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Math Help - Projection of Polynomials

  1. #1
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    Projection of Polynomials

    The following is a worked example:


    In the very first line of working, I don't understand how they've got (g,h)=(x+1,x^2+1)=(1,x^2+1). I mean, how did they get "x+1" equal to "1"?

    This is my own approach by the way:

    proj_{h(x)}g(x)= \frac{(g,h)}{(h,h)}h= \frac{\int^1_{-1}(x^2+1)(x+1)}{\int^1_{-1}(x^2+1)^2}(x^2+1)

    Since \int^1_{-1} (x^2+1)(x+1)dx =4 and \int^1_{-1}(x^2+1)^2 dx= 8, we have:

    \frac{4}{8}(x^2+1)=\frac{1}{4}(x^2+1)

    So, I've got 1/4, but it should've been 5/7 as the provided answer indicates. Could anyone please help?
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  2. #2
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    \left<x+ 1, x^2+ 1\right>= \left<x, x^2+ 1\right>+ \left<1, x^2+ 1\right>

    \left<x, x^2+ 1\right>= \int_{-1}^1 x(x^2+ 1)dx

    But, as they say, x is an odd function and x^2+ 1 is an even function. Their product is an odd function, x^3+ x and so the integral is an even function: \int x^3+ x dx= \frac{1}{4}x^4+ \frac{1}{2}x^2 and that, evaluated at -1 and 1, gives the same thing: \left[\frac{1}{4}x^4+ \frac{1}{2}x^2\right]_{-1}^1=  (\frac{1}{4}+ \frac{1}{2})- (\frac{1}{4}+ \frac{1}{2})= 0.

    That is what is meant by saying that an odd function and an even function are \perp- their inner product is 0. \left<x+ 1, x^2+ 1\right>= \left<x, x^2+ 1\right>+ \left<1, x^2+ 1\right>= 0+ \left<1, x^2+ 1\right>.
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  3. #3
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    Quote Originally Posted by demode View Post
    Since \int^1_{-1} (x^2+1)(x+1)dx =4 and \int^1_{-1}(x^2+1)^2 dx= 8, we have:
    How did you get these? They are both wrong... showing the way would help us tell you where you went wrong.
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