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