a pole's multiplicity, it is usually easy to "cut off" the several expansions at a reasonable small power and thus just focus in what we're interested. – DonAntonio Mar 29 '13 at 18:33
-
Thank you, Don, for enlightening me! Yes, I see why the Laurent series is the best. Btw you missed squaring the $\pi$s in the last lot of equations. – ryang Mar 29 '13 at 20:38
-