1. Recall the Stirling numbers of the 2nd kind are S(n,k) :=...

1. Recall the Stirling numbers of the 2nd kind are $S(n,k) := \# \text{ set partitions of } \{1,2,..., n\} \text{ into }$ \newline $k$ (non-empty) blocks. We saw $F_k(x) := \sum_{n\ge 0} S(n,k)x^n$ satisfies $F_k(x) = \frac{x^k}{(1-x)(1-2x)...(1-kx)}$ \newline for any $k \ge 1$. Find the partial fraction decomposition of $F_k(x)$, i.e., find the numbers $a_j \in \mathbb{Q}$ \newline for which $F_k(x) = a_0 + \frac{a_1}{(1-x)} + \frac{a_2}{(1-2x)} + ... + \frac{a_k}{(1-kx)}$. Conclude that $S(n,k) = \sum_{j=0}^k a_j j^n$. \newline Hint: Clear denominators, and then plug in $x = \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, ..., \frac{1}{k}$ and finally $x = 0$.

Transcript

00:01 In this problem we are going to find the power series of the function 3 divided by x plus 2 times x minus 2 using the partial fractions.

00:16 First of all recall that the power series of a function a divided by 1 minus r when magnitude of r is less than 1 is given by summation n runs from 1 to infinity a times r to the power n in our case observe that the given function is not of this format so we are going to use the partial fraction to convert the given function into this format let's do that the given function will be equal to a divided by x plus 2 plus b divided by x minus 1 we are going to find the constant a and b using partial fraction.

01:05 For that what we will do is so we multiply both sides of the equation by the term x plus 2 times x minus 1.

01:13 This will give us 3 is equal to a times x minus 1 plus b times x plus here observe that if we let x is equal to 1 this term will be equal to 0 so we can find the value of b so which implies 3 is equal to a times 0 plus b times 3 which implies that b is equal to 1 and similarly if we substitute x is equal to minus 2 this expression will become 0 so that we can find the value of a so 3 will be equal to a times minus 2 minus 1 which is minus 3 plus b times minus 2 plus 2 plus 2 plus 2 is 0 which implies that the value of a is equal to minus 1.

02:08 So if we substitute the value of a and b in this expression, we will get our given function as minus 1 divided by x plus 2 plus 1 divided by x minus 1.

02:23 Now observe that this expression is of the form a divided by 1 minus r.

02:29 Now let's consider this expression alone.

02:35 Here observe that we can rewrite this expression as minus 1 divided by 2 times 1 plus x by 2.

02:44 We are taking 2 commonly outside from the denominator and it will be equal to minus 1 divided by 2 divided by 1 minus x divided by 2.

02:56 Here we can consider a as minus 1 by 2 and r as minus x by 2.

03:05 That is if modulus of minus x by 2 is less than 1, then what we can say is the power series will be equal to summation and running from 1 to infinity a times, which is minus 1 by 2 times r to the power n, which is minus x divided.

03:29 By 2 to the power n.

03:32 Here observe that this expression implies that x modulus of x is less than 2...

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