Federico posted a question today about what equals. This is what I tried out, so let me know what you think.

He showed in class how . By differentiating both sides we get . Thus we plug 1 and get that .

Thus we have that

I tried to make sense of this combinatorially and think that it tells us that for a set such that *the number of ways to choose an element of ** and creating a subset containing it *is equal to *the number of ways to choose a subset of ** and picking out an element from it. *

For example take . For the first case say we pick the element 2, then we can create the subsets . Since we can pick 3 elements and do this for each one, we have a total of . Now for the second question we can choose the empty set but have no way of choosing an element ( 0 ways), choose a 1-set and then an element( ways), a 2-subset and then an element( ways), or a 3-subset and then an elements( ways). Thus giving us a total of 0+3+6+3=12 ways.

Cool way to use analysis and algebra to get a not so obvious(at least to me) combinatorial result.

For an algebraic proof we can use the identity $k*\{n\choose k} = n*\{(n-1)\choose(k-1)}$ and then just factor the n out, in case the interval of convergence for such series does not include x=1.

Nice Paco. Dude the latex is a bit different here. To type a formula in WordPress type without the signs .

***To type a formula in WordPress type $ latex Your LaTex code $ without the space between $ and latex.

Three nice proofs! (Notice that there is no issue of convergence here because this is an equality of polynomials, so plugging in is fine.)

I know is kind of late for this but I just saw the video of lecture 2. Another nice interpretation is that is the “weight” of if we assume that the weight of a single element is . Then, if you take any element, the number of subsets containing it is so each element in appears times in . Now just multiply by because there are elements.

Sorry, is in both strange things.