From: mhiroshi@tansei.cc.u-tokyo.ac.jp (H. Murakami)
Newsgroups: sci.math.symbolic
Subject: Test examples for the Computer Algebra System you might interested.
Date: 29 Jul 92 10:18:03 GMT
Organization: Hokkaido Univ. However I am subject to tansei for JUNET.

TO: sci.math.symbolic
Subject: Test examples for the Computer Algebra System you might be interested.

% The following is a small collection of some example problems 
% for the computer algebra systems dated back in 1985.
% I think some problems are fairly easy, and some are not.
% I am especially interested in that which examples are not able
% to be solved by which computer algebra systems.
% I will put this in public domain.
% 			Enjoy!
% 						Hiroshi Murakami
% 						mhiroshi@tansei.cc.u-tokyo.ac.jp

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%                      This is a LaTeX source file.
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%                                                             Nov. 20, 1985
%
%  Some test example problems for the Computer Algebra Systems.
%                                           Written by Hiroshi Murakami.
%
%------------------------------------------------------------------------------
\documentstyle[a4,12pt]{article}

\begin{document}
\subsection{Algebraic Equations.}
\begin{enumerate}

\item Solve numerically
\begin{eqnarray*} 
t^3-3t+1=0.\\
{t_1}=0.347296355, etc.
\end{eqnarray*}

\item Solve algebraically
\[z^5+2z^3+2z^2-15z+10=0.\]
because $(z-1)^{2}(z+2)(z^2+5)=0$\\answer is \[
\begin{array}{l}
z_1=1,\\z_2=1,\\z_3=-2,\\z_4=\sqrt{-5},\\z_5=-\sqrt{-5}
\end{array}
\]

\item Solve numerically
\[ x^3-9x^2+6x-1=0. \]
\[\left\{\begin{array}{l}
x_1=0.2831185828,\\x_2=0.4260220478,\\x_3=8.290859369.\end{array}
\right.\]

\item Consider, \[(x-1)(x-2)(x-3)\cdots(x-20)=\epsilon.\]
Find the perturbation expansion in $\epsilon$ of the root,
\[x=20+{a_1}{\epsilon}+{a_2}{\epsilon}^2+\cdots .\]
\end{enumerate}

\subsection{Indefinite Integration.}
\begin{enumerate}

\item Evaluate \[\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}
=-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,.\]

\item Evaluate \[\int\sqrt{\sqrt{(ax+b)}+c}\,\,dx
=\frac{4}{15a}(3\sqrt{ax+b}-2c)(\sqrt{ax+b}+c)^{\frac{3}{2}}\,.\]

\item Evaluate \[\int\frac{dx}{\sqrt{a\sqrt{x}+b}}
=\frac{4(a\sqrt{x}-2b)}{3a^2}\sqrt{a\sqrt{x}+b}\,.\]

\item Evaluate \[\int\frac{dx}{\sqrt{\sqrt{ax+b}+c}}
=\frac{4}{3a}(\sqrt{ax+b}-2c)\sqrt{\sqrt{ax+b}+c}\,.\]

\item Evaluate \[\int\frac{\sqrt{x+\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}\,dx
=2\sqrt{x+\sqrt{x^2+a^2}}\;\;\;\;[a>0] .\]

\item Evaluate
\begin{eqnarray*}
\int\frac{\sqrt{x+\sqrt{x^2+a^2}}}{x}\,dx &=&2\sqrt{x+\sqrt{x^2+a^2}}\\
&+&\sqrt{a}\log{|\frac{\sqrt{x+\sqrt{x^2+a^2}}-\sqrt{a}}
{\sqrt{x+\sqrt{x^2+a^2}}+\sqrt{a}}}| \\
&-&2\sqrt{a}\arctan{(\sqrt{\frac{x+\sqrt{x^2+a^2}}{a}})}\;\;\;\;[a>0]\,.
\end{eqnarray*}

\item Evaluate \[
\int\frac{x^{(p/q)-1}}{(1+x^p)^{(1-\frac{1}{q})}[(1+x^p)^{1/q}-x^{p/q}]^2}
=\frac{q}{2p}\,\frac{(1+x^p)^{1/q}+x^{p/q}}{(1+x^p)^{1/q}-x^{p/q}}\,.\]

\item Evaluate 
\[\int\frac{dx}{(1+x^{2n})\sqrt{(1+x^{2n})^{1/n}-x^2}}
=\arctan{(\frac{x}{\sqrt{(1+x^{2n})^{1/n}-x^2}})}\,.\]

\item Evaluate 
\[\int\frac{x^{(n/2)-1}}{(1+x^{2n})\sqrt{\sqrt{1+x^{2n}}-x^n}}\,dx
=\frac{2}{n}\arctan{(\frac{x^{n/2}}{\sqrt{\sqrt{1+x^{2n}}-x^n}})}\,.\]

