Modeled PF Trophies

[AlexB23]

Hey guys. I modeled the distribution of PF post trophies using Wolframalpha.com. The formula is 58757.3/x^0.669229 with a fit of 0.999988. It slightly breaks down when the numbers get high.

( data from: https://www.physicsforums.com/members/?key=most_messages )

 

[berkeman]

The formula is 58757.3/x^0.669229

Interesting, so the number of members (call that M) with x posts is reflected by that fit?
$$M = \frac{58757}{x^{0.669}}$$

 

[AlexB23]

Interesting, so the number of members (call that M) with x posts is reflected by that fit?
$$M = \frac{58757}{x^{0.669}}$$

Yep, roughly. It goes to show how real life can be modeled somewhat with math, including social media.

 

[fresh_42]

Yep, roughly. It goes to show how real life can be modeled somewhat with math, including social media.

I have a problem with this formula. $M(x)$ is convex, and your graph is concave.

 

[AlexB23]

I have a problem with this formula. $M(x)$ is convex, and your graph is concave.

The graph is log plotted.

 

[fresh_42]

The graph is log plotted.

That doesn't change the curvature:
https://www.wolframalpha.com/input?i=y=log(x^(-0.669))

 

[AlexB23]

That doesn't change the curvature:
https://www.wolframalpha.com/input?i=y=log(x^(-0.669))

Maybe the axis are flipped on my end. Try the inverse of the formula.

 

[fresh_42]

Maybe the axis are flipped on my end. Try the inverse of the formula.

https://www.wolframalpha.com/input?i=y=e^(60000*(x^(-0.669)))

 

[AlexB23]

https://www.wolframalpha.com/input?i=y=e^(60000*(x^(-0.669)))

No idea then.

 

[fresh_42]

The plot seems correct by comparison with your list. I fed https://elsenaju.eu/Calculator/online-curve-fit.htm with your data to $\log_{10}$ on both scales. It also produces a power law with the wrong curvature: $$M=5.583 x^{−1.09}\quad\text{red}$$
But the polynomial (degree $4$) has a good fit:
$$
M= -0.046x^4+0.418x^3-1.276x^2+0.597x+4.769 \quad\text{blue}
$$

Here is the file that I used:
0;
4.77;
2;
3.446;
2.4;
3.146;
3;
2.67;
3.7;
2.05;
4;
1.785;
4.3;
1.43;
4.6;
0.7;

 

[renormalize]

I have a problem with this formula. $M(x)$ is convex, and your graph is concave.

Hmm, I'm not seeing that. The formula for $M(x)$ is a simple power law and is thus linear on a log-log graph:

And the fit can be significantly improved by introducing a term quadratic in $\text{ln}(x)$:

 

[fresh_42]

Hmm, I'm not seeing that.

The plot in post #1 is clockwise bent, and $y=x^{-2/3}$ anticlockwise. If we take the log then $\log y=-\dfrac{2}{3}\log x$ which is linear, still not curved like the plot is. I first considered only the original function, and then mistakenly took only the log on one scale.

 

[DaveE]

"With four parameters I can fit an elephant, and with five I can make him wiggle his trunk." - John von Neumann

Is there something significant here? What does it mean?

 

[Baluncore]

I expect the 58757.3 coefficient would be a function of the number of members in the population, that will change with time.
But where does the coefficient 0.669 arise from. Could it theoretically be 2/3 = 0.6667 ? Maybe there is a set of PDEs that would reveal how the distribution developed?

 

[Hornbein]

"With four parameters I can fit an elephant, and with five I can make him wiggle his trunk." - John von Neumann

https://arxiv.org/abs/2407.07909

Fitting an Elephant with Four non-Zero Parameters​

? Why non-zero ? Maybe they meant to say "non-Zeno".