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Orthogonal functions

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img55.imageshack.us/img55/8494/67023925dy7.png [Broken]

    3. The attempt at a solution

    All functions orthogonal to 1 result in the fact that: [tex] \int_a^b f(t)\ \mbox{d}t =0 [/tex]

    Now the extra condition is that f must be continous. (because of the intersection).

    But where does the fact that f(a)=f(b)=0 comes from? And why look at the deratives?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 12, 2008 #2


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    Remember way back in first year calc, when you learned that to do that integral you find an antiderivative F(x) and evaluate F(b)-F(a). This is that same problem in disguise.
  4. Nov 12, 2008 #3
    Well I thought of this: [tex] \int_a^b \int_a^t f(s)\ \mbox{d}s \mbox{d}t =0[/tex]
  5. Nov 12, 2008 #4


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    Fine. What are you going to do with it? Why don't you just define [tex]
    F(x)=\int_a^x f(s)\ \mbox{d}s
    What are some of the properties of F(x)?
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