Re: computing derivatives and limits

*To*: mathgroup at smc.vnet.net*Subject*: [mg126080] Re: computing derivatives and limits*From*: Dana DeLouis <dana01 at me.com>*Date*: Sun, 15 Apr 2012 03:18:20 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com

> f[x_] := Cos[Sqrt[x]] > But asking for the derivative at 0 is asking for trouble: > Power::infy: Infinite expression 1/Sqrt[0] encountered. I don't have an answer. In addition to your observation, I wish FullSimplify would try to use the Sinc Function. I've never seen it do this automatically. f[x_]:=Cos[Sqrt[x]] (f[x+d]-f[x]) / d; Limit[%,d->0] //FullSimplify -(Sin[Sqrt[x]]/(2 Sqrt[x])) I wish FullSimplify would simplify this to -Sinc[Sqrt[x ]] / 2 The advantage is that no further limits are required: % /. x-> 0 -(1/2) Otherwise, one needs to use Limit again: Limit[-(Sin[Sqrt[x]]/(2*Sqrt[x])), x -> 0] -(1/2) = = = = = = = = = = Dana DeLouis :>) Mac & Math 8 = = = = = = = = = = On Apr 13, 4:50 am, Andrzej Kozlowski <akozlow... at gmail.com> wrote: > Consider the function: > > f[x_] := Cos[Sqrt[x]] > > This is an analytic function and of course Mathematica knows its Taylor series: > > SeriesCoefficient[f[x], {x, 0, n}] > > Piecewise[{{(-1)^n/(2*n)!, n >= 0}}, 0] > > But asking for the derivative at 0 is asking for trouble: > > f'[0] > > During evaluation of In[203]:= Power::infy: Infinite expression 1/Sqrt[0] encountered. >> > > During evaluation of In[203]:= Infinity::indet: Indeterminate expression 0 ComplexInfinity encountered. >> > > Indeterminate > > This is, in fact, taken from a question that appeared at our recent calculus test and those who gave the same answer as Mathematica got 0 points for it. I understand why this happens but shouldn't Mathematica be a little cleverer about this sort of thing nowadays, particularly in situations when it can find the Taylor series? > > Andrzej Kozlowski