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
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