# Ramanujan's sums

Are the series expansions of arithmetic functions in terms of Ramanujan sums computationally useful? I didn't think they would be, but they seem to be good approximations even when summed with few terms, for example the ramanujan expansion for the divisor function is:

$$\sigma_x(n)=n^x\zeta(x+1)(\frac{C_1(n)}{1^{x+1}}+\frac{C_2(n)}{2^{x+1}}+\frac{C_3(n)}{3^{x+1}}...)=n^x\zeta(x+1)\sum_{k=1}^\infty\frac{C_k(n)}{k^x}$$ $$\text{where the first few } C_k(n) \text{ are:}$$ $$C_1(n)=1$$ $$C_2(n)=\cos(n\pi)$$ $$C_3(n)=2\cos(\frac{2}{3}n\pi)$$ $$C_4(n)=2\cos(\frac{1}{2}n\pi),$$ $$C_5(n)=2\cos(\frac{2}{5}n\pi)+2\cos(\frac{4}{5}n\pi)$$ $$C_6(n)=2\cos(\frac{1}{3}n\pi)$$ ....etc, http://en.wikipedia.org/wiki/Ramanujan

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Is it easier to evaluate the sum of divisors, or a zeta function? Note that if $x$ is even, you cannot use the expression in terms of Bernoulli numbers, and you must then use a numerical algorithm for $\zeta$. This may or may not be simpler than just listing out the divisors procedurally. – J. M. Apr 9 '13 at 1:32
@J.M. I don't know much about Numerical analysis, but I would think for very large integers that it would be easier to approximate a zeta function then to find that integers divisors, sense this would be tantamount to integer factorization. – Ethan Apr 9 '13 at 4:09
Well, then, if $n$ is large, then evaluating the trigonometric functions accurately would be difficult... – J. M. Apr 9 '13 at 4:15
@J.M. You can reduce them modulo $2\pi$ – Ethan Apr 9 '13 at 4:19
That's precisely the problem. If you can do everything in exact arithmetic, then no biggie. The problem is when you do this in inexact arithmetic, the arguments of your trigonometric functions may no longer have any accurate digits after performing the needed subtraction. – J. M. Apr 9 '13 at 4:41