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$$e^ {\pi\sqrt{163}}=262537412640768743.9999999999992\cdots$$

Why does this number run so incredibly close to a whole number? Can I have a logical explanation for why this finding? I know how to calculate it, but I want an answer that would explain it to me by intuition only. Thank you very much for any help. It is not the same as the other similar problem because I am looking for a logical reason and the other is wondering whether or not it is concidence.

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    A simple, though naive explanation: given all possible things you can write in the exponent, sooner or later one of them is bound to look close to a whole number. – Clement C. Jul 14 '15 at 21:15
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    It's not entirely clear what you're asking. First of all, can you rigorously define "incredibly close"? One could argue that 0 is also "incredibly close" to 262537412640768744 (compared to a very large negative number). Second, explain what kind of answer you expect from asking "why" a number is what it is. – chharvey Jul 14 '15 at 21:16
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    https://en.wikipedia.org/wiki/Heegner_number#Almost_integers_and_Ramanujan.27s_constant – Will Jagy Jul 14 '15 at 21:16
  • also, the (logic) tag definitely does not belong on this question. – chharvey Jul 14 '15 at 21:18
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    along with https://en.wikipedia.org/wiki/Heegner_number#Other_Heegner_numbers – Will Jagy Jul 14 '15 at 21:18
  • Yeah, I see what you mean. But look at the number. I mean, come on though, just look at the number. –  Jul 14 '15 at 21:19
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    AAron, if you wish someone to be notified that you have addressed a comment to them, you put an at sign @ followed (no space) by at least the first three letters of their user name. About this particular question, you are allowed to be impressed by the number. Asking for an explanation that will satisfy your intuition is a bit much. – Will Jagy Jul 14 '15 at 21:21
  • Here are some more: $e^\pi - \pi = 19.9990999$, $\pi - \sqrt{40/3 - \sqrt{12}} = 0.00006$, ... As said above, if you try different combinations of $e$ and $\pi$ and square roots of whole numbers you are bound to discover almost-identities like these. – Winther Jul 14 '15 at 21:40
  • Yeah, very true. I just kinda stumbled upon it and found it was cool. –  Jul 14 '15 at 21:51
  • It's more impressive that it's very close to a whole number that is very close to a perfect cube. – mjqxxxx Jul 14 '15 at 21:53
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    I feel like this question could be better worded as "how close to a whole number is $e^{\pi\sqrt{n}}$ with $n = 163$ when compared to other values of $n$?" It would be cool if someone answered with a some kind of graphic illustrating this over a bunch of values of $n$. A related question would be, "what is the smallest natural number $n$ greater than $163$ such that $e^{\pi\sqrt{n}}$ is closer to a whole number than $e^{\pi\sqrt{163}}$?" – Mike Pierce Jul 14 '15 at 21:58
  • @ClementC. : But you cannot say that of all the possible SHORT and SIMPLE expressions you can write down, that will ultimately happen. ${}\qquad{}$ – Michael Hardy Jul 14 '15 at 22:32
  • Why do you think the answers to the other question are not trying to give "logical reasons"? –  Jul 15 '15 at 00:25
  • Similar question: "can someone explain why $2+\sqrt[3]{\frac{1}{e^{10\pi}}}$ is ridiculously close to an integer?" – chharvey Jul 15 '15 at 02:59
  • Dude, your just mocking my question –  Jul 15 '15 at 02:59
  • @AAron — sorry, my mocking is to be taken lightly. I'm just trying to make a point about how this kind of question doesn't belong on the SE network (in my opinion). there are other sites/forums/chatrooms that would welcome your question happily. – chharvey Jul 15 '15 at 03:03
  • Ok, sorry dude, I'm new to this site(only been on a week) and im still trying to understand how everything works –  Jul 15 '15 at 03:04
  • @AAron — understandable. You might want to read "How do I ask a good question?" and "What topics can I ask about here?". – chharvey Jul 15 '15 at 03:12

1 Answers1

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Here's an explanation of these numbers. Say you want $e^{\pi \cdot n}$ to be within a positive error called $\epsilon$ of a natural number $L$, but you want $n$ to be natural as well. In other words...

$$L-\epsilon \lt e^{\pi \cdot n} \lt L+\epsilon$$

You can "solve" for n. Take the natural log and divide by $\pi$ to get...

$${1 \over {\pi}} \cdot \ln{(L-\epsilon)} \lt n \lt {1 \over {\pi}} \cdot \ln{(L+\epsilon)}$$

Now it should be clear that if $L$ ranges through $1$ to $\infty$ that the boundary for $n$ will range $0$ to $\infty$.

Since $L$ must be an natural number, there is little reason to think the values $n$ can take will be any more likely to not contain a natural number.

So now you'd use the above and pick an error tolerance. Values that allowed for both a natural $n$ and $L$ will be within $\epsilon$ of $L$. I'm guessing these are the so called Heegner numbers, you'd just allow for $n$ which are the square root of other numbers.

Zach466920
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