$\frac{7 \cosh(\sqrt 6)}{13} = 3.1415926822 ...$

$\pi = 3.1415926535 ... $

Why are these numbers so close to each other? Is this just a coincidence?


About ten days ago, I saw this question on twitter, but no one has answered it.

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    $\begingroup$ It depends on how we know where the formula $\frac{7 \cosh(\sqrt 6)}{13}$ comes from, I think. $\endgroup$ Commented Aug 3, 2021 at 9:35
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    $\begingroup$ Anyway, an amazing good approximation. At first glance, it is a coincidence, but there might be an expansion (Taylor?) explaining it. $\endgroup$
    – Peter
    Commented Aug 3, 2021 at 9:42
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    $\begingroup$ I just want to make an observation at the high school algebra level. --- Using the well-known Archimedes's approximation for $\pi$, $$\pi≈\dfrac{22}{7}$$ We can write $$\begin{align} \frac{7 \cosh\left(\sqrt x\right)}{13}=\frac {22}{7}\\ \cosh \left(\sqrt x\right)=\frac{286}{49} \\ x≈6,002001...\end{align}$$ I wanted to leave it as a comment. I posted as an answer wrongly. $\endgroup$ Commented Aug 3, 2021 at 10:48
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    $\begingroup$ It gives 7 correct decimals for the cost of 4 digits of numeric input, two choices of function (cosh and square root), and two choices of arithmetic operation (multiplication and division). If we estimate crudely that there's at least as many options for choosing one nice named function as there are digits (and there are obviously 16 ways to choose two basic arithmetic operations), then it shouldn't surprise us that among the expressions of this size one is close to $\pi$. $\endgroup$ Commented Aug 3, 2021 at 10:48
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    $\begingroup$ I would consider this as a coincidence, but I also have some random observations that might possibly related to this topic: $$\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{\cosh x}=\pi, \qquad \frac{\pi}{2}=\prod_{n=1}^{\infty}\left(1-\frac{1}{4n^2}\right),\qquad\cosh(\sqrt{6})=\prod_{n=1}^{\infty}\left(1+\frac{24}{\pi^2(2n-1)^2}\right).$$ $\endgroup$ Commented Aug 3, 2021 at 13:06

4 Answers 4


People still want to have something. This is not really even about $e$, let alone $\cosh(x)=\frac{e^x+e^{-x}}{2}$.

When you unpack $\cosh(x)$ use $e^x=\sum\limits_{k=0}^{\infty}\frac{x^k}{k!}$ you get in the end that is stops being magical at about $M=8$ or $9$


Say we take 8:


after that we start getting slowly away from best possible accuracy (and that is part of the magic, we can add more elements so it would look like $\cosh(\sqrt{6})$). This is giving:

$$\frac{-654104729 + 208208000 \pi}{112112000} \approx -5 \cdot 10^8$$

The part that is actually affecting the precision is $112112000$ while

$$-654104729 + 208208000 \pi \approx -5.78$$

The only surprising element is the fractional structure of:


What you are asking then is why some $\frac{p}{q}\pi$ for small $p$ and $q$ may have a nice series expression that resembles or starts as known functions in form of Taylor polinomial to a high accuracy at some particular integer value.

The only real explanation is that there are way too many such functions to choose from.

But yes your case is surprising since we had to reach $M=8$ before getting close enough and then on, the accuracy of the sum would not change the accuracy of the surprise effect.

Based on this I can create a program that will give anything like that. Notice just how little is $\pi$ involved and what poor accuracy we actually need.



$$\pi = \frac{p}{q}$$

with quite wide and surprise accuracy, i.e. desired one to create a magical effect. There has to be some $a$ so that you can reduce fraction to very small values of $r$ and $s$.


and you have to be able to use some nice Taylor at some integer $t$ for known function and sufficiently large $M$ so that the missing accuracy does not affect our result:


Because the accuracy of $\pi$ is almost of no importance and you can do the same procedure over whatever other constant you want, I call this pure magic. Not even an accident. Just a nice trick.

