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Re: Re: O in Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg95997] Re: [mg95950] Re: O in Mathematica
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Sat, 31 Jan 2009 06:45:10 -0500 (EST)
  • References: <gls1vl$hl1$1@smc.vnet.net> <200901301046.FAA06594@smc.vnet.net>
  • Reply-to: drmajorbob at longhorns.com

>                                              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

I gather (just guessing) that's an approximation to

n Pi + Arctan[n]

for large n?

Bobby

On Fri, 30 Jan 2009 04:46:44 -0600, Scott Hemphill  
<hemphill at hemphills.net> wrote:

> 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



-- 
DrMajorBob at longhorns.com


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