斐波那契双曲函数

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： 斐波那契双曲正弦函数 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} 斐波那契双曲余弦函数 c F h ( x ) = 2 ∗ s i n h ( 2 ∗ x ∗ α ) 5 {\displaystyle cFh(x)={\frac {2*sinh(2*x*\alpha )}{\sqrt {5}}}} 斐波那契双曲正切函数 t F h ( x ) = f s h ( x ) f c h ( x ) {\displaystyle tFh(x)={\frac {fsh(x)}{fch(x)}}} s F h ( − x ) = − s F h ( x ) {\displaystyle sFh(-x)=-sFh(x)} c F h ( − x ) = c F h ( x − 1 ) {\displaystyle cFh(-x)=cFh(x-1)} s F h 2 ( x ) + c F h 2 ( x ) = c F h ( 2 x ) {\displaystyle sFh^{2}(x)+cFh^{2}(x)=cFh(2x)} c F h 2 ( x ) − s F h 2 ( x ) = 1 + s F h ( x ) c F h ( x ) {\displaystyle cFh^{2}(x)-sFh^{2}(x)=1+sFh(x)cFh(x)} s F h ( x ) + s F h ( y ) = ( 5 ) s F h ∗ ( x + y 2 ) c F h ( x − y − 1 2 ) {\displaystyle sFh(x)+sFh(y)={\sqrt {(}}5)sFh*({\frac {x+y}{2}})cFh({\frac {x-y-1}{2}})} c F h ( x ) + c F h ( y ) = 5 c F h ( x + y 2 ) c F h ( x + y − 1 2 ) {\displaystyle cFh(x)+cFh(y)={\sqrt {5}}cFh({\frac {x+y}{2}})cFh({\frac {x+y-1}{2}})} {\displaystyle } {\displaystyle } {\displaystyle } {\displaystyle } {\displaystyle } f s h ( x ) ≈ { ( 4 / 5 5 ln ⁡ ( 1 / 2 + 1 / 2 5 ) x + 8 15 5 ( ln ⁡ ( 1 / 2 + 1 / 2 5 ) ) 3 x 3 + 8 75 5 ( ln ⁡ ( 1 / 2 + 1 / 2 5 ) ) 5 x 5 + 16 1575 5 ( ln ⁡ ( 1 / 2 + 1 / 2 5 ) ) 7 x 7 + O ( x 9 ) ) } {\displaystyle fsh(x)\approx \left\{(4/5\,{\sqrt {5}}\ln \left(1/2+1/2\,{\sqrt {5}}\right)x+{\frac {8}{15}}\,{\sqrt {5}}\left(\ln \left(1/2+1/2\,{\sqrt {5}}\right)\right)^{3}{x}^{3}+{\frac {8}{75}}\,{\sqrt {5}}\left(\ln \left(1/2+1/2\,{\sqrt {5}}\right)\right)^{5}{x}^{5}+{\frac {16}{1575}}\,{\sqrt {5}}\left(\ln \left(1/2+1/2\,{\sqrt {5}}\right)\right)^{7}{x}^{7}+O\left({x}^{9}\right))\right\}} f c h ( x ) ≈ ( 1 / 5 ) ∗ ( 5 ) ∗ ( 5 + ( 5 ) ) / ( ( 5 ) + 1 ) + ( 2 / 5 ) ∗ ( 5 ) ∗ l n ( 1 / 2 + ( 1 / 2 ) ∗ ( 5 ) ) ∗ x + ( 2 / 5 ) ∗ ( 5 ) ∗ ( 5 + ( 5 ) ) ∗ l n ( 1 / 2 + ( 1 / 2 ) ∗ ( 5 ) ) 2 ∗ x 2 / ( ( 5 ) + 1 ) + ( 4 / 15 ) ∗ ( 5 ) ∗ l n ( 1 / 2 + ( 1 / 2 ) ∗ ( 5 ) ) 3 ∗ x 3 + ( 2 / 15 ) ∗ ( 5 ) ∗ ( 5 + ( 5 ) ) ∗ l n ( 1 / 2 + ( 1 / 2 ) ∗ ( 5 ) ) 4 ∗ x 4 / ( ( 5 ) + 1 ) + O ( x 5 ) {\displaystyle fch(x)\approx {(1/5)*{\sqrt {(}}5)*(5+{\sqrt {(}}5))/({\sqrt {(}}5)+1)+(2/5)*{\sqrt {(}}5)*ln(1/2+(1/2)*{\sqrt {(}}5))*x+(2/5)*{\sqrt {(}}5)*(5+{\sqrt {(}}5))*ln(1/2+(1/2)*{\sqrt {(}}5))^{2}*x^{2}/({\sqrt {(}}5)+1)+(4/15)*{\sqrt {(}}5)*ln(1/2+(1/2)*{\sqrt {(}}5))^{3}*x^{3}+(2/15)*{\sqrt {(}}5)*(5+{\sqrt {(}}5))*ln(1/2+(1/2)*{\sqrt {(}}5))^{4}*x^{4}/({\sqrt {(}}5)+1)+O(x^{5})}} f t h ( x ) ≈ 4 ∗ l n ( 1 / 2 + ( 1 / 2 ) ∗ ( 5 ) ) ∗ ( ( 5 ) + 1 ) ∗ x / ( 5 + ( 5 ) ) − 8 ∗ l n ( 1 / 2 + ( 1 / 2 ) ∗ ( 