I'm trying to understand a proof in "Interpolation and Approximation by Polynomials" by Phillips.
Let me quote (page 253):
"For $k\geq 1$ we begin with $$B_{n+k}^{(k)}(f;x)=\frac{(n+k)!}{n!} \sum\limits_{r=0}^n \Delta^k f \left( \frac{r}{n+k} \right) \binom{n}{r} x^r (1-x)^{n-r}$$ and replace $n$ by $n-k$. Then, using $$\frac{\Delta^m f(x_0)}{h^m}=f^{(m)}(\xi)$$ with $h=1/n$, we write $$\Delta^k f \left( \frac{r}{n} \right) = \frac{f^{(k)}(\xi_r)}{n^k},$$ where $r/n<\xi_r<(r+k)/n$. Thus $$B_n^{(k)}(f;x)=\sum\limits_{r=0}^{n-k} c_k f^{(k)}(\xi_r) x^r (1-x)^{n-k-r}$$ [...]."
While he defines $c_0=c_1=1$ and $c_k=\left( 1 - \frac{1}{n} \right)\left( 1 - \frac{2}{n} \right) \cdot ... \cdot \left( 1 - \frac{k-1}{n} \right), 2 \leq k\leq n.$
I don't understand this because:
If i replace $n$ by $n-k$: $$B_{n}^{(k)}(f;x)=\frac{(n)!}{(n-k)!} \sum\limits_{r=0}^{n-k} \Delta^k f \left( \frac{r}{n} \right) \binom{n-k}{r} x^r (1-x)^{n-r}$$
Than insert $\frac{f^{(k)}(\xi_r)}{n^k}$: $$B_{n}^{(k)}(f;x)=\frac{(n)!}{(n-k)!} \sum\limits_{r=0}^{n-k} \frac{f^{(k)}(\xi_r)}{n^k} \binom{n-k}{r} x^r (1-x)^{n-r}$$
Now that his equality would be correct it has to be: $c_k = \frac{n!}{(n-k)!n^k}\binom{n-k}{r}$
Why this should hold? I'm not even seeing why the $r$ shoudl dissapear.
Sincerely, 154085
I am sure that you have seen that :
$$c_k:=\prod_{i=1}^{k-1}(1-\frac{i}{n})=\frac{1}{n^{k-1}}\prod_{i=1}^{k-1}(n-i)=\frac{1}{n^k}\prod_{i=0}^{k-1}(n-i)=\frac{1}{n^k}\frac{n!}{(n-k)!} $$
So if you want :
$$c_k=\frac{n!}{(n-k)!n^k}\begin{pmatrix}n-k\\r\end{pmatrix} $$
This leads to :
$$\begin{pmatrix}n-k\\r\end{pmatrix}=1\Rightarrow r=0\text{ or } r=n-k $$
In particular your last remark is right, there is no reason for the $r$ to disappear. There must be the $\begin{pmatrix}n-k\\r\end{pmatrix}$ term missing in the formula given in the book.