I was given the following problem to do. For $n=1$ distinct nodes $x_{\nu}$, show that $$ [x_0,x_1,...,x_n]f=\sum_{\nu=0}^n \frac{f(x_\nu)}{\prod_{\nu \neq \mu} (x_\nu-x_\mu)} $$ Here is my attempt:
We proceed by induction on $\nu$. Base case: \begin{align*} [x_0]f=\sum_{\nu=0}^0 \frac{f(x_\nu)}{\prod_{\nu \neq \mu} (x_\nu-x_\mu)}=f(x_0). \end{align*} Inductive step: Assume that for any $k$ distinct nodes $$ [x_0,x_1,...,x_{k-1}]f=\sum_{\nu=0}^{k-1} \frac{f(x_\nu)}{\prod_{\nu \neq \mu} (x_\nu-x_\mu)} $$ Then we compute using our assumption \begin{align*} [x_0,x_1,...,x_k]f&=\frac{[x_1,x_2,...,x_k]f-[x_0,x_1,...,x_{k-1}]f}{x_k-x_0} \\ &=\frac{1}{x_k-x_0}\sum_{\nu=1}^{k} \frac{f(x_\nu)}{\prod_{\nu \neq \mu} (x_\nu-x_\mu)}+\frac{1}{x_0-x_k}\sum_{\nu=0}^{k-1} \frac{f(x_\nu)}{\prod_{\nu \neq \mu} (x_\nu-x_\mu)} \end{align*} Since each $[x_1,x_2,...,x_k]f$ and $[x_0,x_1,...,x_{k-1}]f$ have exactly $k$ distinct node then simplifying by factoring out the terms they do not have in common we get \begin{align*} &=\frac{1}{x_k-x_0}\frac{f(x_k)}{\prod_{\nu=1}^{k-1} (x_k-x_\nu)}+\frac{1}{x_k-x_0}\sum_{\nu=1}^{k-1} \frac{f(x_\nu)}{(x_k-x_\nu)\prod_{\nu \neq \mu} (x_\nu-x_\mu)} \\ &+\frac{1}{x_0-x_k}\frac{f(x_0)}{\prod_{ \nu=1}^{k-1} (x_0-x_\nu)}+\frac{1}{x_0-x_k}\sum_{\nu=1}^{k-1} \frac{f(x_\nu)}{(x_0-x_\nu)\prod_{\nu \neq \mu} (x_\nu-x_\mu)} \\ \\ \end{align*} Now here is my problem I need to get $$ =\frac{f(x_k)}{\prod_{\nu=0}^{k-1}(x_k-x_\nu)}+\frac{f(x_0)}{\prod_{ \nu=1}^{k} (x_0-x_\nu)}+\sum_{\nu=1}^{k-1} \frac{f(x_\nu)}{\prod_{\nu \neq \mu,\nu=0}^{k} (x_\nu-x_\mu)} $$ But I don't think my two sums combine to get this by they should. If they do I don't see how or am I doing this wrong. Thanks in advance.
You're pretty close; you should have pulled out $\left(x_{\nu}-x_k\right)$ and $\left(x_{\nu}-x_0\right)$ in the denominators in the last step and you would have been there. Your notation for the products is abysmal and makes it hard to read. If we adopted notation such as $$P^{(k)}_{ij\dots}(x)=\prod_{\begin{array}{c}\mu=0\\\mu\not\in\left\{i,j,\dots\right\}\end{array}}^k\left(x-x_{\mu}\right)$$ Where the index of multiplication (is that the right word?) is implicitly $\mu$ and the lower bound is implicitly $0$ and rewrote your question from there... $$\begin{align}\left[x_0,x_1,\dots,x_k\right]f&=\frac{\left[x_1,x_2,\dots,x_k\right]f-\left[x_0,x_1,\dots,x_{k-1}\right]f}{x_k-x_0}\\ &=\frac1{x_k-x_0}\sum_{\nu=1}^k\frac{f\left(x_{\nu}\right)}{P^{(k)}_{0\nu}\left(x_{\nu}\right)}-\frac1{x_k-x_0}\sum_{\nu=0}^{k-1}\frac{f\left(x_{\nu}\right)}{P^{(k)}_{\nu k}\left(x_{\nu}\right)}\\ &=\frac1{x_k-x_0}\frac{f\left(x_{k}\right)}{P^{(k)}_{0k}\left(x_{k}\right)}+\frac1{x_k-x_0}\sum_{\nu=1}^{k-1}\frac{f\left(x_{\nu}\right)}{P^{(k)}_{0\nu}\left(x_{\nu}\right)}\\ &\quad-\frac1{x_k-x_0}\frac{f\left(x_{0}\right)}{P^{(k)}_{0k}\left(x_{0}\right)}-\frac1{x_k-x_0}\sum_{\nu=1}^{k-1}\frac{f\left(x_{\nu}\right)}{P^{(k)}_{\nu k}\left(x_{\nu}\right)}\\ &=\frac{f\left(x_{k}\right)}{P^{(k)}_{k}\left(x_{k}\right)}+\frac1{x_k-x_0}\sum_{\nu=1}^{k-1}\frac{\left(x_{\nu}-x_0\right)f\left(x_{\nu}\right)}{P^{(k)}_{\nu}\left(x_{\nu}\right)}\\ &\quad+\frac{f\left(x_{0}\right)}{P^{(k)}_{0}\left(x_{0}\right)}-\frac1{x_k-x_0}\sum_{\nu=1}^{k-1}\frac{\left(x_{\nu}-x_k\right)f\left(x_{\nu}\right)}{P^{(k)}_{\nu}\left(x_{\nu}\right)}\\ &=\frac{f\left(x_{k}\right)}{P^{(k)}_{k}\left(x_{k}\right)}+\sum_{\nu=1}^{k-1}\frac{\left(x_{\nu}-x_0-x_{\nu}+x_k\right)f\left(x_{\nu}\right)}{\left(x_k-x_0\right)P^{(k)}_{\nu}\left(x_{\nu}\right)}+\frac{f\left(x_{0}\right)}{P^{(k)}_{0}\left(x_{0}\right)}\\ &=\frac{f\left(x_{k}\right)}{P^{(k)}_{k}\left(x_{k}\right)}+\sum_{\nu=1}^{k-1}\frac{f\left(x_{\nu}\right)}{P^{(k)}_{\nu}\left(x_{\nu}\right)}+\frac{f\left(x_{0}\right)}{P^{(k)}_{0}\left(x_{0}\right)}\end{align}$$