Obvious Observation

[LetYouDown]

Although it seems obvious, just made an observation.

Calculating a C b where (a) is constant and (b) increases from 1 --> (a), the function increases until the midpoint, where it decreases in a mirror fashion.

Calculating a P b where (a) is constant and (b) increases from 1 --> (a), the function increases constantly.

I've never plotted out a graph of this type. Is there any coherent visual pattern from these on a more macro scale? Say (a) = 10000.

[Siegmund]

For large a, a plot of aCb over the possible range of b's approaches the shape of a normal distribution with mean a/2 and variance a/4.

You may also, for studying these with large values of a, want to make use of Stirling's approximation, ln n! ~ n ln n - n + 1. This turns aCb -- or at least ln(aCb) -- into something easy to feed into a graphing calculator or software package and plot as if it were a continuous function.

[LetYouDown]

That's perfect, thank you. You need to relax your brain...it must be stressful coming up with intelligent responses all day.

[pzhon]

[ QUOTE ]

I've never plotted out a graph of this type. Is there any coherent visual pattern from these on a more macro scale? Say (a) = 10000.

[/ QUOTE ]
That depends on what you mean by a coherent visual pattern. People are not used to processing any stream of fairly large numbers.

The collection of numbers (a choose b) is called Pascal's Triangle. As that name suggests, this has been studied for centuries. It's worth studying more.

The Central Limit Theorem says that row n of Pascal's Triangle is roughly a normal distribution after you normalize it by dividing by 2^n.

The even-odd pattern of Pascal's Triangle resembles Sierpinski's gasket.

1
11
1<font color="orange">0</font>1
1111
1<font color="orange">000</font>1
11<font color="orange">00</font>11
1<font color="orange">0</font>1<font color="orange">0</font>1<font color="orange">0</font>1
11111111
1<font color="orange">0000000</font>1
11<font color="orange">000000</font>11
1<font color="orange">0</font>1<font color="orange">00000</font>1<font color="orange">0</font>1
1111<font color="orange">0000</font>1111
1<font color="orange">000</font>1<font color="orange">000</font>1<font color="orange">000</font>1
11<font color="orange">00</font>11<font color="orange">00</font>11<font color="orange">00</font>11
1<font color="orange">0</font>1<font color="orange">0</font>1<font color="orange">0</font>1<font color="orange">0</font>1<font color="orange">0</font>1<font color="orange">0</font>1<font color="orange">0</font>1
1111111111111111

There are more complicated, but related patterns for the last digits in other bases.

[BruceZ]

[ QUOTE ]
Although it seems obvious, just made an observation.

Calculating a C b where (a) is constant and (b) increases from 1 --&gt; (a), the function increases until the midpoint, where it decreases in a mirror fashion.

[/ QUOTE ]

I didn't answer at first because I couldn't figure out what a "C b" was.

These are the coefficients of (x + 1)^a.
In general, (a + b)^n = sum{k = 0 to n} C(n,k)*b^k*a^(n-k).
For example, (a + b)^3 = C(3,0)*a^3 + C(3,1)*a*b^2 + C(3,2)*a^2*b + C(3,3)*b^3.

These form the nth row of Pascal's triangle.

The shallow diagonals of Pascal's triangle sum to the Fibonacci numbers.

When we use the inclusion-exclusion principle to compute the probability that something occurs at least once, the coefficients are 1 and -1. If we want the probability that it occurs exactly n times or at least n+1 times, the coefficients are row n+1 of Pascal's triangle (starting from row 0) with every other entry negated. I discovered this (for myself).