Backward and Central Difference
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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 Comments
Joseph Cheng
on 11 Jun 2021
Edited: Joseph Cheng
on 11 Jun 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
Accepted Answer
J. Alex Lee
on 11 Jun 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
fcpp = cc*fc/dx^2
backward_difference = ((2*fun(x0)-5*fun(x0-dx)+4*fun(x0-2*dx)-fun(x0-3*dx))/dx^2)
central_difference = (-fun(x0+2*dx)+16*fun(x0+dx)-30*fun(x0)+16*fun(x0-dx)-fun(x0-2*dx))/(12*(dx^2))
fpp = funpp(x0)
3 Comments
J. Alex Lee
on 12 Jun 2021
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
fb = fun(xb);
fbpp = cb*fb/dx^2 % This will not be 60.5508
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