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I defined the "function":

$$f(t)=t \delta(t)$$

I know that Dirac "function" is undefined at $t=0$ (see http://web.mit.edu/2.14/www/Handouts/Convolution.pdf).

In Wolfram I get $0 \delta(0)=0$ (http://www.wolframalpha.com/input/?i=0*DiracDelta(0)). Why? I expect $0 \delta(0)=undefined$ (if $\delta(0)=\infty$, thus I will have an indeterminate form $0 \infty$).

Thank you for your time.

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    $\begingroup$ I do not know (and hence would not like to comment much) about f(0) but $\lim_{x \rightarrow 0}f(x) = 0$. You can check that since the Dirac Delta function is zero everywhere except at 0 (where is becomes infinity). So, as the function f(x) approaches 0 (but not equal to zero), your answer will be zero. Maybe f(0) is defined in that way to be zero. Need to check out! $\endgroup$ – Aniruddha Deshmukh Aug 22 '17 at 14:27
  • $\begingroup$ @GennaroArguzzi Define the pairing $\langle T,f \rangle = \int_{-\infty}^\infty T(x) f(x)dx$. Then a distribution $T$ is defined by the value of $\langle T,f \rangle$ for many functions $f$ (the test functions $C^\infty_c$). Then $\delta$ is defined by $\langle \delta, f \rangle = f(0)$, equivalently by $\langle \delta, f \rangle = \lim_{\epsilon \to 0^+} \langle \frac{1_{|x| < \epsilon}}{2 \epsilon}, f \rangle$. Naturally $\langle x \delta ,f \rangle =\lim_{\epsilon \to 0^+} \int_{-\infty}^\infty x \frac{1_{|x| < \epsilon}}{2 \epsilon} f(x)dx = x f(x)|_{x= 0} = 0$ $\endgroup$ – reuns Aug 24 '17 at 21:55
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Look at $\delta$ as distribution: $\langle \delta, f \rangle = f(0)$. Then $\langle t \delta(t), f(t) \rangle = \langle \delta (t), t f(t) \rangle = (tf(t))\mid_{t = 0} = 0 \cdot f(0) = 0$ (see multiplication of distribution by smooth functions).

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  • $\begingroup$ Hi @Cauchy. In my course we do not study distribution. Can you show me some simple webpages on which I can learn it please? $\endgroup$ – Gennaro Arguzzi Aug 22 '17 at 14:32
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    $\begingroup$ @GennaroArguzzi no, sorry. Maybe this could help? math.chalmers.se/~hasse/distributioner_eng.pdf $\endgroup$ – Cauchy Aug 22 '17 at 14:50
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Wolfram Alpha evaluates DiracDelta(1), giving zero. Wolfram Alpha fails to interpret DiracDelta(0), nor does it evaluate 1 * DiracDelta(0), leaving δ(0) unevaluated. Wolfram Alpha believes that the product of 0 and an unevaluated symbol is zero. You can see this with 0*f and with 0*DiracDelta(0), both yielding $0$.

This is not so surprising. From the definition of a field. If we presume that we work in a field large enough to contain $\mathbb{R}$ and $\delta(0)$, so that it is possible to interpret the string "$0 \cdot \delta(0)$", then we must find that $0 \cdot \delta(0) = 0$. It's a standard elementary exercise to show that multiplication of any element of field with the additive identity yields the additive identity.

We may also choose to use elementary limit relations. Suppose we have defined $\delta(t) = \lim_{x \rightarrow 0} f(x,t)$ (a common example is to define $\delta$ to be the limit of Gaussian PDFs as the standard deviation goes to zero), as long as each limit on each RHS in the following exists: \begin{align*} 0 \cdot \delta(0) &\triangleq 0 \cdot \lim_{x \rightarrow 0} f(x,0) \\ &= \lim_{x \rightarrow 0} 0 \cdot \lim_{x \rightarrow 0} f(x,0) \\ &= \lim_{x \rightarrow 0} (0 \cdot f(x,0) ) \\ &= \lim_{x \rightarrow 0} 0 \\ &= 0 \end{align*} The only questionable limit is the first one, from our definition of $\delta$. If we claim $0$ is in the domain of $\delta$, we must allow this limit. Everything else follows. (When one is being careful, one does not claim that $\delta$ has a "domain" because $\delta$ is not a function. At this level of care, one describes delta as an element of some dual space of your space of functions so that the right place for it to appear is as a multiplicand in an integrand. Then one defines $\delta$ in terms of its behaviour in this compound object, not as if it were an independently meaningful thing. This also happens in elementary Calculus : $\mathrm{d}x$ is not a separately intelligible part of $\frac{\mathrm{d}}{\mathrm{d}x}$. The compound object is defined, not its parts. This point of view is described in other answers.)

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You do understand that the Dirac delta "function" isn't a function, since you too put the word in quotes. To justify assertions about it you have to see how those assertions behave in the integrals that involve the delta function. ("Behavior inside integrals" is the idea behind distributions.) That's the essence of @Cauchy 's answer, and what Wolfram knows.

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