The case where adding a point to the data set does not change the interpolating polynomial

Question

Given a set of points $(x_0,f(x_0)),(x_1,f(x_1)),...,(x_n,f(x_n))$, the interpolating polynomial of the function $f(x)$ over the given points is defined as $$\boldsymbol{ L}(x)=\sum_{i=0}^{n}f(x_{i})\boldsymbol{ l}_i\left(x\right)$$

Where $$\boldsymbol{ l}_i\left(x\right)=\prod_{m=0 \\ m \ne i }^{n }\frac{x-x_{m}}{x_{i}-x_{m}}$$

Show that if the point $(x_{n+1},f(x_{n+1}))$ is on the graph of $L(x)$ then $\boldsymbol{ H}(x)=\boldsymbol{ L}(x)$

$$\boldsymbol{ H}(x)=\sum_{i=0}^{n+1}f(x_{i})\boldsymbol{ h}_i\left(x\right)$$

and $$\boldsymbol{ h}_i\left(x\right)=\prod_{m=0 \\ m \ne i }^{n+1 }\frac{x-x_{m}}{x_{i}-x_{m}}$$

Which means the interpolating polynomial won't change if the last added point is itself in the graph of the interpolating polynomial.

I used an algebraic way, but the result was complicated and I could not continue.

Answer 1

The result is false. If $L'(x)=L(x)$ we must have $L(x)= c e^x$, which contradicts the fact that $L(x)$ is a polynomial.

Answer 2

The probably easiest way to prove this is to prove the following statement:

Let there be $n+1$ tuples $\left(x_i,y_i\right)\in\mathbb{C}^2,i\in\left[n+1\right]$.

Then there exists exactly on polynomial $f\left(x\right)=\sum_{i=0}^{n}{a_ix^n}\in\mathbb{C}\left[x\right]$ of degree $\le n$ so that $$f\left(x_i\right)=y_i,\quad i\in\left[n+1\right]$$.

So if you construct the interpolating polynomial $p_1$ for the first $n$ points, then $p_2$ for all $n+1$ points, and find out that $p_1$ and $p_2$ are identical on all $n+1$ points, then from the theorem follows tht they are identical.

To prove the theorem itself, use the fundamental theorem of algebra.