# How to modify \pstInterFF to find the intersection point between a constant m and a function f'(x) without having to explicitly find f'(x) from f(x)?

\documentclass[preview,border=12pt]{standalone}
\usepackage{pst-plot,pst-eucl}

\def\f(#1){((#1)*(#1-5)*(#1-6)/4+1.5*(#1)-5)}
\def\xi{1}
\def\xf{6.5}
\def\m{((\f(\xf)-\f(\xi))/(\xf-\xi))}

\def\fp(#1){}% is f'(x)

% I have to use the following because the newest pst-eucl has not been installed
\pstVerb{/I2P {exec AlgParser cvx exec} def}

\begin{document}
How to find the point $c$ such that
$f'(c) = \frac{f(b)-f(a)}{b-a}$

\small
\begin{verbatim}
\begin{center}
\begin{pspicture}[algebraic,saveNodeCoors](-1,-1)(8,8)
\psaxes[labels=none,ticks=none]{->}(0,0)(-.5,-.5)(7.5,7.5)[$x$,0][$y$,90]
\psplot[linecolor=blue]{.75}{6.75}{\f(x)}
%\pstInterFF[]{{\fp(x)} I2P}{{\m} I2P}{4}{c}% has not been implemented yet.
%\psCoordinate(c)
\end{pspicture}
\end{center}
\end{verbatim}
\end{document}


Shortly speaking, how to modify \pstInterFF to find the intersection point between a constant function m and a function f'(x) without having to explicitly find f'(x) from f(x)?

• Very nice trick to bypass the on-hold barrier! :-) – kiss my armpit Mar 23 '14 at 13:36
• quiet... they may hear you bragging around... – user4686 Mar 23 '14 at 13:59
• @jfbu:...... :-) – kiss my armpit Mar 23 '14 at 14:02

\documentclass[preview,border=12pt]{standalone}

\def\f#1{((#1)*(#1-5)*(#1-6)/4+1.5*(#1)-5)}
\def\A{1}
\def\B{6.5}
\def\M{(\f{\B}-\f{\A})/(\B-\A)}
\begin{document}

\begin{pspicture}[algebraic,saveNodeCoors](-1,-1)(8,8)
\psaxes[labels=none,ticks=none]{->}(0,0)(-.5,-.5)(7.5,7.5)[$x$,0][$y$,90]
\psplot[linecolor=blue,plotpoints=100,linewidth=1.5pt]{.75}{6.75}{\f{x}}
\psplot[linestyle=dashed,linecolor=blue,plotpoints=100,linewidth=1.5pt]{.75}{6.75}{Derive(1,\f{x})}
\pnodes(*{\A} {\f{\A}}){A}(*{\B} {\f{\B}}){B}
\psCoordinates[linestyle=dashed](A)\psCoordinates[linestyle=dashed](B)
\pcline[nodesep=-5mm,linecolor=red](A)(B)
\pstInterFF{\M}{Derive(1,\f{x})}{1}{C}
\pstInterFF{\M}{Derive(1,\f{x})}{4}{D}
\psCoordinates[linestyle=dotted](C)\psCoordinates[linestyle=dotted](D)
\pnodes(*{N-C.x} {\f{x}}){X1}(*{N-D.x} {\f{x}}){X2}
\psCoordinates[linestyle=dashed,linecolor=red](X1)
\psCoordinates[linestyle=dashed,linecolor=red](X2)
\psParallelLine[linecolor=red](A)(B)(X1){0.1}{X}
\pcline[nodesepA=-1,linecolor=red](X1)(X)
\psParallelLine[linecolor=red](A)(B)(X2){0.1}{X}
\pcline[nodesepA=-1,linecolor=red](X2)(X)
\end{pspicture}

