How to evaluate residue of $\cot z/z^4$ at $z=0$?

Question

How to evaluate residue of $\cot z/z^4$ at $z=0$? As we know : $$f(x)=f(0)+f'(0)x+f''(0)x^2/2+...$$ but $\cot(0)\to\infty$ or is undefined? I know that: $$\tan x=x+x^3/3+2x^5/15+...$$

Hint Since $\cot z=\cos z/\sin z$, and since $\sin z$ has a simple zero at $0$, we know $z^{-4}\cot z$ has a pole of order $5$ at $0$. There's a formula to evaluate residues at poles. What you want is a Laurent series, not a Taylor series, and to look at the coefficient of $z^{-1}$. — Pedro, Dec 25 '14 at 07:03
Another approach is to use Bernoulli numbers, but of course this is begging the question, in some sense. Recall (?) that $$\cot z=\frac 1 z+\sum_{\nu\geqslant 0}(-1)^{\nu}\frac{2^{2\nu} B_{2\nu}}{(2\nu)!}z^{2\nu-1}$$ You can derive this from the more familiar $\frac{z}{e^z-1}=\sum_{\nu\geqslant 0}B_\nu\frac{z^{\nu}}{\nu!}$. — Pedro, Dec 25 '14 at 07:12
@PedroTamaroff i know none of B-numbers neither i am familiar with both the formulae you provided; sorry :( — RE60K, Dec 25 '14 at 07:17

score 5 · Answer 1 · answered Dec 25 '14 at 07:18

5

The function has a pole of order $5$ at zero, so it isn't defined there.

Following @PedroTamaroff's hint and using that $\lim_{z\to 0}\frac{\sin z}{z} = 1$ : $$ \frac{\cot z}{z^4} = \frac{\cos z}{z^4\sin z}= \frac{z\cos z}{\sin z}\frac{1}{z^5} $$ Observe that the first fraction is holomorphic in a neighbourhood of $0$. Use Cauchy's theorem to finish.

answered Dec 25 '14 at 07:18

hjhjhj57

4,125

5

How would Cauchy's theorem help to find the residue of $f(z)=\frac{\cot(z)}{z^4}$? – Mussé Redi Jan 19 '17 at 09:46

kobe · Answer 2 · 2020-01-30T16:07:01.597

4

Well,

\begin{align}\cot z = \frac{\cos z}{\sin z} &= \frac{1 - \frac{z^2}{2!} + \frac{z^4}{4!} + O(z^6)}{z - \frac{z^3}{3!} + \frac{z^5}{5!} + O(z^7)}\\ & = \frac{1}{z}\cdot \left(1 - \frac{z^2}{2!} + \frac{z^4}{4!} + O(z^6)\right)\cdot \left(1 + \frac{z^2}{3!} - \frac{z^4}{5!} +\frac{z^4}{3!3!}+ O(z^6)\right)\\ &= \frac{1}{z}\cdot \left(1 + \left(-\frac{1}{2!} + \frac{1}{3!}\right)z^2 + \left(-\frac{1}{2!3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{3!3!}\right)z^4 + O(z^6)\right).\end{align}

Therefore, the coefficient of $\frac{1}{z}$ in the Laurent expansion of $\frac{\cot(z)}{z^4}$ is

$$-\frac{1}{2!3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{3!3!} = -\frac{1}{45}.$$

Hence, $$\text{Res}_{z = 0}\frac{\cot(z)}{z^4} = -\frac{1}{45}.$$

edited Jan 30 '20 at 16:07

answered Dec 25 '14 at 07:42

kobe

41,901

How is the coefficient of $1/z = -1/45$? – Amad27 Jan 12 '15 at 10:03
@Amad27 what do you mean? – kobe Jan 12 '15 at 15:36
I meant how do you derive that the coefficient of the $1/z$ term is $-1/45$? As you say above in the answer? – Amad27 Jan 12 '15 at 15:37
1

