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O in Mathematica
*To*: mathgroup at smc.vnet.net
*Subject*: [mg95893] O in Mathematica
*From*: Francois Fayard <fayard.prof at gmail.com>
*Date*: Thu, 29 Jan 2009 05:55:59 -0500 (EST)
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 ?
Thanks,
Francois Fayard
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