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Math Help - Angle between 1 and x

  1. #1
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    Angle between 1 and x

    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) \int_{0}^{1}f(x)g(x)dx

    Find the angle theta between 1 and x
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  2. #2
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    didnot understand

    Quote Originally Posted by dwsmith View Post
    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) \int_{0}^{1}f(x)g(x)dx

    Find the angle theta between 1 and x
    didnot get your question. could you please elaborate???
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    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) \int_{0}^{1}f(x)g(x)dx

    Find the angle theta between 1 and x
    \left\langle 1,x\right\rangle=\int_0^1 x dx=\frac{1}{2}
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    Quote Originally Posted by dwsmith View Post
    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) \int_{0}^{1}f(x)g(x)dx

    Find the angle theta between 1 and x
    You are thinking of C[0,1] as an inner product space with the standard inner product <f, g>= \int_0^1 f(x)g(x)dx.

    Of course, in any inner product space, <u, v>= |u||v|cos(\theta) where \theta is the angle between u and v.

    The particular calculation for the angle between u= 1 and v= x should be easy.
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    After taking the integral, I obtain 1/2. How does that help me obtain the angle of pi/6?
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  6. #6
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    After taking what integral? The formula I gave you, <u, v>= |u||v|cos(\theta) requires three integrals to determine <u, v>, |u|, and |v|.
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