Compute the triangle of Stirling numbers of the first kind...

Key Concepts

Stirling Numbers of the First Kind

These numbers are a class of special integers that enumerate the number of permutations of n elements with exactly k cycles. They are typically arranged in a lower-triangular array and can be presented in signed or unsigned versions. Their properties make them useful in combinatorics and algebra, particularly in the expansion of falling factorials and in expressing the relationships between elementary symmetric functions and power sums.

Recurrence Relations

A key method to compute Stirling numbers of the first kind is through their recurrence relation, which expresses a number in the array in terms of previous values. The common recurrence relation, s(n, k) = s(n-1, k-1) - (n-1) * s(n-1, k), along with specific base cases such as s(0, 0) = 1 and s(n, 0) = 0 for n > 0, is fundamental for generating the triangle of these numbers.

Triangular Array Representation

The triangular arrangement is a natural way to organize the Stirling numbers of the first kind where the rows represent the number n and the columns represent the parameter k. This layout not only visually captures the recursive structure inherent to these numbers but also facilitates understanding of their interrelationships and combinatorial interpretations.

Transcript

00:01 Okay, so here we're solving the recurrence relation, a .n is equal to a .n minus 1 plus n times n plus 1 over 2.

00:19 Okay, we can actually rewrite this as a.

00:26 N minus 1 plus 1 half n squared plus 1 half n.

00:37 So that sort of separates out that second portion a little bit.

00:42 So if we take the root's characteristic equation where we replace a .n with r, we would get r is equal to an a .n minus 1 will be replaced by 1.

00:54 And then this last part, functions of n equal to 0, we just get r equals 1.

01:00 So we don't have to solve anything.

01:02 Our root is 1.

01:03 And we know that the homogeneous portion, to put a little h up there, is going to be equal to alpha times 1 to the power of n, which of course is just alpha.

01:18 So then we move on to the particular solution, and we will set that equal to our 1 half n squared plus 1 half n, and this is multiplied by 1 to the power of n.

01:39 Okay, and so this is a root.

01:42 Our only root.

01:46 And so that means that our particular solution is going to be equal to n times p2 times n squared plus p1 times n plus p not times one to the power of n, which is just p squared times n cubed, plus p1 times n squared plus p not times n.

02:27 Okay, and so now we need to solve some stuff.

02:32 This must satisfy our recurrence relation, so that means that p squared times n cubed, plus p1 times n squared plus p not times n, must be equal to, and so i'm just using our recurrence relation where a .n is equal to a n minus 1 plus 1 half n squared plus 1 half n and putting this solution into that.

03:05 Okay, so this is our a .n, and that must be equal to then p2 times n minus 1 cubed plus p1 times m minus 1 squared, plus p not times n minus 1.

03:25 So that was our a .n minus 1 portion plus 1 half n squared plus 1 half n.

03:46 Okay, then we can expand these pieces here, and we will get p2 times n cubed minus 3n squared plus 3n squared plus 3n.

04:04 Minus 1 plus p1 times n squared minus 2 n plus 1 plus p not and minus 1 plus 1 half n squared plus 1 half okay then if we subtract some of this stuff from the left hand side into the right hand side we end up with 0 is equal to negative 3 p2 plus 1 1 1 1� n squared plus 3 times p 2 minus 2 p 1 plus 1 half n plus negative p 2 plus p 1 minus p 0.

05:15 Ok, and so now our coefficients have to be equal to 0 in order for this to be equal to 0.

05:26 And so we just have a set of linear equations to solve.

05:32 So we get 0 is equal to negative 3 times p2 plus 1 half.

05:39 0 is equal to 3 times p2 minus 2 times p1 plus 1 1 1�.

05:50 And 0 is equal to negative p2 plus p 1 minus p 0.

06:00 Okay, so then we solve this, and so this one we can solve pretty well, we would minus one half, and then divide by negative three.

06:07 So that gives us that p2 is equal to one -sixth.

06:16 We can put that value into the second equation...