issue with polyval and polyfit measuring time

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Hi!
So I'm working on a code that evaluates how fast a loop can evaluate a varying range of numbers. I put these numbers into an array called x and then the times taken into an array called y. Then I used polyfit and polyval to try to see how long it would take to evaluate numbers between 1 and 1000000000, but when I get my answer, it is very small. I'd expect it to increase like all the others so I think I did something wrong but I am not sure what.
figure(6)
x = [ 1000, 10000, 100000, 1000000]; % represents the tested arrays
y = [ 0.076243, 0.251277, 3.053055, 64.876395]; % represents the time taken
loglog(x,y,'bo--');
title('Time Needed to Calculate the Prime Numbers in an Array');
xlabel('Array length [ - ]');
ylabel('Time taken [ in seconds ]');
grid off;
% d.
xlog = log(x);
tlog = log(y);
coefficients = polyfit(xlog,tlog,1) % which yields [ 0.9874 -9.8979 ]
% Given these coefficients with the polyfit function, the line of best
% fit can be represented by the equation --
% y = 0.9874x - 9.8979 where x is the logarithm of the array length
% and y is the logarithm of the time taken
% e. Polyval may also further be used to find fitted data when given the
% coefficients and the logarithm of 1000000000, which equals 9, as an
% input.
seconds = 10.^polyval(coefficients, 9); % which yields the total seconds
% needed to calculate the logarithm of 1000000000, though this must
% be expressed in hours.
time = seconds./3600 % yields a very small time that doesn't make sense!
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Accepted Answer

Matt J
Matt J on 10 Mar 2020
Edited: Matt J on 10 Mar 2020
xlog = log10(x);
tlog = log10(y);
  2 Comments
Matt J
Matt J on 10 Mar 2020
No problem... You might have gotten there first anyway.

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More Answers (1)

Jon
Jon on 10 Mar 2020
Edited: Jon on 10 Mar 2020
One problem may be that you think that log(x) gives log to the base 10 of x but it is the natural log.
So use instead
xlog = log10(x);
tlog = log10(y);
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