Re: Abs derivative (continued) (was Re: ? ? ? ! ?)

*To*: mathgroup at smc.vnet.net*Subject*: [mg8311] Re: Abs derivative (continued) (was Re: [mg8223] ? ? ? ! ?)*From*: Olivier Gerard <jacquesg at pratique.fr>*Date*: Sun, 24 Aug 1997 13:24:28 -0400*Sender*: owner-wri-mathgroup at wolfram.com

> Original message > > > Can anyone explain what v3.0 of Mathematica thinks it's doing when it > > executes > > > > Plot[Abs'[x], {x, -3/10, 3/10}] > > > > ??? > > > > Note the prime: The first argument of Plot was Abs'[x], not Abs[x]. > > > > The strange behavior I see appears both on a PowerMac 7200/120 running > > Mac OS 7.5.3 and on a Wintel box running Windoze 95, so I presume that > > it isn't platform dependent--or at least, not fully so. > > > > --Lou Talman > Continuing on this subject (please see previous answer), Ok for derivation, it seems ok. What about Integration ? Plotting Integrate[Abs[t], {t,0,x}] gives correct result but this is because Integrate succeeds in giving a symbolic result: 1/2 x Sqrt[ Re[x]^2 + Im[x]^2] Of course, for real x, this is equivalent to 1/2 x Abs[x] but it gives us a good way to enhance the solution I gave for implementing a correct Abs derivative. Let's ask Mma to plot the derivative of the Abs formula: Plot[Sqrt[ Re[#]^2 + Im[#]^2]&'[x], {x, -3/10, 3/10}] It gives us a splendid graph of the Sign function. Why does it seems to works when Abs'[x] gives spurious results ? I suppose it is because the derivative of the Real+Imag formula, Im'[x] Im[x] + Re'[x] Re[x] ------------------------------- Sqrt[ Im[x]^2 + Re[x]^2 ] involves only derivatives of Im and Re. They are certainly calculated by the same algorithm than the one used for Abs'[x] but as they are constant and do not change sign at 0 when going on the real line, the error introduced by the algorithm is a lot less than in the plot that was the subject of Lou's message, as shown for instance by Plot[ Re'[x]-1, {x,-1,1}] where the irregularities are of the order of $MachinePrecision. Besides, these derivatives are on the second order compared to Abs, so the adaptation mechanism of Plot can enter in function. So it would be perhaps a good idea to replace the q&d solution of my previous post by: Unprotect[Abs]; Derivative[1][Abs] ^:= (Re'[#] Re[#] + Im'[#] Im[#]])/Sqrt[Re[#]^2 + Im[#]^2] & ; Protect[Abs]; It has the advantage of giving coherent results in some Complex Number cases while still being not completely rigorous. Olivier Gerard

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