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MathGroup Archive 2000

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Re: Re: Series expansion of ArcSin around 1

  • To: mathgroup at smc.vnet.net
  • Subject: [mg21665] Re: [mg21634] Re: [mg21598] Series expansion of ArcSin around 1
  • From: BobHanlon at aol.com
  • Date: Fri, 21 Jan 2000 04:00:16 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

I forgot to mention that since

ArcSin[-x] == -ArcSin[x]

True

We can add to the definition of f

f[x_?Negative] := -f[-x];

This eliminates most of the divergence and the only appreciable error is then 
for small magnitudes of x

Bob Hanlon

In a message dated 1/18/2000 4:48:15 AM, BobHanlon at aol.com writes:

>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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