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$$f(z)=\frac 1{\cos(z^4)-1}$$

$z=0$ is a pole of $f$, and I believe that the Laurent series centred at $0$ is $-\frac 2{z^8}-\frac 16+...$, which looks like the pole is of order $8$, but why does Wolfram Alpha claim that the pole is of order $2$?

Julián Aguirre
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ryang
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  • Your reasoning sounds correct to me, perhaps it is some misunderstanding between you and Wolfram? Also, I think that if $f(z) = g(z^4)$ with $g$ meromorphic, then all zeros/poles of $f$ have multiplicity divisible by $4$ (normally, $4 \times$ multiplicity of $g$) – Jakub Konieczny Mar 18 '13 at 10:05
  • @feanor Thanks. – ryang Mar 22 '13 at 20:37

2 Answers2

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You are right. And Wolfram seems to agree.

By Taylor expansion $\cos z=1-z^2/2+O(z^4)$ so $$ \cos(z^4)-1=-\frac{z^8}{2}+O(z^{16})=-\frac{z^8}{2}(1+o(1)). $$ Hence $$ f(z)=\frac{1}{\cos(z^4)-1}=\frac{1}{-\frac{z^8}{2}(1+o(1))}=-\frac{2}{z^8(1+o(1))}\sim-\frac{2}{z^8} $$ So $\lim_0 z^8f(z)=-2\neq 0$. This means that $g(z)=z^8f(z)$ is holomorphic at $0$ (in a neighborhood of $0$) with $g(0)=-2$. So a power series expansion of $g$ will yield a Laurent expansion of $f$ starting by $-\frac{2}{z^8}$.

So indeed, $0$ is a pole of $f$ of order $8$.

Julien
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  • Thanks. The query I used was http://www.wolframalpha.com/input/?i=poles+1%2F(cos(z%5E4)-1); you can see how misleading Wolfram Alpha is. – ryang Mar 22 '13 at 20:39
  • @Ryan You're welcome. Misleading it is indeed... – Julien Mar 22 '13 at 20:41
  • @Ryan Actually, when $n=0$, the four formulas they give yield $0$. So this gives an order $4\cdot 2=8$ pole at $0$. – Julien Mar 22 '13 at 20:43
  • Julien, I understand the formal definitions of Big and Little O, but I don't understand the logic in your usage of Little-O above. Could you explain? Alternatively, how else could I obtain the Laurent series for $f$ about 0? – ryang Mar 27 '13 at 01:25
  • The use of $o$ or $O$ is here to get the beginning of the relevant asymptotic expansion. Here the O is really necessary. But then I could have just said $\cos(z^4)-1\sim-\frac{z^8}{2}$. Then it is allowed to take the inverse and get $\sim-\frac{2}{z^8}$. Which is all we need here. Nevertheless, keeping track of the error in the asymptotic with O or o is always a very safe way. Manipulating equivalents solely is very dangerous. You can' just add or subtract them. – Julien Mar 27 '13 at 02:13
  • For instance $f(x)=\frac{1}{x^2}$ and $g(x)=\frac{1}{x^2+x}$ are both equivalent to $\frac{1}{x^2}$ at $+\infty$. But $f(x)-g(x)=\frac{1}{x}$. Would you say it is equivalent to $0$? – Julien Mar 27 '13 at 02:15
  • Isn't $f(x)-g(x)=\frac 1{x^3+x}$? – ryang Mar 27 '13 at 04:04
  • @Ryan Ah, yes, right. That's $\frac{1}{x^3+x^2}$ actually. – Julien Mar 27 '13 at 04:14
  • Actually, I don't understand most of your answer, starting from the little-O notation in line 3 all the way to the whole of the last second paragraph! All I am sure of is that having obtained the Laurent series about $0$ of $f$, we can, by inspection, state that the pole $0$ has order 8. And I guess this resolves my main question as it was posed. – ryang Mar 27 '13 at 04:20
  • Well, it's getting worse each time...! This answer is fairly transparent, though. It proves that $z^8f(z)$ tends to $-2\neq 0$. This proves that $0$ is a pole of order $8$. Maybe you should read about removable singularities and poles here and there... – Julien Mar 27 '13 at 04:33
  • Haha. Well g(0)=0 but you say that it's -2, etc. – ryang Mar 27 '13 at 05:20
  • @Ryan No. It is $g(0)=-2$ by analytic continuation at $0$, where it is not defined in the first place. – Julien Mar 27 '13 at 11:48
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The order of the pole is the number of elements in the finite (!) principal part of the Laurent series of your function (or the highest negative power). You have 4 poles (solve the equation $(\cos \left(z^4\right)-1)$: $\left\{\left\{x\to -\sqrt[4]{2 \pi } \left(\sqrt[4]{n}\right)\right\},\left\{x\to -i \sqrt[4]{2 \pi } \sqrt[4]{n}\right\},\left\{x\to i \sqrt[4]{2 \pi } \sqrt[4]{n}\right\},\left\{x\to \sqrt[4]{2 \pi } \sqrt[4]{n}\right\}\right\}$. If you expand the initial function into Laurent series you will get (for example for the last root): $\frac{5 \left(z-\sqrt[4]{2 \pi }\right)}{32 \sqrt[4]{2} \pi ^{9/4}}+\frac{3}{16\ 2^{3/4} \pi ^{7/4} \left(z-\sqrt[4]{2 \pi }\right)}+\frac{1}{16 \left(\sqrt{2} \pi ^{3/2}\right) \left(z-\sqrt[4]{2 \pi }\right)^2}+\left(-\frac{1}{6}-\frac{19}{128 \pi ^2}\right)+O\left(\left(\text{z}-\sqrt[4]{2 \pi }\right)^2\right)$ The order of this pole is 2. One can conclude that you have 4 poles, each having order 2. So I guess Wolfram Alpha is correct :).

Caran-d'Ache
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    Hi, by your explanation, $\frac {sin z}{z^5}=\frac 1{z^4}-\frac 1{3!z^2}+\frac 1{5!}-\frac{z^2}{7!}+...$ should have an order-$2$ pole at $0$, but my textbook states that the pole is of order $4$. – ryang Mar 18 '13 at 10:39
  • Maybe wikipedia includes terms which are zero when counting the number of elements? – Eckhard Mar 18 '13 at 11:23
  • @Ryan Well I guess my first answer was wrong so I correced myself :). Hope now everything is fine. – Caran-d'Ache Mar 18 '13 at 17:43
  • This would be right for all $n\ne0$, but the OP's question concerns $n=0$, where all four of these expressions coincide. – Greg Martin Mar 18 '13 at 18:52
  • @GregMartin Don't see the problem. Just set $n=0$ (which consequently means that all you poles in the complex plane collide in one) and you'll get the pole of order 8. Isn't it better for understanding to use more general case? – Caran-d'Ache Mar 19 '13 at 04:08
  • @Caran-d'Ache Not usually, no. (That's why we don't teach category theory in kindergarden.) Especially when the original question asks specifically about the particular case. – Greg Martin Mar 19 '13 at 06:12