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Question:

Let $H\in R$,Prove that the transcendental equation $$z\cot{z}+H=0$$ has a countable number of zeros $z_{n}$ and that

$$\lim_{n\to\infty}\left(n+\dfrac{1}{2}\right)\left(z_{n}-\left(n+\dfrac{1}{2}\right)\pi\right)=\dfrac{H}{\pi}$$

My try: we must only prove this

$$z_{n}=\left(n+\dfrac{1}{2}\right)\pi+\dfrac{H}{\left(n+\dfrac{1}{2}\right)\pi}+o\left(\dfrac{1}{n^2}\right)$$ if this problem don't tell this limit reslut,then we how find this limit? Thank you

can you someone help me,Thank you very much!

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  • $\begingroup$ Off the cuff: Showing that there are an infinite number of zeros should be straight forward by geometry. In fact, it will probably be obvious there is a zero in every period of $\cot z$ (hence the $n+1/2$ terms). Given the sequence, try to find $\lim_{n \to \infty} \cot z_n$ using the original equation, which will exist if our original limit exists. Then you should be able to backtrack to get the actual result. $\endgroup$ – abnry Nov 27 '13 at 12:58
  • $\begingroup$ The definition of $z_n$ is needed to do this, i.e. which zero of $z \cot z = H$ is denoted $z_n$. It would seem the left side covers all reals as $z$ is restricted to any interval of the form $(n \pi, (n+1)\pi)$ where $n \ge 1$. Maybe $z_n$ is the solution in this interval, one for each $n \ge 1$. Is that the meaning of $z_n$? $\endgroup$ – coffeemath Nov 27 '13 at 16:28
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Your equation can be rearranged to:

$$\cot(z) = -\frac{H}{z}$$

Put $z = y + x$ where $y = (n+\frac{1}{2})\pi$ and use $\cot(y + x) = -\tan(x)$:

$$\frac{H}{y+x} = \tan(x)$$

We need to show $\lim_{y\rightarrow\inf}xy = H$. Taylor expand both sides:

$$\frac{H}{y}(1 - \frac{x}{y}) = x + O(x^2)$$

$$\left(\frac{H}{y^2} - 1\right)x = -\frac{H}{y} + O(x^2)$$

$$x = \frac{H/y}{1-H/y^2} + O(x^2)$$

From this we can see that $x \in O(1/y)$, so $O(x^2) = O(1/y^2)$, and $xy = H + O(1/y)$ as required.

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  • $\begingroup$ I hope see have methods to find this Gradual estimates?and Thank you $\endgroup$ – china math Nov 27 '13 at 13:55
  • $\begingroup$ I'm sorry, I don't understand. $\endgroup$ – apt1002 Nov 27 '13 at 18:17
  • $\begingroup$ I mean that if I can't tell this limit is $\dfrac{H}{\pi}$,then How can we have this value? $\endgroup$ – china math Nov 27 '13 at 18:19
  • $\begingroup$ $xy = (z_n - (n+1/2)\pi)(n+1/2)\pi$. Note the final $\pi$, which is not present in the limit in the question. If you divide through by that you get the result you want. Is that what you're asking? $\endgroup$ – apt1002 Nov 27 '13 at 18:26

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