Proof that there exist constants $a_i$ such that $\int_0^1 f(x)e^xdx=\sum_{i=1}^na_if(i)$ for polynomial $f(x)$ of degree less than $n$

Question

How do I show that for positive integer $n$ and $f(x)$ all real polynomial functions of degree less than $n$ there exist constants $a_i$ such that $$\int_0^1 f(x)e^xdx=\sum_{i=1}^na_if(i)?$$ I thought maybe I can do something with the basis $(1,x,x^2,...,x^{n-1})$ of the vector space of polynomials, but I'm not sure what exactly?

Edit: $a_i$ have to be chosen independently of $f$

Answer 1

Really, this isn't about integrals. The mapping $f \mapsto \int_0^1 f(x)e^x \, dx$ is a linear functional on the $n$-dimensional space $V$ of polynomials of degree $< n$. The problem asks you to prove that in the dual space $V^{*}$, this functional is in the span of the functionals $$\phi_1 \colon f \mapsto f(1), \quad \phi_2 \colon f \mapsto f(2),\dots, \quad \phi_n \colon f \mapsto f(n).$$ Letting this span be $X \subseteq V^{*}$, it will be enough to prove that $X = V^{*}$. If we had $X \ne V^{*}$, then there would be some nonzero $f \in X^{\bot},$ that is, we'd have $f(1) = f(2) = \dots f(n) = 0$. But this is impossible, since a nonzero polynomial of degree $< n$ cannot have $n$ distinct roots.

Answer 2

A beautifull result establishes that if $F_1,...,F_r, F$ are functionals such that $\bigcap_{i=1}^r\mathrm{Ker}(F_i)\subseteq\mathrm{Ker}(F)$, then there are numbers $a_1,...,a_r$ such that $F=a_1F_1+...+a_rF_r$. The proof can be found here

Let $V=\{\mbox{polynomials with degree }\le n-1\}$, and set $F_i:V\to\mathbb{R}$ given by $F_i(f)=f(i)$, $i\in\{1,...,n\}$. There are $n$ functionals and, by the Identity Theroem, $\bigcap_{i=1}^r\mathrm{Ker}(F_i)=\{0\}\subseteq\mathrm{Ker}(F)$, where $F:V\to\mathrm{R}$ is the functional $F(f)=\int_0^1f(x)e^xdx$.

Thus, there are reals $a_1,...,a_n$ such that $F(f)=a_1F_1(f)+...+a_nF_n(f)$ for all $f\in V$, yielding the desired result.

Answer 3

Posing: $$f(x)=\sum_{j=0}^{n-1}c_jx^j$$ we have: $$\int_0^1 f(x)e^xdx=\int_0^1 \sum_{j=0}^{n-1}c_jx^je^xdx=\sum_{j=0}^{n-1}c_j\int_0^1 x^je^xdx$$ $$\sum_{i=1}^na_if(i)=\sum_{i=1}^na_i\sum_{j=0}^{n-1}c_ji^j=\sum_{j=0}^{n-1}c_j\sum_{i=1}^na_ii^j$$ So, for $0\leq j\leq n-1$: $$\sum_{i=1}^na_ii^j=\int_0^1 x^je^xdx$$ This is a system of $n$ equations in the $n$ variables $a_i$: $$MA=q$$ The rows of $M$ are linearly independent, so the system has but one solution.