How I do evaluate a function handle in other function handle
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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?
1 comentario
Matt J
el 15 de Nov. de 2019
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.
Respuestas (4)
Steven Lord
el 15 de Nov. de 2019
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)
1 comentario
Guillaume
el 15 de Nov. de 2019
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)?
3 comentarios
Guillaume
el 15 de Nov. de 2019
x * 1e-4 or something similar would make more sense to me. That way the delta changes with the magnitude of x.
Matt J
el 15 de Nov. de 2019
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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