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I made a little program that generates Pascal triangles as images :

I first tried it associating to each pixel a color whose intensity was proportional to the number in the Pascal triangle

The colors being 0-255, i used the following function to convert value to colors: $$f(x)=\frac{x-m}{M-m}255$$ where $x$ is the value in the Pascal triangle, $M$ is the max value in the triangle, and $m$ in the min value in the triangle. :

size 50*50 : enter image description here

The axis are like in this picture :

enter image description here

However, as you can see, most of the picture is black due to the numbers being really distant (great distance between highs and lows)

Therefore, i thought it would be good to use a logarithmic scale :

$$f(x)=\frac{\ln(1+x-m)}{\ln(1+M-m)}255$$

Which gives me : size 50*50 : enter image description here

That's way better.

Yet, something was bugging me : as I increased the number of rows, I noticed that some curve was being drawn :

size 50*50 : enter image description here

size 100*100 : enter image description here

size 150*150 : enter image description here

I can't try really high numbers, as my computer isn't good enough, nor is the software I use.

Is there something behind that 'curve' ? If so, what curve would it be ?

Could someone provide explanation why I get such results ?

Thank you.


Progress

We're looking at the level curves of $\ln\binom{N-y}{x},N\in\mathbb{N}^*$

By Stirling, as @TedShifrin remarked, $\ln(n!)\sim n\ln(n)$ therefore $\ln\binom{N-y}{x}\sim y\ln(N-y)-x\ln(x)-(N-y-x)\ln(N-y-x)$ and seem to give us nice curves (cf his answer).

Is there an equation y=f(x) for those curves ?

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    $\begingroup$ Roughly speaking, you're looking at the level curves of $\ln({y\choose x})$. $\endgroup$
    – Jack M
    Jun 21, 2014 at 22:23
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    $\begingroup$ Nice piece of experimental mathematics!!! $\endgroup$ Jun 21, 2014 at 22:24
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    $\begingroup$ An alternative to Stirling here that might be worth looking at is the normal approximation to the binomial distribution. $\endgroup$
    – user21467
    Jun 22, 2014 at 21:57

1 Answer 1

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OK, at long last: For any $N\in\Bbb N$, consider $\binom{N-y}x$, $0\le x,y\le N$. This indicates the intensity at the point $(x,y)$ in the graph, and we are, in fact, as @JackM suggested, looking at its level curves.

Using Stirling's approximation, $\log(n!)\sim n\log n-n$, we consider the plot of level curves of $(N-y)\log(N-y)-x\log x-(N-y-x)\log(N-y-x)$ — note that the linear terms cancel. Here is a Mathematica plot for $N=100$:

Pascal's image

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