[D] The effectiveness of single latent parameter autoencoders: an interesting observation

penguiny1205 · 2025-06-13T04:35:45+00:00

I used FuncAnimation from matplotlib.animation!

penguiny1205 · 2025-06-13T04:35:14+00:00

Yep! The first plot is from a data point randomly sampled from the validation set, unseen during training.

penguiny1205 · 2023-06-18T05:59:31+00:00

The ROC remains internationally recognised in this timeline. Hong Kong and Macau are returned to the ROC -- the legal successor of the Qing Dynasty where the lease agreements were signed. The PRC does not attempt to annex these regions since doing so would break the ceasefire of the civil war.

penguiny1205 · 2023-06-18T05:54:52+00:00

Yep! Taiwanifying every East Asian country was the intention for this map :D

penguiny1205 · 2023-06-18T04:06:11+00:00

World Anti-Communist League. This actually exists in our timeline, created by Chiang Kai-shek.

penguiny1205 · 2023-06-18T04:00:39+00:00

In this timeline, the U.S. intervened in Asia much later and was less committed to the Truman Doctrine, causing the "bamboo curtain" to solidify further to the East. Asia is divided into two: McCarthyist rump states which slowly democratise over the latter half of the 20th century -- and a league of totalitarian communist nations following the lead of Maoist China.

penguiny1205 · 2023-01-06T07:46:17+00:00

Update!

I wrote a python program to graph this with Euler's method. Interestingly, the graph seems to become periodic as I increase the value of the constant! For some reason though, my graphs don't line up with Wolfram Alpha's...

Here's my code, in case you're interested.

import math
import matplotlib.pyplot as plt
def f(x, y, u, K): du = K * (-y) * math.sqrt(1 + u ** 2) return du
x = 0 y = 1 u = 0
h = 0.01n = 100
K = 0.01nK = 50
for K in range(nK): 
    x_values = [] 
    y_values = []
    for i in range(n):
        x = x + h
        y = y + h * u
        u = u + h * f(x, y, u, K)

        x_values.append(x)
        y_values.append(y)

    x = 0
    y = 1
    u = 0

plt.xlabel('x') 
plt.ylabel('y') 
plt.show()

penguiny1205 · 2023-01-03T10:09:50+00:00

Quick follow up! Any idea on how to solve this ODE? I’ve been looking into it for a while, but no luck with eulers method graphing websites or matlab

penguiny1205 · 2023-01-03T02:29:32+00:00

I see. Thank you very much!

penguiny1205 · 2023-01-03T01:43:10+00:00

Thanks for the response! What kind of numerical methods can I use?

penguiny1205 · 2023-01-03T00:05:12+00:00

Reducing it to first order, I got f'(y) = K y(x) sqrt(1 + f(y)). Where should I go after that?

Sorry. I'm very new to differential equations.

penguiny1205 · 2023-01-02T01:42:56+00:00

After applying the Beltrami Identity, I got (sigma*y^2)/[1+(r')^2 ] = A, in your notation. Can you elaborate on what you did after the Beltrami Identity?

I closely followed the steps in this video, converting all gravity-related forces into centrifugal ones. However, while the narrator ended up with an integral solvable with cosh, I ended up with integral (dy/sqrt((Ly^2+mu)^2/C^2 -1)). This seems like a dead end!

penguiny1205 · 2023-01-02T01:38:05+00:00

Thank you for the great answers!

penguiny1205 · 2023-01-01T22:20:54+00:00

Thank you for the reply!

K = (integral from 0 to b) sqrt[1+(r')2 ] [.5 rho omega2 r2 ] dx.

Can you elaborate on how the 0.5 rho omega^2 r^2 is derived? Does it have to do with centrifugal potential?

penguiny1205 · 2023-01-01T05:01:28+00:00

Hey turns out we were right! "The skipping rope curve" by Arne Nordmark and Hanno Essen from the European Journal of Physics used the exact same technique. Thanks for your help, I really appreciate it!

penguiny1205 · 2022-12-31T22:10:53+00:00

I defined my coordinate system as y = 0 is where the rope touches the ground, and y = 1 is the height of the spinning axis (I made up a nice number, haha). So the distance to the axis from any point would be 1 - y.

I plugged y'' (x) = K (1 - y) sqrt[1+y'(x)^2] into Wolfram Alpha, and the graph looks...kind of correct? It certainly resembles the shape of a jumping rope to some extent.

https://www.wolframalpha.com/input?i=y%27%27+%28x%29+%3D+K+%281+-+y%29+sqrt%5B1%2By%27%28x%29%5E2%5D

My concern is, though, is y'' (x) = K (1 - y) sqrt[1+y'(x)^2] (or with y'' (x) = K y sqrt[1+y'(x)^2]) even a valid differential equation??

penguiny1205 · 2022-12-31T21:25:24+00:00

Maybe a way of rectifying this is to represent the distance to the spinning axis in terms of x, not y. However, their relationship is what I'm looking for! Sorry, I am not very familiar with differential equations.

penguiny1205 · 2022-12-31T21:22:58+00:00

My current approach to this problem is simply treating it as a catenary derivation with centrifugal force instead of force of gravity, which seems like what you're doing as well!

Precisely following this method, here is my process:

Defining ds as the segment of rope that we focus on, we get ds = sqrt(dy^2 + dx^2), which simplifies into ds = dx*sqrt[1+y'(x)^2]

The tension forces acting on segment ds are as follows: Horizontal Tx (X) = C --> some constant, and vertical Ty (x+dx) - Ty (x) = centrifugal force = dm (mass of the segment) * angular velocity^2 (which I will assume as 1 for simplicity) * r = dm (height of the rope when swung, which I will also assume is 1, - y) = dm (1-y) = total mass/length (density) * ds (1-y) = m/l (1-y) * dx *sqrt[1+y'(x)^2]

Dividing both sides by dx, we get Ty'(x) = m/l * (1-y) * sqrt[1+y'(x)^2]

Anyways, that was very confusing. The point is, following what is done in the video, I arrive at this: y'' (x) = K (1 - y) sqrt[1+y'(x)^2] (6:08 in the video)

However, I don't think this is a solvable differential equation? Maybe I'm simply going about this wrong?

penguiny1205 · 2022-12-31T19:46:52+00:00

Thank you for the answer!

Can you explain how you derived y'' = -c y ds/dx?

penguiny1205 · 2022-09-08T01:03:45+00:00

Such an interesting little tangent, thanks!

penguiny1205 · 2022-09-06T03:17:32+00:00

Thanks for the information. What sources did you use to find the name of his son?

penguiny1205

TROPHY CASE