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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