Forward, Reverse finite difference question
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%forward
function df1_forward = forward_first(f,x,h)
df1_forward = (f(x(2:end))-f(x(1:end-1)))/h;
end
function df2_forward = forward_second(f,x,h)
df2_forward = (f(x(3:end))-2*f(x(2:end-1))+f(x(1:end-2)))/h^2;
end
function df3_forward = forward_third(f,x,h)
df3_forward = (f(x(4:end))-3*f(x(3:end-1))+3*f(x(2:end-2))-f(x(1:end-3)))/h^3;
end
function df4_forward = forward_fourth(f,x,h)
df4_forward = (f(x(5:end))-4*f(x(4:end-1))+6*f(x(3:end-2))-4*f(x(2:end-3))+f(x(1:end-4)))/h^4;
end
%reverse
function df1_reverse = reverse_first(f,x,h)
df1_reverse = (f(x(2:end))-f(x(1:end-1)))/h;
end
function df2_reverse = reverse_second(f,x,h)
df2_reverse = (f(x(3:end))-2*f(x(2:end-1))+f(x(1:end-2)))/h^2;
end
function df3_reverse = reverse_third(f,x,h)
df3_reverse = (f(x(4:end))-3*f(x(3:end-1))+3*f(x(2:end-2))-f(x(1:end-3)))/h^3;
end
function df4_reverse = reverse_fourth(f,x,h)
df4_reverse = (f(x(5:end))-4*f(x(4:end-1))+6*f(x(3:end-2))-4*f(x(2:end-3))+f(x(1:end-4)))/h^4;
end
I made a forward and backward finite difference code, and the forward and backward codes come out the same. Is this correct?
采纳的回答
x = 0:0.1:1;
f = @(x)x.^2;
derf = (f(x(2:end))-f(x(1:end-1)))/0.1;
hold on
plot(x(2:end),derf); % backward
plot(x(1:end-1),derf); % forward
plot(x,2*x) % analytical
grid on
hold off
更多回答（1 个）
KALYAN ACHARJYA
2024-12-15
1 个投票
In your code, you have written both forward and reverse functions identically, please check it again and make the reverse code.
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