Anonymous Function Differentiation Problem?

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Steve
Steve 2011-9-29
回答： Konstantin Ninidze 2021-4-24
Hello Experts,
I have the following function:
>>f = @(x) x^2-1.
I want to find a function that will do differentiation: g = f' = 2*x so that g will be anonymous too and it will be possible to evaluate for example: g(5) = 10.
Please help me,
Steve

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回答（5 个）

Richard Edmonds
Richard Edmonds 2015-11-5
Yes you can do it. Here is the code to acheive exactly what you want. (too late for you perhaps but this was the first link in my search so I came beack to drop in the best answer I found).
fdash=@(fun,x,epsilon)imag(fun(x(:)+1i*epsilon))/epsilon;
g=@(x)fdash(f,x,10^-20);
Source: http://blogs.mathworks.com/cleve/2013/10/14/complex-step-differentiation/#487a1d22-7fe2-446e-a2a3-69ec283c2d97
Matt Tearle
Matt Tearle 2011-9-29
If you know the independent variable is called 'x':
fprime = str2func(['@(x) ',char(diff(sym(regexprep(char(f),'^@\(x\)',''))))])
fprime(pi)
If not... well, you can do it, but it will be a bit messier.
Walter Roberson
Walter Roberson 2011-9-29
You cannot do that, not without converting the anonymous function in to a character string and doing symbolic differentiation on that, and constructing the appropriate anonymous function from the result.
If you attempt to do this numerically, then because you can only sample at a finite number of numeric points, there will be an infinite number of functions that match the numeric results, leaving you unable to select the correct function to differentiate. You will, for example, be unable to find removable discontinuities numerically (except by luck). And good luck doing a numeric reconstruction of tan(Pi*exp(x)) .
Doug Hull
Doug Hull 2011-9-29
If you know you will be dealing only with polynomials, can you make something like this work?
PolyDer
Konstantin Ninidze
Konstantin Ninidze 2021-4-24
you can try my function: handle_to_sym:
function Fs=handle_to_sym(F)
% input F function_handle
% output Fs symbolic expression
S=char(F);
S=S(5:end);
Fs = evalin(symengine,S);

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