Non-Linear Infinite Resistor Ladder

[Baluncore]

I don't have an immediate answer, but I suspect that this is one of those computations where the initial condition matters a lot, if not most, and so only the 'right' numerical value can give the exact final value, but I'm just speculating...

Using spice to solve the 10 stage ladder.
With a shorted tail = 1.099990
With an open tail = 1.900991
The actual result must lie between those bounds.
Your result, √26 - 4 = 1.099019513592785 is on the wrong side, and slightly outside the bounds.
The tail will look like a straight wire, with resistors to ground of 1e10, 1e11, 1e12, ... = 0.900e10.
You seem to have chosen the wrong tail to the ladder.

 

[WWGD]

Doing that derives the limiting term as $0.9⋅10^n$ discussed in post #8.

I think an obstacle

@lrg
Welcome to PF.


1.099019513592785, is very different from
1.9009910801792881, the numerical result evaluated over more than 100 terms.

Your result is for the CF = 1+1/10+1/10+1/10+1/10+1/10+1/10+1/10+ 1/. . .

The CF we need is = [1, 1, 9, 9, 1, 99, 99, 1, 999, 999, 1, 9999, 9999, 1, ...
According to Google, this number is an example of an irrational number that is extremely well approximated by rational fractions at specific truncation steps.

Is that a Roth- Thue- Siegel result?

 

[Baluncore]

Is that a Roth- Thue- Siegel result?

No idea. I took the numerical solution and then converted it to a CF.
There looked to be a pattern, so I used Google search to confirm it.
If the CF pattern had been cyclic, repeating the same coefficients, then the solution would have been a square root.

 

[bob012345]

Hello everyone,

I think the exact value should be $$R_{\infty} = \sqrt{26} - 4 = 1.099019513592785,$$ that is the solution of the equation $$ x = 1+ \frac{1}{\frac{1}{1}+\frac{10}{x}},$$ or $$x^2+8x-10 = 0,$$ since the equivalent resistance at the right of the first vertical resistor is equal to that at the input divided by 10 (it is an infinite ladder ...). Do you agree?

Welcome to PF!

I believe you are solving a different ladder that goes like this:

The ladder being studied in this thread has the same top resistances but the shunts increase by ten for each stage. In your case where each stage changes both resistors by the same factor, it can be solved exactly and gives the answer you gave.

 

[lrg]

Welcome to PF!

I believe you are solving a different ladder that goes like this: ...

Yes, you are right, very stupid mistake on my side. When I realized it, I decided to simulate the circuit (100 stages) with a Matlab/Octave script, changing the resistance at infinity (far right), and I got this graph:

It shows that the input resistance can go from about 1.1 (short) to about 1.9 (open), as reported also by @Baluncore.
Then, just for fun, I tried to obtain my value (1.099), and I got it, with the help of a 'slightly negative' resistance (~-1.2e-3)! Look at this other graph:

where I extended the range of the resistance at infinity to negative values (here the horizontal scale is linear, not logarithmic). It is even possible to get a short (0 value) as equivalent resistance ...
I think there is something interesting here, namely the infinite ladder cannot be said to have any equivalent input resistance, UNLESS we specify also the behavior at infinity, with a resistance at the far right (kind of boundary condition). Changing that value, we get different equivalent resistances. Of course, negative values are only for fun, but any value between 0 and $+\infty$ (short/open) should be considered 'reasonable', without preferences.

 

[bob012345]

Yes, you are right, very stupid mistake on my side. When I realized it, I decided to simulate the circuit (100 stages) with a Matlab/Octave script, changing the resistance at infinity (far right), and I got this graph:

Which circuit are you simulating with 100 stages? Our original ladder in post #1 or your ladder? Not sure what you mean by “changing the resistance at infinity”? It seems to me simulating a 100 stage ladder, it will make no difference what resistors you add or how you add them at the final stage.

 

[Baluncore]

It seems to me simulating a 100 stage ladder, it will make no difference what resistors you add or how you add them at the final stage.

