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Let $p(x)$ be a polynomial of degree $7$ with real coefficient such that $p(\pi)=\sqrt3.$ and

$$\int_{-\pi}^{\pi}x^{k}p(x)dx=0$$

for $0\leq k \leq 6$. Then find the value of $p(-\pi)$ and $p(0)$?

If I take general polynomial then it is difficult to find and also time consuming so please help to solve. Thanks

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Consider the space $\mathbb{R}_{\le 7}[x]$ of real polynomials of degree at most $7$, equipped with the inner product $\langle f,g\rangle = \int_{-\pi}^\pi fg$.

Your polynomial $p$ satisfies $p \perp \{1, x, \ldots, x^6\}$. We know that a basis for $\mathbb{R}_{\le 7}[x]$ is given by $\{1, x,\ldots, x^7\}$ so applying Gram-Schmidt process on it gives an orthonormal basis

\begin{equation} \frac1{\sqrt{2\pi}}, \frac{\sqrt{\frac{3}{2}} x}{\pi ^{3/2}},\frac{3 \sqrt{\frac{5}{2}} \left(x^2-\frac{\pi ^2}{3}\right)}{2 \pi ^{5/2}},\frac{5 \sqrt{\frac{7}{2}} \left(x^3-\frac{3 \pi ^2 x}{5}\right)}{2 \pi ^{7/2}},\\ \frac{3 \left(35 x^4-30 \pi ^2 x^2+3 \pi ^4\right)}{8 \sqrt{2} \pi ^{9/2}},\frac{\sqrt{\frac{11}{2}} \left(63 x^5-70 \pi ^2 x^3+15 \pi ^4 x\right)}{8 \pi ^{11/2}},\\\frac{\sqrt{\frac{13}{2}} \left(231 x^6-315 \pi ^2 x^4+105 \pi ^4 x^2-5 \pi ^6\right)}{16 \pi ^{13/2}},\frac{\sqrt{\frac{15}{2}} \left(429 x^7-693 \pi ^2 x^5+315 \pi ^4 x^3-35 \pi ^6 x\right)}{16 \pi ^{15/2}} \end{equation}

Hence $p(x) = \frac{\lambda}{16 \pi ^{15/2}}\sqrt{\frac{15}{2}} \left(429 x^7-693 \pi ^2 x^5+315 \pi ^4 x^3-35 \pi ^6 x\right)$ for some constant $\lambda \in \mathbb{R}$. Find $\lambda$ by using the condition $p(\pi) = \sqrt{3}$ and you will have found $p$.

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  • $\begingroup$ I am sorry, may be I misunderstood something but the orthogonal polynomial is $x^7 - \sum_{k=0}^6 \frac{\langle x^7, x^k\rangle}{\langle x^k, x^k\rangle} x^k$, by Gram-Schmidt orthogonalization process, is'nt it? $\endgroup$ – ram ram Sep 9 '18 at 15:15
  • $\begingroup$ @ramram Indeed, I made a stupid mistake, you cannot calculate it without doing the entire Gram-Schmidt process. Have a look now. Turns out this method is still very time consuming. $\endgroup$ – mechanodroid Sep 9 '18 at 15:44
  • $\begingroup$ so is there any better way to do it, if possible give some hint. $\endgroup$ – ram ram Sep 9 '18 at 15:55
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A "cleaner" solution:

Step 1: Let $T_n(x)$ be the polynomial defined in $[-1,1]$ by $T_n(\cos \theta) = \cos(n\theta).$ These are called Chebyshev polynomials. Among their various properties, one of them is the orthogonality: if $n,m \ge 1,$ then

$$ \int_{-1}^{1} T_n(x) T_m(x) dx = \frac{\pi}{2} \delta_{m,n}.$$

Another very important property is that each of those polynomials is of degree $n$. Therefore, they themselves span the space of all polynomials in $[-1,1].$

Step 2: Let $q(x) = p(\pi x).$ By a change of variables, we see that $q$ is a polynomial such that $q(1) = \sqrt{3}$ and

$$ \int_{-1}^{1} x^k q(x) dx = 0, \, k=0,1,...,6. $$

By the considerations above about Chebyshev polynomials, we see that also

$$ \int_{-1}^{1} T_k(x) q(x) dx = 0, \, k=0,1,...,6.$$

Step 3: As we know that $\{T_j\}_{j \le 7}$ spans the space of polynomials of degree $\le 7$ and the orthogonality above, if $q(x) = a_0 T_0(x) + \cdots + a_7 T_7(x),$ then we get that $q(x) = a_7 T_7(x).$ Plugging $x=1$ and using that $T_7(1) = T_7(\cos (2\pi)) = \cos (14 \pi) = 1,$ we get $\sqrt{3} = a_7.$ But $T_7(0) = T_7(\cos(\pi/2)) = \cos(7\pi/2) = 0 \Rightarrow T_7(0) = 0,$ and $T_7(-1) = T_7(\cos(\pi)) = \cos(7\pi) = -1.$ This implies that $q(-1) = -\sqrt{3},$ $q(0) = 0.$ Translating back into $p,$ we have that

$p(0) = 0, p(-\pi) = - \sqrt{3}.$

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