How to develop Lagrange interpolation polynomial basis to exponential basis I have the following question:
"Polynomial interpolation doesn't fit function with fast change. Instead of polynomial interpolation, you can develop an exponential interpolation using the basis 1,exp(x),exp(2x),exp(3x),...,exp(nx). To do so, adjust Lagrange interpolation polynomial to interpolation exponential. Write extensively the base "polynomials" li(x)"
Before posting this I really tried looking everywhere for a similar question.
Therefor, I would love some help with that:)
Thank you!
 A: Let $x_0,..., x_n$ be distinct elements of the field $F$. Let $V$ be the vector space over $F$ with basis $\{1, \exp(x), \exp(2x),..., \exp(nx) \}$
Define $L_i: V \to F$ by
$$L_i(f) = f(x_i), \space\space\space\space\space 0 \le i \le n$$
We want to find a basis for $V$, $\{P_0, P_1,..., P_n\}$, such that $\{L_0, L_1,...,L_n\}$ is its dual basis.
If such a basis exists, it must satisfy:
$$L_j(P_i) = P_i(x_j) = \delta_{ij}$$
Let $$P_i = \prod_{j \ne i} \frac{\exp(x) - \exp(x_j)}{\exp(x_i) - \exp(x_j)}$$
Then $P_i \in V$ for $0 \le i \le n$, and they satisfy the above criterion, so if you can show $P_0,...,P_n$ are linearly independent (an exercise for you), we'd have that $\{L_0,..., L_n\}$ is indeed dual to $\{P_0,...,P_n\}$. This means that $\forall f \in V$,
$$f = \sum_{i = 0}^n f(x_i) P_i$$
I'm not completely sure this is what you are going for, but it is a variant of Lagrange Interpolation using exponentials.
I used techniques here developed in Linear Algebra by Hoffman and Kunze.
A: In Lagrange interpolation, you construct polynomials $l_i(x),i=1,2,\dots,n$ of degree $n-1$, such that $l_i(x_j)=\begin{cases} 1 & i=j \\ 0 & i \neq j \end{cases}$. You do that by defining $L_i(x)=\prod_{j \neq i} (x-x_j)$, which satisfies the second requirement, and then normalizing by taking $l_i(x)=L_i(x)/L_i(x_i)$.
You can do exactly the same thing with exponentials: an exponential function that only vanishes at $x_j$ is $\exp(x)-\exp(x_j)$, so $E_i(x)=\prod_{i \neq j} \exp(x)-\exp(x_j)$ vanishes exactly at the points $i \neq j$. Now normalize that to get something similar to a Lagrange polynomial.
