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Well I find it as a coincidence but how to explain :

$$\frac{1}{1+\frac{a}{1+\frac{a^{2}}{1+\frac{a^{3}}{1+\frac{\cdot\cdot\cdot}{a^{2n}}}}}}-\frac{1}{1+\frac{b}{1+\frac{b^{2}}{1+\frac{b^{3}}{1+\frac{\cdot\cdot\cdot}{b^{2n}}}}}}\simeq 7.33*10^{-5},a=\sqrt{5},b=\sqrt{\pi/4}$$

The difference is very small so : Is there any good reason to explain it ?

My analysis :

From Rogers-Ramanujan continued fraction we have :

Let :

$$\left(a;q\right)_{n}=\prod_{j=0}^{n-1}\left(1-aq^{j}\right)$$ then we have :

$$\frac{\sum_{n=0}^{\infty}\frac{q^{n^{2}}}{\left(q;q\right)_{n}}t^{n}}{\sum_{n=0}^{\infty}\frac{q^{\left(n\left(n-1\right)\right)}}{\left(q;q\right)_{n}}t^{n}}=\frac{1}{1+\frac{t}{1+\frac{tq}{1+\frac{tq^{2}}{1+\frac{tq^{3}}{1+\cdot\cdot\cdot}}}}}$$

Here : $t=q=\sqrt{5}$

Now we can express it as a power series see https://www.sciencedirect.com/science/article/pii/S0723086922000792

Using a very simple algorithm .

Another explanation could be from :

$$R\left(q\right)=\frac{\left(\sqrt{5}-1\right)}{2}\exp\left(-\frac{1}{5}\int_{q}^{1}\frac{\left(1-t\right)^{5}\left(1-t^{2}\right)^{5}\left(1-t^{3}\right)^{5}\cdot\cdot\cdot}{\left(1-t^{5}\right)\left(1-t^{10}\right)\left(1-t^{15}\right)\cdot\cdot\cdot}\frac{dt}{t}\right)=\frac{q^{\frac{1}{5}}}{1+\frac{q}{1+\frac{q^{2}}{1+\cdot\cdot\cdot}}}$$

See for that the Lost's book of Ramanujan.

See also this kind of question A series related to $\pi\approx 2\sqrt{1+\sqrt{2}}$

Remark :

It might be interesting to think at the Gamma's function we have :

$$\left(\frac{1}{\sqrt{5}}\right)!\simeq \sqrt{\frac{\pi}{4}}$$

And again using the work due to Ramanujan $x>0$:

$$\frac{x!}{\sqrt{\pi}\left(\frac{x}{e}\right)^{x}\left(8x^{3}+4x^{2}+x+\frac{1}{30}\right)^{\frac{1}{6}}}\simeq \left(\prod_{n=3}^{\infty}\frac{\left(1-x^{\left(5n-4\right)}\right)\left(1-x^{\left(5n-1\right)}\right)}{\left(1-x^{\left(5n-2\right)}\right)\left(1-x^{\left(5n-3\right)}\right)}\right)^{\frac{1}{144}}$$

We have also :

$$(5^{1/2})!/5^{1/2}-2/\sqrt{\pi}\simeq 0$$

Barackouda
  • 3,879
  • Is there an algorithm ? – Barackouda Mar 17 '23 at 17:19
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    Please show your analysis. – user2661923 Mar 17 '23 at 17:35
  • @user2661923 For the story see https://math.stackexchange.com/questions/4660405/continued-fraction-inequality-with-well-know-constant-as-pi-and-golden-ratio .My analysis is there is a kind of algorithm like Chudnovsky's algorithm with different parameters – Barackouda Mar 17 '23 at 17:41
  • Check your numbers : the difference is $7.33\times 10^{-5}$ – Claude Leibovici Mar 18 '23 at 03:21
  • @ClaudeLeibovici Thanks . – Barackouda Mar 18 '23 at 08:37
  • Do you apply the same number of iterations to each fraction? If so, maybe the only reason would rounding errors but this is strange to me indeed. – NoChance Mar 18 '23 at 09:10
  • For what it's worth, $2/\sqrt{5} - \sqrt{\pi/4} = 0.00820026554$. Their ratio is $1.00925300881$. – Ethan Bolker Mar 19 '23 at 13:34
  • @EthanBolker try $$\frac{\left(\sqrt{5}\right)!}{\sqrt{5}}-\frac{2}{\sqrt{\pi}}+\frac{\frac{1}{6^{3}}4}{\pi}-\frac{\frac{1}{6^{4}}4^{2}}{\pi^{2}}-\frac{\frac{1}{6^{5}}4^{3}}{\pi^{3}}-\frac{\frac{1}{6^{6}}4^{4}}{\pi^{4}}-\frac{\frac{1}{6^{7}}4^{5}}{\pi^{5}}+\frac{\frac{1}{6^{8}}4^{6}}{\pi^{5}}-\frac{\frac{1}{6^{9}}4^{7}}{\pi^{6}}+\frac{\frac{1}{6^{10}}4^{8}}{\pi^{7}}+\frac{\frac{1}{6^{11}}4^{9}}{\pi^{7}}+\frac{\frac{1}{6^{12}}4^{10}}{\pi^{7}}-\frac{\frac{1}{6^{12}}4^{11}}{\pi^{8}}\simeq 0$$ – Barackouda Mar 19 '23 at 13:59

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