%QUESTION3 Implements the central differnce method to find the gradient of
%y=e^x-2sin^2(2x) at the point where x=1.
clear allclcformat longx=1; % defining x
y=1; % defining y, (y=x)
h=0.1 % first value of h
x=x+h; % getting x+h within f(x)
y=y-h; % getting x-h within f(x)
f = @(x) exp(x)-2*(sin(2*x)).^2; % defining f(x)
g = @(y) exp(y)-2*(sin(2*y)).^2; % defining f(y) to later use as f(x-h)
Grad=(f(x)-g(y))/2*h % Central difference method, Grad = Gradient
for n=1:2 % 2 iterations after h=0.1
h=h/10 % new value for h
x=x+h; % to replace f(x) with f(x+h)
y=y-h; % to replace f(x) with f(x-h)
Grad=(f(x)-g(y))/2*h % Central difference method
end
Here is my code, but my output is incorrect. I did the calculations on paper and they're quite different to my answers. My first value of "Grad" is correct but off by a factor of 1/100. My other two answers are completely wrong.
Best Answer