Taylor Series for arctan

I need to write a function that approximates the Taylor Series of arctan(x), with the inputs being x and the error tolerance, and the outputs being the number of terms and the approximation. Here is what I have so far:

function [ARCtanx,n] = arctan_series(x,error)
n = 0;
counter = 0;
while true
    counter = counter + 1;
    series = (-1)^(n)*x^(2*n+1)/(2*n+1);
    terms(counter) = series;
    ARCtanx = sum(terms);
    actualerror = abs((atan(x) - ARCtanx)/(atan(x))*100);
    if actualerror <= error
        break
    end
    n = n +1
end
end

It runs forever, so I know something is wrong with my if statement but I'm not quite sure how to fix it.

Answer 1

[John D'Errico]

There are a few issues here that careful coding would clean up. For example, why do you need variables n AND counter, when both of them do essentially the same thing? But that is just being picky. Code that works is adequate.

By the way, using names like error is a bad idea, as error is a useful function in MATLAB. Again just nitpicking, but it will be important one day to you.

More to the point, what values for x and what error tolerance have you chosen? Have you considered if that series will converge in a reasonable amount of time? If it will ever converge for some x? For what x does it converge?

Note that this is the kind of problem often posed by instructors. Give the student a series that will give them headaches. Then follow that up in the next class to show why. For example, what does this series reduce to when x==1? What do you know about that series?

Does your code converge for small x? For example, how does it do for x = 0.1?

The issue of how to fix the series is easy enough here, but sometimes quite difficult on some other series. For example, for abs(x)>1, is there an identity that would allow you to transform x to a value that DOES have a convergent series? This is typically how such problems are solved. Thus, can you transform the problem to a better one?

[Matrix-Mage]

Yeah, I realized I wasn't too clean with my code, was planning to fix it up once I got a working solution.

Now that I've read your comment, I've tried using small values for x and the code will converge. Normally, I've been doing something like pi/3 and the code will run on for 20,000+ n and keep going. It converges very quickly for small values in radians, largest I've seen being n= 3000 and the smallest being 3 or 4.

Maybe I need to think about the series a bit more, ha ha.

[John D'Errico]

Small values of will show no problems at all. Your code should be working fine then.

That series is convergent only for abs(x)<=1. For larger values of x, things go to hell fast. If you tried it for x=pi/3, that is greater than 1. Those powers of x grow faster than the denominator goes to zero.

And look at the series when x==1. This is in fact a famous series, that converges in theory to pi/4, but it is also well known to converge incredibly slowly.