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Minimisation in Hilbert space

  1. Jan 23, 2012 #1
    1. The problem statement, all variables and given/known data

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

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

    2. Relevant equations

    [itex]x(t) = cos(\frac{\pi t}{2}) [/itex] this is real so [itex]\overline{x} = x [/itex]

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

    3. The attempt at a solution

    [itex]\left|cos(\frac{\pi t}{2}) - p(t) \right|^{2} = \left|x(t) - p(t) \right|^{2} [/itex]

    [itex]= \left\langle x(t) - p(t),x(t) - p(t)\right\rangle [/itex]

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

    [itex]= \int (x\overline{x} - x\overline{p} - \overline{x}p + p\overline{p}) dt [/itex]

    [itex]= \int (x^{2} - x\overline{p} - xp + p\overline{p}) dt [/itex]

    [itex]= \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[/itex]

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

    [itex]= \int (x^{2} - 2x\sum Real(a_{n})t^{n} + \sum \left|a_{n} \right|^{2}t^{2n}) dt[/itex]

    now I don't know what to do. what conditions make this minimal?
    Last edited: Jan 23, 2012
  2. jcsd
  3. Jan 23, 2012 #2
    No need to reply, I have now a solution using the method of least squares and some linear algebra.
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