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$p(x)$ be a polynomial of degree 7 with real coefficients such that $p(π) = √3$ and $$\int_{-π}^{π} x^{k}p(x) = 0, \text{ for} \; 0\leq k \leq 6. $$ I have to find the value of $p(-π)$ and $p(0).$

My initial thoughts were to suppose that $p(x) = a_0 + a_1x + a_2x^2 + \dots + a_7x^7$ and solve it for the coefficients but it was very tedious and like impossible to solve.

Any insight. Thank you.

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Hint: What can be said about similar integrals of polynomial $q(x) = -p(-x)$?

Also helpful: https://en.wikipedia.org/wiki/Legendre_polynomials

Ok, a more thorough solution.

For polynomial $r(x) = \frac12[p(x)+p(-x)]=a_0+a_2x^2+a_4x^4+a_6x^6$, we have: $$ 2\int_{-\pi}^{\pi} x^kr(x)dx = \int_{-\pi}^{\pi} x^kp(x)dx + (-1)^{k+1} \int_{-\pi}^{\pi} x^kp(x)dx = 0,\qquad \text{for }0\le k\le6. $$

Consider integral of non-negative function $r^2(x)$: $$ \int_{-\pi}^{\pi} r^2(x) dx = a_0\int_{-\pi}^{\pi} x^0r(x) dx + \ldots+a_6\int_{-\pi}^{\pi} x^6r(x) dx = 0. $$ It can be only if $r(x)\equiv 0$. So all even coefficients are zero and p(x) is an odd function. Thus, $p(0)=0$ and $p(-\pi)=-p(\pi) = -\sqrt3$

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  • $\begingroup$ Sorry to say but I didn't find the hint useful. Will you elaborate please $\endgroup$ – hiren_garai Sep 18 '18 at 14:29
  • $\begingroup$ Now I get it. Thank you for your time. $\endgroup$ – hiren_garai Sep 18 '18 at 15:05
  • $\begingroup$ I think there will be + where $r(x)$ is defined. $\endgroup$ – hiren_garai Sep 18 '18 at 15:21
  • $\begingroup$ Yeah, you are right. I was thinking of p(x)−q(x) $\endgroup$ – Vasily Mitch Sep 18 '18 at 15:26
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Let's start with $$\int_{-\pi}^\pi p(x)dx = 0$$Note that for all of the terms with $x$ having an odd degree, the integral of the function will have that term with even degree, so substituting $\pi$ for $x$ or $-\pi$ will produce the same value. This gives us our key insight: We solely care about the terms with even degree. Hence, the above integral becomes $$2\pi(a_0+\frac{a_2}3\pi^2+\frac{a_4}5\pi^4+\frac{a_6}7\pi^6)=0\to a_0+\frac{a_2}3\pi^2+\frac{a_4}5\pi^4+\frac{a_6}7\pi^6 =0$$We can similarly construct equations with $k=1,2,...6$ to get $$\frac{a_1}3+\frac{a_3}5\pi^2+\frac{a_5}7\pi^4+\frac{a_7}9\pi^6=0$$$$\frac{a_0}3+\frac{a_2}5\pi^2+\frac{a_4}7\pi^4+\frac{a_6}9\pi^6=0$$$$\frac{a_1}5+\frac{a_3}7\pi^2+\frac{a_5}9\pi^4+\frac{a_7}{11}\pi^6=0$$$$\frac{a_0}5+\frac{a_2}7\pi^2+\frac{a_4}9\pi^4+\frac{a_6}{11}\pi^6=0$$$$\frac{a_1}7+\frac{a_3}9\pi^2+\frac{a_5}{11}\pi^4+\frac{a_7}{13}\pi^6=0$$$$\frac{a_0}7+\frac{a_2}9\pi^2+\frac{a_4}{11}\pi^4+\frac{a_6}{13}\pi^6=0$$And finally, we have one equation due to the condition that $p(\pi)=\sqrt{3}$.$$a_0+a_1\pi+a_2\pi^2+a_3\pi^3+a_4\pi^4+a_5\pi^5+a_6\pi^6+a_7\pi^7=\sqrt{3}$$

After solving this system of equations, what you find is that the only coefficients of the polynomial that are non-zero are those of the odd terms! This means that the function is odd, i.e., $p(x)=-p(-x)$. Moreover, since $a_0=0$, we know that $$p(-\pi)=-\sqrt 3$$$$p(0)=0$$

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    $\begingroup$ A lot of this work was not necessary, but it's a useful exercise to build intuition about odd functions. $\endgroup$ – Don Thousand Sep 18 '18 at 14:35
  • $\begingroup$ Very elaborating answer ! Thank you. $\endgroup$ – hiren_garai Sep 18 '18 at 14:58

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