Vtgyjfy

I found this question in some old exam:

Find 4 reals $a, x_1, x_2, x_3$ such that the equality
$$int_{-1}^1 frac{P(t)}{sqrt{1-t^2}}{rm d}t=a[P(x_1)+P(x_2)+P(x_3)]$$
is true for all polynomial with degree less or equal 3.

My problem is not to compute these reals for a particular example but rather I didn't understand the idea behind this equality. In fact there is a similar question in this exam to prove that $int_{-1}^1 frac{f(t)}{sqrt{1-t^2}}{rm d}t=frac{pi}{3}[f(x_1)+f(x_2)+f(x_3)]$ for all polynomial $f$ with degree less or equal 5; this what make me sure that there is a trick behind these issues. Is there any explanation?

This probably has something to do with Chebyshev polynomials.
begin{align}
T_0(x)=& 1\
T_1(x)=& x\
T_{n+1}(x)=& 2xT_n(x)-T_{n-1}hspace{1em} forall ngeq 1
end{align}

These polynomials have orthogonality property with $1/sqrt{1-x^2}$ as weight.
$$int_{-1}^{1}T_n(x)T_m(x)frac{1}{sqrt{1-x^2}},text{d}x=
leftlbrace
begin{array}{ll}
0 & mneq n\
pi & m=n=0\
pi/2 & m=nneq 0
end{array}
right.$$

For example, in a third order polynomial, $1=T_0(x)T_0(x)$, $x=T_1(x)T_0(x)$, $x^2=T_1(x)T_1(x)$ and $x^3=0.5T_1(x)times(T_2(x)+T_0(x))$.

P.S. I am not allowed to comment given my reputation and hence am posting this as an answer; although, this should have been a comment.

Edit

These polynomials also satisfy what is called "discrete orthogonality" (see this). If $x_i$ are the roots of $T_n(x)$, then for $0leq p,qleq n-1$, the following holds.
begin{align}
sum_i T_p(x_i)T_q(x_i)=leftlbrace
begin{array}{ll}
0 & pneq q\
n & p=q=0\
n/2 & p=qneq 0\
end{array}
right.
end{align}

With this property, I think the question is solved. Let $P_n(x)$ be any $n$-th order polynomial. Then, $P_n(x)$ can be expanded using the Chebyshev polynomial basis.
$$P_n(x)=sum_{i=0}^{n} b_i T_i(x)=sum_{i=0}^{n} b_i T_i(x)T_0(x)$$

By orthogonality, as already pointed out in the comments,
$$int_{-1}^{1} frac{P_n(x)}{sqrt{1-x^2}}=int_{-1}^{1} frac{P_n(x)T_0(x)}{sqrt{1-x^2}}=pi b_0$$

Now, suppose $x_j$, $0leq jleq n-1$ represent roots of $T_n(x)$. Then,
$$sum_{j=0}^{n-1}P_n(x_j)=sum_{j=0}^{n-1}sum_{i=0}^{n} b_i T_i(x_j)T_0(x_j)=sum_{i=0}^{n} b_i sum_{j=0}^{n-1} T_i(x_j)T_0(x_j)=nb_0$$

The last equality follows from discrete orthogonality. Summing up, we have
$$int_{-1}^{1} frac{P_n(x)}{sqrt{1-x^2}}=pi b_0=(pi/n) sum_{j=0}^{n-1}P_n(x_j)$$

Note that $int_{-1}^{1}frac{at^3+bt^2+ct+d}{sqrt{1-t^2}}dt$ is, by symmetry, equal to $int_{-1}^{1}frac{bt^2+d}{sqrt{1-t^2}}dt$. This evaluates to $frac{pi}{2}(b+2d)$. Now we set the first constant to $frac{pi}{2} k$ and try $x_1=0, x_2=z$ and $x_3=-z$, again to exploit the symmetry of the polynomial. The equality the becomes $frac{pi}{2}(b+2d) = pi kbz^2+frac{3}{2}pi kd$. Comparing coefficients you should get $k=frac{2}{3}$ and $z=frac{sqrt{3}}{2}$.
So we have $int_{-1}^{1}frac{at^3+bt^2+ct+d}{sqrt{1-t^2}}dt = frac{pi}{3}[P(-frac{sqrt{3}}{2})+P(0)+P(frac{sqrt{3}}{2})]$.

Surely this is not the most elegant solution but it works just fine for degree 3 or 5 polynomials.