7

Consider $\varphi=\frac{1+\sqrt{5}}{2}$, the golden ratio. Bellow are series $(3)$ and $(6)$ that represent $\varphi$ $$ \begin{align*} \varphi &=\frac{1}{1}+\sum_{k=0}^{\infty}\cdots&(1)\\ \varphi &=\frac{2}{1}+\sum_{k=0}^{\infty}\cdots&(2)\\ \varphi &=\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{(2k)!}{(k+1)!k!2^{4k+3}}&(3)\\ \varphi &=\frac{5}{3}+\sum_{k=0}^{\infty}\cdots&(4)\\ \varphi &=\frac{8}{5}+\sum_{k=0}^{\infty}\cdots&(5)\\ \varphi &=\frac{13}{8}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{(2(k+1))!}{((k+1)+1)!(k+1)!2^{4(k+1)+3}}&(6)\\ \vdots&\\ \end{align*} $$

When looking at the leading terms of $(3)$ and $(6)$ $\;\frac{3}{2}$ and $\frac{13}{8}$ respectively, one is tempted to conjecture that there are similar formulas to fill the holes in the above table.

I'd like to know if such family of formulas exist.

Thanks.

EDIT: Note that both formulas connect the Golden Ratio $\varphi$ to Catalan Numbers $$ C_{k}=\frac{(2k)!}{(k+1)!k!} $$ so for $(3)$ we have $$ \varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{C_{k}}{2^{4k+3}} $$ and for $(6)$ we have $$ \varphi =\frac{13}{8}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{C_{k+1}}{2^{4(k+1)+3}} $$ So, maybe this could be used, somehow, to find the other formulas.

Neves
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2 Answers2

6

Yes, you are right. There is a family. A little experimentation shows that,

$$\varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{C_{k}}{2^{4k+3}}$$

$$\varphi =\frac{13}{2^3}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{C_{k+1}}{2^{4k+7}}$$

$$\varphi =\frac{207}{2^7}+\sum_{k=0}^{\infty}(-1)^{k+2}\frac{C_{k+2}}{2^{4k+11}}$$

$$\varphi =\frac{1657}{2^{10}}+\sum_{k=0}^{\infty}(-1)^{k+3}\frac{C_{k+3}}{2^{4k+15}}$$

$$\varphi =\frac{53019}{2^{15}}+\sum_{k=0}^{\infty}(-1)^{k+4}\frac{C_{k+4}}{2^{4k+19}}$$

and so on. Sorry, but the numerators of the fractions are not Fibonacci numbers. It seems hard to predict what they will be.

Tito Piezas III
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  • very nice, maybe the Fibonaccis are hidden somehow in those fractions... for example $207=3^{2}\cdot 23$. – Neves Dec 28 '14 at 12:20
  • The difference between consecutive fractions is of the form $\pm\frac{1}{2^a}$, except for the last one: $$\frac{3}{2}+\frac{1}{2^3}=\frac{13}{8}$$ $$\frac{13}{2^3}-\frac{1}{2^7}=\frac{207}{2^7}$$ $$\frac{207}{2^7}+\frac{1}{2^{10}}=\frac{1657}{2^{10}}$$ but now we have $$\frac{1657}{2^{10}}-\frac{5}{2^{15}}=\frac{53019}{2^{15}}$$ – Jaume Oliver Lafont Jun 02 '17 at 20:56
  • @JaumeOliverLafont: Interesting observation. – Tito Piezas III Jun 03 '17 at 03:54
5

Let us take the first term out of the summation in equation $(3)$.

$$\begin{align} \varphi&=\frac{3}{2}+\sum_{k=0}^\infty (-1)^k\frac{C_k}{2^{4k+3}}\\ &=\frac{3}{2}+\frac{1}{8}+\sum_{k=1}^\infty (-1)^k\frac{C_k}{2^{4k+3}}\\ &=\frac{13}{8}+\sum_{k=0}^\infty (-1)^{k+1}\frac{C_{k+1}}{2^{4k+7}}\\ \end{align}$$

The result is equation $(6)$. Iterating the same procedure gives all formulas listed in Tito Piezas' answer, so this family is obtained directly from equation $(3)$.