5 ) ) 2 ∗ ( ( 5 ) + 1 ) 2 ∗ x 2 / ( 5 + ( 5 ) ) 2 + ( 2 ∗ ( − ( 8 / 3 ) ∗ l n ( 1 / 2 + ( 1 / 2 ) ∗ ( 5 ) ) 3 + 8 ∗ l n ( 1 / 2 + ( 1 / 2 ) ∗ ( 5 ) ) 3 ∗ ( ( 5 ) + 1 ) 2 / ( 5 + ( 5 ) ) 2 ) ) ∗ ( ( 5 ) + 1 ) ∗ x 3 / ( 5 + ( 5 ) ) + O ( x 4 ) {\displaystyle fth(x)\approx {4*ln(1/2+(1/2)*{\sqrt {(}}5))*({\sqrt {(}}5)+1)*x/(5+{\sqrt {(}}5))-8*ln(1/2+(1/2)*{\sqrt {(}}5))^{2}*({\sqrt {(}}5)+1)^{2}*x^{2}/(5+{\sqrt {(}}5))^{2}+(2*(-(8/3)*ln(1/2+(1/2)*{\sqrt {(}}5))^{3}+8*ln(1/2+(1/2)*{\sqrt {(}}5))^{3}*({\sqrt {(}}5)+1)^{2}/(5+{\sqrt {(}}5))^{2}))*({\sqrt {(}}5)+1)*x^{3}/(5+{\sqrt {(}}5))+O(x^{4})}} s F h ( x ) ≈ { 1 / 5 5 ( ( 5 + 1 ) x ) 2 ( 2 x ) 2 − 1 / 5 5 ( 2 x ) 2 ( ( 5 + 1 ) x ) 2 } {\displaystyle sFh(x)\approx \left\{1/5\,{\frac {{\sqrt {5}}\left(\left({\sqrt {5}}+1\right)^{x}\right)^{2}}{\left({2}^{x}\right)^{2}}}-1/5\,{\frac {{\sqrt {5}}\left({2}^{x}\right)^{2}}{\left(\left({\sqrt {5}}+1\right)^{x}\right)^{2}}}\right\}} c F h ( x ) ≈ { 2 / 5 ( 1 / 4 5 + 1 / 4 ) 5 ( ( 5 + 1 ) x ) 2 ( 2 x ) 2 + 2 / 5 5 ( 2 x ) 2 ( 5 + 1 ) ( ( 5 + 1 ) x ) 2 } {\displaystyle cFh(x)\approx \left\{2/5\,{\frac {\left(1/4\,{\sqrt {5}}+1/4\right){\sqrt {5}}\left(\left({\sqrt {5}}+1\right)^{x}\right)^{2}}{\left({2}^{x}\right)^{2}}}+2/5\,{\frac {{\sqrt {5}}\left({2}^{x}\right)^{2}}{\left({\sqrt {5}}+1\right)\left(\left({\sqrt {5}}+1\right)^{x}\right)^{2}}}\right\}} s F h ( x ) = 4 / 5 5 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 ) ( 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 {\displaystyle sFh(x)=4/5\,{\sqrt {5}}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)\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}} c F h ( x ) = 2 / 5 i 5 ( 2 x + 1 ) ( 5 − 1 ) H e u n B ( 2 , 0 , 0 , 0 , 2 ( − 2 x − 1 ) ( 5 − 1 ) H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 − 1 5 + 1 ) 5 + 1 + 1 / 2 i π ) H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 − 1 5 + 1 ) ( 5 + 1 ) − 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 ) − 1 + 1 / 5 5 π H e u n B ( 2 , 0 , 0 , 0 , 2 ( − 2 x − 1 ) ( 5 − 1 ) H e u n C ( 0 , 1 , 0 , 0 , 1 / 2 , 5 − 1 5 + 1 ) 5 + 1 + 1 / 2 i π ) ( 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 ) − 1 {\displaystyle cFh(x)=2/5\,i{\sqrt {5}}\left(2\,x+1\right)\left({\sqrt {5}}-1\right){\it {HeunB}}\left(2,0,0,0,{\sqrt {2}}{\sqrt {{\frac {\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}}+1/2\,i\pi }}\right){\it {HeunC}}\left(0,1,0,0,1/2,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)\left({\sqrt {5}}+1\right)^{-1}\left({{\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}}}}\right)^{-1}+1/5\,{\sqrt {5}}\pi \,{\it {HeunB}}\left(2,0,0,0,{\sqrt {2}}{\sqrt {{\frac {\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}}+1/2\,i\pi }}\right)\left({{\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}}}}\right)^{-1}} t F h ( x ) = 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 ( − 2 x − 1 ) ( 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 tFh(x)=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(-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}}+1/2\,i\pi }}\right)\right)^{-1}} 反双曲正弦 a r c s F h ( z ) = 1 / 2 a r c s i n h ( 1 / 2 z 5 ) ln ⁡ ( 1 / 2 + 1 / 2 5 ) {\displaystyle arcsFh(z)=1/2\,{\frac {{\it {arcsinh}}\left(1/2\,z{\sqrt {5}}\right)}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}} 满足 s F h ( a r c s F h ( y ) ) = y {\displaystyle sFh(arcsFh(y))=y} 反双曲余弦 a r c c F h ( z ) = − 1 / 2 ln ⁡ ( 1 / 2 + 1 / 2 5 ) − a r c c o s h ( 1 / 2 z 5 ) ln ⁡ ( 1 / 2 + 1 / 2 5 ) {\displaystyle arccFh(z)=-1/2\,{\frac {\ln \left(1/2+1/2\,{\sqrt {5}}\right)-{\it {arccosh}}\left(1/2\,z{\sqrt {5}}\right)}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}} 满足 a r c c F h ( c F h ( z ) ) = z {\displaystyle arccFh(cFh(z))=z} 反双曲正切 a r c t F h ( z ) = 1 / 4 ln ⁡ ( − − z + z 5 + 2 z + z 5 − 2 ) ( ln ⁡ ( 1 / 2 + 1 / 2 5 ) ) − 1 {\displaystyle arctFh(z)=1/4\,\ln \left(-{\frac {-z+z{\sqrt {5}}+2}{z+z{\sqrt {5}}-2}}\right)\left(\ln \left(1/2+1/2\,{\sqrt {5}}\right)\right)^{-1}} 满足 a r c t F h ( t F h ( z ) ) = z {\displaystyle arctFh(tFh(z))=z} 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}} a r c s F h ( z ) ≈ ( 1 / 4 5 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z − 5 96 5 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 3 + 15 512 5 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 5 − 625 28672 5 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 7 + 21875 1179648 5 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 9 + O ( z 11 ) ) > {\displaystyle arcsFh(z)\approx (1/4\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}z-{\frac {5}{96}}\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{3}+{\frac {15}{512}}\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{5}-{\frac {625}{28672}}\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{7}+{\frac {21875}{1179648}}\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{9}+O\left({z}^{11}\right))>} a r c c F h ( z ) ≈ ( − 1 / 2 ln ⁡ ( 1 / 2 + 1 / 2 5 ) + 1 / 2 i c s g n ( i ( 1 / 2 z 5 − 1 ) ) π ln ⁡ ( 1 / 2 + 1 / 2 5 ) + 1 / 4 i c s g n ( i ( 1 / 2 z 5 − 1 ) ) 5 ( ln ⁡ ( 1 2 + 1 / 2 5 ) ) − 1 z + 5 96 i 5 c s g n ( i ( 1 / 2 z 5 − 1 ) ) ( ln ⁡ ( 