\end{document}


• does Derive(1,\f{x}) mean that the derivative is first computed algebraically ? (sorry for not being familiar with pstricks). – user4686 Mar 23 '14 at 17:59
• if I may take this opportunity in something like \psplot{\xa}{\xb}{\f{x}} is there a way to use infix expressions for \xa and \xb. For example \def\xa{\something-1}? I think I tried and it didn't go through. – user4686 Mar 23 '14 at 18:12
• if I understand correctly the user has to help the look for the correct points by launching \psInterFF with suitable starting abscissa, as many times as they are points to be found, right? hence the {1} and {4}? – user4686 Mar 23 '14 at 18:21
• Right, but it is not a big deal to find all intersection points and named it IS1, IS2, ... It is only a question of time :-) – user2478 Mar 23 '14 at 19:26

Here is a naive way.

UPDATE: the update tries to find all points with the given slope.

additional update: this is a code optimization, to detect a local extrema the new version avoids doing a multiplication, it only manipulates signs, hence this is more efficient (this is the part of the code using \xintifboolexpr).

The method is via brute force computation of (by default) 100 intermediate points to identifiy local extrema. Computations are done via xintexpr, which limits the scope (currently) to rational functions and square roots. The code can naturally be modified to do the computations with some other math engine.

\documentclass[multi=pspicture,border=12pt]{standalone}
\usepackage{pst-plot,pst-eucl}
\usepackage{xintexpr}

\def\FindRolleN {100}% number of tested points
\makeatletter
\def\FindRollePt #1#2#3{%
% #1 = function (defined with parentheses as below, not with braces, to
%                conform to OP's code)
% #2 = start of interval
% #3 = end of interval
% the action of this  macro is to set \RolleSlope and \RolleList
\begingroup
\let\ROL@N \FindRolleN
\xintDigits := 6;%
\xintNewFloatExpr \ROL@F [1]{#1(##1)}%
\oodef\ROL@Fa {\ROL@F{#2}}%
\oodef\ROL@Fb {\ROL@F{#3}}%
\oodef\ROL@Dx {\xintfloatexpr #3 - #2 \relax }%
\oodef\ROL@Dy {\xintfloatexpr \ROL@Fb-\ROL@Fa\relax }%
\oodef\ROL@Slope  {\xintfloatexpr \ROL@Dy/\ROL@Dx\relax }%
\oodef\ROL@Stepx  {\xintfloatexpr \ROL@Dx/\ROL@N\relax }%
\oodef\ROL@Stepy  {\xintfloatexpr \ROL@Dy/\ROL@N\relax }%
%
\oodef\ROL@X {\xintfloatexpr #2\relax }%
\let\ROL@Y   \ROL@Fa
%
\oodef\ROL@@X {\xintthefloatexpr \ROL@X + \ROL@Stepx\relax }%
\oodef\ROL@@Y {\xintthefloatexpr \ROL@F{\ROL@@X}\relax }%
% \count0 1 (no need for a count, use \xintiloop!)
\def\ROL@List {}% list of local extrema
\xintiloop [2+1]
\oodef\ROL@@@X {\xintthefloatexpr \ROL@@X + \ROL@Stepx\relax }%
\oodef\ROL@@@Y {\xintthefloatexpr \ROL@F{\ROL@@@X}\relax }%
% commented out, code below avoids a multiplication.
%\xintifSgn {\xintthefloatexpr (\ROL@@Y-\ROL@Y-\ROL@Stepy)*
%                              (\ROL@@@Y-\ROL@@Y-\ROL@Stepy)\relax}
%           {% found a local extrema etc...
%              code as below}
%           {}{}%
\xintifboolexpr
% this is a cleverer way to detect a change of sign. "constancy" compatible, too
{sgn(\ROL@@Y-\ROL@Y-\ROL@Stepy)+sgn(\ROL@@@Y-\ROL@@Y-\ROL@Stepy)}
{}% not an extrema
{% sum of signs=0, hence opposite signs or both zero, hence
% found a local extrema or a "constancy",
% Let's round the coordinates to 4 digits after decimal mark
% Mainly because I don't know how to use floating point notation in pspicture
% point coordinates but surely it can be done.
\edef\ROL@List{\expandafter\unexpanded\expandafter{\ROL@List}%
{{\xintRound{4}{\ROL@@X}}{\xintRound{4}{\ROL@@Y}}}}%
}%
\let\ROL@X\ROL@@X
\let\ROL@Y\ROL@@Y
\let\ROL@@X\ROL@@@X
\let\ROL@@Y\ROL@@@Y
\ifnum\xintiloopindex < \ROL@N\space
\repeat
\global\let\RolleListe\ROL@List
\global\oodef\RolleSlope{\xinttheexpr round(\ROL@Slope,4)\relax}%
\endgroup
}
\makeatother

