What are Stirling Numbers of the Second Kind? [Discrete Mathematics]

[@VitalSine]

What's your favorite application, fact, or observation about Stirling numbers? Let me know in the comments!

[@gamerboyss5310]

nice vid

[@VitalSine]

@@gamerboyss5310 Thank you!

[@Penrose707]

You should mention in the video title or description box that these are specifically Stirling Numbers of the Second Kind. Great video but I was looking for information relating the Stirling numbers of the first kind.

[@VitalSine]

Oh, sorry about that! I will update the title right away, thanks for letting me know. I was going to make a Stirling numbers of the 1st kind video but haven't gotten around to it yet. I hope you find the information you're looking for 👍

[@Penrose707]

@@VitalSine Thank you! I was able to find some information floating around on the web. Are you aware of the relation between partial sums of the harmonic series and the unsigned Stirling Numbers of the First Kind? A partial sum of the series 1/n can be shown to follow the form c(n-1,k) / n! where c(n, k) are the unsigned Stirling number of the first kind with k = 2 for the case of 1/n. Essentially, each partial sum encodes in the numerator a value in the series c(n, 2). I find this to be very fascinating and would like to know more.

[@VitalSine]

@@Penrose707 That is a fascinating relation indeed! There may be more to it than this, but one reason behind this relationship is that when you consider the sequence of numerators of the partial sums of the harmonic series with denominator m!, the first numerator will just be 1. And if you consider any given numerator A_(m) of an (m)th partial sum, with denominator (m)!, the numerator of the (m+1)th partial sum, or A_m+1, will be = (m+1)*A_(m) + (m)!. We can compare this recursive formula to that of Stirling numbers of the 1st kind: C_(n+1, k) = C_(n, k-1) + (n)*C_(n, k), keeping in mind that we are considering the case where k = 2. And now what helps us show the equivalence of these two recurrence relations is the fact that C_(n, 1) = (n-1)!. C_(n+1, 2) = (n-1)! + (n)*C_(n, 2) Now if we let m + 1 = n, the recurrence relation we obtain is the same as that for our numerators in the partial sums of the harmonic series, which was A_m+1 = m! + (m+1)*A_m: C(m+2, 2) = (m)! + (m+1)*C_(m+1,2), and C_(2,2) = A_1 = 1. So the Stirling numbers of the 1st kind C_(n, 2) offset by 1 are the harmonic series' partial sum numerators because they are generated by the same recurrence relation.

[@MyMapTV]

helped me very much, great explanation! thanks alot

[@VitalSine]

Glad it helped!

[@VitalSine]

0:00 Definition 0:44 Objects and Boxes Perspective 1:30 Special cases 3:35 Recursive Formula

[@seal0118]

great video sir

[@ferdy7819]

great video

[@VitalSine]

Thank you! Glad you liked it :)

[@synhegola]

The boxes are distinct too, but their order is not important.

[@RaviSankar-xu9ig]

Dynamic programming brought me here

[@VitalSine]

Awesome! Thanks for watching :)