Finite resistor ladder: n repeating units
Vložit
- čas přidán 28. 08. 2024
- Finding the effective resistance of a ladder-like resistor network with a finite number of repeating units, n. We need to solve a non-linear recurrence relation, and along the way we make use of matrix diagonalization. At the end, we take the limit as n becomes infinite and see that the golden ratio makes an appearance.
Simple derivation in the case of infinite n: • Infinite ladder of res...
About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
My website: benyelverton.com/
#physics #mathematics #electricity #circuits #resistance #effectiveresistance #seriescircuit #parallelcircuit #recurrencerelation #nonlinear #linearalgebra #matrix #matrices #diagonalization #indices #transformations #transformation #powers #eigenvalues #eigenvectors #quadratic #identitymatrix #identity #inverse #determinant #determinants #nthterm #matrixmultiplication #inverseofmatrix #voltage #current #goldenratio #reciprocal #math #science #education
Interesting solution! Nice idea on the ansatz. What i did is the following: find the fixed point of function f(x) = (2x+1)/(x+1) that defines the recurrence , say x_1, x_2 (one of them turns out to be the golden ratio). Then define a new sequence a_n:=(R_n- x_1)/(R_n-x_2), using the recurrence for R_n shows that a_n is a geometric series with common ratio (2-x_1)/(2-x_2).
What a tour de force, bravo!
Thank you!
That's awesome!
Thank you
Great solution using linear algebra! Looks like my previous comment was hidden by CZcams. Perhaps it is because I was trying to write equations. So here I will discuss my method verbally. After obtaining the recurrence relation at 3:05, we let the function of the right-hand side f and then solve for its fixed points, that is values of x such that x=f(x). This equation is quadratic and has two distinct roots, one of the roots being the golden ratio. Then, define a new sequence, say Pn, that equals to the ratio of the differences between Rn and each of the fixed points. Using the recurrence for Rn, we can show that Pn is a geometric sequence and the problem is solved.
I can see both comments in YT studio, apparently there's a known issue with disappearing comments at the moment though. That seems like a nice method but it's not obvious to me why Pₙ defined in that way is a geometric sequence!
@@DrBenYelverton To be honest I got it by a series of trials and errors. I came across the same problem with integer fixed points so it was easier to guess the correct transform. I agree it does seem unmotivated and as of now I dont have an explanation other than it works. The linear algebra method feels less ad hoc. For example, if the fixed points are repeated I have no idea what the appropriate transform is, but linear algebra would have no problem handling that!
I definitely want to revisit this some time and see if I can gain some intuition on why it works! Interestingly, WolframAlpha gives the solution in a different form, involving the coth function, so I wonder if there's yet another way to solve this using some clever hyperbolic trig substitution...
@@DrBenYelverton Wow that is super interesting, thanks for that information! I never knew WolframAlpha can handle this type of nonlinear recurrence to begin with!
Sir, got any approach for the spring problem ? 😅
It's actually very difficult to solve exactly! I was thinking of doing it by finding an effective mass for oscillations of the spring, but it turns out that that's not straightforward unless there is a heavy mass attached to the end of the spring. According to an article on wikipedia, perhaps you can multiply the spring's mass by 4/π² and then use ω = √(k/m) with that effective mass, but you'll need to do some research on where that result comes from and check that it's appropriate to use here! See en.wikipedia.org/wiki/Effective_mass_(spring%E2%80%93mass_system)
Thanks for trying and replying! Looks like it's not a problem to solve as a high school student.. will try later someday. 🙏
These A[2n+1)/A[2n], A[0]=0, A[1]=1, quadratic recursive functions look nice on conformal maps. Magnetic field lines. I guess I should be quiet, but you probably already have a way of reading it. So, no sense deleting.
This sounds interesting - I'm curious but didn't quite understand what you meant!
@@DrBenYelverton I'm going to make a video. Harvard has a competition for amateurs like myself. I'm just trying to figure out how to get Spyder to do Manim animations.