Alternate Representation of Function

  1. Hi all,

    the function that I'm posting about is a piecewise function defined as follows:
    \Delta(x) = \left\{
    1 & \quad x = 0 \\
    0 & \quad x \neq 0

    I decided to call it capital delta because of its similarity to the Dirac delta function. What I'm looking for is another way to represent this function such that it's a single line expression like a limit or a convergent series or something.

    My first attempt at this was to represent it as a Fourier cosine series but I'm not sure that I did this 100% correctly. What I did was consider symmetry on the interval ##[-a, a]## where ##a>0## and consider that ##\Delta(x)=0## at ##x= \pm a##. I used this to find the frequencies of the cosines used in the series. Then, I used Fourier's trick to solve for the coefficient. After all was said and done, I ended up with a series that did not clearly converge to 0 for all ##x \neq 0##. It's been a while since I did a Fourier series so I wasn't too confident in my answer anyways. I'd be interested to see if someone else can make this work.

    For another attempt, I realized the similarity to the Gaussian bell curve. So I defined the function as:
    \Delta(x)=\lim_{\alpha \to \infty}e^{-\alpha x^2}
    For ##|x|>0## this limit converges to zero, for ##x=0##, this limit converges to 1. This next step might be useless, but I'll include it anyways. The next thing I did was represent the function inside the limit as a taylor series. Again, I'm not sure of the convergence of this Taylor series (though I assume it converges everywhere) and I'm not sure how the outer limit interferes with the series or where it converges.
    \Delta(x)=\lim_{\alpha \to \infty}\sum_{n=0}^{\infty}\frac{(-\alpha x^2)^n}{n!}=\lim_{\alpha \to \infty}\sum_{n=0}^{\infty}(-1)^n\frac{\alpha^nx^{2n}}{n!}

    Anyways, the above attempt is the best I could really get at finding a 'single line representation' of ##\Delta##. Like I mentioned though, I'm not sure if the last attempt (with the series) even converges. Any comments about this problem or my methods or whatever are appreciated. Or if you have another attempt I would love to see it.

    Lastly I just wanna mention that this seems eerily familiar to the Dirac delta function so it would indicate that you might want to consider generalized functions when dealing with it. I'm very new to generalized functions but from the bit that I do know I can say this:

    The ##\Delta## function defined above, when multiplied with another function (i.e. ##\Delta(x)f(x)##) is equal to ##f(x)## at zero and 0 everywhere else. Assuming that ##f## is finite at zero, we can say that ##\int_{-\infty}^{\infty}\Delta(x)f(x)dx=0## right? But isn't there an infinite number of functions similar to ##\Delta## which integrate to zero when multiplied by another function? So does this mean the definition of ##\Delta## as a generalized function is not unique?
  2. jcsd
  3. disregardthat

    disregardthat 1,814
    Science Advisor

    A distribution is a uniquely defined functional, but if it is induced by a measurable function f, then f is not unique as such. For example, if f induces a distrubution D by [itex]D(g) = \int^{\infty}_{-\infty} fgdx[/itex], then [itex]f+1_A[/itex] induces the same distrubution where [itex]1_A[/itex] is the unit function on a null set A.

    The distribution induced by your function [itex]\Delta[/itex] is the null-distrubution. Any function which is zero almost everywhere induces the null-distrubution, including the dirac-[itex]\delta[/itex] function. In particular, the dirac-delta distribution (which takes any measurable function to its value in 0) is not induced by the dirac-delta function (which is zero everywhere but in the point 0, where it is [itex]\infty[/itex]).

    As to your attempt to find alternative definitions of [itex]\Delta[/itex], your equation is perfectly fine as the taylor series represents [itex]e^{-\alpha x^2}[/itex] everywhere.
  4. jbunniii

    jbunniii 3,377
    Science Advisor
    Homework Helper
    Gold Member

    You won't be able to make this work. The integral defining the Fourier coefficients cannot distinguish your function from the zero function since they only differ at one point. If calculate the coefficients correctly, they will all be zero, so the Fourier series will converge to the zero function, not to your function.
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