# tvraman/emacspeak

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 % This file holds the original set of examples used to test and demo % \aster{} % in 1992. It is being copied here so we can evolve audio-formatted % maths in emacspeak. \begin{document} %%% {{{ Title \title{Mathematics for computer generated spoken documents.} \author{T. V. Raman\\ Department of Computer Science\\ 4116 Upson Hall\\ Cornell University\\ Ithaca NY 14853--7501\\ \voicemail{(607)255-9202}\\ \email{raman@cs.cornell.edu}} \date{May 27, 1992} \maketitle %%% }}} %%% {{{ simple fractions and expressions \section{simple fractions and expressions. } %% $$a+b+c+d$$ %% $$a+\frac{b}{c} +d$$ %% $$\frac{a+b}{c+d}$$ %% $$\frac{a}{b}+c+d$$ %% $$\frac{a}{b+c+d}$$ %% $$a+\frac{b+c}{d+e}+x$$ %% $$a+bc+d$$ %% $$(a+b)(c+d)$$ %%% }}} %%% {{{ superscripts and subscripts. \section{superscripts and subscripts. } %% $$x^k_1 +x^k_2 + x^k_3 + \cdots + x^k_n = 0$$ %% $$x^{k_1} + x^{k_2} + x^{k_3} + \cdots + x^{k_n} = 0$$ %% $$x_{k^1}+x_{k^2}+x_{k^3}+\cdots+x_{k^n}=0$$ %% $$x^{k^1}+x^{k^2}+x^{k^3}+\cdots+x^{k^n}=0$$ %% $$x_{k_1}+x_{k_2}+x_{k_3}+\cdots+x_{k_n}=0$$ %% $$x +_n y +_n z$$ %%% }}} %%% {{{ Knuth's examples of fractions and exponents \section{Knuth's examples of fractions and exponents. } %% $$x+y^2\over k+1$$ %% $${x+y^2\over k}+1$$ %% $$x+{y^2\over k+1}$$ %% $$x+{y^2\over k}+1$$ %% $$x+y^{2\over k+1}$$ %% $$x^{2^y} \neq {x^2}^y$$ %% $${x^2}^y = x^{2y}$$ %% %%% }}} %%% {{{ Continued fraction \section{A continued fraction. } $1+ {x \over {\scriptstyle 1+ {\scriptstyle x \over {\scriptstyle 1 + {\scriptstyle x \over {\scriptstyle 1 + {\scriptstyle x \over {\scriptstyle 1+ {\scriptstyle x \over 1+{\scriptstyle \atop \ddots }}}}}}}}}}$ %% %%% }}} %%% {{{ Simple School algebra. \section{Simple School algebra. } $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$ %% $$a^3+b^3=(a+b)(a^2-ab+b^2)$$ %% Given $ax^2+bx+c=0$, we have $$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ %% %%% }}} %%% {{{ square roots. \section{square roots. } $$\frac{1+\sqrt{5}}{2}=\phi$$ %% $$\frac{\sqrt{\pi}}{2} \neq \sqrt{\frac{\pi}{2}}$$ %% $$\sqrt{1+\sqrt{2+\sqrt{2+\sqrt {2+\cdots }}}}$$ %% %%% }}} %%% {{{ Trigonometric identities \section{Trigonometric identities. } $$\sin^2x+\cos^2x=1$$ %% $$\sin x^2 + \cos x^2 \neq 1$$ %% $$\sin^{-1}x \neq \sin x^{-1}$$ %% $$\sin (a+b) = \sin a \cos b + \cos a \sin b$$ %% $$\cos (x+y)=\cos x \cos y - \sin x \sin y$$ %% $$\sin 2x = 2 \sin x \cos x$$ %% $$\cos 2x = \cos^2 x -\sin^2 x$$ %% %%% }}} %%% {{{ logs \section{Logarithms. } $$\log^2x\neq2\log