0
$\begingroup$

(2 * 2 * 2 * 3 * 5 * 5 * 5 * 415752385339879101618702506672691517522274861954331757391122938598028498597433207031)^1/5 ~ (very very close to Ramanujan constant)

Would anyone have an idea of why this prime factorisation to the 1/5 power is unbelievably close to the Ramanujan constant, e^(pi*sqrt163) ?. This is precise to 70 digits after the decimal point.

Improve this question
$\endgroup$
8
  • 13
    $\begingroup$ It's extremely unsurprising, because you put more than 70 digits of information into that long number. $\endgroup$
    – Reid Barton
    Commented Nov 9, 2009 at 16:57
  • 5
    $\begingroup$ Agreed. That number is about 90 digits long (I might have miscounted a few.) The spacing between 5th roots of 90 digit numbers is about 10^{-72}. So it's hardly a surprise that you can find some number in that range whose 5th root agrees with a given real number to 70 decimal places. $\endgroup$
    – David E Speyer
    Commented Nov 9, 2009 at 17:05
  • 8
    $\begingroup$ Somewhat surprisingly, exp(Pisqrt(163))^5 is within 10^(-5) of an integer. And this integer is not N^5 where N is the integer closest to exp(Pisqrt(163)). $\endgroup$
    – Michael Lugo
    Commented Nov 9, 2009 at 17:08
  • 1
    $\begingroup$ Either prod Anton to re-open, or post another question. I can prove that c^2 is close to an integer, where c is Ramanujan's constant, and this isn't a formal consequence of c being close to an integer (because it's not close enough, as it were). But my proof doesn't stretch to c^5. $\endgroup$
    – Kevin Buzzard
    Commented Nov 9, 2009 at 18:39
  • 2
    $\begingroup$ Seconding that Michael should ask this as a separate question, unless he wants to work it out himself. $\endgroup$
    – JSE
    Commented Nov 9, 2009 at 19:58

1 Answer 1

Reset to default
7
$\begingroup$

This answer pertains to Michael's comment above.

Let $c$ be $e^{\pi\sqrt{163}}$. It is well-known that $c$ is close to an integer. More precisely, $c$ is about $10^{-12}$ from an integer of size about $10^{17}$, so (squaring $n+\epsilon$) there's no reason for $c^2$ to be close to an integer (just from these facts alone). But it is, and here's a cheap reason why.

Set $q=e^{2\pi i\tau}$ with $\tau=(1+\sqrt{-163})/2$. Then we know from general theory that $j(\tau)=1/q+744+196884q+\ldots$ is an integer, and hence (because $q$ is tiny) that $-1/q$ (which is Ramanujan's constant) is also close to an integer.

But just take this result $1/q+744+196884q+\ldots\in\mathbb{Z}$ and square it. We get $1/q^2+1488/q+ 947304 + 335950912q +\ldots\in\mathbb{Z}$. Now we know $q$ is tiny, and we know $1488/q$ is close to being an integer (but not as close as $q$ is), hence $1/q^2$ is close to being an integer (but not as close as $1/q$ is).

Now rinse and repeat.

But unfortunately this cheap argument doesn't seem to show that $c^5$ is close to an integer, because the power series expansion of $j^5$ involves terms like $1410274829033621720q$, and $q$ is about $10^{-17}$ but this huge integer multiple of $q$ isn't much less than 1 any more so this cheap approach breaks down.

Maybe it's better to do the job properly, and think about what explicit class field theory says about $j(5\tau)$.

Improve this answer
$\endgroup$

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