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Infinite Dirac Series

  1. Nov 4, 2013 #1


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    What is the sum of an infinite Dirac series and why? 1 or infinity?

    [itex]\sum_{n=-\infty}^{\infty}\delta (n)[/itex]

    I can see it being 1 because it's like a series version of the integral:

    [itex]\int_{-\infty}^{\infty}\delta (t)dt = 1[/itex]

    But for the series where n=0,

    [itex]\delta (0) = \infty[/itex]

  2. jcsd
  3. Nov 5, 2013 #2


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    No, it is not "a series version of the integral [itex]\int_{-\infty}^\infty \delta(t)dt[/itex]". The integral has only one delta function.

    Your basic problem is that you don't really understand what the Dirac delta function itself really is. Strictly speaking the "Dirac delta function" is not a function at all, it is a "functional" or "generalized function". The Dirac delta function [itex]\delta(x)[/itex] is the functional that assigns to every function its value at 0: [itex]\delta(x)f(x)= f(0)[/itex].

    What you have written (where did you see it?) makes no sense because "[itex]\delta(0)[/itex]", "[itex]\delta(1)[/itex]" themselves have no meaning. The Dirac delta function, being a functional, is NOT defined at individual values of x.

    If instead you had [itex]\sum_{n=-\infty}^\infty \delta(x- n)[/itex], that would be the functional that assigns, to every function, f, the sum of its values at the integers, [itex]\sum_{n=-\infty}^\infty f(n)[/itex], if that series converges.
  4. Nov 6, 2013 #3


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    Hi HallsofIvy,

    I was solving for the discrete Fourier transform of [itex]δ(n)[/itex] and wondering why is was 1, but I didn't know until now that there are two generalized functions that share the same symbol [itex]δ[/itex], the continuous Dirac delta and the discrete Kronecker delta.

    Also for the Dirac delta, I thought this was true [itex]\int_{-∞}^{∞}δ(x)f(x)dx=f(0)[/itex] instead of [itex]δ(x)f(x)=f(0)[/itex], or are they both the same?
  5. Nov 6, 2013 #4


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    Kroneker delta is not a generalized function, just a function in two variables (both of which need to be integers).

    Edit: So I make this really long post about how that integral isn't really an integral, and then I convince myself that it is an integral. (Well a certain type of integral.)

    Anyway to answer the question, sort of. ##\delta(x)f(x)## can be interpreted as evaluating Dirac on f. Remember linear algebra and how a linear operator L acts on vector v? You wrote it as Lv, right? Same idea, because functional are linear transforms/operators (just with co-domain R). But personally I wouldn't do it because it's more likely to mean "multiply delta by f" with multiplication in the space of distributions.
    Last edited: Nov 6, 2013
  6. Nov 6, 2013 #5


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    Hi pwsnafu,
    I see what you mean about linear operators, but I have never used or seen the Dirac delta in that way. This is the first time I've heard about functionals, so I will probably will have to learn about them properly first before I fully understand how the Dirac delta is a functional.

    The way I thought of the integral [itex]\int_{-∞}^{∞}δ(x-k)f(x)dx=f(k)[/itex] (modified it with a real constant [itex]k[/itex]) before this thread was that the Dirac delta is defined as derivative of the unit step Heaviside function:

    [itex]\delta (x) = \begin{cases}
    +\infty & \text{ if } x= 0\\
    0 & \text{ if } x\neq 0
    \end{cases} [/itex]

    then the multiplication of [itex]δ(x-k)f(x)[/itex] is zero everywhere except at [itex]k[/itex] where it is [itex]δ(0)f(k)[/itex].

    [itex]\int_{-∞}^{∞}δ(x-k)f(x)dx[/itex] then (perhaps informally) becomes [itex]f(k)\int_{-∞}^{∞}δ(x-k)dx=f(k)[/itex]
  7. Nov 6, 2013 #6


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    If you can get a copy of Theories of Generalised Functions: Distributions, Ultradistributions and Other Generalised Functions by Hoskins and Pinto, read it. It is by far the easiest textbook on the topic to read.

    You can't differentiate a function if it not continuous there. Is Heaviside continuous at x=0?

    So what's ##\delta(0)##? And you don't get away with "infinity" because that just means everything becomes infinity, including ##\delta(0)f(k)##.

    Worse, ##\delta(x-k)\,f(x)## is defined, but as a multiplication between a generalized function and a smooth function. The result is a generalized function in its own right, and as Halls has told you, it doesn't have point values.

    Remember, integrating over a point always results in zero: ##\int_0^0 anything(x) \, dx = 0##. But you have
    ##\int_{-\infty}^\infty \delta(x-k) \, dx = \int_{-\infty}^k \delta(x-k)\, dx + \int_{k}^k \delta(x-k)\, dx + \int^{\infty}_k \delta(x-k)\, dx##
    and every term on the right hand side is equal to zero.

    The problem is you were taught a heuristic, which is fine as an introduction. But you haven't been corrected since then. That's the problem.
    Last edited: Nov 6, 2013
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