1 2 + 1 / 2 5 ) ) − 1 z 3 + 15 512 i 5 c s g n ( i ( 1 / 2 z 5 − 1 ) ) ( ln ⁡ ( 1 2 + 1 / 2 5 ) ) − 1 z 5 + 625 28672 i 5 c s g n ( i ( 1 / 2 z 5 − 1 ) ) ( ln ⁡ ( 1 2 + 1 / 2 5 ) ) − 1 z 7 + 21875 1179648 i 5 c s g n ( i ( 1 / 2 z 5 − 1 ) ) ( ln ⁡ ( 1 2 + 1 / 2 5 ) ) − 1 z 9 + O ( z 11 ) ) > {\displaystyle arccFh(z)\approx (-1/2\,{\frac {\ln \left(1/2+1/2\,{\sqrt {5}}\right)+1/2\,i{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right)\pi }{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}+1/4\,i{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right){\sqrt {5}}\left(\ln \left({\frac {1}{2}}+1/2\,{\sqrt {5}}\right)\right)^{-1}z+{\frac {5}{96}}\,i{\sqrt {5}}{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right)\left(\ln \left({\frac {1}{2}}+1/2\,{\sqrt {5}}\right)\right)^{-1}{z}^{3}+{\frac {15}{512}}\,i{\sqrt {5}}{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right)\left(\ln \left({\frac {1}{2}}+1/2\,{\sqrt {5}}\right)\right)^{-1}{z}^{5}+{\frac {625}{28672}}\,i{\sqrt {5}}{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right)\left(\ln \left({\frac {1}{2}}+1/2\,{\sqrt {5}}\right)\right)^{-1}{z}^{7}+{\frac {21875}{1179648}}\,i{\sqrt {5}}{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right)\left(\ln \left({\frac {1}{2}}+1/2\,{\sqrt {5}}\right)\right)^{-1}{z}^{9}+O\left({z}^{11}\right))>} a r c t F h ( z ) ≈ ( 1 / 4 5 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z + 1 / 4 1 / 2 5 ( 5 + 1 ) − 5 / 2 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 2 + 1 / 4 − 5 / 6 5 − 5 / 2 + 1 / 4 5 ( 5 + 1 ) 2 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 3 + 1 / 4 − 5 16 ( 5 + 1 ) 2 + 1 / 8 5 ( 5 + 1 ) 3 − 5 / 8 5 − 25 8 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 4 + O ( z 5 ) ) > {\displaystyle arctFh(z)\approx (1/4\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}z+1/4\,{\frac {1/2\,{\sqrt {5}}\left({\sqrt {5}}+1\right)-5/2}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{2}+1/4\,{\frac {-5/6\,{\sqrt {5}}-5/2+1/4\,{\sqrt {5}}\left({\sqrt {5}}+1\right)^{2}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{3}+1/4\,{\frac {-{\frac {5}{16}}\,\left({\sqrt {5}}+1\right)^{2}+1/8\,{\sqrt {5}}\left({\sqrt {5}}+1\right)^{3}-5/8\,{\sqrt {5}}-{\frac {25}{8}}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{4}+O\left({z}^{5}\right))>} a r c s F h ( z ) ≈ 1 / 2 1 / 2 ln ⁡ ( 5 ) + ln ⁡ ( z ) ln ⁡ ( 1 / 2 + 1 / 2 5 ) + 1 / 10 1 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 2 − 3 100 1 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 4 + 1 75 1 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 