## Sunday, June 28, 2009

## Monday, June 15, 2009

### Bipartite counting , thesis and other things

## Thursday, June 11, 2009

### Today June 11, 2009

## Wednesday, June 10, 2009

### Edward Gorey

Today I read this book by Edward Gorey. Its a bunch of short stories. Some are unsettling like "Loathsome Couple", "Stupid Joke", "Blue ". I guess he is more famous for his dark, melancholy writing. There are some which make little sense in what he was trying to convey like the story about one guy loosing his umbrella later to be found by his dog. I enjoyed the rhymes and his take on alphabets.

## Monday, June 08, 2009

### What I am reading ?

## Sunday, June 07, 2009

### Graphic Witness

## Saturday, June 06, 2009

## Thursday, June 04, 2009

### Some insight

Its 12:26 and I am here at Carbondale library updating this blog. Yesterday I wrote the Gambler's ruin program in Maple and it was good to go over some of the examples that Dr. Clark did in his combinatorics book.

Any course in combinatorics deals with the Stirling numbers, while I understood the recursive proof of 2nd type of Stirling numbers but I was never comfortable with the argument for proving the Stirling numbers of 1st kind until yesterday when I tried writing down all the permutations of c(4) and c(3)and tried to generalize it and at flash of insight every thing was clear. Just to give you a background

c(n,k) = c(n-1,k-1)+(n-1)*c(n-1,k)

The problem term was the factor (n-1) and it is very easy to see that any c(n-1,k) we have (n-1) places where we can put our chosen 'n' element.

Also I was working on derivation of s(n,k) using ordinary generating function. I am comfortable with the proof one gets using mixed generating functions but in ordinary one, one has to deal with partial fraction. Again if you try the trivial example it becomes clear that why one needs to put x = 1/r as a substitution. The recursive definition gives

Bk = x/(1-k*x)*Bk-1 and B0 = 1 and one needs to fish out the coefficient of x^n ie Bk = x^k/((1-x)(1-2x)... (1-kx)) . Thus we need

[x^n] x^k/((1-x)(1-2x)... (1-kx)) and it turns out one has to deal with partial fractions. Where one can write it as

[x^(n-k)] sum(alpha_i/(1-i*x), i = 1.. k)