Explanation of the trick: once you get a very high accuracy for rational approximation of $\pi$ you have a large number of $p$ and $q$ to choose from and reduce your accuracy to a level of surprise. Say you want $10$ digit accuracy. Well take $15$ and then you will have a couple of million values for the pair of $p$ and $q$ to choose from without violating intended $10$ digit accuracy. And then you are free to choose the Taylor expansion over, again, couple of different integer values, anything bellow $20$ you can call a surprise.


$$\pi\simeq 3.141592653589793$$ $$\frac{7}{13}\cosh(\sqrt{6})\simeq 3.141592682218231$$ 8 digits coincide. This is not bad but not sufficient. Clearly both are not equal as already pointed out in comments.

So this is a coincidence. For example, see a lot of formulas approximating $\pi$ with various accuracies in : http://www.contestcen.com/pi.htm

Don't be surprised. Such coincidences are frequent, even with much higher precision. One can easily find numerical coincidences thanks for some software built for this use. A paper for the general public on this subject : https://fr.scribd.com/doc/14161596/Mathematiques-experimentales (in French).

For these kind of search one can also use the ISC : http://wayback.cecm.sfu.ca/projects/ISC/ISCmain.html .

IN ADDITION after Alex Peter's answer.

Citation from https://fr.scribd.com/doc/14161596/Mathematiques-experimentales (Page 2. Rough English translation) :

It is interesting to think about the approach of the researcher confronted with a question which he does not know if the answer will be possible with the actual mathematical knowledge. His exploratory work can make him observe a coincidence or a relationship which is only a conjecture at this stage. The necessary demonstration can be difficult and time-consuming. He can't even know if the game is worth the candle since the conjecture could be wrong. A correct numerical result reinforces the expectation that the conjecture is valid especially as the precision of the calculation is high. It is not a proof but it will avoid wasting efforts on a grossly erroneous conjecture, which is no small thing.

A well known historical example is the infinite series of sum 1/n^2. By an approximate calculation Euler noticed a good proximity with pi^2 / 6 , which reinforced him in the research and finally the discovery of the demonstration of the equality with zeta(2), particular value of the zeta function according to the current writing.

In the same way, let us evoke C.F.Gauss who observed that the number of prime numbers less than n is approximately n/ln(n), which will be confirmed only much later.

End of citation.

The question raised by TOM is an interesting example. In this case it was easy to see that the conjecture is wrong thanks to numerical calculus only. Thus no need to waste time in trying to prove the supposed formula.

  • $\begingroup$ @Ione student. A mathematical coincidence is said to occur when two expressions with no direct relationship show a near-equality which has no apparent theoretical explanation. (From en.wikipedia.org/wiki/Mathematical_coincidence ). $\frac{7 \cosh(\sqrt 6)}{13} $ and $\pi$ are two expressions with no DIRECT relationship since it is prouved that they are not equal. They show a near-equality which has no theoretical explanation. So it is a coincidence. If you don't agree with the Wikipedia definition of coincidence it's up to you to correct the wikipedia article. $\endgroup$
    – JJacquelin
    Commented Aug 7, 2021 at 13:32
  • $\begingroup$ @Ione student. Another definition : A coincidence is a surprising concurrence of events, perceived as meaningfully related, with no apparent causal connection (Diaconis and Mosteller 1989). Which one do you prefer ? $\endgroup$
    – JJacquelin
    Commented Aug 7, 2021 at 13:41
  • $\begingroup$ I'm just a little skeptical. We can never know for sure whether this expression is the result of a mathematical formula involving an infinite series. $\endgroup$ Commented Aug 7, 2021 at 14:14
  • $\begingroup$ I don't understand your comment. What do you mean "this expression" ? $\endgroup$
    – JJacquelin
    Commented Aug 7, 2021 at 14:21
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    $\begingroup$ As an example $$\left|\cos \left(\frac{\pi }{12}\right) \Gamma \left(\frac{5}{12}\right) \Gamma \left(\frac{7}{12}\right)-\pi\right|=4.44 \times 10^{-16}$$ $\endgroup$ Commented Aug 9, 2021 at 8:29

Unfortunately, unless it has some pattern, there is no way to conclude why. Suppose that you find some connection that is short and sweet, but unless you can implement it in at least one or two another cases there is no mathematical conclusion that could be helpful apart from that particular case.

