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);

y_p=f(x,y);

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)

end

v=f(x,y)

end

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?

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);

capj
on 15 Nov 2019 at 17:52

Matt J
on 15 Nov 2019 at 17:58

capj
on 15 Nov 2019 at 19:01

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);

g_n(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);

end

g{10}([1 2 3]) % Effectively this returns [1 2 3] + sum(2:10)

capj
on 15 Nov 2019 at 15:04

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);

end

v=g{n};

end

a=f_dora(f,4)

a(0,0.5)

% -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)?

capj
on 15 Nov 2019 at 16:32

Thanks for answer,

Yes, I made a mistake with the y_p. How do you suggest the delta change would be, more specifically?

Guillaume
on 15 Nov 2019 at 16:49

capj
on 15 Nov 2019 at 17:32

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);

end

v=g{n};

end

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

Matt J
on 8 Dec 2019 at 2:55

Cubic splines have only 3 derivatives, higher order splines have more.

capj
on 8 Dec 2019 at 3:05

Matt J
on 9 Dec 2019 at 13:44

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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## Matt J (view profile)

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https://www.mathworks.com/matlabcentral/answers/491239-how-i-do-evaluate-a-function-handle-in-other-function-handle#comment_767834

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