MHB Can a Unique Polynomial Satisfy Specific Integral Equations?

[evinda]

Hello! (Wave)

Let $\mathbb{R}[x]_{ \leq n}$ be the vector space of the real polynomials of degree $\leq n$, where $n$ a natural number. I want to show that there is a unique $q(x) \in \mathbb{R}[x]_{\leq n}$, with the property that $\int_{-1}^1 p(x) e^x dx=\int_0^1 p(x) q(x) dx$, for each $p(x) \in \mathbb{R}[x]_{\leq n}$. For $n=1$, I want to find the above polynomial $q(x)$.Could you give me a hint how to prove the uniqueness of a polynomial $q$ with the property that $\int_{-1}^1 p(x) e^x dx=\int_0^1 p(x) q(x) dx$ ? (Thinking)

 

[I like Serena]

Hello! (Wave)

Let $\mathbb{R}[x]_{ \leq n}$ be the vector space of the real polynomials of degree $\leq n$, where $n$ a natural number. I want to show that there is a unique $q(x) \in \mathbb{R}[x]_{\leq n}$, with the property that $\int_{-1}^1 p(x) e^x dx=\int_0^1 p(x) q(x) dx$, for each $p(x) \in \mathbb{R}[x]_{\leq n}$. For $n=1$, I want to find the above polynomial $q(x)$.Could you give me a hint how to prove the uniqueness of a polynomial $q$ with the property that $\int_{-1}^1 p(x) e^x dx=\int_0^1 p(x) q(x) dx$ ? (Thinking)

Hey evinda!

Suppose that for a given $p(x)$ with $\int_0^1 p(x)dx\ne 0$ we assume that $q(x)=q_0$.
Then:
$$\int_{-1}^1 p(x) e^x dx=\int_0^1 p(x) q(x) dx
=\int p(x)q_0dx \quad\Rightarrow\quad
q_0=\frac{\int_{-1}^1 p(x) e^x dx}{\int_0^1 p(x)dx}
$$
Similarly we can assume that $q(x)=q_1 x$ and find a different solution, can't we? (Wondering)

It seems to me that $q(x)$ is not unique. (Worried)

 

[Euge]

It seems to me that $q(x)$ is not unique. (Worried)

It is indeed the case that $q(x)$ is unique. Let the vector space $\Bbb R[x]_{\le n}$ be equipped with the inner product defined by setting $\langle p,q\rangle = \int_0^1 p(x)q(x)\, dx$. The mapping $L : \Bbb R[x]_{\le n} \to \Bbb R$ defined by the equation $L(p) = \int_{-1}^1 p(x)e^x\, dx$ is a linear functional on the finite-dimensional normed linear space $(\Bbb R[x]_{\le n}, \langle\cdot,\cdot\rangle)$, so by the Riesz representation theorem there is a unique $q(x) \in \Bbb R[x]_{\le n}$ such that $L(p) = \langle p,q\rangle$ for all $p(x)\in \Bbb R[x]$, i.e., $\int_{-1}^1 p(x)e^x\, dx = \int_0^1 p(x)q(x)\, dx$ for all $p(x)\in \Bbb R[x]$.

In the case $n = 1$, write $p(x) = a + bx$ and set up two equations with $p(x) = 1$ and then $p(x) = x$ to obtain the solution for $p(x)$.

 

[evinda]

It is indeed the case that $q(x)$ is unique. Let the vector space $\Bbb R[x]_{\le n}$ be equipped with the inner product defined by setting $\langle p,q\rangle = \int_0^1 p(x)q(x)\, dx$. The mapping $L : \Bbb R[x]_{\le n} \to \Bbb R$ defined by the equation $L(p) = \int_{-1}^1 p(x)e^x\, dx$ is a linear functional on the finite-dimensional normed linear space $(\Bbb R[x]_{\le n}, \langle\cdot,\cdot\rangle)$, so by the Riesz representation theorem there is a unique $q(x) \in \Bbb R[x]_{\le n}$ such that $L(p) = \langle p,q\rangle$ for all $p(x)\in \Bbb R[x]$, i.e., $\int_{-1}^1 p(x)e^x\, dx = \int_0^1 p(x)q(x)\, dx$ for all $p(x)\in \Bbb R[x]$.

Which version of the Riesz representation theorem do we use? (Thinking)

In the case $n = 1$, write $p(x) = a + bx$ and set up two equations with $p(x) = 1$ and then $p(x) = x$ to obtain the solution for $p(x)$.

For $p(x)=1$ we get that $\int_0^1 q(x) dx=e-\frac{1}{e}$.

For $p(x)=x$ we get that $\int_0^1 x q(x) dx=xe^x-\left( e-\frac{1}{e}\right)$.

How do we find from the above equalities the function $q(x)$ ? (Thinking)

 

[Euge]

Which version of the Riesz representation theorem do we use? (Thinking)

Baby Version: If $L : V \to \Bbb C$ is a linear functional on a finite dimensonal inner product space, then there is a unique $w\in V$ such that $L(v) = \langle v,w\rangle$ for all $v\in V$.

For $p(x)=1$ we get that $\int_0^1 q(x) dx=e-\frac{1}{e}$.

For $p(x)=x$ we get that $\int_0^1 x q(x) dx=xe^x-\left( e-\frac{1}{e}\right)$.

How do we find from the above equalities the function $q(x)$ ? (Thinking)

You are suppose to compute the integrals involving $q$ using the form $q(x) = a + bx$. This will result in a system of two equations in two unknowns $a$ and $b$.