# Question #0ffa1

Dec 22, 2017

A bit of experimenting with linear polynomials of the form $p \left(y\right) = a y + b$ reveals that $p \left(y\right) = \frac{3}{2} y$ is one example that works.

#### Explanation:

To start with, you could hope that a linear polynomial $p \left(y\right) = a y + b$ will work.

The condition that ${\int}_{- 1}^{1} p \left(y\right) \mathrm{dy} = 0$ implies that
$\frac{1}{2} a {y}^{2} + b y {|}_{- 1}^{1} = 0$, or $2 b = 0$, so $b = 0$.

Since $y p \left(y\right) = a {y}^{2} + b y$, the condition that ${\int}_{- 1}^{1} y p \left(y\right) \mathrm{dy} = 1$ implies that $\frac{1}{3} a {y}^{3} + \frac{1}{2} b {y}^{2} {|}_{- 1}^{1} = 1$. This leads to $\frac{2}{3} a = 1$ so that $a = \frac{3}{2}$.

Dec 22, 2017

See below.

#### Explanation:

Calling ${p}_{n} \left(x\right) = {\sum}_{k = 0}^{n} {a}_{k} {x}^{k}$ we have

${\int}_{-} {1}^{1} {\sum}_{k = 0}^{n} {a}_{k} {x}^{k} \mathrm{dx} = {\sum}_{k = 0}^{n} {a}_{k} / \left(k + 1\right) \left(1 - {\left(- 1\right)}^{k + 1}\right) = 0$ or

${\sum}_{k = 0}^{\left\lfloor \frac{n}{2} \right\rfloor} {a}_{2 k} / \left(2 k + 1\right) = 0$

and

${\int}_{-} {1}^{1} x {p}_{n} \left(x\right) \mathrm{dx} = {\int}_{-} {1}^{1} {\sum}_{k = 0}^{n} {a}_{k} {x}^{k + 1} \mathrm{dx} = {\sum}_{k = 0}^{n} {a}_{k} / \left(k + 2\right) \left(1 - {\left(- 1\right)}^{k + 2}\right) = 1$ or

${\sum}_{k = 0}^{\left\lfloor \frac{n}{2} \right\rfloor} {a}_{2 k - 1} / \left(2 k + 1\right) = \frac{1}{2}$

Then, the polynomials obeying the conditions

$\left\{\begin{matrix}{\sum}_{k = 0}^{\left\lfloor \frac{n}{2} \right\rfloor} {a}_{2 k} / \left(2 k + 1\right) = 0 \\ {\sum}_{k = 0}^{\left\lfloor \frac{n}{2} \right\rfloor} {a}_{2 k - 1} / \left(2 k + 1\right) = \frac{1}{2}\end{matrix}\right.$

are solutions.