查看“︁斐波那契双曲函数”︁的源代码

'''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<ref>{{Cite mathworld|urlname=FibonacciHyperbolicFunctions|title=Fibonoacci hyperbolic functions |access-date=2015-05-12 |archive-date=2021-02-24 |archive-url=https://web.archive.org/web/20210224105014/https://mathworld.wolfram.com/FibonacciHyperbolicFunctions.html |dead-url=no }}</ref>。

==斐波那契双曲函数==
定义如下：<ref>Zdzlslaw W. Trzaska,ON FIBONACCI HYPERBOLIC TRIGONOMETRY
AND MODIFIED NUMERICAL TRIANGLES,1994</ref>

;斐波那契双曲正弦函数
<math> sFh(x)=\frac{ 2*sinh(2*x*\alpha)}{\sqrt{5}}        </math>

其中<math>\alpha</math>是[[黄金分割]]的对数：

<math>\alpha=ln(\phi)=ln \frac{1+\sqrt{5}}{2}=0.4812118246</math>



;斐波那契双曲余弦函数
<math> cFh(x)= \frac{2*sinh(2*x*\alpha)}{\sqrt{5}}    </math>





;斐波那契双曲正切函数
<math> tFh(x)=\frac{fsh(x)}{fch(x)}    </math>

===斐氏双曲函数图===
{|
|[[File:Fibonacci hyperbolic sine 01.png|thumb|Fibonacci hyperbolic sine]]
|[[File:Fibonacci hyperbolic sine imaginary part density plot.png|thumb|斐波那契双曲正弦虚数图]]
|[[File:Fibonacci hyperbolic sine imaginary part 3D.png|thumb|斐氏正弦虚数三维图]]
|}
{|
|[[File:Fibonacci hyperbolic cosine.png|thumb|Fibonacci hyperbolic cosine]]
|[[File:Fibonacci hyperbolic cosine imaginary part 3D.png|thumb|斐氏双曲余弦虚数三维图]]
|}
{|
|[[File:Fibonacci hyperbolic tangent.png|thumb|Fibonacci hyperbolic tangent]]
|[[File:Fibonacci hyperbolic tangent imaginary part density plot.png|thumb|斐氏双曲正切虚数图]]
|}

===关系式===
*<math>sFh(-x)=-sFh(x)       </math>
*<math> cFh(-x)=cFh(x-1)      </math>
*<math> sFh^2(x)+cFh^2(x)=cFh(2x)     </math>
*<math> cFh^2(x)-sFh^2(x)=1+sFh(x)cFh(x)     </math>
*<math>sFh(x)+sFh(y)=\sqrt(5)sFh*(\frac{x+y}{2})cFh(\frac{x-y-1}{2}     )      </math>
*<math> cFh(x)+cFh(y)=\sqrt{5}cFh(\frac{x+y}{2}  )cFh(\frac{x+y-1}{2}    )     </math>
*<math>      </math>
*<math>      </math>
*<math>      </math>
*<math>      </math>
*<math>      </math>






===级数展开===
*<math>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\} 
</math>

*<math>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)}</math>

*<math>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)} </math>

===渐近展开===
*<math> 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\} 
</math>


*<math>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\} 
</math>
===Heun函数表示===
*<math>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}
</math>
*<math>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}
</math>
*<math>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}
</math>

==斐波那契双曲反函数==
;反双曲正弦
*<math>arcsFh(z)=1/2\,{\frac {{\it arcsinh} \left( 1/2\,z\sqrt {5} \right) }{\ln 
 \left( 1/2+1/2\,\sqrt {5} \right) }}
          </math>满足<math>sFh(arcsFh(y))=y</math>

;反双曲余弦
*<math> 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) }}
        </math>满足<math>arccFh(cFh(z))=z</math>
;反双曲正切
*<math>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}
         </math>满足<math>arctFh(tFh(z))=z</math>

===Heun函数表示===
<math>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}</math>

<math>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}
</math>

<math>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}
</math>

===级数展开===
*<math>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) )
> 
</math>

*<math>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) )
> 
</math>

*<math>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) )
> 
</math>

===渐近展开===
*<math>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) 
</math>
*<math>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) 
</math>
*<math>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) 
</math>

==反函数图==
{|
|[[File:Fibonacci hyperbolic arcsine 2D.png|thumb|Fibonacci hyperbolic arcsine 2D]]
|[[File:Fibonacci hyperbolic arcsine imaginary part 3D plot.png|thumb|斐氏正弦虚数三维图]]
|[[File:Fibonacci hyperbolic arcsine 2D density plot.png|thumb|Fibonacci hyperbolic arcsine 2D density plot]]
|[[File:Fibonacci hyperbolic sine imaginary part density plot.png|thumb|Fibonacci hyperbolic sine imaginary part density plot]]
|}
{|
|[[File:Fibonacci hyperbolic arccosine 2D.png|thumb|Fibonacci hyperbolic arccosine 2D]]
|[[File:Fibonacci hyperbolic arccosine 2D density plot.png|thumb|Fibonacci hyperbolic arccosine 2D density plot]]
|[[File:Fibonacci hyperbolic arccosine imaginary part 3D plot.png|thumb|斐氏余弦虚数三维图]]
|}
{|
|[[File:Fibonacci hyperbolic tangent imaginary part density plot.png|thumb|Fibonacci hyperbolic tangent imaginary part density plot]]
|[[File:Fibonacci hyperbolic arctan 2D.png|thumb|Fibonacci hyperbolic arctan 2D]]
|[[File:Fibonacci hyperbolic arctan imaginary part 3D plot.png|thumb|斐氏正切虚数三维图]]
|}

==参考文献==
<references/>

[[Category:特殊函数]]