MATLAB Answers


How I do evaluate a function handle in other function handle

Asked by capj
on 15 Nov 2019 at 8:36
Latest activity Edited by Matt J
on 9 Dec 2019 at 14:10
I define the next function:
function [v] = f_deriv(f,x,y,n)
%f is a function of two variables
%x,y is a number
%n is the number of proccess that I need to do
p_y=@(f,x,y) (f(x,y+0.0001) - f(x,y-0.0001))/(2*0.0001);
p_x=@(f,x,y) (f(x+0.0001,y) - f(x-0.0001,y))/(2*0.0001);
for i=1:n
g=@(x,y) f(x,y)*p_y(f,x,y) + p_x(f,x,y);
f=@(x,y) g(x,y)
This function has a input f, like (f=@((x,y) y -x^2 +1), and want to has as output $f^n(x,y)$ (the nth derivative of f, where y depends of x)
The steps that I follow to solve this problem are:
First, I define the function p_y and p_x that is the partial derivate aproximation of f. ()
Next, I define a y_p that is the function f evaluated in (x,y)
And, I define the function g, which is the (p_x + p_y*y_p)
Finally, I want to evaluate n times the function g_k in the same function, something like this:
or something like that:
g_1=@(x,y) f(x,y)*p_y(f,x,y) + p_x(f,x,y);
g_2=@(x,y) g_1(x,y)*p_y(g_1,x,y) + p_x(g_1,x,y);
g_n=@(x,y) g_n-1(x,y)*p_y(g_(n-1),x,y) + p_x(g_(n-1),x,y);
But, when I use the code first I have incorrect results for n> 2. I know the algorithm is correct, because when I manually do the process, the results are correct
A example to try the code, is using f_deriv(@((x,y) y -x^2 +1,0,0.5,n), and for n=1, the result that would be correct is 1.5. For n=2,the result that would be correct is -0.5, for n=3. the result that would be correct is -0.5....
Someone could help me by seeing what the errors are in my code?, or does anyone know any better way to calculate partial derivatives without using the symbolic?

  1 Comment

I know the algorithm is correct, because when I manually do the process, the results are correct.
The algorithm is correct analytically, but is not stable numerically.

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4 Answers

Answer by Matt J
on 15 Nov 2019 at 10:18

I think this line needs to be
g=@(x,y) f(x,y)*p_y(f,x,y) + p_x(f,x,y);


In my case, I need that dy/dx = f(x,y). The code above is only for obtain all possible n-ths derived from f. Next, what I will proceed to do is that with these derivatives I calculate an approximate solution of the differential equation using the Taylor method.
OK, but how do we see that the correct result at 0.,0.5, 3 is -0.5? Do we know the closed form of y(x) somehow?
where y=0.5 and x=0

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Answer by Steven Lord
on 15 Nov 2019 at 14:37

g_1=@(x,y) y_p*p_y(f,x,y) + p_x(f,x,y);
g_2=@(x,y) y_p*p_y(g_1,x,y) + p_x(g_1,x,y);
g_n=@(x,y) y_p*p_y(g_(n-1),x,y) + p_x(g_(n-1),x,y);
Rather than using numbered variables, store your function handles in a cell array. That will make it easier for you to refer to previous functions.
f = @(x) x;
g = cell(1, 10);
g{1} = f;
for k = 2:10
% Don't forget to _evaluate_ the previous function g{k-1} at x
% g{k} = @(x) k + g{k-1} WILL NOT work
g{k} = @(x) k + g{k-1}(x);
g{10}([1 2 3]) % Effectively this returns [1 2 3] + sum(2:10)

  1 Comment

Using your suggestion I wrote the following code,
function [v] = f_doras(f,n)
g = cell(1,n);
g{1} = f
for k=2:n
g{k}=@(x,y) g{k-1}(x,y)*(g{k-1}(x,y+0.001) - g{k-1}(x,y-0.001))/(2*0.001) + (g{k-1}(x+0.001,y) - g{k-1}(x-0.001,y))/(2*0.001);
% -4.5, but the result that would be correct is -0.5
For n=2 and n=3, the results are correct, but for n=4 it's not right.

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Answer by Guillaume
on 15 Nov 2019 at 16:21

Your y_p never changes in your function. It's always the original . Shouldn't y_p be reevaluated at each step?
Your 0.0001 delta is iffy as well. It's ok when you're evaluating the derivative for but for small x (and y) it's not going to work so well. Should the delta be based on the magnitude of x (and y)?


Thanks for answer,
Yes, I made a mistake with the y_p. How do you suggest the delta change would be, more specifically?
x * 1e-4 or something similar would make more sense to me. That way the delta changes with the magnitude of x.
I don't understand your observation very well, for example, seeing the following code (and doing what I think I understood you)
function [v] = f_doras(f,n)
g = cell(1,n);
g{1} = f
for k=2:n
g{k}=@(x,y) (g{k-1}(x,y)*(g{k-1}(x,y+0.0001*y) - g{k-1}(x,y-0.0001*y)) + (g{k-1}(x+0.0001,y) - g{k-1}(x-0.0001,y)))/(2*0.0001*y);
I cannot make the magnitude change, because for example, if x = 0, then x * 0.0001 would be 0. If I do it with y, I think I have worse results than without making the change you suggested.
for example, using a= f_doras(@(x,y) y - x^2 + 1,3), and a(0,0.5) the result that I obtain is worse than using constant delta. Could you explain me more specifically your suggestion?

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Answer by Matt J
on 15 Nov 2019 at 21:27

What you probably have to do is use ode45 or similar to solve the differential equation dy/dx = f(x,y) on some interval. Then, fit f(x,y(x)) with an n-th order smoothing spline. Then, take the derivative of the spline at the point of interest.


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Cubic splines have only 3 derivatives, higher order splines have more.
Is there an implementation in matlab of splines in a higher order,I have not found it, if it exists, you have given me the solution to my problem!
I've never used it, but this File Exchange submission advertises spline fitting of any order
It also appears to contain functions for evaluating spline derivatives.

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