# Lecture Notes

math121b:10.7

### Problem 10.7

Show that $$\frac{d^{l-m}}{dx^{l-m}} (x^2-1)^{l} = \frac{(l-m)!}{(l+m)!} (x^2-1)^m \frac{d^{l+m}}{dx^{l+m}} (x^2-1)^l$$

### Solution

We first rearrange the factorials, so that the differential operators looks nicer $$\tag{*} \frac{1}{(l-m)!} \frac{d^{l-m}}{dx^{l-m}} (x^2-1)^{l} = (x^2-1)^m \frac{1}{(l+m)!} \frac{d^{l+m}}{dx^{l+m}} (x^2-1)^l$$

Next, we open up the parenthesis of $(x^2-1)^l$, we get $$(x+1)^l (x-1)^l = \underbrace{(x+1) \cdots (x+1)}_{ l \z{ times }} \underbrace{(x-1) \cdots (x-1)}_{ l \z{ times }}.$$ Note we have $2l$ factors.

Then, we hit it with $k=l-m$ many $d/dx$s $$(d/dx)^k = \underbrace{(d/dx)}_{(k)} \cdots \underbrace{(d/dx)}_{(1)}.$$ We label them according to the order that they act. The above $2l$ factors $(x\pm 1)$, each of them being linear, can only endure being hit by $d/dx$ once, hence there are $$N=(2l) (2l-1) \cdots (2l-k+1)$$ many ways to matching the $k$ many $d/dx$ to the $2l$ many factors, so that no factors are assigned twice. This is like marking a row of $2l$ boxes by number $1, \cdots k$.

However, in each of the $N$ ways of matching or marking, the result only depends on which boxes are marked, it does not care about the actual number marked on the box. Thus, we can group the $N$ terms into 'bunches', each bunch is of size $k!$, corresponding to the markings that differ only by the numbering of $1,\cdots, k$. Recall $k=l - m$. The prefactor $\frac{1}{(l-m)!}$ can now be used to cancel out the repetition in the bunches.

To summarize, we have interpreted the LHS as $$\frac{1}{(l-m)!} \frac{d^{l-m}}{dx^{l-m}} (x^2-1)^{l} = \z{ sum of all possible ways to delete (l-m) factors out of } \underbrace{(x+1) \cdots (x+1)}_{ l \z{ times }} \underbrace{(x-1) \cdots (x-1)}_{ l \z{ times }}.$$ There are ${2l \choose l-m}$ terms in the sum, each term is a product of $l+m$ linear factors.

Similarly, we can interpret the RHS's partially as $$\frac{1}{(l+m)!} \frac{d^{l+m}}{dx^{l+m}} (x^2-1)^{l} = \z{ sum of all possible ways to delete (l+m) factors out of } \underbrace{(x+1) \cdots (x+1)}_{ l \z{ times }} \underbrace{(x-1) \cdots (x-1)}_{ l \z{ times }}.$$ There are ${2l \choose l+m}$ terms in the sum, each term is a product of $l-m$ linear factors.

Notice that $${n_1 + n_2 \choose n_1} = {n_1 + n_2 \choose n_2}$$ since given a row of people of size $n_1 + n_2$, to pick $n_1$ of them to step forward, is the same as pick $n_2$ of them to step backward.

Thus, we ${2l \choose l+m} = {2l \choose l-m}$, so the two gigantic sums have the same number of terms.

Claim: there is a ways to have a 1-to-1 correspondence of the two sums, so that the term from the sum on RHS, when multiplied by $(x^2-1)^m$ will equal to the corresponding term in the first sum.

I will leave you to verify this claim in the example of $l = 3$, $m=1$. 