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(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.

Mark Thomas
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    It's extremely unsurprising, because you put more than 70 digits of information into that long number. – Reid Barton Nov 09 '09 at 16:57
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    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. – David E Speyer Nov 09 '09 at 17:05
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    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)). – Michael Lugo Nov 09 '09 at 17:08
  • Michael's comment is much more surprising than Mark's. But I suspect Michael's can be explained using the same sorts of ideas which go into the proof that Ramanujan's constant is almost an integer (rather than just the statement that it's almost an integer, which isn't strong enough). – Kevin Buzzard Nov 09 '09 at 17:31
  • buzzard, I agree; I was trying to produce such a proof when this question was closed. – Michael Lugo Nov 09 '09 at 18:18
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    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. – Kevin Buzzard Nov 09 '09 at 18:39
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    Seconding that Michael should ask this as a separate question, unless he wants to work it out himself. – JSE Nov 09 '09 at 19:58
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    As requested by buzzard and JSE, I've asked a related question: http://mathoverflow.net/questions/4775/why-are-powers-of-exppisqrt163-almost-integers – Michael Lugo Nov 09 '09 at 21:27

1 Answers1

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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 $j(5\tau)$.

Kevin Buzzard
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