Let $w_1,w_2$ be Lagrange's interpolating polynomials of functions $f_1,f_2$ using the same nodes.
Prove that polynomial $w(x)=w_1(x)+w_2(x)$ is Lagrange's interpolating polynomial of function $f(x)=f_1(x)+f_2(x)$
Check if the product of these polynomial has analogical property.
So I have noticed that $l_i=\prod_{j=0,\ j\neq i}^n \frac{x-x_j}{x_i-x_j}$ would be the same for poth polynomials, as it depends only on the nodes. $$ w(x)=w_1(x)+w_2(x)=\sum_{i=0}^n f_{1_i} l_i+\sum_{i=0}^n f_{2_i} l_i=\sum_{i=0}^n (f_{1_i}+f_{2_i}) l_i=\sum_{i=0}^n f_i l_i $$ And the last term is just interpolating polynomial of $f(x)=f_1(x)+f_2(x)$, so that concludes the proof.
For 2nd part of the task I'm not sure, but I think that there is no such property, and I just need a counterexample?
Please chceck the validity of proof of the 1st part and answer my question for the 2nd part.
Your response for the first question is correct.
Regarding question 2., the interesting point is that indeed, the product $w_1w_2$ interpolates the function $f_1f_2$. But this is not the Lagrange's interpolating polynomial as it hasn't the right degree.
For example, $w_1(x)=x$ interpolates a (any) function $f_1$ with $f_1(-1)=-1$ and $f_1(1)=1$. In the same manner, $w_2(x)=-x$ interpolates a (any) function $f_2$ with $f_1(-1)=1$ and $f_1(1)=-1$.
The polynomial $w(x) = -x^2 = w_1(x) w_2(x)$ interpolates $f_1f_2$ at $-1$ and $1$. However $w$ is not the Lagrange interpolating polynomial $\mathcal L(x)$ such that $w(-1)=w(1)=-1$. $\mathcal L(x)$ is in fact the constant function equals to $-1$.