Backward and Central Difference

Given that x =10 and delta_x = 0.4,
Is there a better way of writing this code?
x = 10;
delta_x = 0.4;
backward_difference = ((2*f(x)-5*f(x-dx)+4*f(x-2*dx)-f(x-3*dx))/dx^2);
central_difference = (-f(x+2*dx)+16*f(x+dx)-30*f(x)+16*f(x-dx)-f(x-2*dx))/(12*(dx^2));

2 comentarios

Joseph Cheng
Joseph Cheng el 11 de Jun. de 2021
Editada: Joseph Cheng el 11 de Jun. de 2021
Have you already defined "f" as an anonymous function or symbolic function? Otherwise if "f" is an array you would be indexing "f" in a non-integer value
Anna Lin
Anna Lin el 11 de Jun. de 2021
Yes, I have already defined f as an anonymous function.
f=@(x) x.^3+sin(x)

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 Respuesta aceptada

J. Alex Lee
J. Alex Lee el 11 de Jun. de 2021
I guess the answer depends what you want to do with those finite difference approximations. If you want to use it in an algorithm to solve ODEs, your strategy won't work because you don't a priori have a functional form.
This would be a typical matrix math way (assuming your coefficients are correct, i won't check)
cb = [-1,4,-5,2];
cc = [-1,16,-30,16,-1]/12;
fun = @(x) x.^3+sin(x);
funp = @(x) 3*x.^2 + cos(x);
funpp = @(x) 6*x - sin(x);
dx = 0.5;
x0 = 10;
% create stencils on x to define discrete f
xb = x0 - (3:-1:0)'*dx;
xc = x0 + (-2:2)'*dx;
% generate discrete f
fb = fun(xb);
fc = fun(xc);
% execute finite differences
fbpp = cb*fb/dx^2
fbpp = 60.5508
fcpp = cc*fc/dx^2
fcpp = 60.5437
backward_difference = ((2*fun(x0)-5*fun(x0-dx)+4*fun(x0-2*dx)-fun(x0-3*dx))/dx^2)
backward_difference = 60.5508
central_difference = (-fun(x0+2*dx)+16*fun(x0+dx)-30*fun(x0)+16*fun(x0-dx)-fun(x0-2*dx))/(12*(dx^2))
central_difference = 60.5437
fpp = funpp(x0)
fpp = 60.5440

3 comentarios

Anna Lin
Anna Lin el 12 de Jun. de 2021
I'm a bit confused about this 2 lines.
% cb = [-1,4,-5,2];
% xb = x0 - (3:-1:0)'*dx;
% If I wrote it like this?
cb = [2,-5,4,-1];
xb = x0 - (0:3)'*dx;
fbpp = cb*fb/dx^2 % This will not be 60.5508
it is not natural to order it that way (from right node to left note). But it should still work:
fun = @(x) x.^3+sin(x);
dx = 0.5;
x0 = 10;
cb = [2,-5,4,-1];
xb = x0 - (0:3)'*dx
xb = 4×1
10.0000 9.5000 9.0000 8.5000
fb = fun(xb);
fbpp = cb*fb/dx^2 % This will not be 60.5508
fbpp = 60.5508
Anna Lin
Anna Lin el 12 de Jun. de 2021
Thank you.

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