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We know that $$\cosh{x}+\sinh{x}=e^x$$ and that his can be expressed as $$\frac{e^x+e^{-x}}{2}+\frac{e^x-e^{-x}}{2}=\frac{(e^x+e^x)+(e^{-x}-e^{-x})}{2}=e^x$$ and this works out nicely because the $e^{-x}$ cancel. Now consider a "higher order" cosh equation of the form $$C(x)=\frac{e^{\omega^0x}+e^{\omega^1 x}+e^{\omega^2x }}{3}$$ where $\omega^k$ are the 3rd roots of unity; $\omega^k=e^{\frac{2i\pi k}{3}}$. I would like to devise an analagous expression $$C(x)+S_1(x)+S_2(x)=e^x$$ with functions of the form $C$. My first reaction was $$S_1=\frac{e^{\omega^0x}-2e^{\omega^1 x}+e^{\omega^2x }}{3}; S_2=\frac{e^{\omega^0x}+e^{\omega^1 x}-2e^{\omega^2x }}{3}$$ and indeed $$(C+S_1+S_2)(x)=\frac1{3}\left[(3e^{\omega^0 x})+(e^{\omega^1 x}-2e^{\omega^1 x}+e^{\omega^1 x})+(e^{\omega^2 x}+e^{\omega^2 x}-2e^{\omega^2 x})\right]=e^x$$ My intuition was that this was correct; since in the case of $\cosh$ and $\sinh$, the 2nd roots of unity correspond to $1$ and $-1$. But my second thought was that there was no need for subtraction of $2e^{\omega^k x}$ for $k=1,2$. Perhaps there was a way to keep all the signs positive and find an $a,b$ such that $$e^{\omega^1 x}+e^{a\omega^1 x}+e^{b\omega^1 x}=0$$ $$e^{\omega^2 x}+e^{a\omega^2 x}+e^{b\omega^2 x}=0$$ I thought this would keep things a little more "symmetrical" and reduce the need for a $-2$ coefficient around the nontrivial roots of unity. I tried some things to get values for $a,b$ but have been unsuccessful. Should I just be happy with my original $S_1, S_2$ or are there choices for $a,b$ that would keep the symmetry?

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  • $\begingroup$ What is the purpose of this exactly? $\endgroup$ – Matt Samuel Jun 22 '15 at 19:56
  • $\begingroup$ Studying and researching hypergeometric Bernoulli numbers, Appell sequences, came across a paper on compositional Bernoulli numbers and was seeing if there were ways to extend it. $\endgroup$ – Eleven-Eleven Jun 22 '15 at 20:09
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If you want a symmetrical generalization for $n$th roots then define

$$C_{k,n}(z)=\frac{1}{n}\sum_{\zeta^n=1}\zeta^k e^{\zeta z}.$$

Then $C_{0,2}(z)=\cosh z$ and $C_{1,2}(z)=\sinh z$. Note that $C_{k,n}(z)=C_{0,n}^{(k)}(z)$.

In particular, for $n=3$, if $\omega$ is the cube root of unity in the upper half plane then

$$\begin{array}{lll} C_{0,3}(z) & = & \displaystyle \frac{e^z+e^{\omega z}+e^{\omega^2z}}{3} \\ C_{1,3}(z) & = & \displaystyle\frac{e^z+\omega e^{\omega z}+\omega^2 e^{\omega^2 z}}{3} \\ C_{2,3}(z) & = & \displaystyle \frac{e^z+\omega^2 e^{\omega z}+\omega e^{\omega^2z}}{3} \end{array}$$


There is an advanced viewpoint coming from representation theory. If we understand the group $G$ of complex $n$th roots of unity to act on a space of functions $\Bbb C\to\Bbb C$ (where a root of unity $\zeta$ acts by replacing a function $f(z)$ with $f(\zeta^{-1} z)$), then one may apply the isotypical projectors in order to decompose any function into its "constituents," and these are the isotypic components of the exponential function.

If one lets $\zeta$ be a primitive root of unity, then these are precisely the eigenvectors of the operator $f(z)\mapsto f(\zeta^{-1}z)$ (with eigenvalues the $n$th roots of unity) which combine to form the exponential function $\exp(z)$. One obtains these constituent parts with the kind of twisting-averaging operation you see in play in the definition of these $C_{k,n}(z)$s.

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  • $\begingroup$ I was thinking as I was typing that the $-e^{-x}$ was better characterized by $-1\cdot e^{-1\cdot x}$. I like this. Thanks $\endgroup$ – Eleven-Eleven Jun 22 '15 at 20:12
  • $\begingroup$ Real quick, in your summation you have $e^{\zeta x}$. Did you mean $e^{\zeta^k x}$? $\endgroup$ – Eleven-Eleven Jun 22 '15 at 20:32
  • $\begingroup$ @Eleven-Eleven $\zeta$ is an index of summation that varies over all $n$ solutions to $\zeta^n=1$. (Yes, you can do things like that.) $\endgroup$ – whacka Jun 22 '15 at 23:58
  • $\begingroup$ I see. It took a second but it is clear. Thank you for the second elaboration. $\endgroup$ – Eleven-Eleven Jun 23 '15 at 11:39
  • $\begingroup$ In effect, we are taking a discrete Fourier transform here. $\endgroup$ – J. M. is a poor mathematician Jul 11 '15 at 5:55

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