Why is $\log_{10}\pi \simeq 0.5$?

Question

I was looking at my slide rule earlier today, and happened to notice that the dedicated mark for $\pi$ is right around the middle of the scale:

Of course, this is because $\log_{10}\pi \simeq 0.4971498$, which is very close to $1\over2$.

Is there a reason or derivation for this, or is it pure coincidence?

Answer 1

$\pi^2\approx 10$ is not a coincidence. Since $\zeta(2)=\frac{\pi^2}{6}$ we have $$ \pi^2 = 6+\sum_{n\geq 2}\frac{6}{n^2} \leq 6+\sum_{n\geq 2}\frac{6}{n^2-\frac{1}{4}}=10 $$ and the difference between the RHS and the LHS is $$ 10-\pi^2=\sum_{n\geq 2}\frac{6}{n^2(4n^2-1)}\leq \frac{1}{10}+\sum_{n\geq 3}\frac{24}{(4n^2-9)(4n^2-1)}=\frac{29}{210}. $$

Answer 2

Simply because

$$\sqrt{10}=10^{\frac12}\approx 3.16 \approx \pi$$

indeed

$$y=\log_{10}\pi\iff10^y=\pi$$

thus

$$y\approx \frac12$$

Answer 3

$$\pi \approx 3.141592654 \implies log(\pi ) \approx .497149873$$ $$ \sqrt 10 \approx 3.162277660 \implies log \sqrt {10} =.5$$

They are not equal but very close.

Answer 4

I cannot see a very deep reason why this would be so, but the continued fraction of $\sqrt{10}$ is $3 + \frac{1}{6 + \frac{1}{6 + \cdots}}$ and a continued fraction for $\pi$ is $3 + \frac{1}{6 + \frac{9}{6 + \cdots}}$ so they at least agree to $3 + \frac{1}{6}$, which after all is not too bad of an estimate, being off by only 2 in the second decimal place.

I suppose it is worth noting that a general continued fraction for square roots of the form $\sqrt{a^2 + b} = a + \frac{b}{2a + \frac{b}{2a + \cdots}}$ then there is straightforwardly no better approximation to $\pi$ by the square root of an integer since $a$ is constrained to be $3$ and $b$ is constrained to be $1$ right at the start.

I'm not sure how to consider things "coincidental" at this level of navel gazing.