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Recently, I read an article about “almost-identities”. It said, that for every “almost-identity” we have to decide whether it is a coincidence or not. By myself, I discovered that $$ \sum_{k=0}^\infty \frac{\pi^{\frac k2}}{\Gamma{\left(\frac k2+1\right)}}=e^{\pi}\left(1+\text{erf}\left(\sqrt\pi\right)\right)\approx45.9993260894... $$ which is surprisingly close to $46$. So my question is: is this a mere coincidence or can it be “proven” in some sense?

Caleb Stanford
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Redundant Aunt
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    Looks like a coincidence for me. – Peter Dec 25 '14 at 20:31
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    @Peter Like $e^\pi-\pi=19.9990999...$ – Edward Jiang Dec 25 '14 at 20:35
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    Take a look here: 1 and 2 – alexjo Dec 25 '14 at 20:36
  • Could you perhaps provide some examples of why we might deem some near-identies coincidences and others not? As far as I'm concerned, there are no coincidences. – David H Dec 25 '14 at 20:48
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    By discovered you mean you went looking for such almost-identities? Without addressing the question of coincidence we know that the probabillity of finding such almost-identities is very high if you just spend some decent amount of time looking for them (so you should not be surprised finding one). btw here is my favoritte 'coincidence': $\frac{22\pi^4}{2143} = 1.000000001\ldots$ – Winther Dec 25 '14 at 21:29
  • @DavidH "As far as I'm concerned, there are no coincidences." Sorry but I do not know what to do with this statement. At face value it seems obviously false hence you probably mean it in a special sense. Care to elaborate? – Did Dec 30 '14 at 09:03
  • @Did What I meant is that given the axioms of the real numbers plus the relevant definitions of $\pi$, the gamma function, infinite series, etc., there's no other decimal approximation (to five significant figures) the series above could have other than $45.999$. But more importantly, no one has supplied me with a mathematical definition of a coincidence yet, so I don't know how to check if a near-identity has this property or not. – David H Dec 30 '14 at 09:44
  • @EdwardJiang A closed form for $e^\pi-\pi-20$ is given in http://math.stackexchange.com/questions/1598664/why-e-pi-pi-approx-20-and-e2-pi-24-approx-29/1599014#1599014 – Jaume Oliver Lafont Jan 11 '16 at 05:37
  • Curiously: $\exp(\pi)(\text{erf}\sqrt{\pi})+\pi=26.00022611019337959900761019\ldots$ – Anthony Apr 24 '20 at 03:37
  • @AntonioHernandezMaquivar I think this comes from combining the almost-identity in question with the almost identity $e^\pi-\pi\approx 20$, but curious indeed :) – Redundant Aunt Apr 24 '20 at 12:06

2 Answers2

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If you consider the level of simplicity of the expression, you will find that there are at least 10,000 equally simple expressions that you might equally well have checked. For example, instead of $e^\pi$ you might have taken $2^\pi$ or $3^e$ or $\sqrt{2}^{\sqrt{3}}$ or .... If we allow he all combinations using 2 symbols among the numbers from 1 to 10, the constants $e$, $\pi$, and $\gamma$ possibly multiplying those numbers, and square roots (after all, you felt comfortable with a square root in the argument of erf), that one first term has about 50 possibilities.

Similarly, the $\text{erf}(\sqrt{\pi})$ expression could have used any of at least 20 functions about as familiar as erf, and maybe 20 arguments of comparable complexity to the one here.

In the end, the fact that this matches an integer to 3 decimal places is on that basis explainable as a coincidence.

There are other near-equalities that are, in fact, not coincidences. Some are encountered/explained when you study class numbers. But most of those equalities have an astonishing degree of closeness, not just 3 digits.

Mark Fischler
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  • Yes, with this explanation it seems pointless to ask such a question. I just found it astonishing to have an infinite sum involving rational powers of $\pi$ and the reciprocal gamma function and ending up with an almost-integer. It would have been great if there was a deep mathematical reason like with Ramanujans constant, but to think of this is probably way to optimistic. Anyway, thanks for your answer! – Redundant Aunt Dec 26 '14 at 11:12
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It is interesting to note that

$$\sum _{k=0}^{\infty }{\frac {{\pi }^{1/2\,k}}{\Gamma \left( 1/2\,k+a \right) }}={\frac {{{\rm e}^{\pi }} \left( \Gamma \left( a \right) - \Gamma \left( a-1,\pi \right) a+\Gamma \left( a-1,\pi \right) \right) }{\Gamma \left( a \right) {\pi }^{a-1}}}+{\frac {\sqrt {\pi } \left( 2\,{{\rm e}^{\pi }}\Gamma \left( 1/2+a \right) - \left( 2\,a-1 \right) {{\rm e}^{\pi }}\Gamma \left( -1/2+a,\pi \right) \right) }{2\Gamma \left( 1/2+a \right) {\pi }^{-1/2+a}}} $$

Then it is possible to solve the equation

$${\frac {{{\rm e}^{\pi }} \left( \Gamma \left( a \right) - \Gamma \left( a-1,\pi \right) a+\Gamma \left( a-1,\pi \right) \right) }{\Gamma \left( a \right) {\pi }^{a-1}}}+{\frac {\sqrt {\pi } \left( 2\,{{\rm e}^{\pi }}\Gamma \left( 1/2+a \right) - \left( 2\,a-1 \right) {{\rm e}^{\pi }}\Gamma \left( -1/2+a,\pi \right) \right) }{2\Gamma \left( 1/2+a \right) {\pi }^{-1/2+a}}} = 46$$

and the solution is

$$a=0.999987485786729129363813640894$$

which is very near to $1$. According with this the result of @user109899 it is not a coincidence.

Juan Ospina
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