Minimum of polynomial interpolation

Question

We seek a polynomial of rank $2$ that interpolates the function $f(x)=x^3$ in the interval $[-1,1]$ at the points $x=-1,x=0,x=a$ were $0<a<1$. We need find the number $a$ that make the approximation optimal by minimizing the integral $$ \int_{-1}^{1} \left(x^3-p_2(x) \right)^2\ dx. $$

So if I find the Lagrange polynomial for the points $(-1,-1),(0,0),(a,a^3)$, I get: $$ P_L(a) =-1+(x+1)+(a-1)(x+1)x=ax^2+ax-x^2. $$ But I don't know what make now, probably take derivatives of the integral to find its minimum...

Answer 1

The full answer is (continue was Claude Leibovici wrote). We make integral of it and we get:

$2\int\limits_{0}^{1} [a^2x^2 +(a^2-4a+1)x^4] \ dx=\frac{2}{3}a^2+\frac{2}{5}(a^2-4a+1)=g(a)$ Now we need obtain such $a$ that our integral get it minimum, so we need take first derivative on $g(a)$ and make it equal to zero: $g'(a)=0=\frac{2a}{3}+\frac{2a}{5}-\frac{4}{5}$ And the solution is: $a=3/4$

Answer 2

Now that you have $p_a(x)$, you can explicitly compute the error function $$ E(a) = \int_{-1}^1 \left(x^3 - p_a(x)\right)^2dx $$ and then minimize $E(a)$.

You should get $E(a)$ to be a nice quadratic so the minimization is straight-forward even without derivatives... You can check your work with Wolfram Alpha.

Answer 3

By hand, it is quite simple because of the symmetry $$ \left(x^3-p_2(x) \right)^2=a^2 x^2+2 (a-1) a x^3+(a^2-4a+1) x^4+2(1- a) x^5$$ and we do not care about the odd powers.

So, you just need to integrate between the bounds and because of the symmetry $$\int_{-1}^{1} \left(x^3-p_2(x) \right)^2\, dx=2\int_{0}^{1} \Big[a^2 x^2+ (a^2-4a+1) x^4\Big]\,dx$$