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.

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

$$512 \approx e^{2\pi\sqrt1}-24$$ $$512(1+\sqrt2)^3 \approx e^{2\pi\sqrt2}-24$$ and so on.

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 $\pi \approx e^{\pi}-20$?

Update: For example, we have the rather curious,

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

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

With $q = e^{-\pi}$,

$$\pi \approx \beta_1 = e^\pi -20 +\tfrac{1}{48}q-\dots\; (\text{differ:}\; {-4}\times 10^{-22})\\ \pi \approx \beta_2 = e^\pi -20 +\tfrac{1}{48}q-\dots\; (\text{differ:}\; {-3}\times 10^{-22})$$

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 Jan 4 '16 at 4:08
  • $\begingroup$ $e^\pi\approx\frac{5^4}{3^3}$ by Andrew Fraker (2014) en.wikipedia.org/wiki/… $\endgroup$ – Jaume Oliver Lafont Jan 8 '16 at 17:54
  • $\begingroup$ Because coincidences happen. $\endgroup$ – Qudit Oct 28 '17 at 7:23
  • $\begingroup$ @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. $\endgroup$ – Tito Piezas III Oct 28 '17 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 Oct 28 '17 at 8:11

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. :( $\endgroup$ – Tito Piezas III Jan 4 '16 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. $\endgroup$ – Jaume Oliver Lafont Jan 4 '16 at 7:03
  • $\begingroup$ The parametric form is the one by Ferguson, Adamchik, and Wagon and is eq $(55)$ here. $\endgroup$ – Tito Piezas III Jan 4 '16 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 $\endgroup$ – Jaume Oliver Lafont Jan 4 '16 at 7:39

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