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I've recently learned that the cotangent satisfies the following functional equation:

$$\dfrac1{f(z)}=f(z)-2f(2z)$$

(true for $f(z)\neq 0$).

Can we solve this equation for real or complex functions $f?$ Can we give additional conditions such that $\cot$ is the only real or complex function satisfying these conditions and the equation? Or is there perhaps a different functional equation better suited for this purpose?

I'm asking this because I know about such a characterization of the real function $\exp$.

Please note that I know very little about functional equations. I've only seen two examples dealt with in my courses.

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This might be related. The Herglotz trick is essentially the statement that $\pi\cot(\pi z)$ is the unique meromorphic function $f(z)$ satisfying:

$f(z)$ is defined for $z\in\mathbb{C}\backslash\mathbb{Z}$

$f(z+1)=f(z)$

$f(-z)=f(z)$

$-f(z+\frac{1}{2})=f(z)-2f(2z)$

$\lim_{z\to0}\left(f(z)-\frac{1}{z}\right)=0$

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  • $\begingroup$ That's pretty neat that those conditions define a unique function. Do you have a reference for the proof? $\endgroup$ – Antonio Vargas Dec 1 '12 at 3:06
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    $\begingroup$ I first saw this in "Proofs from the book," by Aigner and Ziegler. I also found this on the web www-m9.ma.tum.de/foswiki/pub/WS2010/IMCSeminar/Cotangent.pdf $\endgroup$ – Julian Rosen Dec 1 '12 at 3:13
  • $\begingroup$ Other functions for which functional equations allow the differential equations for the functions to be deduced (this, of course, is much more elementary) are ln, exp, and atan (using the addition formula). $\endgroup$ – marty cohen Dec 1 '12 at 3:59
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Suppose that $\;f(z) = g(rz)\;$ where $\;r\;$ is the residue of the pole at $\;z=0.\;$ Assuming that the Laurent series expansion of $g$ is $\;g(z) = 1/z +\sum_{n=0}^\infty c_n z^n,\;$ then the functional equation rewritten as $\;0 = f(x)(f(x)-2f(2x))-1\;$ gives $\;0=-c_0/x+(-1-c_0^2-3c_1)+O(x)\;$ and solving for the coefficients gives $\;c_0=0,\; c_1=-\frac13,\;c_2=0,c_3=-\frac1{45},\cdots\;$ which gives $g(z)=\cot(z).\;$ Thus the solution is $f(z)=\cot(rz).\;$

There are many other functional equations for $f(x):=a\cot(bx),$ including homogenous quadratic equations in one variable such as $\;0 = f(x)^2-3f(x)f(2x)+f(x)f(3x)+f(2x)f(3x)\;$ and $\;0=f(x)^2-2f(x)f(2x)-f(2x)^2+2f(2x)f(4x).$

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