Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Background to problem (not too important):

enter image description here

My proposed solution:

The infinitely long element,

enter image description here

, however complex, can be represented as a single resistor of resistance $R$. Remembering the initial resistor near $A$, we know $R_{AB}= r+R$. However, as this is an infinitely long element, it is equivalent to a resistor of resistance $R$ attached to the right of two resistors of resistance $r$ (the resistance $R$ is an intrinsic property of the element, so is unaffected by the fact that the further to the right it is, the lower the current passing through it).

Thus, taking $R$ in series with $r$, then the result in parallel with $r$, then in series with $r$: $$R_{AB}= r + \left (\frac{1}{\frac{1}{R+r}+\frac{1}{r}} \right )$$

On the second iteration (moving $R$ further to the right):

$$R_{AB}= r+ \frac{1}{\frac{1}{r}+\frac{1}{r + \left (\frac{1}{\frac{1}{R+r}+\frac{1}{r}} \right )}}$$

Ad infinitum.

I understand this may not be the fastest solution, but I'd like to know a little more about it nonetheless.

The mathematics


$$\large u_{n+1}=r+\frac{1}{\frac{1}{r}+\frac{1}{u_n}}$$

Does $\lim_{n \rightarrow \infty} (u_n)$ exist (important: is the limit a function of $R$?), and, if so, what is it?

First cases





$$\lim _{n \rightarrow \infty} (u_n)\stackrel{?}{=}\varphi r$$

Seems the doing of Fibonnaci. How does one take the limit of this (I assume it requires knowledge of knowledge of the form of $f(n)=F_n$).

Intuitively, why does Fibonnaci appear here? What are the rabbits in this case?

Link to Wolfram's computation.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

If you look at $x_n = \dfrac{u_n}{r}$ you see that

$$ x_{n+1} = 1 + \cfrac{1}{1 + \cfrac{1}{x_n}}$$

This is basically a recurrence for the continued fraction of the golden ratio $\varphi = [1;1,1,\dots]$.

Thus, it is true that $u_n \to r\varphi$

This also explains why you see the Fibonacci numbers. The convergents of the continued fraction are ratios of Fibonacci numbers.

share|improve this answer
Thanks for that. Is $\large \lim _{n \rightarrow \infty}(\frac{F_{n+1}r^2+F_n rR}{F_{n-1}R^2+F_n r})$ computable directly? –  Alyosha Apr 6 '13 at 20:29
@Alyosha: Yes, Divide numerator and denominator by $F_n$, and use the fact that $F_{n+1}/F_n \to \varphi$ (and $F_{n-1}/F_n \to \varphi$). –  Aryabhata Apr 6 '13 at 20:29

I didn't dare look at this earlier, but here's another (transcripted) solution (I prefer Aryabhata's):

Algebraically this equivalence can be written as $$R_{AB}=r+\frac{1}{\frac{1}{r}+\frac{1}{R_{AB}}}$$ Thus $$R_{AB}^2-rR_{AB}-r^2=0$$ This equation has two solutions: $$R_{AB}=\frac{1}{2}(1 \pm \sqrt{5})r$$ The solution corresponding to “-“ in the above formula is negative, while resistance must be positive. So, we reject it. Finally we receive

$$R_{AB}=\frac{1}{2}(1 + \sqrt{5})r= \varphi r$$

share|improve this answer
+1: But, I would suggest you type the answer in, rather than giving an image (which might possibly go away later). –  Aryabhata Apr 6 '13 at 20:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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