0
$\begingroup$

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.

Assumptions:

\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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $a/b \rightarrow \phi$.

$\endgroup$

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.