It makes a difference for a zero ohm termination = tail.
Changing the tail, the starting value for evaluation, of a 100 term ladder.
1.900991080179288 ; Tail = 1e+201 = infinite, open
1.900991080179288 ; Tail = 0.8e+101 = good approximation
1.900991080179287 ; Tail = 1P = 1e15
1.900991080178494 ; Tail = 1T = 1e12
1.900991079385424 ; Tail = 1G = 1e9
1.900990286316557 ; Tail = 1M = 1e6
1.900198002672590 ; Tail = 1k = 1e3
1.502282916827386 ; Tail = 1 ohm
1.099990001000000 ; Tail = 1e-201 = zero, shorted

 

[lrg]

Yes @Baluncore, that is what I found, and I put in the figure above. In fact, the continued fraction that computes the equivalent resistance of the original @bob012345's ladder should be:
$$R = 1 + \cfrac{1}{\frac{1}{1} + \cfrac{1}{b + \cfrac{1}{b + \cfrac{1}{b^2 + \cfrac{1}{b^2 + \cfrac{1}{b^3 + \cfrac{1}{b^3 + \dots}}}}}}}$$
with $b=1/10$.
With that CF, the Seidel-Stern Theorem (search with Google) FAILS: it states that, to have a convergent continued fraction, the sum of the coefficients $1, b, b, b^2, b^2 \dots$ must diverge. Here the sum is 11/9 (when $b=1/10$), so it cannot converge. This is consistent with the numerical result that shows that the truncated ladder fraction (here with 4 stages):
$$R = 1 + \cfrac{1}{\frac{1}{1} + \cfrac{1}{b + \cfrac{1}{b + \cfrac{1}{b^2 + \cfrac{1}{b^2 + \cfrac{1}{b^3 + \cfrac{1}{b^3+ \cfrac{1}{r_{\infty}}}}}}}}},$$
where $r_{\infty}$ is the terminating resistance at the right, is a variable function of $r_{\infty}$ even when the number of stages grows without limit. I understand @bob012345 when he writes in the post #36:

It will make no difference what resistors you add or how you add them at the final stage

​

because I used to think the same thing. But now I have realized that, when the continued fraction of the ladder does not converge, that's wrong: the terminating resistance matters, even when the ladder is infinite!
If you want a 'positive' example, take the 'golden ladder', the one with all the resistors equal to 1. The equivalent resistance is the golden ratio $\frac{1+\sqrt{5}}{2}$, and the continued fraction converges to that value because the sum $1+1+1+\dots$ diverges. In that case, if you simulate the ladder by changing the terminating resistance, you always get the same result (the golden ratio) regardless of the value of that resistance.
It seems paradoxical, but we must remember that, physically, infinite ladders don't exist, so the problem we are discussing is purely mathematical, and in that case the terminating resistance is a kind of boundary condition, a 'hidden' parameter of the infinite ladder that is as important as the parameter $b=1/10$ that defines it.

A very interesting problem indeed.

 

[bob012345]

It makes a difference for a zero ohm termination = tail.

Ok, but then changing the tail to low values is changing the problem significantly. It no longer the ‘tail’ but a different ladder.

 

[bob012345]

I was wrong to say changing the tail doesn’t change the ladder but like I said, it literally changes the ladder to be a different problem if it has no relationship to the actual infinite tail. In the original analysis above our tail value of $0.9 x10^n$ for the $n^{th}$ stage does and was derived from a recursion relation. All it did was make convergence faster. So our original convergence is not wrong.

 

[Baluncore]

I like this problem. It overturns every assumption we would normally make.

 

[bob012345]

If you want a 'positive' example, take the 'golden ladder', the one with all the resistors equal to 1. The equivalent resistance is the golden ratio $\frac{1+\sqrt{5}}{2}$, and the continued fraction converges to that value because the sum $1+1+1+\dots$ diverges. In that case, if you simulate the ladder by changing the terminating resistance, you always get the same result (the golden ratio) regardless of the value of that resistance.
It seems paradoxical, but we must remember that, physically, infinite ladders don't exist, so the problem we are discussing is purely mathematical, and in that case the terminating resistance is a kind of boundary condition, a 'hidden' parameter of the infinite ladder that is as important as the parameter $b=1/10$ that defines it.

A very interesting problem indeed.

If one takes the Golden Ladder equation $R^2 -R -1=0$ and instead of solving it exactly, write it as $R = \sqrt{1+R}$ then start with any terminating resistance under the radical and iterate you will get the Golden Ratio $\phi$. Then the number of iterations you do for a specific desired precision is the number of stages you would need of a finite ladder to converge to.