Re: Series expansion of x_n=Tan[x_n]
- To: mathgroup at smc.vnet.net
- Subject: [mg95948] Re: Series expansion of x_n=Tan[x_n]
- From: Scott Hemphill <hemphill at hemphills.net>
- Date: Fri, 30 Jan 2009 05:46:22 -0500 (EST)
- References: <glpfin$kot$1@smc.vnet.net> <gls1va$hkl$1@smc.vnet.net>
- Reply-to: hemphill at alumni.caltech.edu
Scott Hemphill <hemphill at hemphills.net> writes:
> Francois at news53rd.b1.woo, Fayard at news53rd.b1.woo writes:
>
>> Hello,
>>
>> I'm new to Mathematica and I want to comptute a series expansion of the
>> sequence (x_n) defined by :
>>
>> x_n=Tan[x_n] and n Pi-Pi/2 < x_n < n Pi+Pi/2
>>
>> It's easy to prove that
>>
>> x_n = n Pi + O(1) and x_n = n Pi + ArcTan[x_n]
>>
>>>From these 2 formulas, one could easily compte a series expansion of
>> (x_n) to any order. For example:
>>
>> x_n = n Pi + ArcTan[nPi + O(1)] = nPI + Pi/2 -1/(n Pi) + O(1/n^2)
>>
>> Then we can iterate the Process.
>>
>> I want to do this whith Mathematica, but I have a Few Problems :
>> - How can I enter O(1) ? I've tried O(n,Infinity)^0 but it simplifies to=
>> 1
>> - When I compute ArcTan[n Pi + Pi/2- 1/(Pi n)+O(1/n)^2), it gives me
>> Pi/2-1/(Pi n)+O(1/n)^2. I'm surprised because one could get a better
>> serie expansion from that.
>
> I wouldn't try entering O[1] or O[1/n] because I haven't found a
> useful interaction between that and Series[].
>
> Your basic iteration can be defined this way:
>
> In[1]:= iter[n_,x_] := n*Pi + ArcTan[x]
>
> In[2]:= iter[n,Infinity]
>
> Pi
> Out[2]= -- + n Pi
> 2
>
> This is the series you are looking for, with terms up through the
> constant term.
>
>
> In[3]:= iter[n,%]
>
> Pi
> Out[3]= n Pi + ArcTan[-- + n Pi]
> 2
>
> In[4]:= Series[%,{n,Infinity,1}]
>
> Pi 1 1 2
> Out[4]= Pi n + -- - ---- + O[-]
> 2 Pi n n
>
> Now this series includes the (1/n) term.
>
>
> In[5]:= Normal[%]
>
> 1 Pi
> Out[5]= -(----) + -- + n Pi
> n Pi 2
>
> In[6]:= iter[n,%]
>
> 1 Pi
> Out[6]= n Pi - ArcTan[---- - -- - n Pi]
> n Pi 2
>
> In[7]:= Series[%,{n,Infinity,2}]
>
> Pi 1 1 1 3
> Out[7]= Pi n + -- - ---- + ------- + O[-]
> 2 Pi n 2 n
> 2 Pi n
>
> Now this series includes the (1/n^2) term.
>
> This whole operation can be put together into one expression:
>
> 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
I guess you can solve this in Mathematica the way you want to:
In[1]:= iter[n_,x_] := n*Pi + ArcTan[x]
In[2]:= n*Pi + n*O[n,Infinity]
1 0
Out[2]= Pi n + O[-]
n
In[3]:= iter[n,%]
Pi 1
Out[3]= Pi n + -- + O[-]
2 n
In[4]:= iter[n,%]
Pi 1 1 2
Out[4]= Pi n + -- - ---- + O[-]
2 Pi n n
In[5]:= iter[n,%]
Pi 1 1 1 3
Out[5]= Pi n + -- - ---- + ------- + O[-]
2 Pi n 2 n
2 Pi n
Scott
--
Scott Hemphill hemphill at alumni.caltech.edu
"This isn't flying. This is falling, with style." -- Buzz Lightyear