Re: Series expansion of ArcSin around 1

*To*: mathgroup at smc.vnet.net*Subject*: [mg21634] Re: [mg21598] Series expansion of ArcSin around 1*From*: BobHanlon at aol.com*Date*: Tue, 18 Jan 2000 02:35:19 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

series1 = Normal[Series[ArcSin[x], {x, 1, 4}]] The result is real for x < 1, for example series1 /. x -> .9 2.0218230504180923 The complex factors appear because (x-1) is negative and the powers are radicals. If you wish, you can eliminate the appearance of complex factors by a substitution: series2 = series1 /. (x - 1)^a_ :> ((1 - x)^a*(-1)^a) However, as seen on the Plot below, the result is in the wrong quadrant. Plot[{ArcSin[x], series2}, {x, -1, 1.}, PlotStyle -> {RGBColor[0, 0, 1], RGBColor[1, 0, 0]}]; Both x and Pi-x have the same Sin Plot[Sin[x], {x, 0, 2Pi}]; Consequently, we need to modify the approximation as follows f[x_] := Evaluate[ Pi - Normal[Series[ArcSin[x], {x, 1, 4}]] /. (x - 1)^ a_ :> ((1 - x)^a*(-1)^a)] Plot[{ArcSin[x], f[x]}, {x, -1, 1.}, PlotStyle -> {RGBColor[0, 0, 1], RGBColor[1, 0, 0]}]; As expected, the approximation diverges away from the expansion point. Bob Hanlon In a message dated 1/17/2000 12:12:02 AM, pliszka at fuw.edu.pl writes: >I have the following problem. My x is close to 1 but sligthly >smaller. I want to expand ArcSin[x] around 1 but this is what I get: > >In[53]:= Series[ArcSin[x],{x,1,4}] > > I 3/2 3 I 5/2 > - (-1 + x) --- (-1 + x) > Pi 6 80 >Out[53]= -- - I Sqrt[2] Sqrt[-1 + x] + ------------- - --------------- >+ > 2 Sqrt[2] Sqrt[2] > > 5 I 7/2 > --- (-1 + x) > 448 9/2 >> --------------- + O[-1 + x] > Sqrt[2] > >How to tell Mathematica that my x is real and smaller than 1 >so it will not return all this complex numbers? >