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Re: Series Csc[z]
*To*: mathgroup at smc.vnet.net
*Subject*: [mg77244] Re: Series Csc[z]
*From*: dimitris <dimmechan at yahoo.com>
*Date*: Wed, 6 Jun 2007 06:53:06 -0400 (EDT)
*References*: <f43i9m$3mj$1@smc.vnet.net>
I don't understand completely what you want.
Why don't you take the series expansion at Pi/2
Compare the plots of the following expressions
In[151]:=
o[z_] := Csc[z]
In[152]:=
ser1 = Normal[o[z] + O[z, 0]^4]
Out[152]=
1/z + z/6 + (7*z^3)/360
In[154]:=
ser2 = Normal[o[z] + O[z, Pi]^4]
Out[154]=
(Pi - z)/6 - 1/(-Pi + z) - (7/360)*(-Pi + z)^3
In[176]:=
Plot[{o[z], ser1, ser2}, {z, 0, 2*Pi}, ImageSize -> 600, PlotStyle ->
{Blue, Red, Orange}]
/ wjlee :
> Hello,
>
> The Series[Csc[z], {z, 0, 3}] gives 1/z + z/6 + 7/360*z^3 + O[z]^4
>
> However the first few terms of the Laurent series expansion
> approximate well for z<=pi/2. it will need many terms (perhaps >300)
> for good approximation for region pi/2 < z < pi.
>
> Mathematica has a "Pade" function but depending on where the "centered
> around" point is, the solution is still rather complicated.
>
> Is there other Mathematica functions (or way to do it) that could give
> us nice Laurent series form but using less than 20 terms?
>
> Thanks.
>
> wj
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