# Backward and Central Difference

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Anna Lin 2021 年 6 月 11 日
コメント済み: Anna Lin 2021 年 6 月 12 日 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));
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Anna Lin 2021 年 6 月 11 日
Yes, I have already defined f as an anonymous function.
f=@(x) x.^3+sin(x)

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### 採用された回答

J. Alex Lee 2021 年 6 月 11 日
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
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Anna Lin 2021 年 6 月 12 日
Thank you.

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R2021a

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