What is the optimal polynomial of degree 2 to minimize the given expression?

[sunrah]

Homework Statement

$P_{2} \subset L_{2}$ is the set of all polynomials of degree $n \leq 2$. Complete the following approximation. In other words find the polynomial of degree 2 that minimises the following expression:

$\int \left|cos(\frac{\pi t}{2}) - p(t)\right|^{2}dt = min$ with -1 <= t >= 1

Homework Equations

$x(t) = cos(\frac{\pi t}{2})$ this is real so $\overline{x} = x$

$p(t) = \sum a_{n}t^{n}$ because polynomial max 2nd degree: 0 <= n >= 2 (we do not know whether it has imaginary terms or not)

The Attempt at a Solution

$\left|cos(\frac{\pi t}{2}) - p(t) \right|^{2} = \left|x(t) - p(t) \right|^{2}$

$= \left\langle x(t) - p(t),x(t) - p(t)\right\rangle$

$= \int (x(t) - p(t))( \overline{x(t) - p(t)})dt$ scalar product defined in set of polynomial functions

$= \int (x\overline{x} - x\overline{p} - \overline{x}p + p\overline{p}) dt$

$= \int (x^{2} - x\overline{p} - xp + p\overline{p}) dt$

$= \int (x^{2} - x\sum\overline{a_{n}}t^{n} - x\sum a_{n}t^{n} + \sum a_{n}\overline{a_{n}}t^{2n}) dt$

$= \int (x^{2} - x\sum(\overline{a_{n}} + a_{n})t^{n} + \sum \left|a_{n} \right|^{2}t^{2n}) dt$

$= \int (x^{2} - 2x\sum Real(a_{n})t^{n} + \sum \left|a_{n} \right|^{2}t^{2n}) dt$now I don't know what to do. what conditions make this minimal?

 

[sunrah]

No need to reply, I have now a solution using the method of least squares and some linear algebra.