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I am starting to study linear algebra, in the first pages of the book appeared a problem that I could not tackle. What polynomial $p$ satisfies $\int_{-1}^{1}p(y)\,dy=0$ and $\int_{-1}^{1}yp(y)\,dy=1$. I have some knowledge of differential and integral calculus, but I never learned how to solve equations with integrals. How are they solved and when are you supposed to learn to solve them ?

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  • $\begingroup$ Take a polynomial and compute explicitly the two integrals. Nothing else is needed. $\endgroup$
    – John B
    Dec 22 '17 at 22:04
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The integral equations you've mentioned aren't that hard and can be just solved and conclusions can be made just by using brute force :

Let $p(y)$ be a polynomial, such:

$$p(y) = a_ny^n + a_{n-1}y^{n-1} + \dots + a_1y + a_0$$

Then the $2$ given integral equations will be :

$$\int_{-1}^1p(y)dy = \int_{-1}^1(a_ny^n + a_{n-1}y^{n-1} + \dots + a_1y + a_0)dy = 0 $$ $$\Rightarrow$$ $$\bigg[\frac{a_n}{n+1}y^{n+1} + \frac{a_{n-1}}{n}y^{n} + \dots + \frac{a_1}{2}y^2 + a_0 y \bigg]_{-1}^1 = 0$$

$$\text{and}$$

$$\int_{-1}^1yp(y)dy = \int_{-1}^1 y(a_ny^n + a_{n-1}y^{n-1} + \dots + a_1y + a_0)dy =1$$

$$\Leftrightarrow$$

$$\int_{-1}^1 (a_ny^{n+1} + a_{n-1}y^{n} + \dots + a_1y^2 + a_0y)dy=1$$

$$\Leftrightarrow$$

$$\bigg[\frac{a_n}{n+2}y^{n+2} + \frac{a_{n-1}}{n+1}y^{n+1} +\dots +\frac{a_1}{3}y^3 + \frac{a_0}{2}y^2 \bigg]_{-1}^1=1$$

Finally, for the polynomial $p(y)$ and it's coefficients and order, you'll get the system :

$$\begin{cases} \bigg[\frac{a_n}{n+1}y^{n+1} + \frac{a_{n-1}}{n}y^{n} + \dots + \frac{a_1}{2}y^2 + a_0 y \bigg]_{-1}^1 = 0 \\ \bigg[\frac{a_n}{n+2}y^{n+2} + \frac{a_{n-1}}{n+1}y^{n+1} +\dots +\frac{a_1}{3}y^3 + \frac{a_0}{2}y^2 \bigg]_{-1}^1=1 \end{cases}$$

Can you derive a conclusion from now on ?

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  • $\begingroup$ From this equations, we can see that not only one polinomy satisfy this problem, there are infinite polinomies. We have arbritrary long number of variables $a_i$ but only 2 equations. $\endgroup$
    – Daniel Pol
    Dec 22 '17 at 23:00
  • $\begingroup$ @DanielPol There may be infinite polynomials satisfying the concluded system of equations, but there are a specific subset of the set of all polynomials over $\mathbb R$. Specifically, you can derive a relation for $n$. $\endgroup$
    – Rebellos
    Dec 22 '17 at 23:02
  • $\begingroup$ Yes. Only a specific set of polinomies. $\endgroup$
    – Daniel Pol
    Dec 22 '17 at 23:04
  • $\begingroup$ @DanielPol Then the question is answered, as you find a polynomial that satisfies the equations given. The term "polynomial" can mean a general form, which will satisfy the elements of the polynomial group that you'll have reduced to. $\endgroup$
    – Rebellos
    Dec 22 '17 at 23:41
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Since the question doesn't ask for all polynomials, we can just find one, so it makes sense to try looking for the smallest. A constant can't work (why not?), so let's try a linear function. $x$ is odd, so something of the form $p(y) = ay$ seems likely, since the constant term must be zero as well.

Putting this into the second integral gives $$1=\int_{-1}^1 ay^2 \, dy =a\left[\frac{y^3}{3}\right]_{-1}^1=\frac{2a}{3}.$$ So, $p(y) = \frac{3}{2} y$ satisfies the given equation.

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The first equation is true for any odd function, so you could try $\;p(x)=ax+bx^3\;,\;\;a,b\in\Bbb R\;$, or something of the like.

But if $\;p\;$ is an odd polynomial, then $\;yp(y)\;$ is an even one, so

$$\int_{-1}^1 yp(y)\,dy=2\int_0^1 yp(y)\,dy$$

Observe carefully the above and I bet you'll be able to come up with some ideas...

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