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Hyperbolic Functions.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Hyperbolic Functions} \begin{align*} \text{\bf Revision:}\quad&\text{Consider circle }x^2+y^2=1.\quad \text{The line segment from the origin to a point on the circle $(x,y)$ form}\\ &\text{an angle $\alpha$ with the $x$-axis. The area bound by this line segment, the $x$-axis and the circle is $\tfrac{~\alpha~}{2}$}.\\ &\text{We define: $\cos\alpha=x$\quad and\quad$\sin\alpha=y$.}\\ \\ \text{\bf Definition:}\quad&\text{Consider hyperbola }x^2-y^2=1.\quad \text{Since there is no ``angle at the centre'' for a hyperbola, we define}\\ &\text{$\alpha$ such that $\tfrac{~\alpha~}{2}$ equals to the area bound by the $x$-axis, the hyperbola and the line segment}\\ &\text{joining the origin and a point $(x,y)$ on the hyperbola. We define: $\cosh\alpha=x$\quad and\quad$\sinh\alpha=y$.}\\ \\ &\text{Let us consider the first quadrant only, where the hyperbola can be expressed by }y=\sqrt{x^2-1}.\\ &\text{The area }\frac{~\alpha~}{2}=\frac{~1~}{2}xy-\int_1^x y~dx=\frac{~1~}{2}xy-\int_1^x\sqrt{x^2-1}~dx\\ &=\frac{~1~}{2}xy-\left[\frac{~1~}{2}x\sqrt{x^2-1}-\frac{~1~}{2}\ln\left|x+\sqrt{x^2-1}\right|\right]_1^x\\ &=\frac{~1~}{2}x\sqrt{x^2-1}-\left(\frac{~1~}{2}x\sqrt{x^2-1}-\frac{~1~}{2}\ln\left|x+\sqrt{x^2-1}\right|\right)\\ &=\frac{~1~}{2}\ln\left|x+\sqrt{x^2-1}\right|\\ &\therefore\quad\alpha=\ln\left(x+\sqrt{x^2-1}\right)\quad\text{for }x,y\geq 1.\qquad e^{\alpha}=x+\sqrt{x^2-1}.\\ &(e^{\alpha}-x)^2=x^2-1,\quad e^{2\alpha}-2e^{\alpha}x+x^2=x^2-1,\quad \cosh\alpha=x=\frac{e^{2\alpha}+1}{2e^{\alpha}}=\frac{e^{\alpha}+e^{-\alpha}}{2}.\\ &\sinh\alpha=y=\sqrt{x^2-1}=\sqrt{\frac{(e^{\alpha}+e^{-\alpha})^2}{4}-1}=\sqrt{\frac{e^{2\alpha}+e^{-2\alpha}+2-4}{2}} =\sqrt{\frac{(e^{\alpha}-e^{-\alpha})^2}{4}}=\frac{e^{\alpha}-e^{-\alpha}}{2}.\\ &\therefore\quad\boxed{\cosh\alpha=\frac{e^{\alpha}+e^{-\alpha}}{2},\quad \sinh\alpha=\frac{e^{\alpha}-e^{-\alpha}}{2}}\text{ which can be easily extended to all four quadrants.}\\ &\text{Likewise, }\tanh\alpha=\frac{\sinh\alpha}{\cosh\alpha}=\frac{e^{\alpha}-e^{-\alpha}}{e^{\alpha}+e^{-\alpha}},\quad \coth\alpha=\frac{\cosh\alpha}{\sinh\alpha}=\frac{e^{\alpha}+e^{-\alpha}}{e^{\alpha}-e^{-\alpha}},\quad\text{etc.}\\ \\ \text{\bf Imagine\ldots}\quad&\text{It follows that }\cosh^2 x-\sinh^2 x=1,\quad\text{which links to the ``circular'' functions: }\cos^2 x+\sin^2 x=1.\\ &\text{Since $i^2=-1$, you may imagine that the hyperbolic and ``circular'' functions are related through $i$.}\\ &\text{Consider }\cosh(i\alpha)=\frac{e^{i\alpha}+e^{-i\alpha}}{2}=\frac{(\cos\alpha+i\sin\alpha)+(\cos\alpha-i\sin\alpha)}{2}=\cos\alpha,\\ &\text{and }\sinh(i\alpha)=\frac{e^{i\alpha}-e^{-i\alpha}}{2}=\frac{(\cos\alpha+i\sin\alpha)-(\cos\alpha-i\sin\alpha)}{2}=i\sin\alpha,\\ &\text{We define: }\boxed{\cos(i\alpha)=\cosh\alpha\quad\text{and}\quad\sin(i\alpha)=i\sinh\alpha} \quad\text{so }\cosh\alpha=\cos(i\alpha)\quad\text{and}\quad\sinh\alpha=-i\sin(i\alpha).