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 $\pi$.
$\endgroup$ 86 Answers
$\begingroup$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 $\pi$ ?
$\endgroup$ 5 $\begingroup$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 $$\int_{-\infty}^\infty e^{\frac{-x^2}{2t}}dx=\sqrt{2\pi t}.$$
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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!
$\endgroup$ 4 $\begingroup$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.
$\endgroup$ 7 $\begingroup$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. $$\pi \approx \frac{2\times38^{\log(38)}}{355205} \approx 3.1415926$$
$\endgroup$
