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\pstInterFF を変更して、f(x) から f'(x) を明示的に見つけることなく、定数 m と関数 f'(x) の交点を見つけるにはどうすればよいでしょうか?

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\pstInterFF を変更して、f(x) から f'(x) を明示的に見つけることなく、定数 m と関数 f'(x) の交点を見つけるにはどうすればよいでしょうか?

ここに画像の説明を入力してください

\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}

簡単に言えば、から明示的に検索せずに、定数関数と関数\pstInterFFの交差点を見つけるために変更するにはどうすればよいでしょうか。mf'(x)f'(x)f(x)

答え1

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

\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}

ここに画像の説明を入力してください

答え2

素朴な方法は次のとおりです。

アップデート: アップデートは全て指定された傾斜を持つ点。

追加アップデート: これはコードの最適化です。局所的な極値を検出するために、新しいバージョンでは乗算を行わず、符号のみを操作するため、より効率的です (これは を使用しているコードの部分です\xintifboolexpr)。

ロール複数

ロール複数B

この方法は、（デフォルトでは）中間点を力ずくで計算して100局所的極値を特定する。計算は次のように行われる。xintexpr、これは（現時点では）有理関数と平方根に範囲を限定します。当然、他の数学エンジンで計算を行うようにコードを変更することもできます。

\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}

この回答の最初のバージョン:

ロール

\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
      \advance\count0 1
      \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}

答え3

計算機 (および xpicture) を使用した可能な解決策。平均値の例

\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}}{%
  \ADD{\XZERO}{\currentamplitude}{\XMED}
  \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}

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