Understanding a proof for divided differences

Question

suppose you are given the following proof of the statement:
$[x_0, …, x_n; f \cdot g]=\sum_{k=0}^n[x_0,…,x_k; f]\cdot [x_k,…,x_n; g]$.

Some preliminary steps:

1. Let $[x_0,…,x_n; f]=\sum_{i=0}^n\frac{f(x_i)}{(x_i-x_0)\cdot … \cdot (x_i-x_{i-1})(x_i-x_{i+1})…(x_i-x_n)}$

Then

2. $[x_0, …, x_n; L(x_0, …, x_n; f)\cdot g] = [x_0, …, x_n; f\cdot g]$. Where $L(x_1,…,x_n;f)$ is the Langrange interpolating polynomial of the Function $f$ with respect to $x_0, …, x_n$.
3. $$[x_0, …, x_i, y_0, …, y_j; (t-x_0)…(t-x_i)f]=[y_0,…,y_j; f]$$
   Appereantly the step 3) can also be deduced by identifying the coefficients at $x^n$ in the following immediate identity
4. $$L(x_0, …, x_i, y_0,…,y_j; (t-x_0)…(t-x_i)f(t))(x)=$$
   $$=(x-x_0)…(x-x_i)L(y_0,…,y_j; f)(x)$$
   Using divided differences, we can rewrite the lagrange polynomial in the normal Newton form
5. $L(x_0,…,x_n; f)(x)=\sum_{k=0}^n(x-x_0)…(x-x_{k-1})[x_0,…,x_n; f]$

The meat of the proof (using 2., 3. and 5.):

$$[x_0,…,x_n; f\cdot g]=[x_0,…,x_n; L(x_0, …, x_n; f)\cdot g]=$$
$$[x_0,…,x_n; \sum_{k=0}^n (t-x_0)…(t-x_{k-1})[x_0,…,x_k; f]\cdot g(t)]$$
$$=\sum_{k=0}^n [x_0, …, x_k; f]\cdot [x_0,…,x_n; (t-x_0)…(t-x_{k-1})g(t)]=$$
$$=\sum_{k=0}^n [x_0,…,x_k; f]\cdot [x_k,…,x_n; g]$$

How do you explain step 3 using simple calculation (ignoring step 4)? Using 1 I only get:

$$[x_0, …, x_i, y_0, …, y_j; (t-x_0)…(t-x_i)f]=$$
$$\sum_{i=0}^n\frac{(t-x_0)…(t-x_i)f(x_i)}{(x_i-x_0)\cdot … \cdot (x_i-x_{i-1})(x_i-x_{i+1})…(x_i-x_n)}$$
how do I get $\sum_{i=0}^j\frac{f(y_i)}{(y_i-y_0)\cdot … \cdot (y_i-y_{i-1})(y_i-y_{i+1})…(y_i-y_j)}=[y_0,…,y_n; f]$?

Kind regards

Answer 1

You did not consequently translate between the symbols/variables of 1. and 3. It should be \begin{align} &[x_0,...,x_i,y_0,...,y_j;(t-x_0)...(t-x_i)f(t)]\\ &=\sum_{m=0}^i\frac{(x_m-x_0)...(x_m-x_i)f(x_m)}{(x_m-x_0)...(x_m-x_{m-1})(x_m-x_{m+1})...(x_m-x_i)(x_m-y_0)...(x_m-y_j)}\\ &~~~~+\sum_{n=0}^j\frac{(y_n-x_0)...(y_n-x_i)f(y_n)}{(y_n-x_0)...(y_n-x_i)(y_n-y_0)...(y_n-y_{n-1})(y_n-y_{n+1})...(y_n-y_j)}\\ &=\sum_{n=0}^j\frac{f(y_n)}{(y_n-y_0)...(y_n-y_{n-1})(y_n-y_{n+1})...(y_n-y_j)}\\ &=[y_0,...,y_j;f] \end{align}

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You can also prove 4. without 3. if you use the characterization of the Lagrange interpolation polynomial as the (unique) minimum-degree polynomial that passes through the sampling points. That both sides in 4. have the same values in the $x_m$ and $y_n$ is then almost trivial. That the degrees have to be equal requires a little more thought.