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Thread: sin(x) as a vector

  1. #1
    Nov 2018

    sin(x) as a vector


    I could do with some help on an assignment for my calculus class!
    This is the info given at the beginning of the assignment:

    Fourier series can be seen as a generalisation of the idea of the scalar product. Consider on the one hand the ordinary vectors
    v1 = (1, 1) and v2 = (1,°1), and on the other hand the functions w1 = sin(x), w2 = sin(2x), w3 = sin(3x), ..., which we shall think
    of as a kind of “vectors” as well.

    And this is the subquestion I am at now:

    Find the normalised vectors w1* , w2* , w3* , ... (above you may have used Pythagoras’s theorem for this step, but now you will need to formulate the normalisation purely in terms of scalar products, using the usual relation between scalar products and lengths).

    For the previous subquestion I indeed used pythagoras to get the normalised vectors for v1 and v2. What I did sofar for this question is the following:

    sin(x)∙sin(x)=sin(x)^2= |w1|^2
    |w1|=sqrt(sin(x)^2)= sin⁡(x)
    w1*=w1/|w1| =sin⁡(x)/sin⁡(x)=1

    I also got 1 as the answer for w2* and w3*, using the same method. However, the next assignment requires me to check that these vectors are orthogonal so I figured my answers cannot be correct, as the scalar product will not be 0 if they are all 1.

    Can someone please explain to me what I should be doing instead? Thanks for your help!
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  2. #2
    Senior Member
    Jan 2009

    Re: sin(x) as a vector

    The square root of a square is the absolute value. I will use abs() to denote absolute value.
    In your case |w1| = abs(sin(x))

    This means that w1* = w1/|w1| = sin(x) / abs(sin(x)), which is equal to the piece-wise function shown here:
    1 for x in (2*pi*n, pi + 2*pi*n)
    -1 otherwise

    w2*, w3* have different periods for which it is 1 and -1. I'm not sure what to do next, but hope this helps.
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