# The most complex formula for the golden ratio $\varphi$ that I have ever seen. How was it achieved?

I am fascinated by the following formula for the golden ratio $\varphi$: $$\Large\varphi = \frac{\sqrt{5}}{1 + \left(5^{3/4}\left(\frac{\sqrt{5} - 1}{2}\right)^{5/2} - 1\right)^{1/5}} - \frac{1}{e^{2\pi\,/\sqrt{5}}}\,\mathop{\LARGE \mathrm{K}}_{n=0}^{\infty}e^{-2\pi n\sqrt{5}}$$ such that $$\Large{\mathop{\LARGE\mathrm K}_{n=0}^{\infty}e^{-2\pi n\sqrt{5}} = \cfrac{1}{1 + \frac{e^{-2\pi\sqrt{5}}}{1 + \frac{e^{-4\pi\sqrt{5}}}{1 + \cdots}}}}.$$ This formula is just a rearrangement of a formula giving the value of the continued fraction, and was to no surprise created by Srinivasa Ramanujan.

My question is, just how on Earth did he create something like this? Is there some explanation? Does anybody know?? I did some research and he had three other very similar formulae, where in each of them, he showed the values of $$\large\mathop{\LARGE \mathrm K}_{n=0}^\infty e^{-2\pi n}\quad\text{and}\quad e^{-2\pi/5}\mathop{\LARGE \mathrm K}_{n=0}^\infty e^{-2\pi n}\quad\text{and}\quad e^{-\pi/5}\mathop{\LARGE\mathrm K}_{n=0}^\infty e^{-\pi n}$$ and so many other summations regarding Gelfond's constant $e^\pi$. I apologise if, in the event you know how he created his formula, you might be sitting at your desk for hours writing a long long answer with the workings out.

Thank you in advance.

My motive for finding things like these originated from here.

• The closed formulas for that continued fraction is naturally obtained by solving certain q-difference equations which are analogues of classical differential equations. This is explained in George Andrews' book on the theory of partitions. – Mariano Suárez-Álvarez Feb 18 '18 at 3:15
• Huh? ${}{}{}{}$ – Mariano Suárez-Álvarez Feb 18 '18 at 5:01
• Well, there is this new thing called google. I am sure it would be a great help in finding information about both! – Mariano Suárez-Álvarez Feb 18 '18 at 5:23
• See this blog post for the proof by Ramanujan. Ramanujan was systematically investigating the properties of Rogers Ramanujan continued fraction $R(q)$ and he found that it was also a modular function which allowed its values to be computed as algebraic numbers for $q=\pm e^{-\pi\sqrt{n}}$ where $n$ is a positive rational number. – Paramanand Singh Feb 24 '18 at 8:02
• Well you will perhaps need to study the theories of theta functions as developed by Ramanujan to get a complete picture. You may search for Ramanujan in my archives page paramanands.blogspot.com/p/archives.html and read the posts at your leisure. Also note that this is a difficult topic and may require reasonable amount of time. – Paramanand Singh Feb 24 '18 at 8:09

## 1 Answer

Ramanujan's ability to prove theorem's and define new functions was remarkable. In his very short life (1887-1920), he manage to rediscover all elliptic and theta functions theory that take centuries to developed. His work is in research until nowdays. He did not leave us writing proof's, because he has no time to do this. For example in [BerIII] one can see that firstly he defined the basic ''null'' theta functions and then construct from them a huge number of theorem's, from the very simple ones, to modular equations of very high degree say 255. This may seen as: If $$$$K(x)=\int^{\pi/2}_{0}\frac{d\theta}{\sqrt{1-x^2\sin^2(\theta)}}=\frac{\pi}{2}{}_2F_1\left(\frac{1}{2},\frac{1}{2};1;x^2\right),$$$$ is the complete elliptic integral of the first kind. The elliptic singular modulus $$k_r$$ is defined from the equation (I use the traditional notation) $$\frac{K\left(\sqrt{1-x^2}\right)}{K\left(x\right)}=\sqrt{r}\Leftrightarrow x=k_r\in(0,1)$$ and $$z_n:={}_2F_1\left(\frac{1}{2},\frac{1}{2};1;k^2_{n^2r}\right)=K(k_{n^2r})=K[n^2r]\textrm{, }n=1,2,3,\ldots,$$ then what is $$m_n=m_n(r)=\frac{z_n}{z_1}=\frac{K[n^2 r]}{K[r]}=?$$
(In general if $$r$$ is a positive rational, then both $$k_r$$and $$m_n(r)$$ are algebraic numbers).

In his notes, Hardy commented that Ramanujan had produced groundbreaking new theorems, including some that "defeated me completely; I had never seen anything in the least like them before".

Returning to Berndt, in his book says (litle rephrased by me): ''We don't have proof's. Our proof's are perhaps more aptly may called verifications''. He then uses computer program MACSYMA to verify "prove" these results, using not the theory of tranformation of theta functions and $$q-$$series, but rather results from the theory of modular forms developed much later. Berndt uses as general references books of Rankin in 1977 and Pettersson 1974 (see [BerIII] pg. 326 and its related references).

