Vtyjkyuk

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

$$int_-pi^pix^kp(x)dx=0$$

for $0leq k leq 6$. 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

Consider the space $mathbbR_le 7[x]$ of real polynomials of degree at most $7$, equipped with the inner product $langle f,grangle = int_-pi^pi fg$.

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

beginequation
frac1sqrt2pi, fracsqrtfrac32 xpi ^3/2,frac3 sqrtfrac52 left(x^2-fracpi ^23right)2 pi ^5/2,frac5 sqrtfrac72 left(x^3-frac3 pi ^2 x5right)2 pi ^7/2,\
frac3 left(35 x^4-30 pi ^2 x^2+3 pi ^4right)8 sqrt2 pi ^9/2,fracsqrtfrac112 left(63 x^5-70 pi ^2 x^3+15 pi ^4 xright)8 pi ^11/2,\fracsqrtfrac132 left(231 x^6-315 pi ^2 x^4+105 pi ^4 x^2-5 pi ^6right)16 pi ^13/2,fracsqrtfrac152 left(429 x^7-693 pi ^2 x^5+315 pi ^4 x^3-35 pi ^6 xright)16 pi ^15/2
endequation

Hence $p(x) = fraclambda16 pi ^15/2sqrtfrac152 left(429 x^7-693 pi ^2 x^5+315 pi ^4 x^3-35 pi ^6 xright)$ for some constant $lambda in mathbbR$. Find $lambda$ by using the condition $p(pi) = sqrt3$ and you will have found $p$.

A "cleaner" solution:

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

$$ int_-1^1 T_n(x) T_m(x) dx = fracpi2 delta_m,n.$$

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 $q(1) = sqrt3$ 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 $T_7(1) = T_7(cos (2pi)) = cos (14 pi) = 1,$ we get $sqrt3 = a_7.$ But $T_7(0) = T_7(cos(pi/2)) = cos(7pi/2) = 0 Rightarrow T_7(0) = 0,$ and $T_7(-1) = T_7(cos(pi)) = cos(7pi) = -1.$ This implies that $q(-1) = -sqrt3,$ $q(0) = 0.$
Translating back into $p,$ we have that

$p(0) = 0, p(-pi) = - sqrt3.$