This might be a little long so please bear with me.

The Golden Ratio $\phi$ is defined as the single positive root of the polynomial $p(t) = t^2 - t - 1$. One can think of it as a line divided into two segments $a$ and $b$ such that $\frac{a+b}{a} = \frac{a}{b} \equiv \phi$ with $a > b$.

Now imagine we take not a line (1d) but a block (2d) such that $\frac{(a+b)a}{a^2} = \frac{a^2}{ab} = \phi$ (i.e. $a$ and $b$ are in the golden ratio) and assume that $(a+b)a = 1$. Then we have for the two blocks $a^2 = \frac{1}{\phi} \equiv \Phi$ and $ab = 1 - \Phi = \Phi^2$ thus

\begin{eqnarray*} a &=& \sqrt{\Phi} \\ b &=& \sqrt{\Phi^3} \end{eqnarray*}

Extending this line of thought to $n$-dimensions and keeping $(a+b)a^{n-1} = 1$ we have

\begin{eqnarray*} a &=& \sqrt[n]{\Phi} \\ b &=& \sqrt[n]{\Phi^{n+1}} \end{eqnarray*}

Letting $n \rightarrow \infty$ we have $a=1$ and $b=\Phi$ and thus $a + b = \phi$. A few notes

  • if $(a+b)a^{n-1} = \phi$ then $a = 1$ and $b = \Phi$ for all $n \in \mathbb{N}$
  • if $(a+b)a^{n-1} = x$ then $a = \sqrt[n]{x\Phi}$, $b = \sqrt[n]{x\Phi^{n+1}}$, which both converge to $1$ and $\Phi$ respectively if $n \rightarrow \infty$
  • if $(a+b)a^{n-1} = x^n$ then $a = x\sqrt[n]{\Phi}$, $b = x\sqrt[n]{\Phi^{n+1}}$, which both converge to $x$ and $x\Phi$ respectively if $n \rightarrow \infty$
  • if $(a+b)a^{n-1} = x^{-n}$ then $a = \frac{1}{x}\sqrt[n]{\Phi}$, $b = \frac{1}{x}\sqrt[n]{\Phi^{n+1}}$, which both converge to $\frac{1}{x}$ and $\frac{\Phi}{x}$ respectively if $n \rightarrow \infty$

Now to my question.


\begin{eqnarray*} (a+b)a^{n-1} &=& x^{-n^2}\\ \frac{(a+b)a^{n-1}}{a^n} = \frac{x^{-n^2}}{a^n} &=& \frac{a}{b} = \phi. \end{eqnarray*}

We then calculate $a$ and $b$:

$$ a^n = \Phi x^{-n^2} \rightarrow a = \Phi^{1/n} x^{-n^{2^{1/n}}} \\ a^{n-1}b = \Phi^2 x^{-n^2} \rightarrow b = \Phi^{1 + 1/n} x^{-n^{2^{1/n}}} $$

In the limit $n \rightarrow \infty$ it seems like $a = b = 0$. But then $\frac{a}{b} = \frac{0}{0}$. Where is the error?

  • $\begingroup$ Compute the quotient between $a$ and $b$ before taking $n \to \infty$ $\endgroup$ – FormerMath Oct 21 '14 at 23:00
  • $\begingroup$ If I take the ratio beforehand, then I don't need $n \rightarrow \infty$ anymore since $a/b = \phi$. In the above cases both $a$ and $b$ had converging values $\neq 0$. Why is it that in the last case it would converge to 0? I would expect it to have an analogue convergence as in the case $x^{-n}$. $\endgroup$ – David Seres Oct 21 '14 at 23:41

Redoing your last lines:

$$a^n = \Phi x^{-n^2} \rightarrow a = \Phi^{1/n} x^{-n^2\over{n}}=\Phi^{1/n}x^{-n}= \frac{1}{x^n}\sqrt[n]{\Phi} \\ a^{n-1}b = \Phi^2 x^{-n^2} \rightarrow b = \Phi^{1 + 1/n} x^{-n}= \frac{1}{x^n}\sqrt[n]{\Phi^{n+1}}$$

So, yes, in the limit, as $n \rightarrow \infty$ both $a \rightarrow 0$ and $b \rightarrow 0$ (for $x>1$),

But, clearly, $a > 0$ and $b > 0$ (for $x>0$) for all $n$ before the limit,

So you can safely do the division $a/b = \phi$ without dividing by zero,

So $a/b \rightarrow \phi$.


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