\item Evaluate \[
\int\frac{(\sqrt{x}-\sqrt{a})\sqrt{\sqrt{x}+\sqrt{x-a}}}{\sqrt{x}(x-a)^{3/2}}
\,dx=\frac{2}{a^{1/4}}[-\sqrt{\frac{2\sqrt{a}}{\sqrt{x}+\sqrt{a}}}
+\arcsin{(\frac{\sqrt{x}-\sqrt{a}}{\sqrt{x-a}})}]\;\;\;\;[a>0]\,.\]
\end{enumerate}

\subsection{Definite Integration.}
\begin{enumerate}

\item Evaluate \[\int_0^{\infty} \frac{x}{1+x^2}\sin{wx}\,dx
=\frac{\pi}{2}e^{-w}\,.\]

\item Evaluate \[
\int_0^{\infty} \frac{\sin^2{(\frac{x}{2})}}{x}\,\sin{wx}\,dx
=\left \{\begin{array}{lll}
\frac{\pi}{4} &;& w < 1,\\
\frac{\pi}{8} &;& w = 1,\\
0             &;& w > 1.
\end{array}
\right.\]

\item Evaluate 
\[\int_{0}^{1}\,\int_{0}^{1}\frac{dx\,dy}{1-xy}=\frac{{\pi}^2}{6}\,.\]

\item Evaluate 
\[\int_{0}^{\frac{\pi}{2}}\frac{x^2\,dx}{\sin^{2}{x}}=\pi\ln{2}\,.\]

\item Evaluate 
\[\int_{0}^{\frac{\pi}{2}}\frac{x^3\cos{x}}{\sin^3{x}}\,dx
=-\frac{{\pi}^3}{16}+\frac{3}{2}\pi\ln{2}\,.\]

\item Evaluate 
\[\int_{0}^{1}\arcsin{x}\cdot\ln{x}\,dx=2-\ln{2}-\frac{\pi}{2}\,.\]

\item Evaluate 
\[\int_{0}^{1}\arccos{x}\cdot\ln{x}\,dx=\ln{2}-2\,.\]

\item Evaluate 
\[\int_{0}^{\infty}\frac{x\tan(ax)}{x^2+b^2}\,dx
=\frac{\pi}{e^{2ab}+1}\;\;,\;\;[a>0,b>0]\,.\]

\item Evaluate 
\[\int_0^{\infty}\frac{x\cot{(ax)}}{x^2+b^2}\,dx
=\frac{\pi}{e^{2ab}-1}\;\;,\;\;[a>0,b>0]\,.\]

\item Evaluate 
\[\int_0^{\infty}\frac{\ln{x}}{x+a}\,\frac{dx}{x-1}
=\frac{{\pi}^2+(\ln{a})^2}{2(a+1)}\;\;,\;\;[a>0]\,.\]

\item Evaluate 
\[\int_0^{\infty}\ln{(\frac{a^2+b^2x^2}{x^2})}\cdot\frac{dx}{c^2+g^2x^2}
=\frac{\pi}{cg}\ln{(\frac{ag+bc}{c})}\;\;,\;\;[a,b,c,g>0]\,.\]

\item Evaluate 
\[\int_{0}^{\infty}\frac{x^{2}e^{x}\,dx}{\sqrt{(e^{x}-1)^3}}=8\pi\ln{2}\,.\]

\item Evaluate 
\[\int_{0}^{\infty}\frac{x^{3}e^{x}\,dx}{\sqrt{(e^x-1)^3}}
=24\pi[(\ln{2})^2+\frac{{\pi}^2}{12}]\,.\]

\item Evaluate 
\[\int_{0}^{\infty}\frac{x^2\,dx}{\cosh{x}}=\frac{{\pi}^3}{8}\,.\]

\item Evaluate 
\[\int_0^{\infty}\frac{dx}{(x+\sqrt{1+x^2})^{\alpha}}
=\frac{\alpha}{{\alpha}^2-1}\,.\]
\end{enumerate}

\subsection{Infinite Summation.}
\begin{enumerate}

\item Evaluate 
\[\sum_{n=1}^{\infty}\frac{n^3}{n!}=5e\,.\]

\item Evaluate 
\[\sum_{n=1}^{\infty}\frac{(n!)^2}{(2n+1)!}=\frac{2\pi}{3\sqrt{3}}\,.\]

\item Evaluate 
\[\sum_{n=1}^{\infty}\frac{1}{n^2(n+1)}=\frac{{\pi}^2}{6}-1\,.\]

\item Evaluate 
\[\sum_{n=1}^{\infty}\frac{n}{(4n^2-1)^2}=\frac{1}{8}\,.\]

\item Evaluate 
\[\sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}=10-{\pi}^2\,.\]

\item Evaluate 
\[\sum_{n=1}^{\infty}\frac{1}{(3n+1)(3n+4)}=\frac{1}{3}\,.\]

\item Evaluate 
\[\sum_{n=1}^{\infty}\frac{1}{n^2+a^2}
=-\frac{1}{2a^2}+\frac{\pi}{2a}\coth{a\pi}\,.\]