\begin{document}
% How to find the point or points $c$ such that
% $% f'(c) = \frac{f(b)-f(a)}{b-a} %$

\def\f(#1){((#1)*(#1-5)*(#1-6)/4+1.5*(#1)-5)}
\def\xi{1}
\def\xf{6.5}

% \FindRollePt {\f}{\xi}{\xf}%
% \show\RolleListe % debugging

\begin{pspicture}[algebraic,saveNodeCoors](-1,-1)(8,8)
\psaxes[labels=none,ticks=none]{->}(0,0)(-.5,-.5)(7.5,7.5)[$x$,0][$y$,90]
\psplot[linecolor=blue]{.75}{6.75}{\f(x)}
\FindRollePt {\f}{\xi}{\xf}%
\psplot[linecolor=green]{.75}{6.75}{\RolleSlope*(x-\xi)+\f(\xi)}
\psCoordinates[linestyle=dashed](*\xi\space {\f(\xi)})
\psCoordinates[linestyle=dashed](*\xf\space {\f(\xf)})
\xintFor* #1 in {\RolleListe}\do {%
\xintAssign  #1\to\RolleX\RolleY
\psdot(\RolleX,\RolleY)
\psplot[linecolor=red]{.75}{6.75}{\RolleSlope*(x-\RolleX)+\RolleY}
}
\end{pspicture}

\def\f(#1){2.5*(#1-2.5)*((#1-2.5)^2-1)*((#1-2.5)^2-2)}
\def\xi{1}
\def\xf{4}
\psset{unit=2cm}

\begin{pspicture}[algebraic,saveNodeCoors](-.25,-2.5)(5,3)
\rput(2.5,2.5){\parbox {10cm}{How to find the point or rather the points $c$
such that $f'(c) = \frac{f(b)-f(a)}{b-a}$}}
\psaxes[labels=none,ticks=none]{->}(0,0)(-.25,-2)(5,2)[$x$,0][$y$,90]
\psplot[linecolor=blue,plotpoints=200]{.95}{4.05}{\f(x)}
\psCoordinates[linestyle=dashed](*\xi\space {\f(\xi)})
% NOTE: how to use * with negative coordinate? I had to shift
% everything to be positive
\psCoordinates[linestyle=dashed](*\xf\space {\f(\xf)})
\FindRollePt {\f}{\xi}{\xf}%
\psplot[linecolor=green]{\xi}{\xf}{\RolleSlope*(x-\xi)+\f(\xi)}
\xintFor* #1 in {\RolleListe}\do {%
\xintAssign  #1\to\RolleX\RolleY
\psdot(\RolleX,\RolleY)
\oodef\xa {\xinttheexpr round(\RolleX-.5,4)\relax}%
\oodef\xb {\xinttheexpr round(\RolleX+.5,4)\relax}%
\psplot[linecolor=red]{\xa}{\xb}
{\RolleSlope*(x-\RolleX)+\RolleY}
}
\end{pspicture}

\end{document}


\documentclass[preview,border=12pt]{standalone}
\usepackage{pst-plot,pst-eucl}
\usepackage{xintexpr}