The general formula for the fractions is therefore $$\frac{3}{2}+\sum_{k=0}^N (-1)^{k}\frac{C_k}{2^{4k+3}}$$

Egyptian fraction versions of $(2)$, $(4)$ and $(6)$ are given by

$$\varphi=2-\sum_{k=0}^\infty \frac{1}{F(2^{k+2})}$$

$$\varphi=\frac{5}{3}-\sum_{k=0}^\infty \frac{1}{F(2^{k+3})}$$

$$\varphi=\frac{13}{8}-\sum_{k=0}^\infty \frac{1}{F(3·2^{k+2})}$$

and additional series involving Fibonacci numbers are given by

$$\varphi=\frac{34}{21}-\sum_{k=0}^\infty \frac{1}{F(2^{k+4})}$$

$$\varphi=\frac{89}{55}-\sum_{k=0}^\infty \frac{1}{F(5·2^{k+2})}$$

$$\varphi=\frac{233}{144}-\sum_{k=0}^\infty \frac{1}{F(3·2^{k+3})}$$

$$\varphi=\frac{610}{377}-\sum_{k=0}^\infty \frac{1}{F(7·2^{k+2})}$$

The general pattern is

$$\varphi = \frac{F(2n+1)}{F(2n)}-\sum_{k=0}^\infty \frac{1}{F(n2^{k+2})}$$

Alternatively, we can write formulas (1), (2), (4) and (5) as a linear combination of formulas (3) and (6), from relationships

$$1=5\left(\frac{3}{2}\right)-4\left(\frac{13}{8}\right)$$

$$1=-3\left(\frac{3}{2}\right)+4\left(\frac{13}{8}\right)$$

$$\frac{5}{3}=-\frac{1}{3}\left(\frac{3}{2}\right)+\frac{4}{3}\left(\frac{13}{8}\right)$$

$$\frac{8}{5}=\frac{1}{5}\left(\frac{3}{2}\right)+\frac{4}{5}\left(\frac{13}{8}\right)$$

After some algebra, the results are

$$\begin{align}\varphi &= 1 + 3\sum_{k=0}^\infty \frac{(-1)^k(2k)!(4k+7)}{k!(k+2)!16^{k+1}}\tag{1}\\ \\ \varphi &= 2 - \sum_{k=0}^\infty \frac{(-1)^k(2k)!(8k+13)}{k!(k+2)!16^{k+1}}\tag{2}\\ \\ \varphi &= \frac{5}{3}-\frac{1}{3}\sum_{k=0}^\infty \frac{(-1)^k(2k)!(4k+5)}{k!(k+2)!16^{k+1}}\tag{4}\\ \\ \varphi &= \frac{8}{5}+\frac{3}{5}\sum_{k=0}^\infty \frac{(-1)^k(2k)!}{k!(k+2)!16^{k+1}}\tag{5} \end{align}$$

Two observations: the procedure used has little to do with these particular fractions, similar results could be expected for any fraction; the simplest formula, (5), is obtained from a true interpolation, with both coefficients positive.

Since all these fractions lie between $1$ and $2$, we may take equations (1) and (2) as a basis and parameterize the family of approximations $\varphi\approx \frac{p}{q}$ according to

$$\frac{p}{q}=2\alpha+1(1-\alpha)=\alpha + 1$$

so the general equation may be written

$$\varphi=\frac{p}{q}+\sum_{k=0}^\infty \frac{(-1)^k(2k)!((3-5\alpha)4k+21-34\alpha)}{k!(k+2)!16^{k+1}}$$

where $\alpha=\dfrac{p}{q}-1$.

When $\dfrac{p}{q}=\dfrac{3}{2}$ then $\alpha=\dfrac{1}{2}$ and the multiplying polynomial $(12-20\alpha)k+21-34\alpha$ reduces to $(2k+4)=2(k+2)$. This cancels the largest factor of $(k+2)$! in the denominator and equation (3) is obtained.

Jaume Oliver Lafont
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