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At the end of a Physics examination, I decided to play around with my calculator, as I always do when I have time left over, and found that:

$$\frac{1}{100} \cdot 11^{\ln(11)} \approx 3.14159789211,$$ where $\ln(x)$ is the natural logarithm of $x$, gives $\pi$ correct to five decimal places, where $\pi \approx 3.14159265359$.

Does anybody know of any reason why this may be, or if this is simply a coincidence?

Edit

I must thank @Shailesh for providing me with the link to the following Reddit page, for it also begs the same question as to whether or not there is a relationship between $11, \ln(11),$ and $\pi$.

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    $\begingroup$ looks like a coincidence to me. Unless you find a recurring pattern of incresingly accurate approximations, it's usually just a coincidence. $\endgroup$ – AlexR Jun 13 '15 at 15:29
  • $\begingroup$ @AlexR Ah, okay. Well, I will look to see if there is any sort of pattern. $\endgroup$ – Taylor Jun 13 '15 at 15:30
  • $\begingroup$ As math goes, you can always find some more or less convincing justification for each of those. There must be a very strange series or an integral that evaluates to the difference between this and the true $\pi$. $\endgroup$ – orion Jun 13 '15 at 15:32
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    $\begingroup$ Interestingly the same question was there posted 14 Feb 2011 on reddit. Here is the link. m.reddit.com/r/math/comments/fl35c/… $\endgroup$ – Shailesh Jun 13 '15 at 15:43
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    $\begingroup$ What is a reason and what is a coincidence? This is realted to the mathematical phenomenon of almost integers, calculations involving seemingly random irrationals that turn out to be almost an integer (your choice could be rearranged to say $\frac{\ln^2 11}{\ln 100 \pi} \sim 1$). A reasonable 'why' answer might come from equations that involve floors (like powers of the Fibonacci constant). $\endgroup$ – Mitch Jun 13 '15 at 21:16
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Though Srinivasa Ramanujan would probably find a deep explanation, you can reason as follows: taking two small integers, say in range $1$ to $100$ and combining them in $100$ different ways using simple expressions $(+,-,\times,\div,\sqrt{},x^y,\log_yx\cdots)$, what is the probability that the six leading digits of the expression will be those of $\pi$ ?

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  • $\begingroup$ I haven't the foggiest! What would you say? $\endgroup$ – Taylor Jun 13 '15 at 16:56
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    $\begingroup$ With similar rules, $74^2\tan(74)=-\color{green}{31415.9}3542\cdots$ (among $4843142$ other expressions evaluated). $\endgroup$ – Yves Daoust Jun 13 '15 at 17:21
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    $\begingroup$ Also, $\ln(20)/\sqrt{\sin(2)}=\color{green}{3.14159}3698\cdots$ among $8227233$ expressions tried. $\endgroup$ – Yves Daoust Jun 13 '15 at 17:31
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    $\begingroup$ Interestingly, the magical number $\tan(74)$ also appears in $\tan(74)\tan(35)=-\color{green}{2.71828}5730\cdots$. $\endgroup$ – Yves Daoust Jun 13 '15 at 17:58
  • $\begingroup$ $\tan(74)$ is interesting. Since it could give approximation to six digits of $\pi$ and $e$, is there an expression with it to produce $\phi$ ? $\endgroup$ – KKZiomek Apr 2 '16 at 5:28
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Maybe not a complete coincidence. Your approximation can be rewritten as $$e^{\frac{\ln^2 11}{2}}\approx\sqrt{100\pi},$$ and it may be an approximation of the normal distribution integral $$\int_{-\infty}^\infty e^{\frac{-x^2}{2t}}dx=\sqrt{2\pi t}.$$

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  • $\begingroup$ It's doubtful, that there's anything special about $11$... $\endgroup$ – DVD Jun 17 '15 at 17:11
  • $\begingroup$ Looks like the residue theorem for the contour integral: en.wikipedia.org/wiki/Residue_theorem#Example $\endgroup$ – DVD Jun 17 '15 at 17:45
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It seems to point towards an expression of the form $\bigg[\dfrac{(1+x)^{\ln(1+x)}}{x^2}\bigg]_{x=10}\simeq\pi$, which, together with $\pi^2\simeq10$, indicate $\dfrac{\ln^211}{\ln\pi}\simeq5$ as a starting point.

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  • $\begingroup$ That's a cool find! Thank you! $\endgroup$ – Taylor Jun 13 '15 at 16:53
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Reversing your equation and using an approximate form of $\pi$,

$$\frac{1}{100} \cdot 11^{\ln(11)} \approx \pi \implies 11^{\ln(11)} \approx 100\pi$$

$$\implies \log_{11}(100\pi) \approx \ln(11)$$

$$\implies {\ln(100\pi)\over\ln11} \approx \ln(11)$$

$$\implies \ln(100\pi)\approx \ln^2(11)$$

but I've no idea why that might be!

