Recent content by bob012345

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...

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...

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

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.

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...

Here are recursion relations that give the exact polynomials for the numerator and denominator for any ladder with ##N## stages and a scaling factor ##q##. In all cases the first stage resistors are both 1 unit. For the ##N^{th}## stage, the shunt resistor is ##q^n## where ##n=N-1##. For example...

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

I believe if it’s transcendental it is also irrational.

And it may be rational, irrational and if irrational it may be expressible as a compact, exact quantity involving a square root or it may not be such as ##\pi##. Such a number is transcendental. Our ladder may be the latter.

For the original infinite ladder in post #1 I have found that the number of digits in ##p## and ##q## goes as ##1+n(n-1)## as in post #5 above.

BTW, the reverse ladder, where the horizontal resistors are large and the vertical resistors are small, converges super fast with no assumptions about the terminal stage since that is the one that vanishes. I get ##R_∞=1.90990909917198…##

I’m not sure what you mean. It makes a big difference what value I pick for the infinite termination. The value of ##0.9⋅10^n## is not arbitrary.

I have found that this continued fraction converges rather slowly. As ##n→∞##, the horizontal resistors become vanishingly small and the rest of the ladder collapses to ##0.9 ⋅10^n##. Substituting that for the last term in the series for each ##n## yields much faster convergence.

Thanks. In looking at the exact solutions for the first few terms it seems the fractions quickly get gigantic. These seem also to have no common factors so each fraction cannot be reduced. I suspect this pattern continues as ##n→∞## we get an infinitely large fraction which cannot be reduced...

Thanks but do you agree it’s converging to an irrational number?