Question:
Let H∈R,Prove that the transcendental equation zcotz+H=0 has a countable number of zeros zn and that
limn→∞(n+12)(zn−(n+12)π)=Hπ
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!
Answer
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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