       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 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'
>
> During evaluation of In:= Power::infy: Infinite expression 1/Sqrt encountered. >>
>
> During evaluation of In:= 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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