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RE : Bug in O[x]

• To: mathgroup at smc.vnet.net
• Subject: [mg48264] RE : [mg48233] Bug in O[x]
• From: "Florian Jaccard" <florian.jaccard at eiaj.ch>
• Date: Thu, 20 May 2004 04:03:39 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

Hello Boris,

No, everything is alright !

You have to consider O[x] as somethig of the form a*x , O[x]^2 as something
of the form a*x^n , etc.
Rule :	O[x]+O[x^2]=O[x]  etc.

(1+O[x])^2=1+2*O[x]+O[x]^2=1+2*O[x]=1+O[x]

(1/x+O[x])^2=1/x^2+2*1/x*O[x]+O[x]^2=1/x^2+2*1/x*O[x]=x^(-2)+O[x]^0
(because 1/x * O[x]  is something behaving like a constant...)

(1+O[x])^3=1+3*O[x]+3*O[x]^2+O[x]^3=1+O[x]

Nothing is wrong... you just d'd'nt understand the meaning of O[x]...

O[x]^n represents a term of order x^n. O[x]^n is generated to represent \
omitted higher-order terms in power series.

Regards

F.Jaccard


-----Message d'origine-----
De : Boris_Hollas [mailto:No.hollas at informatik.uni-ulm.de.Spam] 
Envoyé : mercredi, 19. mai 2004 08:42
À : mathgroup at smc.vnet.net
Objet : [mg48233] Bug in O[x]

Have a look at this:

Mathematica 5.0 for Linux
Copyright 1988-2003 Wolfram Research, Inc.
 -- Motif graphics initialized --

In[1]:= (1+O[x])^2

Out[1]= 1 + O[x]

In[2]:= (1/x+O[x])^2

         -2       0
Out[2]= x   + O[x]

In[3]:= (1+O[x])^3

Out[3]= 1 + O[x]


Obviously something is wrong with the O function in Mathematica 5.


Boris Hollas

• Follow-Ups:
  • Re: RE : Bug in O[x]
    • From: Murray Eisenberg <murray@math.umass.edu>