\def\FindRolleN {100}% number of tested points
\makeatletter
\def\FindRollePt #1#2#3{%
% #1 = function (defined with parentheses as below, not with braces, to
% confirm to OP)
% #2 = start of interval
% #3 = end of interval
% the action of this  macro is to set \RolleX, \RolleY, \RolleSlope
\begingroup
\let\ROL@N \FindRolleN
% I use floating point numbers with some hesitation, it is not always faster
% than exact evaluations when the manipulated numbers don't have many digits
\xintDigits := 6;%
% I define this for the parsing of the algebraic expression to be done once and
% for all.
% subtle detail: if \ROL@F was defined with \xintNewExpr, rather than
% \xintNewFloatExpr, its output would be in a/b[n] form, its use in the next
% expressions would have to be mandatorily within a brace pair (see manual)
% other subtle detail: \ROL@F is a macro not able to parse an argument like
% \xintexpr knows how to do. For this, explicit \xinttheexpr ..\relax, or
% \xintthefloatexpr..\relax must be used in this argument.
\xintNewFloatExpr \ROL@F [1]{#1(##1)}%
\oodef\ROL@Fa {\ROL@F{#2}}%
\oodef\ROL@Fb {\ROL@F{#3}}%
\oodef\ROL@Slope {\xintfloatexpr (\ROL@Fb-\ROL@Fa)/(#3-#2)\relax }%
\oodef\ROL@Step  {\xintfloatexpr (#3-#2)/\ROL@N\relax }%
\def\ROL@U {0}%
\count0 0
\def\ROL@J {0}%
\xintloop
\oodef\ROL@X {\xintthefloatexpr #2+\count0*\ROL@Step\relax }%
\oodef\ROL@V {\xintthefloatexpr
abs(\ROL@N*\ROL@F{\ROL@X}-(\ROL@N-\count0)*\ROL@Fa-\count0*\ROL@Fb)
\relax }%
\xintifGt\ROL@V\ROL@U {\let\ROL@U\ROL@V\odef\ROL@J{\the\count0}}{}%
\ifnum\count0 < \ROL@N\space
\repeat
% Now converting to fixed point format with 4 digits after decimal mark
% Mainly because I don't know how to use floating point notation in pspicture
% but surely can be done.
\oodef\ROL@Result       {\xinttheexpr round(#2+\ROL@J*\ROL@Step,4)\relax}%
\global\let\RolleX\ROL@Result
\global\oodef\RolleY    {\xinttheexpr round(\ROL@F{\RolleX},4)\relax}%
\global\oodef\RolleSlope{\xinttheexpr round(\ROL@Slope,4)\relax}%
\endgroup
}
\makeatother

\def\f(#1){((#1)*(#1-5)*(#1-6)/4+1.5*(#1)-5)}
\def\xi{1}
\def\xf{6.5}

\begin{document}
How to find the (rather, a') point $c$ such that
$f'(c) = \frac{f(b)-f(a)}{b-a}$

\begin{center}
\begin{pspicture}[algebraic,saveNodeCoors](-1,-1)(8,8)
\psaxes[labels=none,ticks=none]{->}(0,0)(-.5,-.5)(7.5,7.5)[$x$,0][$y$,90]
\psplot[linecolor=blue]{.75}{6.75}{\f(x)}
\FindRollePt {\f}{\xi}{\xf}%
\psdot(\RolleX,\RolleY)
\psplot[linecolor=red]{.75}{6.75}{\RolleSlope*(x-\RolleX)+\RolleY}
\psplot[linecolor=green]{.75}{6.75}{\RolleSlope*(x-\xi)+\f(\xi)}
% I don't know how to create a dot, using algebraic expressions for the
% coordinates, I don't know anything about pstricks
% I have tried various things, for example this:
\psCoordinates[linestyle=dashed](*\xi\space {\f(\xi)})
\psCoordinates[linestyle=dashed](*\xf\space {\f(\xf)})
% user manual is big.
% \rput(*\xf\space {\f(\xf)}){\rule{3pt}{3pt}}
% \rput(*\xi\space {\f(\xi)}){\rule{3pt}{3pt}}
\end{pspicture}
\end{center}

