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 commentaires
Joseph Cheng
le 11 Juin 2021
Modifié(e) : Joseph Cheng
le 11 Juin 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
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J. Alex Lee
le 11 Juin 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 commentaires
J. Alex Lee
le 12 Juin 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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