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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






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