If you use Inverse symbolic calculator you can find at least $10$ expressions for the precision that you require. As you can see if you try, these algorithms require $10$ or more digits in order to even start making some assumptions, which is to say that our current mathematical notation is rich enough that you can easily find hundreds or thousands of connections with a small number of digits.

$3.14159$ is giving $92$ simple expressions, $3.141592$ is giving $23$, $3.1415926$ is giving $15$, $3.14159265$ is giving $11$. Some of these with higher accuracy are indeed $\pi$, but not all for smaller accuracy.

However, notice that this is a direct connection, without fractions. Imagine that you take that adding a small fraction, say with denominator and numerator up to about $10$, for example one we allow to be $1$ to $5$ and another $1$ to $20$ could still be called surprising. That is $200$ variants. If our inverse symbolic algorithms are able to provide just $5$ connections, that would be $1000$ simple connections for whatever small constant you like in your form.

Your number of decimals is somewhere about having way too many connections in similar form to decide if this is accidental or coming from something substantial.

And even if there is some shorter and sweeter compression of this connection, unless it is more generalized, it would remain just a curiosity.

If we all provide such a compressed explanation, you would not be able to tell which one is more right. So it would be more of a game than anything to do with math itself.

  • $\begingroup$ I don't agree with your conclusion : "So it would be more of a game than anything to do with math itself." On the contrary I think this is a very usefull field of research. This would need a comment too long to be typed here. That is why I edited my comment in addition to my main answer. Apart this, I agree with a large part of what you wrote. $\endgroup$
    – JJacquelin
    Commented Aug 7, 2021 at 11:04
  • $\begingroup$ @JJacquelin No. In this form it is bad one. Very bad one There are researches around this in a serious form that lead to very serious consequences in math. This, one equation out of nowhere is a serious waste of time and it leads nowhere in 99% of the cases. If you say ok, now I have this for 3,4,6,9,12,14... I would say ok, but a function crossing rational lattice at one point is of no interest whatsoever. It is of no use to remind me that there are researches, THIS particular question is similar to billion others. And I explained why. $\endgroup$
    – Alex Peter
    Commented Aug 7, 2021 at 11:05
  • $\begingroup$ @JJacquelin if you look at the equation $7$ and $13$ and $6$ and $2$ (square root) are totally arbitrary. If you take say $3 \cdot 10 \cdot 10 \cdot 5$ combinations for these that would look "surprising" that would make $1500$ possible combinations. That is a lot. Quite some number of functions would fit this pattern. Just by trying out couple I have found other funny examples of this sort. And I can assure you that you can construct them. It is just magical, nothing else, in this case. $\endgroup$
    – Alex Peter
    Commented Aug 7, 2021 at 11:14
  • $\begingroup$ Well I understand your point. But I don't agree anyways. Even if one equation out of nowhere is a serious waste of time and it leads nowhere in 99% of the cases (You are right) , If 1% leads to something of interest and even less leads to a discovery, I say this is not a game only. $\endgroup$
    – JJacquelin
    Commented Aug 7, 2021 at 11:17
  • $\begingroup$ @JJacquelin Do not waste your time on this one :) I did. $\endgroup$
    – Alex Peter
    Commented Aug 7, 2021 at 11:28

My answer is $\color{brown}{\mathbf{coincidence}}.$

Just now, using the approach kitten on the keys, I have got $$e\,\sqrt[\large21]{21}\approx3.142,$$ $$e\,\sqrt[\large29]{\dfrac{133}2}\approx3.1415964,$$ $$e\sqrt[\large40]{\dfrac{1307}4}\approx3.1415926472,$$ when $$\pi=3.141592653589793\dots$$

I hope I've stopped in time :)

  • $\begingroup$ @lonestudent I've tried too... $\endgroup$ Commented Aug 8, 2021 at 18:14

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