       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>

```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 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:= iter[n_,x_] := n*Pi + ArcTan[x]
>
> In:= iter[n,Infinity]
>
>         Pi
> Out= -- + n Pi
>         2
>
> This is the series you are looking for, with terms up through the
> constant term.
>
>
> In:= iter[n,%]
>
>                       Pi
> Out= n Pi + ArcTan[-- + n Pi]
>                       2
>
> In:= Series[%,{n,Infinity,1}]
>
>                Pi    1       1 2
> Out= Pi n + -- - ---- + O[-]
>                2    Pi n     n
>
> Now this series includes the (1/n) term.
>
>
> In:= Normal[%]
>
>            1      Pi
> Out= -(----) + -- + n Pi
>           n Pi    2
>
> In:= iter[n,%]
>
>                        1     Pi
> Out= n Pi - ArcTan[---- - -- - n Pi]
>                       n Pi   2
>
> In:= Series[%,{n,Infinity,2}]
>
>                Pi    1        1        1 3
> Out= 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:= f[k_] := FixedPoint[Simplify[Series[n*Pi + ArcTan[Normal[#1]],
>         {n, Infinity, k}]] & , Infinity]
>
> In:= f
>
>                                              2         2
>                Pi    1        1      8 + 3 Pi    8 + Pi       1 5
> Out= 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:= iter[n_,x_] := n*Pi + ArcTan[x]

In:= n*Pi + n*O[n,Infinity]

1 0
Out= Pi n + O[-]
n

In:= iter[n,%]

Pi     1
Out= Pi n + -- + O[-]
2      n

In:= iter[n,%]

Pi    1       1 2
Out= Pi n + -- - ---- + O[-]
2    Pi n     n

In:= iter[n,%]

Pi    1        1        1 3
Out= 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

```

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