6 − 7 1000 1 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 8 + O ( z − 10 ) {\displaystyle arcsFh(z)\approx 1/2\,{\frac {1/2\,\ln \left(5\right)+\ln \left(z\right)}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}+1/10\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{2}}}-{\frac {3}{100}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{4}}}+{\frac {1}{75}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{6}}}-{\frac {7}{1000}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{8}}}+O\left({z}^{-10}\right)} a r c c F h ( z ) ≈ − 1 / 2 ln ⁡ ( 1 / 2 + 1 / 2 5 ) − 1 / 2 ln ⁡ ( 5 ) − ln ⁡ ( z ) ln ⁡ ( 1 / 2 + 1 / 2 5 ) − 1 / 10 1 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 2 − 3 100 1 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 4 − 1 75 1 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 6 − 7 1000 1 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 8 + O ( z − 10 ) {\displaystyle arccFh(z)\approx -1/2\,{\frac {\ln \left(1/2+1/2\,{\sqrt {5}}\right)-1/2\,\ln \left(5\right)-\ln \left(z\right)}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}-1/10\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{2}}}-{\frac {3}{100}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{4}}}-{\frac {1}{75}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{6}}}-{\frac {7}{1000}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{8}}}+O\left({z}^{-10}\right)} a r c t F h ( z ) ≈ 1 / 4 ln ⁡ ( 5 − 1 5 + 1 ) + i π ln ⁡ ( 1 / 2 + 1 / 2 5 ) + 1 / 4 2 + 2 5 − 1 5 + 1 ( 5 − 1 ) ln ⁡ ( 1 / 2 + 1 / 2 5 ) z + 1 / 4 8 5 ( 5 + 1 ) 2 ( 5 − 1 ) − 2 5 ( 2 + 2 5 − 1 5 + 1 ) ( 5 + 1 ) ( 5 − 1 ) 2 ln ⁡ ( 1 / 2 + 1 / 2 5 ) z 2 + O ( z − 3 ) {\displaystyle arctFh(z)\approx 1/4\,{\frac {\ln \left({\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)+i\pi }{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}+1/4\,{\frac {2+2\,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}}{\left({\sqrt {5}}-1\right)\ln \left(1/2+1/2\,{\sqrt {5}}\right)z}}+1/4\,{\frac {8\,{\frac {\sqrt {5}}{\left({\sqrt {5}}+1\right)^{2}\left({\sqrt {5}}-1\right)}}-2\,{\frac {{\sqrt {5}}\left(2+2\,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)}{\left({\sqrt {5}}+1\right)\left({\sqrt {5}}-1\right)^{2}}}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{2}}}+O\left({z}^{-3}\right)} Weisstein, Eric W. (编). Fibonoacci hyperbolic functions. at MathWorld--A Wolfram Web Resource. Wolfram Research, Inc. [2015-05-12]. （原始内容存档于2021-02-24） （英语）.  Zdzlslaw W. Trzaska,ON FIBONACCI HYPERBOLIC TRIGONOMETRY AND MODIFIED NUMERICAL TRIANGLES,1994