斐波那契双曲函数

✍ dations ◷ 2025-06-08 11:28:51 #特殊函数

斐波那契双曲函数(Fibonoacci hyperbolic functions)是一个与黄金分割有关的特殊函数

定义如下：

$sFh(x)={\frac {2*sinh(2*x*\alpha )}{\sqrt {5}}}$ $sFh(x)={\frac {2*sinh(2*x*\alpha )}{\sqrt {5}}}$

其中 $\alpha$ $\alpha$ 是黄金分割的对数：

$\alpha =ln(\phi )=ln{\frac {1+{\sqrt {5}}}{2}}=0.4812118246$ $\alpha =ln(\phi )=ln{\frac {1+{\sqrt {5}}}{2}}=0.4812118246$

$cFh(x)={\frac {2*sinh(2*x*\alpha )}{\sqrt {5}}}$ $cFh(x)={\frac {2*sinh(2*x*\alpha )}{\sqrt {5}}}$

$tFh(x)={\frac {fsh(x)}{fch(x)}}$ $tFh(x)={\frac {fsh(x)}{fch(x)}}$

$arcsFh(z)=1/2\,z{\sqrt {5}}\left({\sqrt {5}}+1\right){\it {HeunC}}\left(0,1/2,0,0,1/4,5\,{\frac {{z}^{2}}{5\,{z}^{2}+4}}\right){\frac {1}{\sqrt {5\,{z}^{2}+4}}}\left({\sqrt {5}}-1\right)^{-1}\left({\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)\right)^{-1}$ $arcsFh(z)=1/2\,z{\sqrt {5}}\left({\sqrt {5}}+1\right){\it {HeunC}}\left(0,1/2,0,0,1/4,5\,{\frac {{z}^{2}}{5\,{z}^{2}+4}}\right){\frac {1}{\sqrt {5\,{z}^{2}+4}}}\left({\sqrt {5}}-1\right)^{-1}\left({\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)\right)^{-1}$

$arccFh(z)=-1/2+\left(1/2\,{\sqrt {-\left(-2+z{\sqrt {5}}\right)^{2}}}z{\sqrt {5}}\left({\sqrt {5}}+1\right){\it {HeunC}}\left(0,1/2,0,0,1/4,5/4\,{\frac {{z}^{2}}{5/4\,{z}^{2}-1}}\right)\left(-2+z{\sqrt {5}}\right)^{-1}{\frac {1}{\sqrt {-5\,{z}^{2}+4}}}\left({\sqrt {5}}-1\right)^{-1}-1/4\,{\frac {{\sqrt {-\left(-2+z{\sqrt {5}}\right)^{2}}}\pi \,\left({\sqrt {5}}+1\right)}{\left(-2+z{\sqrt {5}}\right)\left({\sqrt {5}}-1\right)}}\right)\left({\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)\right)^{-1}$ $arccFh(z)=-1/2+\left(1/2\,{\sqrt {-\left(-2+z{\sqrt {5}}\right)^{2}}}z{\sqrt {5}}\left({\sqrt {5}}+1\right){\it {HeunC}}\left(0,1/2,0,0,1/4,5/4\,{\frac {{z}^{2}}{5/4\,{z}^{2}-1}}\right)\left(-2+z{\sqrt {5}}\right)^{-1}{\frac {1}{\sqrt {-5\,{z}^{2}+4}}}\left({\sqrt {5}}-1\right)^{-1}-1/4\,{\frac {{\sqrt {-\left(-2+z{\sqrt {5}}\right)^{2}}}\pi \,\left({\sqrt {5}}+1\right)}{\left(-2+z{\sqrt {5}}\right)\left({\sqrt {5}}-1\right)}}\right)\left({\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)\right)^{-1}$

$arctFh(x)=z=4\,x\left({\sqrt {5}}-1\right){\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right){\it {HeunB}}\left(2,0,0,0,2\,{\sqrt {\frac {x\left({\sqrt {5}}-1\right){\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)}{{\sqrt {5}}+1}}}\right){{\rm {e}}^{1/2\,{\frac {-4\,{\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)x{\sqrt {5}}+4\,{\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)x-2\,{\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right){\sqrt {5}}+2\,{\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)+i\pi \,{\sqrt {5}}+i\pi }{{\sqrt {5}}+1}}}}\left({\sqrt {5}}+1\right)^{-1}\left({{\rm {e}}^{2\,{\frac {x\left({\sqrt {5}}-1\right){\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)}{{\sqrt {5}}+1}}}}\right)^{-1}\left({\frac {2\,i\left(2\,x+1\right)\left({\sqrt {5}}-1\right){\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)}{{\sqrt {5}}+1}}+\pi \right)^{-1}\left({\it {HeunB}}\left(2,0,0,0,{\sqrt {2}}{\sqrt {{\frac {\left(-1-2\,x\right)\left({\sqrt {5}}-1\right){\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)}{{\sqrt {5}}+1}}+1/2\,i\pi }}\right)\right)^{-1}$ $arctFh(x)=z=4\,x\left({\sqrt {5}}-1\right){\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right){\it {HeunB}}\left(2,0,0,0,2\,{\sqrt {\frac {x\left({\sqrt {5}}-1\right){\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)}{{\sqrt {5}}+1}}}\right){{\rm {e}}^{1/2\,{\frac {-4\,{\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)x{\sqrt {5}}+4\,{\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)x-2\,{\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right){\sqrt {5}}+2\,{\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)+i\pi \,{\sqrt {5}}+i\pi }{{\sqrt {5}}+1}}}}\left({\sqrt {5}}+1\right)^{-1}\left({{\rm {e}}^{2\,{\frac {x\left({\sqrt {5}}-1\right){\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)}{{\sqrt {5}}+1}}}}\right)^{-1}\left({\frac {2\,i\left(2\,x+1\right)\left({\sqrt {5}}-1\right){\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)}{{\sqrt {5}}+1}}+\pi \right)^{-1}\left({\it {HeunB}}\left(2,0,0,0,{\sqrt {2}}{\sqrt {{\frac {\left(-1-2\,x\right)\left({\sqrt {5}}-1\right){\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)}{{\sqrt {5}}+1}}+1/2\,i\pi }}\right)\right)^{-1}$

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