斐波那契双曲函数

✍ dations ◷ 2025-04-04 20:47:24 #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:

s F h ( x ) = 2 s i n h ( 2 x α ) 5 {\displaystyle sFh(x)={\frac {2*sinh(2*x*\alpha )}{\sqrt {5}}}}

其中 α {\displaystyle \alpha } 是黄金分割的对数:

α = l n ( ϕ ) = l n 1 + 5 2 = 0.4812118246 {\displaystyle \alpha =ln(\phi )=ln{\frac {1+{\sqrt {5}}}{2}}=0.4812118246}


c F h ( x ) = 2 s i n h ( 2 x α ) 5 {\displaystyle cFh(x)={\frac {2*sinh(2*x*\alpha )}{\sqrt {5}}}}



t F h ( x ) = f s h ( x ) f c h ( x ) {\displaystyle tFh(x)={\frac {fsh(x)}{fch(x)}}}





a r c s F h ( z ) = 1 / 2 z 5 ( 5 + 1 ) H e u n C ( 0 , 1 / 2 , 0 , 0 , 1 / 4 , 5 z 2 5 z 2 + 4 ) 1 5 z 2 + 4 ( 5 1 ) 1 ( H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 1 5 + 1 ) ) 1 {\displaystyle 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}}

a r c c F h ( z ) = 1 / 2 + ( 1 / 2 ( 2 + z 5 ) 2 z 5 ( 5 + 1 ) H e u n C ( 0 , 1 / 2 , 0 , 0 , 1 / 4 , 5 / 4 z 2 5 / 4 z 2 1 ) ( 2 + z 5 ) 1 1 5 z 2 + 4 ( 5 1 ) 1 1 / 4 ( 2 + z 5 ) 2 π ( 5 + 1 ) ( 2 + z 5 ) ( 5 1 ) ) ( H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 1 5 + 1 ) ) 1 {\displaystyle 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}}

a r c t F h ( x ) = z = 4 x ( 5 1 ) H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 1 5 + 1 ) H e u n B ( 2 , 0 , 0 , 0 , 2 x ( 5 1 ) H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 1 5 + 1 ) 5 + 1 ) e 1 / 2 4 H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 1 5 + 1 ) x 5 + 4 H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 1 5 + 1 ) x 2 H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 1 5 + 1 ) 5 + 2 H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 1 5 + 1 ) + i π 5 + i π 5 + 1 ( 5 + 1 ) 1 ( e 2 x ( 5 1 ) H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 1 5 + 1 ) 5 + 1 ) 1 ( 2 i ( 2 x + 1 ) ( 5 1 ) H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 1 5 + 1 ) 5 + 1 + π ) 1 ( H e u n B ( 2 , 0 , 0 , 0 , 2 ( 1 2 x ) ( 5 1 ) H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 1 5 + 1 ) 5 + 1 + 1 / 2 i π ) ) 1 {\displaystyle 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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