\\ &\text{It follows: }\tan(i\alpha)=\frac{\sin(i\alpha)}{\cos(i\alpha)}=\frac{i\sinh\alpha}{\cosh\alpha}=i\tanh\alpha \quad\text{and}\quad \cot(i\alpha)=\frac{\cos(i\alpha)}{\sin(i\alpha)}=\frac{\cosh\alpha}{i\sinh\alpha}=-i\coth\alpha\\ &\text{Now the equivalence is clear: }1=\cos^2(i\alpha)+\sin^2(i\alpha)=\cosh^2\alpha+(i\sinh\alpha)^2=\cosh^2\alpha-\sinh^2\alpha.\\ \\ &\text{You may imagine a rectangular hyperbola with its foci on $x$-axis being a circle with an imaginary $y$-axis}\\ &\text{(conceptually at $x=\pm\infty$). The angle at its (far away) ``centre'' can be represented by $2i$ times the area}\\ &\text{bound by the $x$-axis, the hyperbola and line segment $(0,0)-(x,y)$. The $y$ coordinate of this imaginary}\\ &\text{circle (given by $\cos(i\alpha)$) is $iy$ (or $i\sinh\alpha$ in the context of the hyperbola). On this notion, hyperbolic}\\ &\text{functions are equivalent to their ``circular'' counterparts on imaginary $y$ and angle measurements.}\\ \end{align*} % % % \begin{align*} \text{\bf To take }&\text{\bf it further}\\ &\text{Formulae of hyperbolic functions can be easily derived from their ``circular'' counterparts.}\\ &\text{Given }\cosh x=\cos(ix),\quad\sinh x=-i\sin(ix),\quad\tanh x=-i\tan(ix),\\ &\coth x=i\cot(ix),\quad\mathrm{sech~}x=\sec(ix),\quad\text{and }\mathrm{csch~}x=i\csc(ix),\quad\text{we have\ldots}\\ \\ &\sinh(-x)=-i\sin(-ix)=i\sin(ix)=-\sinh x,\\ &\cosh(-x)=\cos(-ix)=\cos(ix)=\cosh x,\\ &\tanh(-x)=-i\tan(-ix)=i\tan(ix)=-\tanh x,\\ \\ &\cosh^2 x-\sinh^2 x=\cos^2(ix)-(-i\sin(ix))^2=\cos^2(ix)+\sin^2(ix)=1,\\ &1-\mathrm{sech}^2 x=1-\sec^2(ix)=-\tan^2(ix)=-(i\tanh x)^2=\tanh^2 x,\\ &1+\mathrm{csch}^2 x=1+(i\csc(ix))^2=1-\csc^2(ix)=-\cot^2(ix)=-(-i\coth x)^2=\coth^2 x,\\ \\ &\sinh(x+y)=-i\sin(ix+iy)=-i[\sin(ix)\cos(iy)+\cos(ix)\sin(iy)]=\sinh x\cosh y+\cosh x\sinh y,\\ &\cosh(x+y)=\cos(ix+iy)=\cos(ix)\cos(iy)-\sin(ix)\sin(iy)=\cosh x\cosh y+\sinh x\sinh y,\\ &\tanh(x+y)=-i\tan(ix+iy)=-i\frac{\tan(ix)+\tan(iy)}{1-\tan(ix)\tan(iy)}=\frac{-i\tan(ix)-i\tan(iy)}{1+(-i)\tan(ix)(-i)\tan(iy)} =\frac{\tanh x+\tanh y}{1+\tanh x\tanh y},\\ \\ &\frac{d}{dx}\cosh x=\frac{d}{dx}\cos(ix)=-i\sin(ix)=\sinh x,\\ &\frac{d}{dx}\sinh x=\frac{d}{dx}(-i\sin(ix))=-i(i\cos(ix))=\cos(ix)=\cosh x,\\ &\frac{d}{dx}\tanh x=\frac{d}{dx}(-i\tan(ix))=-i(i\sec^2(ix))=\mathrm{sech}^2 x.\\ \\ &\text{For inverse hyperbolic functions, let us try $\tanh^{-1}x$~:}\\ &\text{Let }u=\tanh^{-1}x.\quad x=\tanh u=\frac{e^u-e^{-u}}{e^u+e^{-u}},\quad x(e^u+e^{-u})=e^u-e^{-u},\quad e^u(1-x)=e^{-u}(1+x),\\ &u+\ln(1-x)=-u+\ln(1+x),\quad 2u=\ln(1+x)-\ln(1-x),\quad\tanh^{-1}x=u=\frac{~1~}{2}\ln\left|\frac{1+x}{1-x}\right|.\\ &\text{Let }v=\tan^{-1}(ix).\quad ix=\tan v=\tan(i(-iv))=i\tanh(-iv),\quad x=\tanh(-iv),\\ &\text{so }\tanh^{-1}x=-iv=-i\tan^{-1}(ix).\\ \\ &\text{This can be used in integration. For example, given }\int\frac{1}{1+x^2}~dx=\tan^{-1}x+C,\quad\text{find }\int\frac{1}{1-x^2}~dx:\\ &\int\frac{1}{1-x^2}~dx=\int\frac{1}{1+(ix)^2}~dx=\frac{~1~}{i}\tan^{-1}(ix)+C=-i\tan^{-1}(ix)+C\\ &=\tanh^{-1}x+C=\frac{~1~}{2}\ln\left|\frac{1+x}{1-x}\right|+C.\\ \\ &\text{No wonder in integration formulae, negating the sign of $x^2$ often causes the inverse trigonometric}\\ &\text{function in the result to change to a logarithm function.}\\ \end{align*} \end{document}