For several days, I've been wondering how it would be possible to compute the sine of huge numbers like 100000! (radians). I obviously don't use double but cpp_rational from the boost multiprecision library. But I can't simply do 100000! mod 2pi and then use the builtin function sinl (I don't need more than 10 decimal digits..) as I'd need several million digits of pi to compute this accurately.

Is there any way to achieve this?

  • 4
    $\begingroup$ Why? Why is this necessary? $\endgroup$ – Jared Jul 5 '16 at 9:06
  • 1
    $\begingroup$ You chould try to use the following relationship: $\Gamma(n+0.5)=\frac{2n!}{n!4^n}\sqrt{\pi}$ $\endgroup$ – MrYouMath Jul 5 '16 at 9:08
  • 3
    $\begingroup$ $100000!=31415*100000*99999!/31415≒pi*k$. So $sin100000!≒0.000$. $\endgroup$ – Takahiro Waki Jul 5 '16 at 9:14
  • 1
    $\begingroup$ Applying the taylor-series won't work. $1)$ The alternating signs lead to numerical instability $2)$ we need too many terms. It seems that we need a program with high precision-calculation and cannot compute , for example $sin(10^{10^{10^{10}}})$ $\endgroup$ – Peter Jul 5 '16 at 9:21
  • 1
    $\begingroup$ $(10^5)!$ has $456,574$ digits. Do you need the sine for even larger numbers ? You need less tha a million digits for $sin((10^5)!)$ $\endgroup$ – Peter Jul 5 '16 at 9:27

I believe you may be able to calculate this without obscene numbers of digits of $\pi$ if you take advantage of the fact that these are factorials. To simplify the algebra, we can calculate $a_n=e^{i(n!)}$ instead (you want the imaginary part). Then $$a_{n+1}=e^{i(n!)(n+1)}=a_n^{n+1},$$ and it's perfectly reasonable to calculate $a_{100000}$ recursively with a high-precision library.

The downside is that to start the recursion you need a very good approximation of $e^i$, and I don't know if the error dependence works out any differently than in the $\pmod{2\pi}$ approach.

But to answer your actual question, Mathematica doesn't even break a sweat with the mere million digits needed for this:

> Block[{$MaxExtraPrecision = 1000000}, N[Sin[100000!], 10]]


takes about 15 ms on my computer.

For calculating the sine or cosine of a large arbitrary precision real number $x$, the gains of this method (which are tuned for $\sin n$ for integer $n$) are mostly lost, so I would recommend your original idea of reducing the argument $\bmod 2\pi$. As has been noted, the main bottleneck is a high-precision estimation of $\pi$. Your answer will be useless unless you can at least calculate $\frac{x}{\pi}$ to within $1$ (otherwise you may as well answer "somewhere between $-1$ and $1$"), so you need at least $\log_2(x/\pi^2)$ bits of precision for $\pi$. With $x\approx100000!$, that's about $1516701$ bits or $456572$ digits. Add to this the number $a$ of bits of precision you want in the result, so about $1516734$ digits of $\pi$ to calculate $33$ bits ($\approx 10$ digits) of $\sin x$ in the range $x\approx 100000!$.

Once you have an integer $n$ such that $y=2\pi n$ is close to $x$ (ideally $|x-2\pi n|\le2\pi$, it doesn't have to be perfectly rounded), calculate $\pi$ to precision $a+\log_2(n)$, so that $y$ is known to precision $a$, and then $x-y$ is precision $a$ and $\sin x=\sin (x-y)$ can be calculated to precision $a$ as well.

  • $\begingroup$ If I am right, the relative error on a power is the relative error on the base times that power. So you will need to evaluate $e^i$ with $n!$ decimals. $\endgroup$ – Yves Daoust Jul 5 '16 at 9:37
  • 3
    $\begingroup$ @YvesDaoust Actually, you only need error within $\frac1{n!}$ of the target error, which means about $\log(n!)$ decimals. $n!$ decimals would truly be obscene. $\endgroup$ – Mario Carneiro Jul 5 '16 at 9:40
  • $\begingroup$ you are right, my bad. $\endgroup$ – Yves Daoust Jul 5 '16 at 9:46
  • $\begingroup$ I don't actually compute the sine of a factorial but of a number as big as 100000! - but anyway: how is Mathematica actually capable of computing the sine that fast?! $\endgroup$ – Oliver Borchert Jul 5 '16 at 18:04
  • $\begingroup$ @OliverBorchert Actually, nothing about this method really depends on it being $n!$. I'm just taking $e^i$ to a very large integer power, which you could do just as well with other numbers by exponentiation by squaring. The important part is that to calculate $e^{in}$ you need about $a+\log_2(n)$ bits of precision on $e^i$ to get $a$ bits at the end. $\endgroup$ – Mario Carneiro Jul 5 '16 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.