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 2*sinh(3^h*asinh(x/2)) is the superfunction of (...) ? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/11/2013, 10:49 AM (This post was last modified: 09/11/2013, 10:51 AM by mike3.) Letting $P_n(x) = \sinh((2n+1) \mathrm{arsinh}(x))$ (so your $f_{2n+1}(x) = 2 P_n(x/2)$), I noticed $P_0(x) = x$ $P_1(x) = 4x^3 + 3x$ $P_2(x) = 16x^5 + 20x^3 + 5x$ $P_3(x) = 64x^7 + 112x^5 + 56x^3 + 7x$ $P_4(x) = 256x^9 + 576x^7 + 432x^5 + 120x^3 + 9x$ ... Now look at the Chebyshev polynomials $T_n(x)$ for odd $n$... $T_1(x) = x$ $T_3(x) = 4x^3 - 3x$ $T_5(x) = 16x^5 - 20x^3 + 5x$ $T_7(x) = 64x^7 - 112x^5 + 56x^3 - 7x$ $T_9(x) = 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x$ ... So $P_n(x) = \frac{T_{2n+1}(ix)}{i} = -i T_{2n+1}(ix)$. Then, $f_{2n+1}(x) = 2 \sinh((2n+1) \mathrm{arsinh}(x/2)) = -2i T_{2n+1}\left(\frac{ix}{2}\right)$. This, I suppose, is as "close as we can get to something in the 'usual' toolbox", at least if your "usual" toolbox has enough to include well-known named sequences of polynomials like the Chebyshev polynomials. Note also that $T_n(x) = \cos(n \mathrm{arccos}(x)) = \cosh(n \mathrm{arcosh}(x))$ and, for $x > 0$, \begin{align} \mathrm{arcosh}(ix) &= \log(ix + \sqrt{(ix)^2 - 1})\\ &= \log(ix + \sqrt{-x^2 - 1})\\ &= \log(ix + \sqrt{-(x^2 + 1)})\\ &= \log(ix + i \sqrt{x^2 + 1}) \\ &= \log(i(x + \sqrt{x^2 + 1})\\ &= \log(i) + \log(x + \sqrt{x^2 + 1})\ (\mathrm{note\ that\ }-\pi < \arg(i) + \arg(x + \sqrt{x^2 + 1}) < \pi\ \mathrm{so\ the\ log\ identity\ is\ OK})\\ &= \log(i) + \mathrm{arsinh}(x) \end{align} . Now $T_{2n+1}(ix) = \sinh((2n+1) (\log(i) + \mathrm{arsinh}(x))) = \cosh((2n+1) \log(i)) \sinh((2n+1) \mathrm{arsinh}(x)) + \sinh((2n+1) \log(i)) \cosh((2n+1) \mathrm{arsinh}(x))$ Taking $\log(i) = \frac{i \pi}{2}$ and using the correspondence between hyperbolic and trigonometric functions gives $\sinh\left((2n+1)\frac{i \pi}{2}\right) = \sinh\left(i\left(n + \frac{1}{2}\right) \pi\right) = i \sin\left(\left(n + \frac{1}{2}\right)\pi\right) = i (-1)^n$. Also, for cosh we get $\cosh\left((2n+1)\frac{i \pi}{2}\right) = \cos\left(\left(n + \frac{1}{2}\right)\pi\right) = 0$. Thus the result above simplifies to $T_{2n+1}(x) = i \sinh((2n+1) \mathrm{arsinh}(x))$ (the $(-1)^n$ drops out due to the evenness of cosh) and so we have a formal proof of the relation to the Chebyshev polynomials we just gave. « Next Oldest | Next Newest »

 Messages In This Thread 2*sinh(3^h*asinh(x/2)) is the superfunction of (...) ? - by Gottfried - 09/10/2013, 12:23 PM RE: 2*sinh(3^h*asinh(x/2)) is the superfunction of (...) ? - by tommy1729 - 09/10/2013, 09:50 PM RE: 2*sinh(3^h*asinh(x/2)) is the superfunction of (...) ? - by tommy1729 - 09/10/2013, 10:06 PM RE: 2*sinh(3^h*asinh(x/2)) is the superfunction of (...) ? - by mike3 - 09/11/2013, 10:49 AM RE: 2*sinh(3^h*asinh(x/2)) is the superfunction of (...) ? - by Gottfried - 09/11/2013, 06:30 PM RE: 2*sinh(3^h*asinh(x/2)) is the superfunction of (...) ? - by Gottfried - 09/11/2013, 08:32 PM

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