7
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ very nice, maybe the Fibonaccis are hidden somehow in those fractions... for example $207=3^{2}\cdot 23$. $\endgroup$ – Neves Dec 28 '14 at 12:20
  • $\begingroup$ 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 $$\frac{1657}{2^{10}}-\frac{5}{2^{15}}=\frac{53019}{2^{15}}$$ $\endgroup$ – Jaume Oliver Lafont Jun 2 '17 at 20:56
  • $\begingroup$ @JaumeOliverLafont: Interesting observation. $\endgroup$ – Tito Piezas III Jun 3 '17 at 3:54
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.