Own mathematical function f(x)

5,308

Solution 1

If this works generally, I just got lucky. EDITED to do derivatives.

EDITED To be more true to math mode. EDITED to allow different function names with use of optional argument (default \f). EDITED to use more natural syntax \f(3) rather than \f{3}. EDITED to provide \listfunc macro. EDITED to work with amsmath.

Finally, EDITED to allow a more general syntax that can include primes, subscripts etc. in the function name itself.

\documentclass{article}
\usepackage{amsmath}% BREAKS ORIGINAL CODE; REQUIRES \[email protected] IN \setfunc
\makeatletter
\newcommand\setfunc[2][f]{\expandafter\[email protected]\csname#1\endcsname(##1){#2}}
\makeatother
\def\func#1(#2){\csname#1\endcsname(#2)}
\def\listfunc#1(#2){#1(#2)=\func#1(#2)}
\newcommand\x{(##1)}
\begin{document}
\setfunc{\sin\x} I can list the function: $\listfunc f(3)$\par
or I can just print out $\f(x+y)$.\par
or with a general input syntax: $\func f(x+y)$\par
\setfunc[g'_y]{\ln\x + 3\x^2} Now we can have $\listfunc g'_y(7)$\par
\medskip
Derivatives:\par
\setfunc[y]{4\x^5 - 2\x^2 +3}
\setfunc[y']{20\x^4 - 4\x}
\setfunc[y'']{80\x^3 - 4}
\setfunc[y''']{240\x^2}
\setfunc[y^{iv}]{480\x}
$\listfunc y(2)$\par
$\listfunc y'(2)$\par
$\listfunc y''(2)$\par
$\listfunc y'''(2)$\par
$\listfunc y^{iv}(2)$\par
\end{document}

enter image description here

NOTE: Joel noted that the method can get confused if the evaluation value itself contains a term in parentheses, for example, $\f ( \ln(a + 1.5) )$. The workaround for this is to embrace the inner argument, such as $\f({\ln(a + 1.5)})$ or $\listfunc y''({\ln(a + 1.5)})$.

Solution 2

This raises more complications than it solves, in my opinion, but here's an idea:

\documentclass{article}

\makeatletter
\newcommand{\setfunc}[4]{%
  \@namedef{[email protected]}##1{#1}%
  \@namedef{[email protected]'}##1{#2}%
  \@namedef{[email protected]''}##1{#3}%
  \@namedef{[email protected]'''}##1{#4}%
}
\def\f#1#{\@nameuse{[email protected]#1}}
\makeatother

\begin{document}

\setfunc{\sin(#1)}{\cos(#1)}{-\sin(#1)}{-\cos(#1)}
$\f{x}$ $\f'{1}$ $\f''{\pi}$ $\f'''{\pi/2}$

\setfunc{\log(#1)+1}{}{}{}
$\f{3}$

\end{document}

enter image description here

If you just want the function and not the derivatives, it's much simpler:

\newcommand{\setfunc}[1]{\renewcommand\f[1]{#1}}

Full example:

\documentclass{article}

\newcommand{\f}[1]{} % initialize
\newcommand{\setfunc}[1]{\renewcommand{\f}[1]{#1}}

\begin{document}

\setfunc{\sin(#1)}
$\f{x}+\f{3}$

\setfunc{\log(#1)+1}
$\f{3}$

\end{document}

enter image description here

A different implementation, where you can set any name you like (but beware of not redefining already existing commands)

\documentclass{article}

\newcommand{\setfunc}[2][\f]{\def#1##1{#2}}

\begin{document}

\setfunc{\sin(#1)}
$\f{x}+\f{3}$

\setfunc[\g]{\log(#1)+1}
$\g{3}$

\end{document}
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Updated on March 25, 2020

Comments

  • Joel Duscha
    Joel Duscha over 2 years
    • I'd like to define an custom function f(x).

    • For example \f{3} should print ln(3) + 3 if the function is set to ln(x) + 3.

    • One should be able change the function: \setfunc{sin(\x}}.

    • This should only affect future uses of \f{...}

    • And it should be possible to define the first three derivatives..

    The commands do not have do be this way. There may be a more elegant/practical way. Warning: It should work in this environment: https://tex.stackexchange.com/a/299720/101053

    Edit: Changed derivate to derivative; It's unclear what "define the derivative" should mean. I tried to say that I can simply add other functions (whether derivative or nor).

