- #1
sunrah
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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}) dtnow I don't know what to do. what conditions make this minimal?
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