Backward and Central Difference
显示 更早的评论
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 个评论
Joseph Cheng
2021-6-11
编辑：Joseph Cheng
2021-6-11
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
2021-6-11
采纳的回答
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)
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
