Re: Problem with "Which"
- To: mathgroup at smc.vnet.net
- Subject: [mg122212] Re: Problem with "Which"
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Fri, 21 Oct 2011 06:22:38 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201110201144.HAA05724@smc.vnet.net>
- Reply-to: drmajorbob at yahoo.com
In your Which statement, when x has not yet been substituted, neither f2 >
0 nor f2 < 0 is True, so "f2 = 0" is reached.
(You meant f2 == 0.)
The value of f2 = 0 is zero, not True, so all three conditions have failed
and Which remains unevaluated.
The value of ew (which probably should have been a localized Module
variable but isn't) is
ew
{{x -> 0}, {x -> 2}}
When x -> 0, all three conditions fail again, since f2 = 0 still resolves
to 0, not True.
When x -> 2, the condition f2 > 0 is True, so Which[...] becomes "T", and
that's what you're getting.
In the If statement, f2 is still 0, having been set that way in the Which
statement, so Which resolves to "T".
The code did what you told it to do.
Bobby
On Thu, 20 Oct 2011 06:44:54 -0500, mbmb <sb at 9y.com> wrote:
> Who can check my module?
>
> KD[f0_, a0_, b0_] := Module[{f = f0, a = a0, b = b0},
> f2 = D[f, {x, 2}];
> ew = Solve[D[f, x] == 0, x, Reals];
> Print["Extremwerte: ", {Which[f2 > 0, "T", f2 < 0, "H", f2 = 0, "S" ],
> x, f} /. ew];
> Print["Extremwerte: ", {If[f2 > 0, "T", "H" ], x, f} /. ew];
> ]
>
> When I enter: KD[x^3 - 3 x^2, -1, 4] the output is
>
> Extremwerte: {{Which[-6+6 x<0,H,f2=0,S],0,0},{T,2,-4}}
>
> whereas the IF-line gives
>
> Extremwerte: {{H,0,0},{T,2,-4}}
>
> Why can't I use Which in this case. Why doesn't Mathematica evaluate f2
> in all cases?
>
--
DrMajorBob at yahoo.com
- References:
- Problem with "Which"
- From: mbmb <sb@9y.com>
- Problem with "Which"