斐波那契双曲函数

定义如下：^[2]

斐波那契双曲正弦函数

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

其中${\displaystyle \alpha }$ 是黄金分割的对数：

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

斐波那契双曲余弦函数

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

斐波那契双曲正切函数

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

斐氏双曲函数图

关系式

• ${\displaystyle sFh(-x)=-sFh(x)}$ 
• ${\displaystyle cFh(-x)=cFh(x-1)}$ 
• ${\displaystyle sFh^{2}(x)+cFh^{2}(x)=cFh(2x)}$ 
• ${\displaystyle cFh^{2}(x)-sFh^{2}(x)=1+sFh(x)cFh(x)}$ 
• ${\displaystyle sFh(x)+sFh(y)={\sqrt {(}}5)sFh*({\frac {x+y}{2}})cFh({\frac {x-y-1}{2}})}$ 
• ${\displaystyle cFh(x)+cFh(y)={\sqrt {5}}cFh({\frac {x+y}{2}})cFh({\frac {x+y-1}{2}})}$ 
• ${\displaystyle }$
• ${\displaystyle }$
• ${\displaystyle }$
• ${\displaystyle }$
• ${\displaystyle }$

级数展开

• ${\displaystyle fsh(x)\approx \left\{(4/5\,{\sqrt {5}}\ln \left(1/2+1/2\,{\sqrt {5}}\right)x+{\frac {8}{15}}\,{\sqrt {5}}\left(\ln \left(1/2+1/2\,{\sqrt {5}}\right)\right)^{3}{x}^{3}+{\frac {8}{75}}\,{\sqrt {5}}\left(\ln \left(1/2+1/2\,{\sqrt {5}}\right)\right)^{5}{x}^{5}+{\frac {16}{1575}}\,{\sqrt {5}}\left(\ln \left(1/2+1/2\,{\sqrt {5}}\right)\right)^{7}{x}^{7}+O\left({x}^{9}\right))\right\}}$ 

• ${\displaystyle fch(x)\approx {(1/5)*{\sqrt {(}}5)*(5+{\sqrt {(}}5))/({\sqrt {(}}5)+1)+(2/5)*{\sqrt {(}}5)*ln(1/2+(1/2)*{\sqrt {(}}5))*x+(2/5)*{\sqrt {(}}5)*(5+{\sqrt {(}}5))*ln(1/2+(1/2)*{\sqrt {(}}5))^{2}*x^{2}/({\sqrt {(}}5)+1)+(4/15)*{\sqrt {(}}5)*ln(1/2+(1/2)*{\sqrt {(}}5))^{3}*x^{3}+(2/15)*{\sqrt {(}}5)*(5+{\sqrt {(}}5))*ln(1/2+(1/2)*{\sqrt {(}}5))^{4}*x^{4}/({\sqrt {(}}5)+1)+O(x^{5})}}$ 

• ${\displaystyle fth(x)\approx {4*ln(1/2+(1/2)*{\sqrt {(}}5))*({\sqrt {(}}5)+1)*x/(5+{\sqrt {(}}5))-8*ln(1/2+(1/2)*{\sqrt {(}}5))^{2}*({\sqrt {(}}5)+1)^{2}*x^{2}/(5+{\sqrt {(}}5))^{2}+(2*(-(8/3)*ln(1/2+(1/2)*{\sqrt {(}}5))^{3}+8*ln(1/2+(1/2)*{\sqrt {(}}5))^{3}*({\sqrt {(}}5)+1)^{2}/(5+{\sqrt {(}}5))^{2}))*({\sqrt {(}}5)+1)*x^{3}/(5+{\sqrt {(}}5))+O(x^{4})}}$ 

渐近展开

• ${\displaystyle sFh(x)\approx \left\{1/5\,{\frac {{\sqrt {5}}\left(\left({\sqrt {5}}+1\right)^{x}\right)^{2}}{\left({2}^{x}\right)^{2}}}-1/5\,{\frac {{\sqrt {5}}\left({2}^{x}\right)^{2}}{\left(\left({\sqrt {5}}+1\right)^{x}\right)^{2}}}\right\}}$ 

• ${\displaystyle cFh(x)\approx \left\{2/5\,{\frac {\left(1/4\,{\sqrt {5}}+1/4\right){\sqrt {5}}\left(\left({\sqrt {5}}+1\right)^{x}\right)^{2}}{\left({2}^{x}\right)^{2}}}+2/5\,{\frac {{\sqrt {5}}\left({2}^{x}\right)^{2}}{\left({\sqrt {5}}+1\right)\left(\left({\sqrt {5}}+1\right)^{x}\right)^{2}}}\right\}}$ 

Heun函数表示

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

反双曲正弦

• ${\displaystyle arcsFh(z)=1/2\,{\frac {{\it {arcsinh}}\left(1/2\,z{\sqrt {5}}\right)}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}}$ 满足${\displaystyle sFh(arcsFh(y))=y}$ 

