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On nascent delta 'function'

  1. Feb 1, 2012 #1
    greetings . i have two questions regarding the sinc function in the week limit , where it can be used as a nascent delta function.
    the definition :
    [tex] \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}} \phi(x)dx=\phi(x_{0}) [/tex] is said to be valid for any smooth function [itex]\phi(x)[/itex] with compact support . does that mean that the following is not valid :
    [tex] \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}}xdx=x_{0} [/tex]

    moreover . if we expand the sine function, we get :
    [tex] \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\phi(x)\sum_{n=0}^{\infty}\frac{(-1)^n(x-x_{0})^{2n}}{(2n+1)!(\varepsilon)^{2n+1}}dx =\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}} \int_{-\infty}^{\infty}(x-x_{0})^{2n}\phi(x) =\phi(x_{0})

    is it legit to perform the integration term by term ??
    Last edited: Feb 1, 2012
  2. jcsd
  3. Feb 1, 2012 #2


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    You're missing a factor of epsilon. Your integral should be

    [tex]\lim_{\epsilon \rightarrow 0} \frac{1}{\pi}\int_{-\infty}^{\infty} dx \frac{\sin\left(\frac{x-x_0}{\epsilon}\right)}{\frac{x-x_0}{\epsilon}} \phi(x)[/tex]

    If you make the change of variables [itex]y = (x-x_0)/\epsilon[/itex] you can show (under the assumption that you can exchange the limit and the integral) that the result is [itex]\phi(x_0)[/itex]. Whether or not you can legitimately choose [itex]\phi(x) = x[/itex] depends on whether or not you can actually exchange the limit with the integral.

    I don't think expanding the sine as a series will really get you anywhere. The key property of the nascent delta function is that you pick a function that integrates to 1 when you scale out the epsilon.
  4. Feb 2, 2012 #3
    no , i'am not ... check (37) here "delta function"

    as for the expansion of the sine , here is what i was trying to do :

    [tex] \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}} (x-x_{0})^{2n} = \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}}\left(\sum_{k=0}^{2n}\binom{2n}{k}x^{2n-k} (-1)^{k} (x_{0})^{k} \right)

    now define [itex]\Phi(x_{0j})[/itex] as a delta distribution such that :
    [tex] \Phi(\textbf{X})=\sum_{j=1}^{m} \delta(x-x_{0j}) [/tex]

    [tex] \Phi(\textbf{X})=\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}}\left(\sum_{k=0}^{2n}\binom{2n}{k}x^{2n-k} (-1)^{k} \sum_{j=1}^{m} (x_{0j})^{k} \right)
    now, for a certain situation , we are able to compute [itex]\sum_{j=1}^{m} (x_{0j})^{k}[/itex] for any [itex]k[/itex], but not the [itex]x_{0j}[/itex] themselves. so i was hoping to extract the [itex]x_{0j}[/itex] by :
    [tex] x_{0j}= \int x\Phi(\textbf{X}) dx [/tex] over some period that contains [itex]x_{0j}[/itex]
    Last edited: Feb 2, 2012
  5. Feb 2, 2012 #4
    mathematica is giving me some encouraging results . to spice things up, for a polynomial [itex] f(x) [/itex] , we can find [itex] \sum_{j=1}^{m} (x_{0j})^k [/itex] where [itex]x_{0j}[/itex] are the roots of the polynomial :
    [tex]\sum_{j=1}^{m} (x_{0j})^k = \lim_{x \rightarrow 0}\frac{1}{q(0)k!}\left( \sum_{j=0}^{k}\binom{k+1}{j+1} \left( p(x) \hat{q}(x)^{j}\right)^{(k)}\right)[/tex]
    where :
    [tex] p(x)=x^{m-1}f^{'}\left(\frac{1}{x} \right) [/tex]
    [tex] q(x)=x^{m}f\left(\frac{1}{x} \right) [/tex]
    [tex] \hat{q}(x)=-\frac{q(x)}{q(0)}[/tex]
    [tex] m= deg[f(x)] [/tex]

    so , for a root [itex]z[/itex] contained in a small enough contour [itex]\kappa [/itex] :

    [tex] z(\kappa )=\lim_{\varepsilon \rightarrow 0}\lim_{s \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}}\left(\sum_{k=0}^{2n}\binom{2n}{k}\frac{(-1)^{k}}{{q(0)}k!}\left(\oint_{\kappa } {x}^{2n-k+1}dx\right)\sum_{j=0}^{k}\binom{k+1}{j+1}\left(p(s)\hat{q}(s)^{j}\right)^{(k)}\right) [/tex]

    i was able to calculate the roots of some polynomials this way . and i know i can take it to the next step , and come up with a general formula . furthermore, this technique could be easily extended to study the roots of some complex functions .
    Last edited: Feb 2, 2012
  6. Feb 3, 2012 #5


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    You're right, my mistake.

    It's still not clear to me what you're trying to accomplish here. Are you expanding the sine because you want to compute the integral

    [tex]\int_{-\infty}^\infty dx (x-x_0)^{2n}\phi(x)[/tex]

    and then evaluate the infinite sum and see if it reduces to [itex]\phi(x_0)[/itex]? If so, interchanging the integral with the infinite sum is only going to be valid in certain cases. If you chose [itex]\phi(x) = x[/itex] the integrals aren't well defined. Even if you replace the infinite limits with finite limits [itex]\pm T[/itex] which you take to infinity at the end, you'll have to be careful with the order of the epsilon and T limits. I think you probably get different results for functions [itex]\phi(x)[/itex] which are not smooth.
  7. Feb 3, 2012 #6
    i am expanding the sine , and then sum over [itex]x_{0j}[/itex] in order to isolate the term
    [tex] \sum_{j=1}^{m} (x_{0j})^{k} [/tex]

    which i can calculate via :

    for some polynomial [itex] f(x) [/itex]

    if we plug this term into the expansion we get :

    [tex] \Phi(\mathbf{X})=\sum_{j=1}^{m}\delta (x-x_{0j})=\lim_{\varepsilon \rightarrow 0}\lim_{s \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}}\left(\sum_{k=0}^{2n}\binom{2n}{k}\frac{(-1)^{k}}{{q(0)}k!} {x}^{2n-k}\sum_{j=0}^{k}\binom{k+1}{j+1}\left(p(s)\hat{q}(s)^{j}\right)^{(k)}\right) [/tex]

    if we integrate, we get to extract the roots of the polynomial :

    [tex] \oint _{\kappa} x\Phi(\mathbf{X})dx = x_{i} [/tex]

    where [itex] x_{i} [/itex] is contained in the contour [itex] \kappa [/itex] .
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