Dgdrxrt

Consider the set of polynomials $phi_n$built using the Gramm-Schmidt method with respect to the inner product:

$$(f,g) = int_a^b f(x)g(x)w(x)dx$$

with $w(x)>0$.

Prove that if $x_0$ is a zero of $phi_n$, then:

(1)
$$x_0 = fracint_a^bx[fracphi_nx-x_o]^2 w(x)dxint_a^b[fracphi_nx-x_o]^2 w(x)dx(1)$$

From that,I've got to this:

$$int_a^b phi_n(x)frac phi_n(x)x-x_0w(x)dx = 0$$
I suspect this is the solution but I don't know how to argue it. The second term inside the integral is of one degree less than $phi_n$, by the properties of the polynomials, would that make it orthogonal to $phi_n$?.

Also, the problem then says: find an analogue formula for $x_0^2$ pertaining to (1), which I don't understand, maybe someone gets it.

You are on the right track. All you need to say is that $fracphi_nx-x_0$ is a polynomial of degree $n-1$. Since $x_0$ is a root, you have $phi_n(x)=(x-x_0)P_n-1(x)$. Now we can write $P_n-1$ in terms of the orthogonal basis, and we only need the polynomials with degree less or equal to $n-1$:$$P_n-1(x)=sum_j^n-1c_jphi_j(x)$$Here $c_j$ are just some constants. When you plug in this expression into the left hand side of your last formula, you get $$sum_j^n-1c_jint_a^bphi_n(x)phi_j(x)w(x)dx$$
Since the $phi_n$ polynomials are orthogonal and $jne n$, the last integral is always $0$.

Now for the last question, you want to find a formula for $x_0^2$. Start from your last formula.
$$int_a^b phi_n(x)frac phi_n(x)x-x_0w(x)dx = 0$$
Now multiply and divide by $(x-x_0)^2=x^2-2xx_0+x_0^2$:

$$int_a^b phi_n(x)frac phi_n(x)(x-x_0)^3(x^2-2xx_0+x_0^2)w(x)dx = 0$$
Expanding the parentheses, and moving the $x_0^2$ part to the other side, we get

$$x_0^2int_a^bphi_n(x)frac phi_n(x)(x-x_0)^3w(x)dx=int_a^b(2xx_0-x^2)phi_n(x)frac phi_n(x)(x-x_0)^3w(x)dx$$
Now use $2xx_0-x^2=x^2-2x(x-x_0)$, and you get
$$x_0^2int_a^bphi_n(x)frac phi_n(x)(x-x_0)^3w(x)dx=int_a^bx^2phi_n(x)frac phi_n(x)(x-x_0)^3w(x)dx-2int_a^bxphi_n(x)frac phi_n(x)(x-x_0)^2w(x)dx$$