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Background to problem (not too important):

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My proposed solution:

The infinitely long element,

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, 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

$$u_{1}=r+R$$

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

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

First cases

$$u_{2}=\frac{3r^2+2rR}{R+2r}$$

$$u_{3}=\frac{8r^2+5rR}{3R+5r}$$

$$u_{4}=\frac{21r^2+13rR}{8R+13r}$$

$$u_{4}=\frac{55r^2+34rR}{21R+34r}$$

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

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.

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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.

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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
1  
@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

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

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+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

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