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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?
>
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