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Math Help - linear algebra Gram-Schmidt in Function Spaces problem

  1. #31
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    Oh ok. So for the problem 6.1 is that really all I need to show? Since H_2n(x) would be even and H_2n+1(x) would be odd?
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  2. #32
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    I didn't really follow on that part so I would rather wait for Ackbeet to return and give you a definitive answer
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  3. #33
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    Alright, thank you. Do you by any chance understand any other parts of this problem? I'm really hurrying to get it because my exam is tomorrow and this is one of the practice problems that I just can't get
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  4. #34
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    Okay, at a second glance it's not that difficult:

    For 6.1 you will also need to show that:

    1) The product of 2 even functions is even
    2) The product of an odd function and an even function is odd
    3) The derivative of an odd function is even (similar to what you just proved)

    I can understand the other questions, but it's been a long time since I've done anything similar so there's a big chance that I will make stupid mistakes. Which you don't want
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  5. #35
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    ok, thank you. I think i understand 6.1 for the most part. Now to move onto the rest, hopefully in time for the exam
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  6. #36
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    So for 6.2, you can actually generalize a bit more and say that since you're integrating an even function times an odd function (which is what, even or odd?) over a symmetric interval, the result is...

    For 6.3, just turn the crank and plug things into the expression for the inner product.

    For 6.4, just turn the crank and plug things into the Gram-Schmidt procedure. Follow your footnote 11 - it makes good sense and will save you some work.
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  7. #37
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    6.2 even times an odd is odd, but I don't see where this is going

    6.3 so I just plug it into the inner product to normalize it?
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  8. #38
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    The integral of an odd function over a symmetric (symmetric about the origin) interval is like \int_{-R}^{R}x\,dx. It's going to be zero, because any positive area on one side of the origin will be canceled by negative area on the other side since the function is odd. Make sense? The exponential allows you to take the limit as R\to\infty, since you know the integrand vanishes at \pm\infty.

    For 6.3, you have to find the norm, or length, of H_{0} and H_{1}. You're trying to normalize them, which in this context means that you make them of length 1. To make any vector of length 1, you divide it by what?
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  9. #39
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    Oh! Yeah the symmetric integral makes sense. I don't see how this is proving 6.2 though.

    6.3 divide by the magnitude. I tried setting up H_0 as [1 0 0 0]^T and H_1 as [0 2 0 0]^T so normalized (assuming this is correct) would be the same for H_0 and H_1 = [ 0 2/sqrt(2) 0 0]^T
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  10. #40
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    I don't see how this is proving 6.2 though.
    Ah, but what is the inner product? You know that if the inner product of two functions is zero, they're "orthogonal", at least with respect to that inner product. That is, two functions f and g are orthogonal, by definition, if f\cdot g=0, or

    \displaystyle{\int_{-\infty}^{\infty}f(x)\,g(x)\,e^{-x^{2}}\,dx=0.}

    And now, hopefully, you see why we're pretty much done, right?

    For 6.3, I'm not sure I would represent your functions as vectors. You can do that - there's nothing stopping you. But the Hermite polynomials get longer and longer. Eventually, you're going to have to expand the size of your vectors. In fact, the Hermite polynomials are of infinite dimension! I would just stick to the ol' H_{0}(x)=1 style of representation. Keep things simple. So what is the magnitude of H_{0}? And what is the magnitude of H_{1}?
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  11. #41
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    oh ok I think I understand it.

    6.3 - Wouldn't the magnitude of H_0 be 1 and H_1 is sqrt(2)?
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  12. #42
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    No. You have to compute the integral. Remember that the magnitude of a function f is |f|=\sqrt{f\cdot f}=\sqrt{\int_{-\infty}^{\infty}f\,f\,e^{-x^{2}}\,dx}. That's what you have to compute.

    I think your vector representation is confusing the issue. I would ditch it, if I were you.
    Last edited by Ackbeet; August 18th 2010 at 10:59 AM. Reason: Forgot the integral's square root.
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  13. #43
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    Ok, I'll forget about the vector representation.
    |H_{0}|=\sqrt{H_{0}\cdot H_{0}}=\sqrt{\int_{-\infty}^{\infty}H_{0}\,H_{0}\,e^{-x^{2}}\,dx} = \sqrt{\int_{-\infty}^{\infty}1\,(1)\,e^{-x^{2}}\,dx}

    and for |H_{1}| =  \sqrt{\int_{-\infty}^{\infty}2x\,(2x)\,e^{-x^{2}}\,dx}

    I feel really stupid saying this but I honestly cannot remember how to do the integral of those.
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  14. #44
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    @Ackbeet - \lim_{r \to \infty} \int_{-r}^r f(x)dx is actually the Cauchy principal value of the integral, and IIRC it's not always equal to the value of the integral.
    A neat example can be found here -- Cauchy principal value - Wikipedia, the free encyclopedia
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  15. #45
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    SpiffyEh @ 43:

    Hehe. You shouldn't feel stupid. Those are not elementary integrals. You might possibly have seem them in a multi-variable calculus course, or a statistics course. It's been said that you are a mathematician if you know that \int_{-\infty}^{\infty}e^{-x^{2}}\,dx=\sqrt{\pi}. To derive this result, you do a dirty trick: set I equal to the value of the integral, and then square it. Combine the two integrals together, one with a change of variable to y. Then you switch the integral over to polar coordinates, at which point you can do one or more different u substitutions, which gives you the result. Finally, you have to take the square root again.

    The reason all of that is necessary is that there is no known elementary antiderivative of that exponential function. The fact that we seek a definite integral makes all the difference. To get the H_{1} integral, integrate by parts twice to reduce it to the integral I just gave you.

    Defunkt @ 44:

    I would agree with you that you have to be careful, and also that we're dealing with the Cauchy PV integral. However, in this case, our functions are so well-behaved that I don't think that sort of behavior is going to show up. The integrals in the examples are all ones whose convergence is dodgy at best. We're dealing with all L^{2} functions here. The exponential is enough to squash all the Hermite polynomials of any finite degree fast enough into the x-axis to make the integrals converge quite quickly. My instincts (I freely admit, not at all backed up by rigor) are that it doesn't matter how R goes to infinity. We're not dealing with an \infty-\infty situation.
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