Show uniqueness of polynomial

Solving this system, you will get the unique polynomial $q(x)$. In summary, we discussed the problem of finding a unique polynomial $q(x)$ with the property that $\int_{-1}^1 p(x) e^x dx=\int_0^1 p(x) q(x) dx$ for each $p(x) \in \mathbb{R}[x]_{\leq n}$, and how to prove its uniqueness using the Riesz representation theorem. We also explored the case of $n=1$ and how to find the polynomial $q(x)$ by setting up two equations involving $q(x)$ and solving for its coefficients.
  • #1
evinda
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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)
 
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  • #2
evinda said:
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)
 
  • #3
I like Serena said:
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 $p(x)$.
 
  • #4
Euge said:
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)
Euge said:
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)
 
  • #5
evinda said:
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 $v\in V$.
evinda said:
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 $b$.
 

1. What is a polynomial?

A polynomial is a mathematical expression that consists of variables and coefficients, combined using only the operations of addition, subtraction, and multiplication. The variables can only have non-negative integer exponents.

2. How do you show uniqueness of a polynomial?

To show uniqueness of a polynomial, you must prove that it has a unique set of coefficients for each term and that the variables have unique exponents. This can be done by setting up a system of equations and solving for the coefficients, or by using mathematical induction.

3. Why is it important to show uniqueness of a polynomial?

Showing uniqueness of a polynomial is important because it ensures that there is only one possible solution for the polynomial. This is crucial in mathematical proofs and in applications of polynomials in various fields such as physics and engineering.

4. What are some common methods used to show uniqueness of a polynomial?

Some common methods used to show uniqueness of a polynomial include mathematical induction, proof by contradiction, and solving a system of equations. Other techniques such as using the Fundamental Theorem of Algebra or the Rational Root Theorem may also be employed depending on the specific polynomial.

5. Can a polynomial have more than one unique solution?

No, a polynomial can only have one unique solution. This is because the variables and coefficients are defined to have specific values and cannot be changed. If a polynomial has more than one solution, it is not considered unique.

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