Recent content by bob012345

That’s like saying I’m a billionaire but I just happen to have zero billion dollars at the moment.

Perhaps, but we are dealing with a gamma ray and a radio wave in the same reference frame. Consider what happens if both interact with a microscopic primordial black hole of comparable energy to the gamma ray for example?

I wonder if that is just an approximation? I would think that in principle, the energy difference between a gamma Ray and a radio wave would change the spacetime curvature even if in the most minuscule amount to change to geodesic.

Well something needs to have mass in order to curve spacetime.

So what effective mass is used when starlight is bent by the sun? Isn’t it just ##m=\frac{E}{c^2}##?

I do not know what typo you are referring to? Perhaps it’s the way I specified the termination term. I meant if the ##n^{th}## stage has two resistors ##10^{-n}## and ##10^n##, one can change the latter to ##0.9⋅10^n## or one can leave that alone and add another resistor in parallel ##9⋅10^n##...

Take the limit of ##\frac{R_n R}{R_n+R}## as ##R_n→0## which gives zero. So if one shorts the first stage the minimum resistance is 1 unit.

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.