Given n+1 data pairs $(x_0,y_0)...(x_n,y_n)$ for j=0,1,2...,n we have $p_j=\prod_{i\neq j}(x_j-x_i)$ and $\psi(x)=\prod_{i=0}^n(x-x_i)$.

Question

Given n+1 data pairs $(x_0,y_0)...(x_n,y_n)$ for j=0,1,2...,n we have $p_j=\prod_{i\neq j}(x_j-x_i)$ and $\psi(x)=\prod_{i=0}^n(x-x_i)$.

I am having trouble determining what $\psi(x_j)$ is and what $\psi'(x_j)$ would be.

I feel like $\psi(x_j)= 0$ because it would contain the $x_j-x_j$ term... But I feel like I am missing something...

Answer 1

You aren't missing something; $\psi(x_j)=0$ and the reason is the one you wrote. As for finding $\psi'(x_j)$ one way is to "differenciate logarithmically"; $$ \log\psi(x)=\sum_{i=0}^n\log(x-x_i); $$ $$ \frac{d}{dx}\left[\log\psi(x)\right]=\sum_i\frac{d}{dx}\log(x-x_i)=\sum_i\frac{1}{x-x_i}. $$ Now, in the other hand we know that $\frac{d}{dx}\log\psi(x)=\frac{\psi'(x)}{\psi(x)}$, so $$ \psi'(x)=\psi(x)\sum_i\frac{1}{x-x_i}=\sum_ip_i(x). $$ Now plug in $x=x_j$.

Answer 2

Hint: $\psi'(x) = \sum_{j=0}^n \left(\prod_{i \ne j} (x-x_i)\right)$ by the product rule.