This was inspired by this post. Let $q = e^{2\pi\,i\tau}$. Then,

$$\alpha(\tau) = \left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24} = \frac{1}{q} - 24 + 276q - 2048q^2 + 11202q^3 - 49152q^4+ \cdots\tag1$$

where $\eta(\tau)$ is the Dedekind eta function and coefficients are A007191.

For example, let $\tau =\sqrt{-1}$. Then $\alpha(\tau) = 512 = 2^9$ and the series $(1)$ "explains" why,

\begin{align} 512 &\approx e^{2\pi}-24\\[4pt] 512(1+\sqrt2)^3 &\approx e^{2\pi\sqrt2}-24 \end{align}

and so on for other $\tau$.

Q: Let $q = e^{-\pi}$. Can we find a relation, $$\pi = \frac{1}{q} - 20 +c_1 q + c_2 q^2 +c_3 q^3 +\cdots\tag2$$ where the $c_i$ are well-defined integers or rationals such that $(2)$ "explains" why $\pi \approx e^{\pi}-20$?

For example, we have the rather curious functions,

$$\beta_1(q) := \frac{1}{q} - 20 +\tfrac{1}{48}q - \tfrac{1}{300}q^3 -\tfrac{1}{972}q^5 +\tfrac{1}{2160}q^7+\tfrac{1}{\color{brown}{2841}}q^9-\tfrac{1}{\color{brown}{2369}}q^{11}-\cdots\tag3$$

$$\beta_2(q) := \frac{1}{q} - 20 +\tfrac{1}{48}q - \tfrac{1}{300}q^3 -\tfrac{1}{972}q^5 +\tfrac{1}{2160}q^7+\tfrac{1}{\color{brown}{2842}}q^9-\tfrac{1}{\color{brown}{2810}}q^{11}-\cdots\tag4$$

Let $q = e^{-\pi}$, then,

$$\beta_1(q) \approx \pi,\quad (\text{difference:}\; {-4}\times 10^{-22})\\ \beta_2(q) \approx \pi,\quad(\text{difference:}\; {-3}\times 10^{-22})$$

However, there seems to be an indefinite number of formulas, where the choice of a coefficient (say, $2841$ or $2842$) determines an ever-branching tree of formulas. But there might be a subset where the coefficients have a nice closed-form.

  • $\begingroup$ Ah, yes, the eternal question. :-$)$ $\endgroup$
    – Lucian
    Commented Jan 4, 2016 at 4:08
  • $\begingroup$ $e^\pi\approx\frac{5^4}{3^3}$ by Andrew Fraker (2014) en.wikipedia.org/wiki/… $\endgroup$ Commented Jan 8, 2016 at 17:54
  • $\begingroup$ Because coincidences happen. $\endgroup$
    – Qudit
    Commented Oct 28, 2017 at 7:23
  • 1
    $\begingroup$ @Qudit: No one is saying it is not coincidence. It could be or it could not be. But the well-known "coincidence" McKay noticed about $196883,196884$, and Monstrous Moonshine should give one pause. $\endgroup$ Commented Oct 28, 2017 at 7:33
  • $\begingroup$ @TitoPiezasIII If someone could give a proof that provides insight into why it is true, that would be interesting. However, these sorts of problems typically seem to be solved by expanding power series or similar arguments. These may be an entertaining exercise for some, but I don't see how they are any more enlightening than approximating the expression with my calculator. $\endgroup$
    – Qudit
    Commented Oct 28, 2017 at 8:11

1 Answer 1


This does not follow your proposal exactly but it is built on series with rational terms only.

From expansions

$$ e^\pi=\sum_{k=0}^\infty\frac{3\left(e^{3\pi}-\left(-1\right)^ke^{-3\pi}\right)\Gamma\left(\frac{k}{2}+3i\right)\Gamma\left(\frac{k}{2}-3i\right)}{2 \pi k!}=\sum_{k=0}^\infty a(k) $$ http://oeis.org/A166748


$ \pi=3+\sum_{k=1}^\infty\frac{1}{4·16^k}\left(-\frac{40}{8k+1}+\frac{56}{8k+2}+\frac{28}{8k+3}+\frac{48}{8k+4}+\frac{10}{8k+5}+\frac{10}{8k+6}-\frac{7}{8k+7}\right)=3+\sum_{k=1}^\infty b(k) $


the following representation is obtained

$$e^\pi-\pi-20 = \frac{1201757159}{10580215726080}+\sum_{k=16}^\infty a(k) -\sum_{k=2}^\infty b(k) \approx -0.00090002 \approx -\left(\frac{3}{100}\right)^2$$

Cancellation comes from the first three decimal digits: $$ \sum_{k=0}^{15} a(k) = \frac{991388824265291953}{42849873690624000}\approx 23.136(2) $$ $$ b(1) = \frac{49087}{360360} \approx .136(1) $$ [EDIT] Three digit cancellation may also be obtained by taking 14 terms from the series for $e^\pi$ and 3 terms from the simpler $$\pi-3=\sum_{k=1}^{\infty}\frac{3}{(1+k)(1+2k)(1+4k)}$$

(Lehmer, http://matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27121.pdf, pag 139-140)

[EDIT] Another expression with a "wrong digit" that leads to higher precision when corrected is given by $$ e - \gamma-\log\left(\frac{17}{2}\right) \approx 0.0010000000612416$$ $$ e \approx \gamma+\log\left(\frac{17}{2}\right) +\frac{1}{10^3} +6.12416·10^{-11}$$

Why is $e$ close to $H_8$, closer to $H_8\left(1+\frac{1}{80^2}\right)$ and even closer to $\gamma+\log\left(\frac{17}{2}\right) +\frac{1}{10^3}$?

[EDIT] $$\sum_{k=0}^\infty \frac{2400}{(4 k+9) (4 k+15)} = 100 \pi-\frac{58480}{231} \approx 60.99909$$

  • 1
    $\begingroup$ The large fraction $\displaystyle \frac{1201757159}{10580215726080}$ does not seem to be particularly inspiring. :( $\endgroup$ Commented Jan 4, 2016 at 3:33
  • $\begingroup$ The equation for $\pi$ comes from a parametric form that can be adjusted to match whatever initial fraction. There should be one that cancelled the fraction in $e^\pi-\pi-20$, maybe with larger coefficients in the $\pi$ expression. $\endgroup$ Commented Jan 4, 2016 at 7:03
  • $\begingroup$ The parametric form is the one by Ferguson, Adamchik, and Wagon and is eq $(55)$ here. $\endgroup$ Commented Jan 4, 2016 at 7:07
  • $\begingroup$ Yes. The idea to get 22/7 as the first term comes from page 8 by Adamchik and Wagon in cs.cmu.edu/~adamchik/articles/pi/pi.pdf $\endgroup$ Commented Jan 4, 2016 at 7:39

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