Timeline for Does the derivative of log have a Dirac delta term?
Current License: CC BY-SA 3.0
28 events
| when toggle format | what | by | license | comment | |
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| Jun 29, 2020 at 22:15 | answer | added | Pulcinella | timeline score: 7 | |
| Jun 29, 2020 at 20:18 | answer | added | Tom Copeland | timeline score: 4 | |
| Nov 2, 2016 at 13:30 | history | edited | Andrés E. Caicedo |
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| Jan 12, 2016 at 16:46 | vote | accept | Mikhail Katz | ||
| Oct 8, 2015 at 18:11 | answer | added | Tron | timeline score: 4 | |
| Jan 24, 2014 at 12:12 | comment | added | Tom Copeland | See page 508 in Masters of Theory: Cambridge and the Rise of Mathematical Physics by Andrew Warwick for a note on Heaviside's influence on Dirac. | |
| Apr 16, 2013 at 10:11 | comment | added | Tom Copeland | @Liviu, anticipating? Check the dates. Dirac was famous for his terseness, originality, and ingenuity, as reflected in his presentation. | |
| Apr 16, 2013 at 9:24 | comment | added | Liviu Nicolaescu | Like several other people here, I believe Diracthat was anticipating Sokhotski-Plemelj formula. In any case have a look at the first volume of the wonderful treatise "Generalized Functions" by Gelfand and Shilov. This volume is full of very useful computations hard to find elsewhere. Dirac-Sokhotski-Plemelj formula is Example 6, Sect. 2.2 of Chapter I. | |
| Apr 15, 2013 at 22:58 | comment | added | Terry Tao | Yes, there is a d/dx missing on the left-hand side, thanks for the correction! | |
| Apr 15, 2013 at 22:44 | comment | added | Tom Copeland | Dirac would have known all these results from electrostatics to corroborate his assertion. | |
| Apr 15, 2013 at 22:39 | comment | added | Tom Copeland | Terry, you're missing a derivative. d/dz log(z)=1/z then look at limits as in the Poisson kernel. | |
| Apr 15, 2013 at 21:50 | comment | added | Terry Tao | I would write Dirac's formula as $\log(x+i0^+)= p.v.\frac{1}{x}−i\pi \delta(x)$, and this is now a perfectly rigorous equation in the space of distributions (using whatever branch of the logarithm one wishes which does not cut ${\bf R}+i0^+$), indeed it is just the Plemelj formula given in the answers below, written in distributional form. | |
| Apr 15, 2013 at 21:34 | answer | added | Bazin | timeline score: 0 | |
| Apr 15, 2013 at 21:24 | comment | added | Tom Copeland | Dirac's bachelor's degree was in electrical engineering at a British university, so I'm sure he was familiar with and influenced by Heaviside's style of mathematics. | |
| Apr 15, 2013 at 21:05 | comment | added | Tom Copeland | For those who may be a little confused about the absolute value sign, just switch to polar coordinates and restrict to the real line. | |
| Apr 15, 2013 at 19:07 | comment | added | Ryan Budney | @katz: Scott is right, you probably just missed this detail. If $x<0$, the argument of $log(-|x|)$ is still negative. You want to write $log(x) = log(-x) + \pi i H(-x)$. | |
| Apr 15, 2013 at 18:36 | answer | added | jbc | timeline score: 17 | |
| Apr 15, 2013 at 12:48 | comment | added | Mikhail Katz | What is spurious about the absolute value in Tom Copeland's formula as well as mine? Note that Dirac spoke of both $H$ and $\delta$ as having LOCAL values. | |
| Apr 15, 2013 at 12:40 | comment | added | S. Carnahan♦ | Your information seems to be out of date. We mathematicians have been satisfied with $H$ and its derivatives for over 60 years. Also, the last equation in your comment has a spurious absolute value sign. | |
| Apr 15, 2013 at 11:59 | comment | added | Mikhail Katz | More precisely, for negative $x$, $\log x = \log(-|x|)=\log(e^{\pi i}|x|)$ which Dirac seems to replace by $\log(e^{\pi i})+\log|x|=\pi i+\log |x|=\pi i H(-x)+\log |x|$. But not all mathematicians are satisfied by $H$ and especially not with $\frac{d}{dx}H$ :-) | |
| Apr 15, 2013 at 11:21 | comment | added | Igor Khavkine | @Tom, in the text surrounding that formula, Dirac states precisely that (though less explicitly). | |
| Apr 15, 2013 at 10:27 | comment | added | Tom Copeland | I'm sure Dirac was thinking that ln(x)=ln|x|+i H(-x)π, where H(x) is the Heavside step function. | |
| Apr 15, 2013 at 9:04 | answer | added | Carlo Beenakker | timeline score: 19 | |
| Apr 15, 2013 at 8:58 | comment | added | Mikhail Katz | That's part of my "formalizing" query--but I am pretty sure Dirac wasn't speaking of $x>0$. | |
| Apr 15, 2013 at 8:55 | comment | added | Marc Palm | how is log defined for negative $x$? | |
| Apr 15, 2013 at 8:47 | comment | added | Mikhail Katz | $\log x$ for all real $x$, as stated. | |
| Apr 15, 2013 at 8:37 | comment | added | Marc Palm | $x>0$ or $\log |x|$? | |
| Apr 15, 2013 at 8:31 | history | asked | Mikhail Katz | CC BY-SA 3.0 |