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