Look at the last line (line 3) for the expansion of $\cot z$. If you divide through by $z^4$, you find that $$\frac{\cot z}{z^4} = \cdots + \left(-\frac{1}{2!3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{3!3!}\right)\frac{1}{z} + O(z).$$ So the coefficient of $\frac{1}{z}$ is $$-\frac{1}{2!3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{3!3!} = -\frac{1}{45}.$$ – kobe Jan 12 '15 at 15:52
Fantastic, this deserves much more than $(+1)$ – Amad27 Jan 12 '15 at 15:58
Wow, thank you @Amad27! – kobe Jan 12 '15 at 16:00
2

@kobe How did you arrive at your second equality from the first one? I'm talking about the second factor, which seems to refer to the denominator. – Mussé Redi Jan 19 '17 at 09:35
1

@MusséRedi using $\frac{1}{1-x} = 1 + x + x^2 + \cdots$. – kobe Jan 19 '17 at 14:49
@kobe What is $x$ is in this case? – Mussé Redi Jan 19 '17 at 14:53
1

@MusséRedi $x = \frac{z^2}{3!} - \frac{z^4}{5!} + O(z^6)$. – kobe Jan 19 '17 at 15:03
@kobe I guess that's a valid approach since $|z|<1$. – Mussé Redi Jan 19 '17 at 15:10
at the third equality, the coefficient of $z^2$ should be $(-\frac{1}{2!} + \frac{1}{3!})$,shouldn’t it? – glimpser Jan 30 '20 at 10:17
@glimpser yes, I had evaluated $3! = 6$, but accidentally left the factorial symbol next to $6$. Thanks for the comment. – kobe Jan 30 '20 at 16:08

RE60K · Accepted Answer · 2014-12-25T20:09:33.693

0

$$\frac{\cos z}{z^4\sin z}$$

$$\newcommand{\f}[2]{\frac{#1}{#2}} \newcommand{\r}[1]{\frac{1}{#1}} \newcommand{\array}[2]{\begin{array}{#1}#2\end{array}} \newcommand{\b}[1]{\left(#1\right)} \newcommand{\s}[1]{\left[#1\right]} \newcommand{\d}[0]{\ldots} \newcommand{\ul}[1]{\underline{#1}} \array{r|lll}{ &&\r{z^5}&-\b{\r{3!}-\r{2!}}\r{z^3}&+\s{\b{\r{4!}-\r{5!}}+\r{3!}\b{\r{3!}-\r{2!}}}\r{z}&+\d\\\hline z^5-\f{z^7}{3!}+\f{z^9}{5!}+\d&&1&-\f{z^2}{2!}&+\f{z^4}{4!}&+\d\\ &&1&-\f{z^2}{3!}&+\f{z^4}{5!}&+\d\\\hline &&&\b{\r{3!}-\r{2!}}z^2&+\b{\r{4!}-\r{5!}}z^4&+\d\\ &&&\b{\r{3!}-\r{2!}}z^2&+\r{3!}\b{\r{3!}-\r{2!}}z^4&+\d\\\hline &&&&\s{\b{\r{4!}-\r{5!}}+\r{3!}\b{\r{3!}-\r{2!}}}z^4&+\d\\ &&&&\s{\b{\r{4!}-\r{5!}}+\r{3!}\b{\r{3!}-\r{2!}}}z^4&+\d\\\hline &&&&&+\d }\\ \begin{align} R&=\r{4!}\b{1-\r{5}}+\r{3!}\s{\r{2!}\b{\r{3}-1}}\\ &=\r{24}\b{\f45}+\r{12}\b{\f{-2}3}\\ &=\r{6}\b{\r{5}-\r{3}}\\ &=\f{-2}{6\times15}\\ &=-\r{45} \end{align}$$

edited Dec 25 '14 at 20:09

answered Dec 25 '14 at 07:50

RE60K

17,716

@PedroTamaroff I tried but can't make up the long-division in latex, the bottom portion i can easily write, i need help, anyways will try my best to make it readablle. – RE60K Dec 25 '14 at 17:22
@PedroTamaroff done – RE60K Dec 25 '14 at 20:09
You really need to find a better way to code this. Doesn't \frac work for you? You don't need to use the long division symbol. – Pedro Dec 25 '14 at 22:28

How to evaluate residue of $\cot z/z^4$ at $z=0$?

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