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。

## 斐波那契双曲函数

定义如下：

斐波那契双曲正弦函数

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

其中$\alpha$是黄金分割的对数：

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

斐波那契双曲余弦函数

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

斐波那契双曲正切函数

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

### 斐氏双曲函数图

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### 关系式

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

### 级数展开

- $$fsh(x) \approx \left\{ (4/5\,\sqrt {5}\ln \left( 1/2+1/2\,\sqrt {5} \right) x+{ \frac {8}{15}}\,\sqrt {5} \left( \ln \left( 1/2+1/2\,\sqrt {5} <pre> \right) \right) ^{3}{x}^{3}+{\frac {8}{75}}\,\sqrt {5} \left( \ln \left( 1/2+1/2\,\sqrt {5} \right) \right) ^{5}{x}^{5}+{\frac {16}{ </pre> 1575}}\,\sqrt {5} \left( \ln \left( 1/2+1/2\,\sqrt {5} \right) <pre> \right) ^{7}{x}^{7}+O \left( {x}^{9} \right) ) \right\} </pre>$$
- $$fch(x) \approx {(1/5)*\sqrt(5)*(5+\sqrt(5))/(\sqrt(5)+1)+(2/5)*\sqrt(5)*ln(1/2+(1/2)*\sqrt(5))*x+(2/5)*\sqrt(5)*(5+\sqrt(5))*ln(1/2+(1/2)*\sqrt(5))^2*x^2/(\sqrt(5)+1)+(4/15)*\sqrt(5)*ln(1/2+(1/2)*\sqrt(5))^3*x^3+(2/15)*\sqrt(5)*(5+\sqrt(5))*ln(1/2+(1/2)*\sqrt(5))^4*x^4/(\sqrt(5)+1)+O(x^5)}$$
- $$fth(x) \approx {4*ln(1/2+(1/2)*\sqrt(5))*(\sqrt(5)+1)*x/(5+\sqrt(5))-8*ln(1/2+(1/2)*\sqrt(5))^2*(\sqrt(5)+1)^2*x^2/(5+\sqrt(5))^2+(2*(-(8/3)*ln(1/2+(1/2)*\sqrt(5))^3+8*ln(1/2+(1/2)*\sqrt(5))^3*(\sqrt(5)+1)^2/(5+\sqrt(5))^2))*(\sqrt(5)+1)*x^3/(5+\sqrt(5))+O(x^4)}$$

### 渐近展开

- $$sFh(x) \approx \left\{ 1/5\,{\frac {\sqrt {5} \left( \left( \sqrt {5}+1 \right) ^{x } \right) ^{2}}{ \left( {2}^{x} \right) ^{2}}}-1/5\,{\frac {\sqrt {5} <pre> \left( {2}^{x} \right) ^{2}}{ \left( \left( \sqrt {5}+1 \right) ^{x} \right) ^{2}}} \right\} </pre>$$
- $$cFh(x) \approx \left\{ 2/5\,{\frac { \left( 1/4\,\sqrt {5}+1/4 \right) \sqrt {5} <pre> \left( \left( \sqrt {5}+1 \right) ^{x} \right) ^{2}}{ \left( {2}^{x} \right) ^{2}}}+2/5\,{\frac {\sqrt {5} \left( {2}^{x} \right) ^{2}}{ \left( \sqrt {5}+1 \right) \left( \left( \sqrt {5}+1 \right) ^{x} \right) ^{2}}} \right\} </pre>$$

### Heun函数表示

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

## 斐波那契双曲反函数

反双曲正弦

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

反双曲余弦

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

反双曲正切

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

### Heun函数表示

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

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

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

### 级数展开

- $$arcsFh(z) \approx (1/4\,{\frac {\sqrt {5}}{\ln \left( 1/2+1/2\,\sqrt {5} \right) }}z-{ \frac {5}{96}}\,{\frac {\sqrt {5}}{\ln \left( 1/2+1/2\,\sqrt {5} <pre> \right) }}{z}^{3}+{\frac {15}{512}}\,{\frac {\sqrt {5}}{\ln \left( 1 </pre> /2+1/2\,\sqrt {5} \right) }}{z}^{5}-{\frac {625}{28672}}\,{\frac { \sqrt {5}}{\ln \left( 1/2+1/2\,\sqrt {5} \right) }}{z}^{7}+{\frac { 21875}{1179648}}\,{\frac {\sqrt {5}}{\ln \left( 1/2+1/2\,\sqrt {5} <pre> \right) }}{z}^{9}+O \left( {z}^{11} \right) ) </pre> >$$
- $$arccFh(z) \approx (-1/2\,{\frac {\ln \left( 1/2+1/2\,\sqrt {5} \right) +1/2\,i{\it csgn } \left( i \left( 1/2\,z\sqrt {5}-1 \right) \right) \pi }{\ln <pre> \left( 1/2+1/2\,\sqrt {5} \right) }}+1/4\,i{\it csgn} \left( i \left( 1/2\,z\sqrt {5}-1 \right) \right) \sqrt {5} \left( \ln \left( {\frac {1}{2}}+1/2\,\sqrt {5} \right) \right) ^{-1}z+{\frac { </pre> 5}{96}}\,i\sqrt {5}{\it csgn} \left( i \left( 1/2\,z\sqrt {5}-1 <pre> \right) \right) \left( \ln \left( {\frac {1}{2}}+1/2\,\sqrt {5} \right) \right) ^{-1}{z}^{3}+{\frac {15}{512}}\,i\sqrt {5}{\it csgn} \left( i \left( 1/2\,z\sqrt {5}-1 \right) \right) \left( \ln \left( {\frac {1}{2}}+1/2\,\sqrt {5} \right) \right) ^{-1}{z}^{5}+{ </pre> \frac {625}{28672}}\,i\sqrt {5}{\it csgn} \left( i \left( 1/2\,z\sqrt {5}-1 \right) \right) \left( \ln \left( {\frac {1}{2}}+1/2\,\sqrt { 5} \right) \right) ^{-1}{z}^{7}+{\frac {21875}{1179648}}\,i\sqrt {5}{ \it csgn} \left( i \left( 1/2\,z\sqrt {5}-1 \right) \right) \left( \ln \left( {\frac {1}{2}}+1/2\,\sqrt {5} \right) \right) ^{-1}{z}^{9 }+O \left( {z}^{11} \right) ) >$$
- $$arctFh(z) \approx (1/4\,{\frac {\sqrt {5}}{\ln \left( 1/2+1/2\,\sqrt {5} \right) }}z+1/ 4\,{\frac {1/2\,\sqrt {5} \left( \sqrt {5}+1 \right) -5/2}{\ln <pre> \left( 1/2+1/2\,\sqrt {5} \right) }}{z}^{2}+1/4\,{\frac {-5/6\,\sqrt </pre> {5}-5/2+1/4\,\sqrt {5} \left( \sqrt {5}+1 \right) ^{2}}{\ln \left( 1/ 2+1/2\,\sqrt {5} \right) }}{z}^{3}+1/4\,{\frac {-{\frac {5}{16}}\, <pre> \left( \sqrt {5}+1 \right) ^{2}+1/8\,\sqrt {5} \left( \sqrt {5}+1 \right) ^{3}-5/8\,\sqrt {5}-{\frac {25}{8}}}{\ln \left( 1/2+1/2\, </pre> \sqrt {5} \right) }}{z}^{4}+O \left( {z}^{5} \right) ) >$$

### 渐近展开

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

## 反函数图

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