\end{document}

• naturally, until xint has supplanted maple and mathematica you may switch to another (latex) math engine to do the computations, particularly if your function has non-rational components, apart from square root which xint knows how to handle. – user4686 Mar 23 '14 at 13:40

A possible solution, using calculator (and xpicture).

\documentclass[preview,border=12pt]{standalone}
\usepackage{xpicture}
\usepackage{amsmath,ifthen}

\begin{document}
How to find the point $c$ such that
$f'(c) = \frac{f(b)-f(a)}{b-a}$

\newcpoly{\fI}{0}{30}{-11}{1}                      % fI(x)=30x-11x^2+x^3
\newlpoly{\fII}{-5}{1.5}                           % fII(x)=-5+1.5x
\LINEARCOMBINATIONfunction{0.25}{\fI}{1}{\fII}{\F} % F(x)=(1/4)fI(x)+fII(x) (this is our function)

Our function is
$F(x)=\frac{x(x-5)(x-6)}{4}+1.5x-5$

\F{0}{\solZero}{\DsolZero}
\F{7}{\solSeven}{\DsolSeven}
Values of $F$ and $F'$ at $0$ and $7$ are
$\begin{gathered} f(0)=\solZero\qquad f'(0)=\DsolZero \\ f(7)=\solSeven\qquad f'(7)=\DsolSeven \end{gathered}$
\SUBTRACT{\solSeven}{\solZero}{\meanvalue}
\DIVIDE{\meanvalue}{7}{\meanvalue}
And the mean value is
$\frac{f(7)-f(0)}{7-0}=\meanvalue$
So, we search $c\in[0,7]$ such that $f'(c)=\meanvalue$.
We apply the bisection strategy
(in fact, this strategy is not secure, because the sign of the derivative
changes sign several times).

\COPY{0}{\XZERO}
\COPY{7}{\XONE}
\COPY{3.5}{\currentamplitude}
\COPY{0.0001}{\tolerance}

\whiledo{\lengthtest{\currentamplitude pt>\tolerance pt}}{%
\F{\XZERO}{\solzero}{\Dsolzero}
\F{\XMED}{\solmed}{\Dsolmed}
\F{\XONE}{\solone}{\Dsolone}
\SUBTRACT{\Dsolzero}{\meanvalue}{\derzero}
\SUBTRACT{\Dsolmed}{\meanvalue}{\dermean}
\MULTIPLY{\derzero}{\dermean}{\derproduct}
\ifthenelse{\lengthtest{\derproduct pt<0 pt}}{\COPY{\XMED}{\XONE}}{\COPY{\XMED}{\XZERO}}
\DIVIDE{\currentamplitude}{2}{\currentamplitude}
}
\medskip

Having applied this method we have obtained that
the derivative equals mean value of function $F$ at $c=\XMED$.

\setlength{\unitlength}{1cm}
\begin{Picture}(-1,-6)(8,11)
\cartesiangrid(0,-5)(7,10)
\pictcolor{red}
\PlotFunction[10]{\F}{0}{7}
\pictcolor{blue}
\xLINE(0,\solZero)(7,\solSeven)
\Put(\XMED,\solmed){\xLINE(-1,-\Dsolmed)(1,\Dsolmed)}
\Polyline(\XMED,0)(\XMED,\solmed)(0,\solmed)
\Put[S](\XMED,0){$c$}
\Put[W](0,\solmed){$f(c)$}
\Put[E](\XMED,\solmed){\scriptsize$f'(c)=\frac{f(b)-f(a)}{b-a}$}
\end{Picture}

\end{document}
`