[mg21598] Series expansion of ArcSin around 1

[Andrzej Kozlowski]

Is the following satisfactory?

In[1]:=
Simplify[ComplexExpand[Normal[Series[ArcSin[x], {x, 1, 4}]]], x < 1]

Out[1]=
2 3
6720 Pi + Sqrt[2 - 2 x] (14887 - 1849 x + 477 x - 75 x )
---------------------------------------------------------
13440

or if you prefer:

In[2]:=
Expand[%]

Out[2]=
Pi 14887 Sqrt[2 - 2 x] 1849 Sqrt[2 - 2 x] x
-- + ------------------- - -------------------- +
2 13440 13440

2
159 Sqrt[2 - 2 x] x 5 3
-------------------- - --- Sqrt[2 - 2 x] x
4480 896

> From: Jacek Pliszka <pli...@fuw.edu.pl>
> Date: Sun, 16 Jan 2000 22:43:46 -0500 (EST)
> To: math...@smc.vnet.net
> Subject: [mg21598] Series expansion of ArcSin around 1
>
> Hi!
>
> 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?
>
> Thanks for any help,
>
> Jacek
>
>
>

[BobH...@aol.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

[Mark Harder]

Jacek,
I don't know how to get Mathematica to modify its expansion for the
restricted case x=real; but if you need to compute with Series, you can
convert to Normal form and replace x with values, or you can extract the
real part of the normal form and then replace. i.e.:

In[30]:=ts=Simplify[Series[ArcSin[x],{x,1,4}] ];
Normal[ts]/.x->.99
Re[%]

Out[31]=1.71233580011662844` +0.` I
Out[32]=1.71233580011662844`

or

In[33]:=Re[Normal[ts]];
%/.x->.99
Out[34]=1.71233580011662844`

-mark

-----Original Message-----
From: Jacek Pliszka <pli...@fuw.edu.pl>
Subject: [mg21598] Series expansion of ArcSin around 1

>Hi!
>

>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?
>

[Hugh Walker]

>Hi!
>
>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?
>
>Thanks for any help,
>

==========
This works OK:

Series[ArcSin[1 - q], {q, 0, 3}]

==

Hugh Walker
Gnarly Oaks