He also had no book's to read. He did not pass in the exams in the university (because he was bad at English). The Rogers-Ramanujan continued fraction and the specific evaluation you mention, was only a very-very small touch οn a very-very large board. While many great mathematicians of the 20th century were enjoying reading papers of Euler, Gauss, Dirichlet, Jacobi, Eisenstein, Weber, Kronecker, Hurwitz, Riemann and others, Ramanujan open this way alone and passed them.

About the Rogers-Ramanujan Continued Fraction

Ramanujan have first discovered that:

For $$|q|<1$$ and $$|z|<|q|^{-1}$$ a more general version of Rogers Ramanujan continued fraction is $$$$R(z;q):=\frac{q^{1/5}}{1+}\frac{qz}{1+}\frac{q^2z}{1+}\frac{q^3z}{1+}\ldots$$$$ For $$|q|<1$$ we also define $$$$(a;q)_n:=\prod^{n-1}_{k=0}{(1-aq^k)}\textrm{, } (a;q)_0:=1$$$$ and $$$$(q)_n:=(q;q)_n\textrm{, }(q)_0:=1.$$$$ Set also $$f(-q):=(q;q)_{\infty}$$

Theorem 1.(See [Berndt] chaper 16, Entry 15) If $$|q|<1$$ and $$$$H(z;q):=\sum^{\infty}_{n=0}\frac{q^{n(n+1)}z^n}{(q)_n}\textrm{, } G(z;q):=\sum^{\infty}_{n=0}\frac{q^{n^2}z^n}{(q)_n},$$$$ then $$$$R(z;q)=q^{1/5}\frac{H(z;q)}{G(z;q)}.$$$$

Having in mind the above he proceeded and find

Theorem 2. (The Rogers-Ramanujan identities) (see [Andrews]: Sec. 14-3,14-4) If $$|q|<1$$, then $$$$H(1;q)=f(-q)^{-1}\sum^{\infty}_{n=-\infty}(-1)^nq^{5n^2/2+3n/2}=\prod^{\infty}_{n=0}\frac{1}{(1-q^{5n+2})(1-q^{5n+3})}$$$$ and $$$$G(1;q)=f(-q)^{-1}\sum^{\infty}_{n=-\infty}(-1)^nq^{5n^2/2+n/2}=\prod^{\infty}_{n=0}\frac{1}{(1-q^{5n+1})(1-q^{5n+4})}.$$$$

Theorem 3. For all $$|q|<1$$, we have (we denote $$R(q):=R(1;q))$$: $$$$\frac{1}{R(q)}-1-R(q)=\frac{f(-q^{1/5})}{q^{1/5}f(-q^5)}$$$$ and $$$$\frac{1}{R(q)^5}-11-R(q)^5=\frac{f^6(-q)}{q f^6(-q^5)}.$$$$

Theorem 4. (see [BerIII] Chapter 16, Entry 39). If $$a,b>0$$ and $$ab=\pi^2$$, then $$$$\left\{\frac{\sqrt{5}+1}{2}+R(e^{-2a})\right\}\cdot\left\{\frac{\sqrt{5}+1}{2}+R(e^{-2b})\right\}=\frac{5+\sqrt{5}}{2}.$$$$

Remarks. From Jacobi's theory of elliptic functions [W,W] pg. 488 we have the next exersize $$f(-q^2)^6=2\pi^{-3}q^{-1/2}k_rk'_rK^3\textrm{, this was known to Ramanujan,}$$ where $$q=e^{-\pi\sqrt{r}}$$, $$r>0$$, $$k'_r:=\sqrt{1-k_r^2}$$.

Also the lower modular equations and its multipliers are

$$K=K(k_r)=z_1\textrm{, }m_2=\frac{z_2}{z_1}=\frac{1+k'_r}{2}\textrm{, }k_1=\frac{1}{\sqrt{2}},$$ $$k_{4r}=\frac{1-k'_r}{1+k'_r}\textrm{, the modular equation of degree 2.}$$ (Multipliers of degree 3 and 5): $$$$27m^4_3-18m^2_3-8(1-2k^2_r)m_3-1=0$$$$ $$$$(5m_5-1)^5(1-m_5)=256k^2_r(1-k^2_r)m_5$$$$ As an example Ramanujan have calculated (see [Vil]) $$k_{210}^2=(4-\sqrt{15})^4(8-3\sqrt{7})^2(2-\sqrt{3})^2(6-\sqrt{35})^2(\sqrt{10}-3)^4 (\sqrt{7}-\sqrt{6})^4\times$$ $$\times(\sqrt{15}-\sqrt{14})^2(\sqrt{2}-1)^2.$$

REFERENCES

[Andrews]: G.E. Andrews, ''Number Theory''. Dover Publications, New York. 1994.

[BerIII]: B.C.Berndt, ''Ramanujan`s Notebooks Part III''. Springer Verlag, New York (1991)

[Vil]: Mark B. Villarino. ''Ramanujan's Most Singular Modulus''. arXiv:math/0308028v4 [math.HO] 17 Jul 2005.

[W,W]: E.T.Whittaker and G.N.Watson, ''A course on Modern Analysis''. Cambridge U.P. (1927)

• +1 you have presented a very nice survey. – Paramanand Singh May 23 at 9:45
• Also nice paper by Villarino. It is an interesting read. – Paramanand Singh May 23 at 9:50
• @ParamanandSingh. Thank you very much. – Nikos Bagis May 23 at 13:19