# Backward and Central Difference

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Anna Lin on 11 Jun 2021
Commented: Anna Lin on 12 Jun 2021 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 on 11 Jun 2021
Yes, I have already defined f as an anonymous function.
f=@(x) x.^3+sin(x)

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
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
Anna Lin on 12 Jun 2021
Thank you.

R2021a

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