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:

slide rule with pi circled

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?

  • 4
    $\begingroup$ Wow, that brings back memories. I haven't seen a slide rule for years. $\endgroup$ – Robert Israel Feb 13 '18 at 21:57
  • $\begingroup$ @RobertIsrael I've never used it! $\endgroup$ – user Feb 13 '18 at 21:59
  • $\begingroup$ According to Wikipedia, "This coincidence was used in the design of slide rules, where the "folded" scales are folded on $\pi$ rather than $\sqrt{10}$ because it is a more useful number and has the effect of folding the scales in about the same place" $\endgroup$ – David Quinn Feb 13 '18 at 22:27
  • $\begingroup$ That page on Mathematical coincidences also notes that this is the reason the gravitational constant $g\approx10 m/s^2$, because a meter was originally defined as the length of a pendulum with a half period of 1 second, and $T_{1\over2}\approx\pi\sqrt{L\over{g}} \implies g\approx\pi^2$ $\endgroup$ – kgutwin Feb 16 '18 at 20:41

$\pi^2\approx 10$ 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 $$ 10-\pi^2=\sum_{n\geq 2}\frac{6}{n^2(4n^2-1)}\leq \frac{1}{10}+\sum_{n\geq 3}\frac{24}{(4n^2-9)(4n^2-1)}=\frac{29}{210}. $$

  • $\begingroup$ How is the $\sum\limits_{n \geq 2} \frac{6}{n^2-1/4}=4$? $\endgroup$ – Botond Feb 13 '18 at 22:14
  • $\begingroup$ Why it should demonstrate that it is not a coincidence? I don't see the reason. $\endgroup$ – user Feb 13 '18 at 22:15
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    $\begingroup$ @Botond: that is a telescopic series. $\endgroup$ – Jack D'Aurizio Feb 13 '18 at 22:15
  • $\begingroup$ @JackD'Aurizio Thanks, I'm going to check it. $\endgroup$ – Botond Feb 13 '18 at 22:16
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    $\begingroup$ @Botond: you may also have a look at the first section of my notes where the identity $\pi^2=\sum_{n\geq 1}\frac{18}{n^2\binom{2n}{n}}\approx 10$ is proved through creative telescoping. $\endgroup$ – Jack D'Aurizio Feb 13 '18 at 22:17

Simply because

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




$$y\approx \frac12$$

  • $\begingroup$ I don't feel like this answer addresses OP's core question, "Is there a reason or derivation for this, or is it pure coincidence?" $\endgroup$ – Chris Culter Feb 13 '18 at 22:03
  • $\begingroup$ @ChrisCulter Interesting point! I'll think about it, post also your own answer if you have some idea to share! $\endgroup$ – user Feb 13 '18 at 22:04
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    $\begingroup$ @ChrisCulter I think gimusi showed its reason: $\pi^2 \approx 10$. Maybe the $\Delta \log(a) < \Delta a$ could be added. $\endgroup$ – Botond Feb 13 '18 at 22:05
  • $\begingroup$ @ChrisCulter Anyway I think it is just a coincidence, indeed it is expressed in base-ten (for counting) which is an arbitrary choice. I'm wondering how it should be in base-2 or others. $\endgroup$ – user Feb 13 '18 at 22:09
  • $\begingroup$ @Botond Well, why is $\pi^2\approx10$? A satisfactory explanation would involve some argument beyond rearranging the equation and performing a numerical evaluation. If someone asks you why $\pi\approx3$, you don't evaluate some series and conclude that $3.14\approx3$; you should be drawing a regular hexagon inscribed in a circle. If someone asks why $\pi\approx\frac{22}{7}$, you should start talking about $\int_0^1 \frac{x^4\left(1-x\right)^4}{1+x^2} \, dx$. For $\pi^2\approx10$, Jack D'Aurizio suggests that the answer is implied by $\zeta(2)$; and there could be other explanations as well. $\endgroup$ – Chris Culter Feb 14 '18 at 8:53

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.


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

They are not equal but very close.

  • $\begingroup$ Your value of pi has a typo. 3.1*4* $\endgroup$ – JTP - Apologise to Monica Feb 14 '18 at 1:09
  • $\begingroup$ @JoeTaxpayer Thanks for the comment. I fixed it per your advice. $\endgroup$ – Mohammad Riazi-Kermani Feb 14 '18 at 1:28

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