x$$ %% $$\log x^2=2\log x$$ %% $$\frac{\log x}{\log a} = \log_a x$$ %% $$\log_{a^2} x = \frac{1}{\log_x a^2}= \frac{1}{2\log_x a} = \frac{\log_a x}{2}$$ %% %%% }}} %%% {{{ Series \section{Series. } $$1+x+x^2+x^3+x^4+\cdots+x^{n-1}+\cdots = \frac{1}{1-x}$$ %% $$1+x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots +\frac{x^n}{n}+\cdots$$ %% $$x - \frac{x^2}{2} +\frac{x^3}{3} -\frac{x^4}{4}+\frac{x^5}{5} \pm \cdots = \log(1+x)$$ %% $$\gamma = 1+\frac{1}{2}+\frac{1}{3} +\frac{1}{4} +\cdots +\frac{1}{n} -\log n$$ %% $$\log (1+x) - \log (1-x) = \log \frac{1+x}{1-x} = \sum_{i=1}^\infty \frac{x^{2i-1}}{2i -1}$$ %%% }}} %%% {{{ Integrals \section{Integrals. } %% $$\int\frac{\dx}{x} =\log x$$ $$\int_1^a \int_1^b\int_1^c e^{x+y+z}\dx\dy\dz$$ %% $$\int_1^\infty e^{x^2-x-1}\dx$$ %% $$\int_1^\infty e^{x^{2-x}-1}\dx$$ %% $$\int_0^1\int_0^{\sqrt{1 -y^2}}1\dx\dy= \int_0^{\pi/2}\int_0^1 r\varint{r}\varint{\theta}$$ $$s=\int_a \int_b f \dx\dy+1$$ %% %%% }}} %%% {{{ Summation \section{Summations. } $$\sum_{i=1}^n a_i =1$$ %% $$\sum_{1\leq i\leq n}a_i =1$$ %% $$\sum_{i=1}^n a_i + b_i = 1$$ %% %%% }}} %%% {{{ Limits \section{Limits. } $$\lim_{x \to \infty}\int_0^x e^{-y^2}\dy = \frac{\sqrt{\pi}}{2}$$ %% $$\lim_{x\to 0}\frac{\sin x}{x} =1$$ %% %%% }}} %%% {{{ Cross referenced equations \section{Cross referenced equations. } \begin{equation} \cosh x = \frac{e^x + e^{-x} }{2} \label{eq:cosh} \end{equation} \begin{equation} \sinh x = \frac{e^x-e^{-x}}{2} \label{eq:sinh} \end{equation} Squaring~\ref{eq:cosh} and~\ref{eq:sinh} and computing their difference gives $$\cosh^2x -\sinh^2 x = 1$$ %% %%% }}} %%% {{{ Distance formula \section{Distance formula. } %% Given $x=(x_1,x_2), y=(y_1,y_2)$ the distance between the two points is given by: $$d(x,y) = \sqrt{(x_1-y_1)^2 +(x_2-y_2)^2}$$ This is the distance formula. %%% }}} %%% {{{ Quantified expression \section{Quantified expression. } %% $$\forall x \in X: \exists y \in Y : x=y$$ %%% }}} %%% {{{ Exponentiation \section{Exponentiation} Consider the expression: $$e^{e^{e^x}}$$ Differentiating with respect to $x$ gives: $$e^{e^{e^x}} e^{e^x} e^x$$ Simplifying this expression gives: $$e^{(e^{e^x} + e^x + x)}$$ %%% }}} %%% {{{Matrix \section{A generic matrix} Notice the use of vertical and diagonal dots in the generic matrix shown below. $$A=\pmatrix{a_{1 1}&a_{1 2}&\ldots&a_{1 n}\cr a_{2 1}&a_{2 2}&\ldots&a_{2 n}\cr \vdots&\vdots&\ddots&\vdots\cr a_{m 1}&a_{m 2}&\ldots&a_{m n}\cr}$$ %%% }}} %%% {{{ Faa de Bruno's formula \section{Faa de Bruno's formula } \uselongsummation %% Let $D^k_xu$ represent