Differences

This shows you the differences between two versions of the page.

--- math121b:prob9-1 [2020/03/14 20:38]
pzhou created
+++ math121b:prob9-1 [2026/02/21 14:41] (current)
@@ Line 23: / Line 23: @@
 \end{aligned}
 $$
+In one of the steps, we used
+$\frac{d^{n-1}}{dx^{n-1}} (x^2-1)^n$ equals 0 when $x=\pm 1$.
+Now, we need to figure out, what is  $\frac{d^{n-1}}{dx^{n-1}} (x^2-1)^n (0)$. This is looking for $x^{n-1}$'s coefficient in $(x^2-1)^n$. Thus, we see this is only nonzero if $n-1=2m$ is an even number. That would be the binomial coefficient of $(x^2-1)^n$'s $m$-th term
+$$ (x^2-1)^n = (x^2)^n + {n \choose n-1} (x^2)^{n-1} (-1)^1 + \cdots + {n \choose m} (x^2)^{m} (-1)^{n-m} + \cdots $$
+Hence, we have (using $n-1=2m$)
+$$ (d/dx)^{2m} (x^2-1)^n (0) = (2m)! {n \choose m} (-1)^{n-m} =  (2m)! {2m+1 \choose m} (-1)^{m+1} $$
+Hence, we get, for $n=2m+1$)
+$$ \int_{-1}^1 f(x) P_n (x) dx = \frac{1}{2^n n!} (-2) (2m)! {2m+1 \choose m} (-1)^{m+1} = \frac{(-1)^{m}}{2^{2m} (2m+1)}  {2m+1 \choose m} $$
+All in all, we have coefficient
+$$ c_n = \frac{2n+1}{2} \cdot \begin{cases} 0 & n \text{ even } \cr
+\frac{(-1)^{m}}{2^{2m} (2m+1)}  {2m+1 \choose m}  & n = 2m+1
+\end{cases}
+$$
+By the way, if you can find the first few terms, that would be enough.

Lecture Notes