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Homework Help: Inner product

  1. Mar 26, 2010 #1
    In C[0,1], with inner product defined by (3), consider the vectors 1 and x.

    Find the angle theta between 1 and x.

    (3)[tex]\int_{0}^{1}f(x)g(x)dx[/tex]

    Find the angle theta between 1 and x

    I don't know what to do with polynomial inner product vector space
     
  2. jcsd
  3. Mar 27, 2010 #2

    Dick

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    Can't you compute the inner product of 1 and x given the definition?
     
  4. Mar 27, 2010 #3
    Yes but how is that going to find the angle?
     
  5. Mar 27, 2010 #4

    Mark44

    Staff: Mentor

    How would you find the angle between two ordinary vectors? Wouldn't you use something like this?

    [itex]u \cdot v = |u| |v| cos(\theta)[/itex]

    This idea can be generalized to any inner product space.
     
  6. Mar 27, 2010 #5
    After taking the integral, I obtain 1/2. How does that help me obtain the angle of pi/6?
     
  7. Mar 27, 2010 #6

    Dick

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    Didn't you see Mark44's suggestion? Set u=1 and v=x and then figure out u.v, |u|, |v| and then what cos(theta) is. Remember |v|=sqrt(v.v). It's a number, not a function.
     
  8. Mar 27, 2010 #7
    By definition, <1,1> is the integral from 0 to 1 of 1^2 which is 1/2.
    And <x,x> is the integral from 0 to 1 of x^2 which is 1/3.

    Now the equation is 1/2=1/6 cos theta so theta is arccos 3 which isn't pi/6.
     
  9. Mar 27, 2010 #8

    Dick

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    The integral of 1*1 from 0 to 1 is 1/2??? And I already warned you that |x|=sqrt(<x,x>).
     
  10. Mar 27, 2010 #9
    1 sorry. The definition of inner product on this space is (3)

    <u, v>=[tex]\int_{0}^{1}f(x)g(x)[/tex] which implies <x, x>=[tex]\int_{0}^{1}x^{2}=1/3[/tex].


    So now we have 1/2=1/3 cos theta and now 3/2=cos theta which is greater than 1 so it doesn't exist.
     
  11. Mar 27, 2010 #10

    Dick

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    <x,x> is 1/3. That doesn't mean |x|=1/3.
     
  12. Mar 27, 2010 #11
    Ok I understand now.
     
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