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Difference of a function and a finite sum

  1. Jun 6, 2014 #1
    Hi everybody,

    I am looking for some help with a problem that has been nagging me for some time now.
    I'm going to give you the gist of it, but I can provide more details if needed.
    So, after some calculations I am left with a function of the following form
    $$
    F_L(y) = f(y) -S_L(y), \quad S_L(y) = \sum_{k=1}^{L/2-1} g_L(y_k,y) \Theta(y-y_k)
    $$
    where ## \Theta(y) ## is the Heaviside function limiting the sum, ##y\in[-1,1]##, and the ##y_k## are related to the roots of the unity, ##y_k = -\cos(2 \pi/L k)##.

    The problem is that ##S_L(y)## is very close to the function ##f(y)##, so that ##F_L(y)## is much smaller than both ##f(y)## and ##S_L(y)##.

    I tried to trade ##S_L(y)## for a (definite) integral, but I'm not sure that's the way to go.
    I actually find that, at the leading order, ##S_L(y)=f(y)##, but I am sort of stuck in evaluating the sub-leading terms.

    From other considerations I am pretty sure that ##F_L(y) \approx C_L (1-y^2)^{L+\ell} ## where ##\ell## is a number of the order of 1 ( which could be ignored for large ##L##), and ##C_L## is a constant exponentially decreasing with ##L##. I'm also able to verify this guess numerically, for not too large ##L##.

    What I'd like is to actually prove that ##F_L## has the form I expect.

    I hope someone can provide some insight.

    Thanks a lot in advance
    Franz
     
    Last edited: Jun 6, 2014
  2. jcsd
  3. Jun 6, 2014 #2

    mfb

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    How can we possibly find out anything without knowing gL and f(y)?
     
  4. Jun 7, 2014 #3
    Hi mfb,

    sorry about that. I thought there might be a general answer. As I mention in my question, of course I can provide details. So

    $$ f_L(y) = (y+1)^{L-2} $$

    and

    $$ g_L(x,y) = x (y-x)^{L-2} +(L-2)(1-x^2) (y-x)^{L-3} $$

    If I'm not making obvious mistakes, when one trades the sum with an integral, the density of ##y_k## makes it so that the integrand is the derivative of a function

    $$ S_L \approx -\int_{y_1}^y dx\, \frac{d}{dx} \left[\sqrt{1-x^2} (y-x)^{L-2}\right] $$

    The problem is that the ultimate result is the difference of ##f_L(y)## and, ##S_L(y)##, which basically is what one usually throws away by turning ##S_L## into an integral.

    I have not made much progress with that, but I'm in the process of giving a reliable estimate of the constant ##C_L##.

    I was thinking that perhaps I can show that the derivative of ##F_L(y)## is "the same" as that of the guess function.

    Thanks for your input
     
    Last edited: Jun 7, 2014
  5. Jun 7, 2014 #4

    mfb

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    A general answer for the difference of two unknown, arbitrary functions? No.

    Putting everything together:

    $$F_L(y) = (y+1)^{L-2} -S_L(y)$$
    $$S_L(y) = \sum_{k=1}^{L/2-1} \left(y_k (y-y_k)^{L-2} +(L-2)(1-y_k^2) (y-y_k)^{L-3}\right) \Theta(y-y_k)$$
    $$y_k = -\cos(2 \pi k/L)$$

    Simplify a bit:
    $$S_L(y) = \sum_{k=1}^{L/2-1} \left(yy_k +L -2\right) (y-y_k)^{L-3} \Theta(y-y_k)$$

    Looks still complicated. I tested it in WA with L=10, and the functions have a factor of 2 as difference. If I fix that, the agreement looks good but the difference has an odd shape. A larger difference for larger y is not surprising, as you have more elements to sum over then.

    Hmm, I don't know.
     
  6. Jun 8, 2014 #5
    Hi mfb,

    and thanks for your help!

    No, I was referring to the problem in general. That is: I was wondering whether there exists a general procedure for evaluating the sub-leading contribution or, if you like, the leading contribution in the difference between the sum and the corresponding integral.

    Uhm... I think there something weird here. You can sure collect a factor ##(y-y_k)^{L-3}##, but it seems to me that you're missing a couple of terms in the remaining factor. I think it should be ##y y_k +L-2 +(1-L) y_k^2##. Unless there is some reason I'm not seeing for discarding the missing terms.

    By the way, I just realized that I forgot an overall factor 2 in ##g_L##. That is, the problem is
    $$
    F_L(y) = (1+y)^{L-2}-S_L(y), \qquad S_L(y) = 2 \sum_{k=1}^{L/2-1}
    \left[ y_k (y-y_k)^{L-2} + (L-2)(1 - y_k^2)(y-y_k)^{L-3}\right]
    \Theta(y-y_k)
    $$

    Here you can see a plot of ##F_{24}(y)/F_{24}(0)## (circles), compared to ##(1-y^2)^{23}## (solid gray).

    As you can see the agreement is pretty good... I'm not really sure that the exponent should be ##L-1##. Of course, for sufficiently large ##L## it should not really matter.
    I'd really like to prove convincingly that ##F(y)## tends to ##C_L (1-y^2)^{L-q}##

    Thanks again for your attention, and for any ideas you can share.

    Franz
     
    Last edited: Jun 8, 2014
  7. Jun 8, 2014 #6

    mfb

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    The google link wants a login (?).

    Hmm, somehow I made a mistake in the simplification. You are right.
     
  8. Jun 8, 2014 #7
    I was sure I had shared the image with anyone having the link, but you're right, I checked again and it was not. Sorry about that.

    Now it should be shared.

    I tried to include the image directly in the post, but for some reason it does not show (perhaps it is the form of the google link).
    Let me know if you still have problems... I'll try something different.
     
    Last edited: Jun 8, 2014
  9. Jun 8, 2014 #8

    mfb

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    Right. Google does not give the image directly, just a page that creates the image.

    Sorry, I don't see an easy way to estimate the error.
    Those integrals usually give deviations in regions with large second derivatives of the integrand and at the edges. Maybe you can check this by comparing the sum of 3 adjacent elements with the integral part in this range. If you can set an upper bound on that, you can probably get an upper bound for the whole (sum-integral) difference.
     
  10. Jun 9, 2014 #9
    I see... I'll try what you suggest.
    I'll update the post with any progress, in case anyone is interested.
    Thanks a lot again for your help :)
    Franz
     
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