Re: O in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg95950] Re: O in Mathematica*From*: Scott Hemphill <hemphill at hemphills.net>*Date*: Fri, 30 Jan 2009 05:46:44 -0500 (EST)*References*: <gls1vl$hl1$1@smc.vnet.net>*Reply-to*: hemphill at alumni.caltech.edu

Francois Fayard <fayard.prof at gmail.com> writes: > Hello, > > At first, thanks for your help, but I've found what I was asking for. > To input a O(1) in mathematica, you juste have to write > > n O[n,Infinity] > > which gives you O[1/n]^0 which is not simplified to 0. > > Now, I've got another question around O. Let's first explain what I > call a O, or big O (in France). A O(f(x)) around zero is a function > that can be written B(x)f(x) where B(x) is bounded around 0. I just > want to make sure everyone speaks about the same thing. > With that definition x = O(x) (around 0), but x Log[x] is not a O(x) > (around 0) as x Log[x]/x=Log[x] is not bounded around 0. But when I > write in Mathematica > > Log[x] O[x,0]^1 > > It is simplified to O[x,0]^1 which is obviously wrong. I've seen that > if you multiply O[x,0]^1 by a fonction g(x) that is negligeable > compared to x^epislon around 0 for a epsilon>0, the result is > simplified to O[x,0]^1 which is wrong form a mathematical point of view. > > Do I have to understand that O[x,0]^n (in Mathematica) should be > considered as a O[x,0]^(n-epsilon) (in mathematics) for whatever > epsilon>0 you want ? If we consider this definition, are the results > from Mathematica "certified" ? > Another question should be : Why does Mathematica behave like that ? Hi Francois, Did you see my response to one of your other posts? In it I develop the following expression, which gives the expansion you want to the number of terms you desire: In[8]:= f[k_] := FixedPoint[Simplify[Series[n*Pi + ArcTan[Normal[#1]], {n, Infinity, k}]] & , Infinity] In[9]:= f[4] 2 2 Pi 1 1 8 + 3 Pi 8 + Pi 1 5 Out[9]= Pi n + -- - ---- + ------- - --------- + -------- + O[-] 2 Pi n 2 3 3 3 4 n 2 Pi n 12 Pi n 8 Pi n Scott -- Scott Hemphill hemphill at alumni.caltech.edu "This isn't flying. This is falling, with style." -- Buzz Lightyear

**Follow-Ups**:**Re: Re: O in Mathematica***From:*DrMajorBob <btreat1@austin.rr.com>