\item Evaluate 
\[\sum_{n=1}^{\infty}\frac{1}{{n^2}{2^n}}
=\frac{\pi^2}{12}-\frac{1}{2}(\log{2})^2\,.\]

\end{enumerate}

\subsection{Resolvent.}
\begin{enumerate}
\item
Let $x^4+a_1x^3+a_2x^2+a_3x+a_4
=(x^2+\alpha_1x+\alpha_2)^2-\lambda(x+\beta_1)^2\,.$

If we equate the coefficient of the both hand side,
the simultaneous algebraic relations be obtained:
\[\begin{array}{lll}
x^4&:&1=1\\
x^3&:&a_1=2{\alpha_1}\\
x^2&:&a_2=-{\lambda}+{\alpha_1}^2+2{\alpha_2}\\
x^1&:&a_3=2({\alpha_1}{\alpha_2}-{\lambda}{\beta_1})\\
x^0&:&a_4={\alpha_2}^2-{\lambda}{\beta_1}^2
\end{array}\]

Elimination of ${\alpha_1}$, ${\alpha_2}$, ${\beta}$ from above equations will 
produce the ${\lambda}$'s algebraic equation of the third degree.

For the special case ${a_1}=0\;({\alpha_1}=0)$ we will get:
\[{\lambda}^3+2{a_2}{\lambda}^2+({a_2}^2-4a_4){\lambda}-{a_3}^2=0\,.\]

\item
Similaly, for the case of the equation of degree 6:
\[x^6+{a_2}x^4+{a_3}x^3+{a_4}x^2+{a_5}x+{a_6}
=(x^3+{\alpha_2}x+{\alpha_3})^2-{\lambda}(x^2+{\beta_1}x+{\beta_2})^2\,.\]

Eliminate ${\alpha}$'s and ${\beta}$'s and find the algebraic equation of 
$\lambda$ !

\item
Similaly, consider the case of the equation of degree 8:
\[x^8+{a_2}x^6+\cdots+{a_8}=(x^4+{\alpha_2}x^2+\cdots+{\alpha_4})^2
-{\lambda}(x^3+{\beta_1}x^2+\cdots+{\beta_3})^2\,.\]

\item
How far can we go on this kind of computations?

\end{enumerate}

\subsection{Newton's problem.}
Consider a curve 
\[y^3-xy^{\frac{1}{2}}+x^{\frac{4}{3}}=0\,.\]
The origin $(0,0)$ is on the curve.

Find the Puiseux expansion of y to as higher degree as possible:
\[y=x^{\frac{2}{3}}+2x^{\frac{4}{3}}+13x^2+\cdots.\]

\subsection{Functional Equation.}
\begin{enumerate}
\item
Solve the following functional equation by power series:

\begin{eqnarray*}
\frac{p^2(x^2)+q^2(x^2)}{2}&=&\{\frac{p(x)+q(x)}{2}\}^2 \\
\frac{p^2(x^2)-q^2(x^2)}{2}&=&\{\frac{p(x)-q(x)}{2}\}^2 \\
p(0)=q(0)=1,&&{\rm and\,analytic\,at\,} 0.
\end{eqnarray*}

Assume the series expansions:
\begin{eqnarray*}
p(x)=1+2\sum_{j=1}^{\infty} p_j x^j\,,\\
q(x)=1+2\sum_{j=1}^{\infty} q_j x^j\,.
\end{eqnarray*}

Equating the coefficients, we can get the simultaneous equation of
$p_j$ and $q_j$\,.

The right answer should be if we assume $p_1-q_1 \ne 0$,
\begin{eqnarray*}
p(x)&=1+2\sum_{k=1}^{\infty}  x^{k^2} =&1+2x+2x^{4}+2x^{9}+2x^{16}+\cdots,\\
q(x)&=1+2\sum_{k=1}^{\infty}(-x)^{k^2}=&1-2x+2x^{4}-2x^{9}+2x^{16}-\cdots \\
    &=p(-x).
\end{eqnarray*}

\item
Similary, find out the following functional equation's analytic solutions.\\
Note: ${\omega}{\ne}1,\;{\omega}^3=1$.
\begin{eqnarray*}
\frac{p^3(x^3)+q^3(x^3)+r^3(x^3)}{3} &=&\{\frac{p(x)+q(x)+r(x)}{3}\}^3 \\
\frac{p^3(x^3)+{\omega}q^3(x^3)+{\omega^2}r^3(x^3)}{3}
&=&\{\frac{p(x)+{\omega}q(x)+{\omega^2}r(x)}{3}\}^3 \\
\frac{p^3(x^3)+{\omega^2}q^3(x^3)+{\omega}r^3(x^3)}{3} 
&=&\{\frac{p(x)+{\omega^2}q(x)+{\omega}r(x)}{3}\}^3 \\
p(0)=q(0)=r(0)=1,&& p'(0),\,q'(0),\,r'(0)\,{\rm are\,not\,all\,zeros\,}.
\end{eqnarray*}
\end{enumerate}
\end{document}
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Newsgroups: sci.math.symbolic