The fact that Ramanujan's Constant $e^{\pi \sqrt{163}}$ is almost an integer ($262 537 412 640 768 743.99999999999925...$) doesn't seem to be a coincidence, but has to do with the $163$ appearing in it. Can you explain why it's almost-but-not-quite an integer in layman's terms (I'm not a mathematician)?

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    $\begingroup$ mathoverflow.net/questions/30787 and mathoverflow.net/questions/4775 are relevant, but let's see somebody try to explain it without saying "modular". ;) $\endgroup$ – J. M. is a poor mathematician Sep 13 '10 at 16:01
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    $\begingroup$ This is not really the kind of phenomenon with a layman's-terms explanation... $\endgroup$ – Qiaochu Yuan Sep 13 '10 at 16:01
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    $\begingroup$ @Qiaochu, indeed! $\endgroup$ – Mariano Suárez-Álvarez Sep 13 '10 at 16:02
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    $\begingroup$ similar question: "can you explain why $2+\sqrt[3]{\frac{1}{e^{10\pi}}}$ is almost-but-not-quite an integer?" $\endgroup$ – chharvey Jul 15 '15 at 3:08
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    $\begingroup$ @chharvey The answers below seem to indicate that there is a mathematical explanation for why the expression is almost integral. $\endgroup$ – Strants Jul 15 '15 at 5:39

I do not think "why" has a reasonable layman's-terms answer, but let me at least explain "how" with a simpler example of such a numerical coincidence. If one takes powers of the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$ it is not hard to see that they are close to integers. For example, $\phi^{20} = 15126.999934...$. One might ask an analogous question about why these numbers are close to integers. The answer is that

$$\phi^n + \varphi^n = L_n$$

where $L_n$ is the $n^{th}$ Lucas number (an integer), and where $\varphi = \frac{1 - \sqrt{5}}{2}$ has absolute value less than $1$. As $n$ gets larger, $\varphi^n$ gets smaller, so $\phi^n$ becomes a better and better approximation to the integer $L_n$.

A similar, but much more complicated, phenomenon is happening here. The reason $e^{\pi \sqrt{163}}$ is so close to an integer is that it, plus a small error term, gives a special formula for that integer. But "why" this formula exists is a rather long and complicated story (as Robin Chapman hints at) and I do not think there is any reasonable way to talk about it in layman's terms.

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    $\begingroup$ For a generalization of the above example, see en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number . $\endgroup$ – Qiaochu Yuan Sep 13 '10 at 18:02
  • $\begingroup$ accepted as answer for explaining how some numbers approach integers, albeit with a completely different example. I understand Rumanujan's Constant is way beyond my understanding of mathematics. $\endgroup$ – stevenvh Sep 14 '10 at 14:56
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    $\begingroup$ The above explanation is not correct. The approximation property of powers of the golden ratio are due to the fact that it is a Pisot number, i.e., it is > 1 and all its conjugates have absolute value < 1. While interesting, it has applications to Fourier analysis, the Salem-Zygmund theorem characterizing a class of sets of uniqueness, this is not at all what is going on here. $\endgroup$ – ilan vardi Jul 17 '13 at 11:53
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    $\begingroup$ @ilan: I never claimed that it was. I claimed that it was similar (in that there is a formula for an integer which has a bulk term and a small error term). $\endgroup$ – Qiaochu Yuan Jul 17 '13 at 19:08

This is quite a challenge to express in "layman's terms", but the reason is that $$j\left(\frac{1+\sqrt{-163}}{2}\right)$$ is an integer where $j$ is the $j$-function. When you substitute $(1+\sqrt{-163})/2$ into the $q$-expansion (see the wikipedia page) of $j$, all terms save the first two are small, and the first two equal $$-\exp(\pi\sqrt{163})+744.$$

The reason that this $j$-value is an integer is due to the quadratic field $\mathbb{Q}(\sqrt{-163})$ having class number one, or equivalently that all positive-definite integer binary quadratic forms of discriminant $-163$ are equivalent.

Added I'll try to explain the connection with binary quadratic forms. Consider a quadratic form $$Q(x,y)=ax^2+bxy+cy^2$$ with $a$, $b$ and $c$ integers. I'll only consider forms $Q$ which are primitive, so that $a$, $b$ and $c$ have no common factor $ > 1$, and positive-definite, that is $a > 0$ and the discriminant $D=b^2-4ac < 0$. There is a notion of equivalence of quadratic forms, and two primitive positive-definite forms $Q$ and $Q'(x,y)=a'x^2+b'xy+c'y^2$ (necessarily also of discriminant $D$) are equivalent if and only if $$j\left(\frac{b+\sqrt{-D}}{2a}\right) =j\left(\frac{b'+\sqrt{-D}}{2a'}\right).$$ For each possible discriminant there are only finitely many equivalence classes. Thus we get a finite set of $j$-values for each discriminant, and the big theorem is that they are the solutions of a monic algebraic equation with integer coefficients. When there is only one class the equation has the form $x-k=0$ where $x$ is an integer, and the $j$-value must be an integer.

My recommended reference for this is David Cox's book Primes of the form $x^2+ny^2$. But these results appear towards the end of this 350-page book.


As QY pointed out this cannot be explained in layman terms: Here are the papers which you would like to see: www.isibang.ac.in/~sury/ramanujanday.pdf


  • $\begingroup$ Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. $\endgroup$ – Guy Apr 1 '14 at 5:30

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