反双曲余弦

• ${\displaystyle arccFh(z)=-1/2\,{\frac {\ln \left(1/2+1/2\,{\sqrt {5}}\right)-{\it {arccosh}}\left(1/2\,z{\sqrt {5}}\right)}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}}$ 满足${\displaystyle arccFh(cFh(z))=z}$ 

反双曲正切

• ${\displaystyle arctFh(z)=1/4\,\ln \left(-{\frac {-z+z{\sqrt {5}}+2}{z+z{\sqrt {5}}-2}}\right)\left(\ln \left(1/2+1/2\,{\sqrt {5}}\right)\right)^{-1}}$ 满足${\displaystyle arctFh(tFh(z))=z}$ 

Heun函数表示

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

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

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

级数展开

• ${\displaystyle arcsFh(z)\approx (1/4\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}z-{\frac {5}{96}}\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{3}+{\frac {15}{512}}\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{5}-{\frac {625}{28672}}\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{7}+{\frac {21875}{1179648}}\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{9}+O\left({z}^{11}\right))>}$ 

• ${\displaystyle arccFh(z)\approx (-1/2\,{\frac {\ln \left(1/2+1/2\,{\sqrt {5}}\right)+1/2\,i{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right)\pi }{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}+1/4\,i{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right){\sqrt {5}}\left(\ln \left({\frac {1}{2}}+1/2\,{\sqrt {5}}\right)\right)^{-1}z+{\frac {5}{96}}\,i{\sqrt {5}}{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right)\left(\ln \left({\frac {1}{2}}+1/2\,{\sqrt {5}}\right)\right)^{-1}{z}^{3}+{\frac {15}{512}}\,i{\sqrt {5}}{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right)\left(\ln \left({\frac {1}{2}}+1/2\,{\sqrt {5}}\right)\right)^{-1}{z}^{5}+{\frac {625}{28672}}\,i{\sqrt {5}}{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right)\left(\ln \left({\frac {1}{2}}+1/2\,{\sqrt {5}}\right)\right)^{-1}{z}^{7}+{\frac {21875}{1179648}}\,i{\sqrt {5}}{\it {csgn}}\left(i\left(1/2\,z{\sqrt {5}}-1\right)\right)\left(\ln \left({\frac {1}{2}}+1/2\,{\sqrt {5}}\right)\right)^{-1}{z}^{9}+O\left({z}^{11}\right))>}$ 

• ${\displaystyle arctFh(z)\approx (1/4\,{\frac {\sqrt {5}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}z+1/4\,{\frac {1/2\,{\sqrt {5}}\left({\sqrt {5}}+1\right)-5/2}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{2}+1/4\,{\frac {-5/6\,{\sqrt {5}}-5/2+1/4\,{\sqrt {5}}\left({\sqrt {5}}+1\right)^{2}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{3}+1/4\,{\frac {-{\frac {5}{16}}\,\left({\sqrt {5}}+1\right)^{2}+1/8\,{\sqrt {5}}\left({\sqrt {5}}+1\right)^{3}-5/8\,{\sqrt {5}}-{\frac {25}{8}}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}{z}^{4}+O\left({z}^{5}\right))>}$ 

渐近展开

• ${\displaystyle arcsFh(z)\approx 1/2\,{\frac {1/2\,\ln \left(5\right)+\ln \left(z\right)}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}+1/10\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{2}}}-{\frac {3}{100}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{4}}}+{\frac {1}{75}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{6}}}-{\frac {7}{1000}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{8}}}+O\left({z}^{-10}\right)}$ 
• ${\displaystyle arccFh(z)\approx -1/2\,{\frac {\ln \left(1/2+1/2\,{\sqrt {5}}\right)-1/2\,\ln \left(5\right)-\ln \left(z\right)}{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}-1/10\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{2}}}-{\frac {3}{100}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{4}}}-{\frac {1}{75}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{6}}}-{\frac {7}{1000}}\,{\frac {1}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{8}}}+O\left({z}^{-10}\right)}$ 
• ${\displaystyle arctFh(z)\approx 1/4\,{\frac {\ln \left({\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)+i\pi }{\ln \left(1/2+1/2\,{\sqrt {5}}\right)}}+1/4\,{\frac {2+2\,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}}{\left({\sqrt {5}}-1\right)\ln \left(1/2+1/2\,{\sqrt {5}}\right)z}}+1/4\,{\frac {8\,{\frac {\sqrt {5}}{\left({\sqrt {5}}+1\right)^{2}\left({\sqrt {5}}-1\right)}}-2\,{\frac {{\sqrt {5}}\left(2+2\,{\frac {{\sqrt {5}}-1}{{\sqrt {5}}+1}}\right)}{\left({\sqrt {5}}+1\right)\left({\sqrt {5}}-1\right)^{2}}}}{\ln \left(1/2+1/2\,{\sqrt {5}}\right){z}^{2}}}+O\left({z}^{-3}\right)}$ 

1. ^ 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）. 
2. ^ Zdzlslaw W. Trzaska,ON FIBONACCI HYPERBOLIC TRIGONOMETRY AND MODIFIED NUMERICAL TRIANGLES,1994