the $k$th derivative of a function $u$ with respect to $x$. The chain rule states that $D^1_xw = D^1_uw D^1_xu$. If we apply this to second derivatives, we find $D^2_xw = D^2_uw (D^1_xu)^2+D^1_uw D^2_xu$. Show that the {\em general formula} is %% \begin{equation}\label{eq:faa-de-bruno} D^n_xw = \sum_{0\le j\le n} \sum_{\scriptstyle k_1+k_2+\cdots+k_n=j \atop {\scriptstyle k_1+2k_2+\cdots+nk_n=n \atop {\scriptstyle k_1,k_2,\ldots,k_n\ge0 }}} D^j_u w \frac{n! {(D^1_x u)}^{k_1} \cdots {(D^n_x u)}^{k_n} } {k_1!{(1!)}^{k_1} \cdots k_n!{(n!)}^{k_n}} \end{equation} %% \endlongsummation %%% }}} %%% {{{ Variable substitution. \section{Using variable substitutions.} \activatevariablesubstitution The following examples demonstrate the effectiveness of using the {\em variable substitution\/} reading style. Applying variable substitution to Faa de Bruno's formula shown in equation~\ref{eq:faa-de-bruno} results in: \begin{equation} \label{eq:faa-de-bruno-subst} D^n_xw = \sum_{0\le j\le n} \sum_{\underbrace{{\scriptstyle k_1+k_2+\cdots+k_n=j \atop {\scriptstyle k_1+2k_2+\cdots+nk_n=n \atop {\scriptstyle k_1,k_2,\ldots,k_n\ge0 }}}}_{\mbox{\em lower constraint 1\/}}} D^j_u w \frac{\overbrace{n! {(D^1_x u)}^{k_1} \cdots {(D^n_x u)}^{k_n}} ^{\mbox{\em numerator 1\/}}} {\underbrace{k_1!{(1!)}^{k_1} \cdots k_n!{(n!)}^{k_n}}_{\mbox{\em denominator 1\/}}} \end{equation} Consider the expression: \begin{equation} e^{x+e^{x+e^{x+e^x}}} \label{eq:exponentsum} \end{equation} Differentiating expression~\ref{eq:exponentsum} with respect to $x$ gives: $$e^{x + e^{x + e^{x + e^x}}} (1 + e^{x + e^{x + e^x}} (1 + e^{x + e^x} (1 + e^x)))$$ $$\int_{5u+t}^t \frac{ \int_{5u+t}^x \frac{e^{-x^2} + e^{x^3} } {sx^2 + c^{2 }x } \dy} {s x^2 + c^2 x} \dx$$ $$\int_1^\infty (a+b+c)(e^{-x^2} + e^{x^3})\dx$$ $$\int \frac{\exp{-x^2} + \exp {x^3} }{\sin x^2 + \cos^2 x} \dx$$ $$\frac{x\sin(\log x)}{2} - \frac{x\cos(\log x)}{2}$$ $$\frac{2x^5}{5} - \sqrt{\frac{2x^5 \log x}{5}} + \frac{x^5\log^2 x}{5}$$ Consider the expression: $e^{\tan(e^{\arctan(e^x)})}$ Differentiating with respect to $x$ gives: ${e^{\left( \tan(e^{\arctan(e^x)}) + \arctan(e^x) + x \right)} \over (e^{(2 x)} + 1) \cos^2(e^{\arctan(e^x)})}$ ${3 x^2 \sin{x} + x^3 \cos{x} + e^{\sin{x}} \cos{x} \over e^{\cos{x}} - \tan\left( {x \over 5} + 1 \right)} + {\left( {0.2 \over \cos^2 \left( {x \over 5} + 1 \right)} + e^{\cos{x}} \sin{x} \right) \times \left( e^{\sin{x}} + x^3 \sin{x} \right) \over \left( e^{\cos{x}} - \tan\left( {x \over 5} + 1 \right) \right)^2}$ \deactivatevariablesubstitution %%% }}} \end{document}