- #1
peripatein
- 880
- 0
Hi,
I'd sincerely appreciate it if someone were willing to review the few lines of MATLAB code below and indicate why they don't quite yield the expected output.
I am asked to generate using MATLAB approximated values of f(x)=cos(x) at nodes x+h,x-h with random errors <=5*10-6 (using rand) for h=10-8,10-7,...,10-1. Hence,
f~(x+h)=f(x+h)+e(x+h)
f~(x-h)=f(x-h)+e(x-h)
where |e(x)|<=5*10-6
in order to then find approximation for f'(1.2) by using the approximation:
f'(x)=[f(x+h)-f(x-h)]/2h
I am finally asked to plot the error with respect to the value of h.
Below is my code. I am not really sure why it yields one line across the y-axis and another across the x axis.
x=1.2;
fminush=cos(x-h)+(5e-6)*rand(1,1);
fplush=cos(x+h)+(5e-6)*rand(1,1);
fder=(fplush-fminush)./(2*h);
plot(h,abs(-sin(x)-fder))
I'd sincerely appreciate it if someone were willing to review the few lines of MATLAB code below and indicate why they don't quite yield the expected output.
Homework Statement
I am asked to generate using MATLAB approximated values of f(x)=cos(x) at nodes x+h,x-h with random errors <=5*10-6 (using rand) for h=10-8,10-7,...,10-1. Hence,
f~(x+h)=f(x+h)+e(x+h)
f~(x-h)=f(x-h)+e(x-h)
where |e(x)|<=5*10-6
in order to then find approximation for f'(1.2) by using the approximation:
f'(x)=[f(x+h)-f(x-h)]/2h
I am finally asked to plot the error with respect to the value of h.
Below is my code. I am not really sure why it yields one line across the y-axis and another across the x axis.
Homework Equations
The Attempt at a Solution
h=(10^-1).^[1:8];x=1.2;
fminush=cos(x-h)+(5e-6)*rand(1,1);
fplush=cos(x+h)+(5e-6)*rand(1,1);
fder=(fplush-fminush)./(2*h);
plot(h,abs(-sin(x)-fder))