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
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