    • Werner
      Werner over 6 years
      Your final point can also be done manually, right? Say, \func{n} for the function, \func[1]{n} for the first or derivative, \func[2]{n} for the second order derivative, ...
    • cfr
      cfr over 6 years
      Please post a complete example. What's a derivate? Do you mean derivative? With Werner's suggestion, you could define the function using 4 arguments, say, \setfunc{}{}{}{}. I don't see how TeX can be expected to figure out the derivatives automatically. (Unless you restrict the set of possible functions so that they can be handled by an automatic algorithm, say.)
    • Werner
      Werner over 6 years
      Do you want to evaluate the functions, or just print their algebraic representation?
    • Joel Duscha
      Joel Duscha over 6 years
      @Werner is right. This could be done.cfr Yes I mean derivatives. Or in other words just other functions. They should be entered manually. But 4 arguments would be difficult to use because sometimes I may only use one or two functions.
  • Joel Duscha
    Joel Duscha over 6 years
    Really impressive. Is there a way to add more functions like setfunc[2]{}?
  • Steven B. Segletes
    Steven B. Segletes over 6 years
    @JoelDuscha I've been tweaking the answer so I can name the target function \f or \g for example. But I'm not sure what functionality you mean by \setfunc[2]{}?
  • Steven B. Segletes
    Steven B. Segletes over 6 years
    @JoelDuscha Please see my revision. The syntax has changed slightly (for the better).
  • Joel Duscha
    Joel Duscha over 6 years
    Naming them /g /f was exactly what I needed. New syntax is even better. EDIT works now.
  • Joel Duscha
    Joel Duscha over 6 years
    Unfortunately, I get this error only in my document: "Use of \\setfunc doesn't match its definition. \@ifnextchar ... \[email protected] =#1\def \[email protected] { #2}\def \[email protected] {#3}\f... l.21 \setfunc{\sin(\x)}"
  • Steven B. Segletes
    Steven B. Segletes over 6 years
    @JoelDuscha Did you update your syntax to use parens instead of braces? If you can't locate the error in your syntax, maybe edit your question and post a small MWE at the end. I could then have a look.
  • Joel Duscha
    Joel Duscha over 6 years
    The error is caused by \usepackage{amsmath}. Is this helpful or do you still need a small MWE?
  • Steven B. Segletes
    Steven B. Segletes over 6 years
    @JoelDuscha That helps. Let me consider it more.
  • Steven B. Segletes
    Steven B. Segletes over 6 years
    @JoelDuscha Fixed!
  • Joel Duscha
    Joel Duscha over 6 years
    This got somehow more upvotes than the other answers. I don't quite understand why. For my purpose this would make thinks messy I guess. But Maybe I overlook something here.
  • egreg
    egreg over 6 years
    @JoelDuscha Wasn't you who asked for supporting also setting the first three derivatives?
  • Steven B. Segletes
    Steven B. Segletes over 6 years
    @JoelDuscha One last improvement... Function names can now have primes, subscripts, etc., and can be invoked, as in my MWE as \setfunc[g'_y]{\ln(\x) + 3\x^2} and displayed with \func.
  • Joel Duscha
    Joel Duscha over 6 years
    Which is just another function. I see I confused everyone. Sorry for that.
  • Steven B. Segletes
    Steven B. Segletes over 6 years
    @JoelDuscha Derivates are now possible, with latest improvements.
  • Joel Duscha
    Joel Duscha over 6 years
    \f ( \ln(a + 1.5) ) causes the definition to end at the first ) Maybe it would make more sense to implement \f {...} or some special brackets to avoid this. But I agree that \f(3) is a more natural syntax.
  • Steven B. Segletes
    Steven B. Segletes over 6 years
    @JoelDuscha I could implement a braced format, if that were the case. However, the function definition is done with \setfunc{} which already uses a braced format. Your example, \f (\ln(a + 1.5)), is it really plausible? What it is asking is to take a function defined in \setfunc{}, and replace every instance of \x with \ln(a + 1.5). Correct me if I am mistaken, but that does not seem like a plausibly desired use of the method, is it?
  • Joel Duscha
    Joel Duscha over 6 years
    Yes it is exactly what I want. The value of the function at x = ln(a + 5). I'm doing function group. Every value for an 'a' gives another function.
  • Steven B. Segletes
    Steven B. Segletes over 6 years
    @JoelDuscha You can use the existing code, if you embrace the argument inside the parens: $\listfunc y''({\ln(a + 1.5)})$