Why is it that this gives a good approximation of $\pi$?

pi approximation

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 $\pi$.

Solution 1:

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 $\pi$ ?

Solution 2:

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 $$\int_{-\infty}^\infty e^{\frac{-x^2}{2t}}dx=\sqrt{2\pi t}.$$

Solution 3:

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.