Program for printing a Pascal Triangle:

Using Mathematica ( what else? ):

Pascal[n_]:=MatrixExp[Table[If[i == (j+1),j+1,0],{i,0,n},{j,0,n}]]

Usage: Pascal[n]//MatrixForm

Pascal's Triangle as a Matrix Exponential

For me, this is an amazing result. When I read about it in "Accessing Bernoulli-Numbers by Matrix-Operations, Gottfried Helms 3'2006 Version 2.3" I immediately started Mathematica and tried it myself. The picture above is the result. It's what I call some deep mathematics, although the formula for calculating the Matrix Exponential is easily understood.

Dangerous Knowledge

Dangerous Knowledge is a BBC Four documentary about the lives and work of Georg Cantor, Kurt Godel and Alan Turing. ( Find it on Google Video. ) Cantor, Godel and Turing worked on paradoxical stuff. The paradox about the documentary is that it is about mathematics without showing any of it. I suppose it is a nice documentary if you are familiar with Cantor, Godel and Turing. The documentary in that respect becomes a " complementary ".

Sequences

Sequence
Closed form(ula)
Generating function

{1,1,1,1,...}
a(n)=1
f(x)=1/(1-x)

{1,2,4,8,...}
a(n)=2^n
f(x)=1/(1-2x)

{0,0,0,0,1,1,...}
if (n<4) then a(n)=0 else a(n)=1
f(x)=x^4/(1-x)

See also the Mathematica package RSolve

Generating Functions

Generating functions are one of the most surprising, useful, and clever inventions in discrete mathematics.

Generating functions transform problems about sequences into problems about functions.
For example:
* sequence {1, 4, 9, 16, 25, ... }
* closed form a(n) = n^2
* generating function: F(x)=x(1+x)/(1-x)^3.
Interested? An introduction to generating functions (pdf document) can be found [ here ].

Difference sequences

Rather early in our math education we learn about functions and derivative functions.

  
              f(x+h)-f(x)
f'(x) = lim   -----------
         h-> 0    h 

For example: 
f(x) = x^n
f'(x) = n*x^(n-1)

Something similar can be defined for sequences, in that case the 'derative' is called the difference sequence.

a_n = {1, 16, 81, 256, ... }
f(n) = n^4
f'(n) = 1 + 4n + 6n^2 + 4n^3
f''(n) = 14 + 24n + 12n^2
f(3)(n) = 36 + 24n
f(4)(n) = 24

For an arbitrary sequence f:
f(m)(n) = Sum[(-1)^(k)*Binomial[m,k]*f[n-k+m],{k,0,m}]

Here also we see that Pascal's Triangle has a crucial meaning.

Binomial Theorem

(x + y)^n = Sum(n,k) Choose(n,k) * x^(n-k) * y^k

So what? I can now -prove- it. ( By induction. )

Literary Mathematics

"... A generating function is a clothesline on which we hang up a sequence of numbers for display. ..." ( by Herbert S. Wilf in Generatingfunctionology )