Plotting log x and y

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Francis Wendell
Francis Wendell on 12 Feb 2021
format long;
syms i imin n error y x h emin;
n = 30;
x = 0.5;
h = 1;
emin = 1;
for i =1:n
h = h*0.25;
y = [sin(x+h)-sin(x)]/h;
error(i) = abs(cos(x)-y);
i;
h;
y;
error;
TruncationError(i) = ((-1/6)*h^2*cos(x));
TotalError = ((-1/6)*h^2*cos(x)+error);
if error < emin
emin = error;
imin = i;
end
end
imin;
emin;
%hold on
%a = -log10(abs(error));
%b = log10(h);
%plot (a,b)
%plot (error);
%plot (TruncationError);
%plot (TotalError);
I'm new to matlab and am not sure how all the commands work. I am trying to plot error, truncation error, and total error all on the same graph. The x and y axis need to be what a and b are both log. It should look similar to the picture below numbers are different. Any guidance in the correct direction will be extremely helpful. Thanks.
  1 Comment
dpb
dpb on 12 Feb 2021
Edited: dpb on 12 Feb 2021
You've not saved the values of h in the loop to use to plot against -- but also note that at the end of your loop that h will be (1/4)^n --> (1/4)^30
>> (1/4^30)
ans =
8.6736e-19
>> log10(ans)
ans =
-18.0618
>>
so your log10(h) axis would go from 0 --> -19 instead of 0 --> 6
So unless the plot is mislabeled, it shows h went from ~10 to 10^6 instead.
There is also no need for symbolic variables here...

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Answers (1)

Rafael Hernandez-Walls
Rafael Hernandez-Walls on 12 Feb 2021
Why don't you treat like this:
n = 30;
x = 0.5;
h(1) = 1;
emin = 1;
for i =1:n
if i==1
h(1)=0.5;
else
h(i) = h(i-1)*0.25;
end
y = [sin(x+h(i))-sin(x)]/h(i);
error(i) = abs(cos(x)-y);
TruncationError(i) = ((-1/6)*h(i)^2*cos(x));
TotalError(i) = ((-1/6)*h(i)^2*cos(x)+error(i));
if error(i) < emin
emin = error;
imin = i;
end
end
loglog(h,TotalError,'r')
figure
loglog(h,TruncationError,'k')

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