How I do evaluate a function handle in other function handle

Question

capj 2019-11-15

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编辑： Matt J 2019-12-9

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?

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Matt J 2019-11-15

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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Answer 1

Matt J 2019-11-15

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在 MATLAB Online 中打开

I think this line needs to be

g=@(x,y) f(x,y)*p_y(f,x,y) + p_x(f,x,y);

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Matt J 2019-11-15

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?

capj 2019-11-15

编辑：capj 2019-11-15

where y=0.5 and x=0

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Answer 2

Steven Lord 2019-11-15

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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)

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capj 2019-11-15

编辑：capj 2019-11-15

在 MATLAB Online 中打开

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 3

Guillaume 2019-11-15

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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)?

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Guillaume 2019-11-15

x * 1e-4 or something similar would make more sense to me. That way the delta changes with the magnitude of x.

capj 2019-11-15

在 MATLAB Online 中打开

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 4

Matt J 2019-11-15

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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.