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I'm trying to expand cosine function by using the Mittag-Leffler theorem which is introduced in Arfken 7ed.:

$$f(z)=f(0)\exp\left(\frac{zf'(0)}{f(0)}\right)\prod_{n=1}^{\infty}\left(1-\frac{z}{z_n}e^{z/z_n}\right) \tag{11.88}$$

Where $z_n$ is simple point.

First I found $z_n=\frac{2n-1}{2}\pi$.

And the result on the book was

$cosz=\prod_{n=1}^{\infty}(1-\frac{z^2}{(n-1/2)^2\pi^2})$

So I could guess $e^{z/z_n}$ was deformed as $1+\frac{z}{z_n}$.

But if we expand $e^{z/z_n}$, it becomes $1+\frac{z}{z_n}+\frac{1}{2!}(\frac{z}{z_n})^2+\frac{1}{3!}(\frac{z}{z_n})^3+\cdots$

Should I study the gamma function? I didn't study it yet.

I wonder why the higher order disappeared. Can you give me some advice?

Thomas Andrews
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Chan J.
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  • Why do you introduce a question about studying the gamma function? – FShrike Oct 01 '22 at 13:35
  • Because the last part of the section introduces the concept of the gamma function, I was wondering if it is a problem that can only be solved by knowing the concept of the gamma function. – Chan J. Oct 01 '22 at 16:40
  • You figured out that the zeros of cosine are $\frac{2n - 1}2\pi$. but note that that is for every integer $n$. But in the result in the book, notice that the product is only for positive integers $n$. Can you figure out how the negative roots of the cosine fit into this? – Paul Sinclair Oct 02 '22 at 03:48
  • Also, please review the theorem. I don't have a reference available, but I'm pretty sure the exponential factor multiplies the entirety of $1 - z/z_n$ , not just the last term. I also vaguely remember it being more general than you are claiming. – Paul Sinclair Oct 02 '22 at 04:17
  • Thanks for your advice! I didn't consider the case of negative integers. I will try it again! – Chan J. Oct 02 '22 at 12:39
  • @ChanJ. - FYI: if you want to reply to someone, you need to add @ followed by their username (preferably without spaces) to the comment. Then they will be notified of your reply. I just happened to look in again and noticed your reply. I was not notified. There are some exceptions to this - in particular, any comment on a post is automatically notified to the poster, which is why I didn't actually need to do this for you. I only included it as an example. – Paul Sinclair Oct 02 '22 at 13:35
  • @PaulSinclair Oh, sorry. I didn't know that. And thank you so much for noticing me. – Chan J. Oct 03 '22 at 12:06

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