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mathematician, composer, photographer, fiddler

27 Nov 2017 | categories: Mathematics

A Dynamical Proof That the Ratio of Successive Fibonacci Numbers Approaches the Golden Ratio

In his 1611 essay On the Six-Cornered Snowflake, Johannes Kepler observed that the ratio of successive Fibonacci numbers $F_n/F_{n-1}$ approaches the golden ratio $\tfrac12(1+\sqrt5)$. [Recall that $F_1=F_2=1$ and $F_{n+1}=F_n+F_{n-1}$ for $n\ge2$.]

Many proofs have since been found. Here is a quick, dynamical one I thought of which I have been unable to find in the literature.

Proof: Consider the function $f(x)=(1+x)/x$. The sequence of ratios of successive Fibonacci numbers is the orbit of $F_2/F_1=1$ under $f$, i.e. $$ f(F_n/F_{n-1})=F_{n+1}/F_n. $$ On the other hand, $f$ has precisely two fixed points, the roots of $x^2-x-1$, one positive (the golden ratio), the other negative. Since $f'(x)=-1/x^2$, $f$ is a decreasing function for $x>0$. And $f(2)=3/2$ while $f(3/2)=5/3$. So $f$ maps the interval $[3/2,2]$ into itself. In fact, $f$ is a contracting mapping on $[3/2,2]$ (with Lipschitz constant $4/9$) since $$ |f(x)-f(y)|=|y-x|/|xy|\le\tfrac49|y-x| $$ for $x,y\in[3/2,2]$. By the Banach fixed point theorem, then, all orbits in $[3/2,2]$ approach the golden ratio. Although $1$ is not in $[3/2,2]$, $f(1)=2$ is, so the orbit of $1$ approaches the golden ratio. ∎

This proof can be extended to show that the conclusion holds even if the first two terms $F_1,F_2$ of the Fibonacci sequence are altered, provided their ratio $F_2/F_1$ is not $\tfrac12(1-\sqrt5)$ (the negative, unstable fixed point of $f$).

In a sense this explains why the limit is the golden ratio: being a fixed point of $f$ is the defining property of the golden ratio.

11 Jul 2017 | categories: Photographs

Chimney Bluff

A drumlin eroding into Lake Ontario.

Chimney Bluff (2017)

15 Mar 2017 | categories: Photographs

Der Himmel über Rochester

Der Himmel über Rochester (2017)

27 Mar 2016 | categories: Photographs

Letchworth Lower Falls

Letworth Lower Falls (2016)

17 Dec 2014 | categories: Mathematics

How to Recognize Generators for String Bordism

A key result of my paper:

The Cayley Plane and String Bordism, Geometry & Topology 18-4 (2014), 2045–2078.

unmentioned in the abstract or introduction—and likely of independent interest—is the following characteristic-number-theoretic criterion for a set to generate the String bordism ring (with 6 inverted).

Theorem 4. A set $S$ generates the ring $\pi_*\mathrm{MO}\langle8\rangle[1/6]$ if:

  1. For each integer $n>1$, there is an element $M^{4n}$ of $S$ such that for any prime $p>3$: $$\begin{align*} \mathrm{ord}_p \big( \mathrm{s}_n[M^{4n}] \big) = \begin{cases} 1 & \text{if $2n=p^i-1$ or $2n=p^i+p^j$ for some integers $0 \le i \le j$} \\ 0 & \text{otherwise} \end{cases} \end{align*}$$ where $\mathrm{s}_n[M]$ is the characteristic number corresponding to the $n$th power sum symmetric polynomial $\sum x_i^n$ in the Pontrjagin roots of the tangent bundle of $M$, and $\mathrm{ord}_p(-)$ its $p$-adic order, i.e. the heighest power of $p$ which divides it.
  2. For each prime $p>3$ and each pair of integers $0<i<j$, there is an element $N^{2(p^i+p^j)}$ of $S$ such that: $$\begin{align*} \mathrm{s}_{(p^i+p^j)/2}[N^{2(p^i+p^j)}]&=0 \\ \mathrm{s}_{(p^i+1)/2,(p^j-1)/2}[N^{2(p^i+p^j)}] &\not\equiv 0 \mod p^2 \end{align*}$$ where $\mathrm{s}_{m,n}[N]$ is the characteristic number corresponding to the symmetric polynomial $\sum_{i\ne j} x_i^mx_j^n$ in the Pontrjagin roots of the tangent bundle of $N$.

This is a consequence of Hovey’s calculation of:

The homotopy of $\mathrm{MString}$ and $\mathrm{MU}\langle6\rangle$ at large primes, Algebr. Geom. Topol. 8 (2008), 2401–2414.

which itself builds on Hopf-ring-theoretic work of Ravenel and Wilson. It should be compared with the analogous result for the oriented bordism ring (with 2 inverted):

Theorem (Novikov, cf. Stong p. 180). A sequence $\{M^{4n}\}_{n\ge1}$ generates the ring $\pi_*\mathrm{MSO}[1/2]$ if and only if:

  • For any integer $n>0$ and any odd prime $p$: $$\begin{align*} \mathrm{ord}_p \big( \mathrm{s}_n[M^{4n}] \big) = \begin{cases} 1 & \text{if $2n=p^i-1$ for some integer $i>0$} \\ 0 & \text{otherwise} \end{cases} \end{align*}$$


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