Re: UnitStep

*To*: mathgroup at smc.vnet.net*Subject*: [mg52383] Re: UnitStep*From*: DrBob <drbob at bigfoot.com>*Date*: Thu, 25 Nov 2004 05:50:03 -0500 (EST)*References*: <20041124100658.160$aO@newsreader.com>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

>> I don't know why you think it's wrong. Because THESE are the right answers: f[x_] = (1/x)*UnitStep[x - 1]; g[x_] = D[f[x], {x}] h[x_] = Simplify[Integrate[g[x], x]] DiracDelta[-1 + x]/x - UnitStep[-1 + x]/x^2 UnitStep[-1 + x]/x The x==0 case in Piecewise is a red herring, BTW. It's removable, yet Mathematica lets it become a problem. f[x_] = Piecewise[{{0, x == 0}, {(1/x)*UnitStep[x - 1], True}}]; g[x_] = D[f[x], {x}] h[x_] = Integrate[g[x], x] Piecewise[{{0, x < 1}, {-(1/x^2), x > 1}}, Indeterminate] Piecewise[{{0, x <= 1}}, -1 + 1/x] Both definitions above for f are identical, point by point and in form at the only discontinuity. I'm not saying Mathematica should have gotten this right, but I _am_ saying it got it wrong. (That's life, when we use a CAS.) The OP's original problem should not have occurred at all. UnitStep[x-c] f[x] is a well-behaved function if f[x] is well-behaved for x >= c. The behavior of f on x < c is (or should be) irrelevant. Bobby On Wed, 24 Nov 2004 10:06:58 -0500 (EST), David W. Cantrell <DWCantrell at sigmaxi.org> wrote: > DrBob <drbob at bigfoot.com> wrote: > [snip] >> In version 5.1 we have a different choice: >> >> (inputs) >> f[p_] = Piecewise[{{0, p == 0}, {(1/p) UnitStep[p - 1], True}}]; >> Integrate[f@x, {x, -5, 5}] >> D[f@x, {x}] >> Integrate[%, x] >> Plot[f@x, {x, -5, 5}] >> >> (outputs) >> Log[5] >> Piecewise[{{0, x < 1}, {-(1/x^2), x > 1}}, Indeterminate] >> Piecewise[{{0, x <= 1}}, -1 + 1/x] >> (and a plot) >> >> This time D followed by Integrate is evaluated, but WRONG. > > I don't know why you think it's wrong. But in fact, it is not only right, > but also clever. Note that the original function is discontinuous, as is > its derivative. Then when we Integrate that, we need to get an > antiderivative of the derivative of f. But due to the discontinuity, an > antiderivative of the derivative of f need not be f itself. To confirm that > the result of Integrate is correct, just differentiate that result > > Piecewise[{{0, x <= 1}}, -1 + 1/x] > > in your head. Don't you get > > Piecewise[{{0, x < 1}, {-(1/x^2), x > 1}}, Indeterminate] > > just as desired? > > The antiderivative is clever because Mathematica chose -1 as the constant > of integration in the part for x > 1, thereby nicely making the > antiderivative continuous. > > David Cantrell > > > -- DrBob at bigfoot.com www.eclecticdreams.net