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  • $\begingroup$ That is rather interesting. Thank you for your answer, and I hope that we may find a reason for why this is. $\endgroup$ – Taylor Jun 13 '15 at 15:31
  • $\begingroup$ @G.Sassatelli Well, no, of course not, but I do believe that it is a rather interesting, and intriguing mathematical fact. $\endgroup$ – Taylor Jun 13 '15 at 15:37
  • $\begingroup$ Faster: $$\frac{1}{100}\cdot 11^{\ln(11)}\approx \pi\iff 11^{\ln(11)}\approx 100\pi\stackrel{\ln(\cdot)}\iff\ln(11)^2\approx \ln(100\pi) $$ $\endgroup$ – user26486 Jun 13 '15 at 15:50
  • $\begingroup$ To me this looks just like another, equivalent, coincidence ^^ $\endgroup$ – AlexR Jun 13 '15 at 16:07
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Here's how I think of your result:

Let's look for integers $n$, such that the beginning of the decimal expansion of $n^{\log n}$ agrees with that of $\pi$ (up to some point). Using a for loop, I found the following approximations for $n<100,000$:

$$ \pi \approx \frac{11^{\log 11}}{10^2},\frac{53599^{\log{53599}}}{10^{51}},\frac{59546^{\log{59546}}}{10^{52}}.$$

Note that the last two only approximate $\pi$ to 4 digits after the decimal point.

It seems that for $n<1,000,000$, $n=11$ gives the best approximation of the form $\frac{n^{\log{n}}}{10^{d(n^{\log n})-1}}$ where $d(m)$ is the number of digits of $m$ left to the decimal point.

I'm stil trying to find better approximations though...

EDIT:

In Mathematica I used something of the form

For[n = 1, n < 100000, n++,If[Floor[n^Log[n]/10^(IntegerLength[Floor[n^Log[n]]] - 5)] == 31415,Print[N[n^Log[n], 10], " ", n]]]

This will give you approximations good to 4 decimal places in the range $n<100000$.

EDIT2:

Using a longer loop for finer approximations I found

$$\pi \approx \frac{3214471^{\log 3214471}}{10^{97}},\frac{3745521^{\log 3745521}}{10^{99}} $$

both to 6 decimal places.


Overall, it seems that the case $n=11$ is extraordinarily good for small values of $n$. I still can't see why nonetheless.

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  • $\begingroup$ Thank you for your answer, as this is interesting. On another note, you said that you used a "for loop," which I only know to be a feature in programming. If I am correct, may I ask what programming language did you use? $\endgroup$ – Taylor Jun 13 '15 at 16:11
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    $\begingroup$ @Taylor Such things can be done nicely in SAGE and Mathematica, to name two relatively easy to use arbitraty-precision programming languages. $\endgroup$ – AlexR Jun 13 '15 at 16:12
  • $\begingroup$ Also, you have put $\log,$ instead of $\ln,$ which is what I used. $\endgroup$ – Taylor Jun 13 '15 at 16:13
  • $\begingroup$ @AlexR Ah, thank you! I am not very familiar with either of the two, but I will be sure to check them out. $\endgroup$ – Taylor Jun 13 '15 at 16:13
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    $\begingroup$ @Taylor In Mathematica Log means $\ln$. Also, I'm running a loop looking for 6 decimal places approximations for $n<10,000,000$ as we speak. Nothing has been found yet. $\endgroup$ – user1337 Jun 13 '15 at 16:19
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We can write the approximation as the following equivalent almost-integer.

$$\frac{11^{\log(11)}}{\pi}\approx 100.0001667\approx 10^2+\frac{1}{6000}$$

Two similar ones with three zeros after the decimal point also involve multiples of Heegner numbers.

$$\frac{(4\times43)^{\log(4\times43)}}{\pi}\approx 102381257746.0007223$$

$$\frac{(20\times163)^{\log(20\times163)}}{\pi} \approx 8374566425664827543971729687.0002623$$

Yet another Heegner number, $19$, appears in the almost-integer $$\frac{2(2\times 19)^{\log(2\times 19)}}{\pi} \approx 355204.9991 \approx5\left(41+71\times10^3\right)=4^2(12^2+5)^2-11$$

which gives an approximation to $\pi$ with seven correct decimal digits. $$\pi \approx \frac{2\times38^{\log(38)}}{355205} \approx 3.1415926$$

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