# Golden Ratio and a Ramanujan-Type Integral

## Abstract

**:**

## 1. Introduction

## 2. Proof of Equation (3)

- Step I: Rewriting the sum side of Equation (7)Our goal is to show that the left-hand side of Equation (7) is the same as$$\sum _{n=-\infty}^{\infty}\frac{x{q}^{n}}{{(1-x{q}^{n})}^{2}}-\frac{z{q}^{n}}{{(1-z{q}^{n})}^{2}}$$Indeed, let us consider the sum involving x in Equation (8). We break it up according to $n=0$, $n>0$ and $n<0$:$$\begin{array}{ccc}\hfill \sum _{n=-\infty}^{\infty}\frac{x{q}^{n}}{{(1-x{q}^{n})}^{2}}& =& \frac{x}{{(1-x)}^{2}}+\sum _{n=1}^{\infty}\frac{x{q}^{n}}{{(1-x{q}^{n})}^{2}}+\frac{x{q}^{-n}}{{(1-x{q}^{-n})}^{2}}\hfill \\ & =& \frac{x}{{(1-x)}^{2}}+\sum _{n=1}^{\infty}\frac{x{q}^{n}}{{(1-x{q}^{n})}^{2}}+\frac{{x}^{-1}{q}^{n}}{{(1-{x}^{-1}{q}^{n})}^{2}}\hfill \\ & =& \frac{x}{{(1-x)}^{2}}+\sum _{j,n=1}^{\infty}j{x}^{j}{q}^{jn}+j{x}^{-j}{q}^{jn}\hfill \\ & =& \frac{x}{{(1-x)}^{2}}+\sum _{j=1}^{\infty}\frac{j{x}^{j}{q}^{j}}{1-{q}^{j}}+\frac{j{x}^{-j}{q}^{j}}{1-{q}^{j}}\hfill \end{array}$$
- Step II: Identifying the poles of the infinite product in Equation (7)Let us denote by $F(x,z)$ the right-hand side of Equation (7). First we treat $F(x,z)$ as a function of x. At the same time, we will treat z as a parameter that is not an integral power of q. We claim that $F(x,z)$ has poles of order two at$$x={q}^{n}$$Indeed, the denominator ${(1-x)}^{2}$ in $F(x,z)$ implies $x=1$ is a pole of order 2. Similarly, the denominator ${(1-{x}^{\pm 1}{q}^{j})}^{2}$ implies $x={q}^{\pm j}$ is a pole of order 2. This proves our claim.Next we want to find the partial fraction expansion of $F(x,z)$. To this end, we need to determine the symmetries of $F(x,z)$.
- Step III: Exploring the symmetries of $F(x,z)$Readers can easily verify the following:$$\begin{array}{ccc}\hfill F(x,z)& =& -F(z,x)\hfill \end{array}$$$$\begin{array}{ccc}\hfill F(x,z)& =& F(qx,z)\hfill \end{array}$$$$\begin{array}{ccc}& =& F({q}^{-1}x,z)\hfill \end{array}$$
- Step IV: Finding the partial fraction expansion of $F(x,z)$Our goal is to prove that$$F(x,z)=\sum _{n\in \mathbb{Z}}\frac{x{q}^{n}}{{(1-x{q}^{n})}^{2}}+H(x,z):=G\left(x\right)+H(x,z)$$Here $H(x,z)$ is a “remainder” term that has a Laurent expansion in x. Note that, by comparing with Step I above, Equation (12) implies that we have “half” of Equation (7).To prove Equation (12), we start with the observation that Step II implies$$F(x,z)=\frac{{a}_{0}\left(x\right)}{{(1-x)}^{2}}+\sum _{n=1}^{\infty}\frac{{a}_{n}\left(x\right)}{{(1-x{q}^{n})}^{2}}+\frac{{a}_{-n}\left(x\right)}{{(1-{x}^{-1}{q}^{n})}^{2}}+H(x,z)$$First we show that ${a}_{0}\left(x\right)=x$. The part of $F(x,z)$ that contributes to ${a}_{0}\left(x\right)$ comes solely from the overall prefactor (note that the infinite product becomes 1 as $x\to 1$). Regarding this prefactor, we note that$$\frac{(x-z)(1-xz)}{{(1-x)}^{2}{(1-z)}^{2}}=\frac{x}{{(1-x)}^{2}}-\frac{z}{{(1-z)}^{2}}$$This implies the principal part at $x=1$ is$$\frac{x}{{(1-x)}^{2}}$$For $n>0$, we have$${a}_{\pm n}\left(x\right)={x}^{\pm 1}{q}^{n}$$
- Step V: Determining $H(x,z)$This is the final step: let us show that$$H(x,z)=-G\left(z\right)$$We recall that both H and G were defined in Equation (12). Previously, $H(x,z)$ represented what cannot be determined by understanding the pole structure of $F(x,z)$. What Equation (15) says is that, $H(x,z)$ can indeed be written as something known (i.e., G)—but there is a catch: the argument of G on the right-hand side of Equation (15) is z, not x. In fact, the same equation tells us the $H(x,z)$ is independent of x. Let us turn to the proof.Since $H(x,z)$ is a Laurent expansion in x (cf. the sentence right after Equation (12)), we can write it as$$H(x,z)=\sum _{n\in \mathbb{Z}}{b}_{n}\left(z\right){x}^{n}$$First we note that $G\left(x\right)$ satisfies$$G\left(x\right)=G\left(xq\right)$$$$H(x,z)=H(xq,z)$$Indeed,$$\begin{array}{ccc}\hfill H(x,z)& =& F(x,z)-G\left(x\right)\hfill \\ & =& F(xq,z)-G\left(xq\right)\hfill \\ & =& H(xq,z)\hfill \end{array}$$Equation (17) implies that only ${b}_{0}\left(z\right)$ survives in the expansion in Equation (16):$$H(x,z)={b}_{0}\left(z\right)$$This, with Equation (12), implies$$F(x,z)=G\left(x\right)+{b}_{0}\left(z\right)$$By Equations (9) and (18), we have$$G\left(x\right)+{b}_{0}\left(z\right)=-G\left(z\right)-{b}_{0}\left(x\right)$$$$G\left(x\right)+{b}_{0}\left(x\right)=-(G\left(z\right)+{b}_{0}\left(z\right))$$$${b}_{0}\left(z\right)=\lambda -G\left(z\right)$$$$F(x,z)=G\left(x\right)-G\left(z\right)+\lambda $$$$0=F(x,z)+F(z,z)=2\lambda $$$$F(x,z)=G\left(x\right)-G\left(z\right)$$$$BIGSTEP2:IntegratingEquation\text{Equation}\left(4\right))$$

I had never seen anything in the least like them before. A single look at them is enough to show that they could only be written down by a mathematician of the highest class. They must be true because, if they were not true, no one would have had the imagination to invent them.

## 3. Final Remarks

## Acknowledgements

## References

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Chan, H.-C.
Golden Ratio and a Ramanujan-Type Integral. *Axioms* **2013**, *2*, 58-66.
https://doi.org/10.3390/axioms2010058

**AMA Style**

Chan H-C.
Golden Ratio and a Ramanujan-Type Integral. *Axioms*. 2013; 2(1):58-66.
https://doi.org/10.3390/axioms2010058

**Chicago/Turabian Style**

Chan, Hei-Chi.
2013. "Golden Ratio and a Ramanujan-Type Integral" *Axioms* 2, no. 1: 58-66.
https://doi.org/10.3390/axioms2010058