Re: Ungleich

["Udo und Susanne Krause" <su.krause@XXXXXXX.ch>]

Hallo Frank und Stefan,

> das kann am Ausdruck liegen, denn:

Allenfalls noch zu tippen

In[7]:= FullSimplify[d^3/Abs[d], Assumptions -> d != 0]
Out[7]= d^2 Sign[d]

In[10]:= FullSimplify[d^3/Abs[d], Assumptions -> d < 0 || d > 0]
Out[10]= d Abs[d]

In[16]:= Refine[d^3/Abs[d], Assumptions -> d > 0 || d < 0]
Out[16]= d^3/Abs[d]

In[24]:= Refine[d^3/Abs[d], Assumptions -> d != 0]
Out[24]= d^3/Abs[d]

und letzteres veranlasst zu einer Eselsbrueckenfunktion

In[33]:= Remove[richterRefine]
richterRefine[f_, a_] := If[FreeQ[a, Unequal], Refine[f, a],
  If[Head[a] == Unequal,
   {{#, Refine[f, Assumptions -> #]} &[
     Less[First[Apply[List, a]] , Last[Apply[List, a]]]],
    {#, Refine[f, Assumptions -> #]} &[
     Greater[First[Apply[List, a]] , Last[Apply[List, a]]]]},
   Refine[f, a]
   ]
  ]

In[35]:= richterRefine[d^3/Abs[d], d != 0]
Out[35]= {{d < 0, -d^2}, {d > 0, d^2}}

In[38]:= richterRefine[d^3/Abs[d], 0 != d]
Out[38]= {{0 < d, d^2}, {0 > d, -d^2}}

In[39]:= richterRefine[(x - y)^3/Abs[x - y], x != y]
Out[39]= {{x < y, (x - y)^3/(-x + y)}, {x > y, (x - y)^2}}

In[40]:= richterRefine[d^3/Abs[d], d < 0 || d > 0]
Out[40]= d^3/Abs[d]

Gruss && Schönen Sonntag
Udo.