Re: Maclaurin series for ArcCosh[x]

• To: mathgroup at smc.vnet.net
• Subject: [mg73620] Re: Maclaurin series for ArcCosh[x]
• From: "dimitris" <dimmechan at yahoo.com>
• Date: Fri, 23 Feb 2007 04:34:55 -0500 (EST)
• References: <erjoou\$p9a\$1@smc.vnet.net>

```
In[47]:=
\$Version

Out[47]=
"5.2 for Microsoft Windows (June 20, 2005)"

I think you have encountered something I would say is not a bug but
rather a feature.

Anyway, I believe that, the problematic behavior is due to the
presence of Floor function in the series expansion
not only of ArcCosh but also of ArcCot[x], ArcCoth[x], ArcCsc[x],
ArcCsch[x], ArcSec[x] nad ArcSech[x].

The following commands will demontrate that

In[48]:=
ToExpression[Names["Arc*"]]
Through[%[x]]
({#1, Series[#1, {x, 0, 11}]} & ) /@ %
({#1, #1 + O[x]^12} & ) /@ %%

Out[48]=
{ArcCos, ArcCosh, ArcCot, ArcCoth, ArcCsc, ArcCsch, ArcSec, ArcSech,
ArcSin, ArcSinh, ArcTan, ArcTanh}

Out[49]=
{ArcCos[x], ArcCosh[x], ArcCot[x], ArcCoth[x], ArcCsc[x], ArcCsch[x],
ArcSec[x], ArcSech[x], ArcSin[x], ArcSinh[x], ArcTan[x], ArcTanh[x]}

Out[50]=
{{ArcCos[x], SeriesData[x, 0, {Pi/2, -1, 0, -1/6, 0, -3/40, 0, -5/112,
0, -35/1152, 0, -63/2816}, 0, 12, 1]},
{ArcCosh[x], (-1)^Floor[Arg[x]/(2*Pi)]*SeriesData[x, 0, {(I/2)*Pi, -
I, 0, -I/6, 0, (-3*I)/40, 0, (-5*I)/112, 0, (-35*I)/1152,
0, (-63*I)/2816}, 0, 12, 1]}, {ArcCot[x], (1/2)*(-1)^Floor[(Pi +
2*Arg[x])/(2*Pi)]*Pi +
SeriesData[x, 0, {-1, 0, 1/3, 0, -1/5, 0, 1/7, 0, -1/9, 0, 1/11},
1, 12, 1]},
{ArcCoth[x], (-I)*((1/2)*(-1)^Floor[Arg[x]/Pi]*Pi + SeriesData[x, 0,
{I, 0, I/3, 0, I/5, 0, I/7, 0, I/9, 0, I/11}, 1, 12,
1])}, {ArcCsc[x], (1/2)*I*(-1)^Floor[Arg[x]/
Pi]*(-2*I*Pi*Floor[Arg[x]/Pi] +
SeriesData[x, 0, {(-I)*Pi - Log[4] + 2*Log[x], 0, 1/2, 0, 3/16,
0, 5/48, 0, 35/512, 0, 63/1280}, 0, 12, 1])},
{ArcCsch[x], (-(1/2))*(-1)^Floor[(Pi + 2*Arg[x])/
(2*Pi)]*(-2*I*Pi*Floor[(Pi + 2*Arg[x])/(2*Pi)] +
SeriesData[x, 0, {-Log[4] + 2*Log[x], 0, -1/2, 0, 3/16, 0, -5/48,
0, 35/512, 0, -63/1280}, 0, 12, 1])},
{ArcSec[x], Pi/2 - (1/2)*I*(-1)^Floor[Arg[x]/
Pi]*(-2*I*Pi*Floor[Arg[x]/Pi] +
SeriesData[x, 0, {(-I)*Pi - Log[4] + 2*Log[x], 0, 1/2, 0, 3/16,
0, 5/48, 0, 35/512, 0, 63/1280}, 0, 12, 1])},
{ArcSech[x], (-(1/2))*I*(-1)^Floor[Arg[x]/Pi]*Pi +
(1/2)*(2*I*Pi*Floor[Arg[x]/Pi] + SeriesData[x, 0, {I*Pi + Log[4] -
2*Log[x]}, 0, 12, 1]) +
SeriesData[x, 0, {-1/4, 0, -3/32, 0, -5/96, 0, -35/1024, 0,
-63/2560}, 2, 12, 1]},
{ArcSin[x], SeriesData[x, 0, {1, 0, 1/6, 0, 3/40, 0, 5/112, 0,
35/1152, 0, 63/2816}, 1, 12, 1]},
{ArcSinh[x], SeriesData[x, 0, {1, 0, -1/6, 0, 3/40, 0, -5/112, 0,
35/1152, 0, -63/2816}, 1, 12, 1]},
{ArcTan[x], SeriesData[x, 0, {1, 0, -1/3, 0, 1/5, 0, -1/7, 0, 1/9,
0, -1/11}, 1, 12, 1]},
{ArcTanh[x], SeriesData[x, 0, {1, 0, 1/3, 0, 1/5, 0, 1/7, 0, 1/9, 0,
1/11}, 1, 12, 1]}}

Out[51]=
{{ArcCos[x], SeriesData[x, 0, {Pi/2, -1, 0, -1/6, 0, -3/40, 0, -5/112,
0, -35/1152, 0, -63/2816}, 0, 12, 1]},
{ArcCosh[x], ArcCosh[x] + SeriesData[x, 0, {}, 12, 12, 1]},
{ArcCot[x], ArcCot[x] + SeriesData[x, 0, {}, 12, 12, 1]},
{ArcCoth[x], ArcCoth[x] + SeriesData[x, 0, {}, 12, 12, 1]},
{ArcCsc[x], ArcCsc[x] + SeriesData[x, 0, {}, 12, 12, 1]},
{ArcCsch[x], ArcCsch[x] + SeriesData[x, 0, {}, 12, 12, 1]},
{ArcSec[x], ArcSec[x] + SeriesData[x, 0, {}, 12, 12, 1]},
{ArcSech[x], ArcSech[x] + SeriesData[x, 0, {}, 12, 12, 1]},
{ArcSin[x], SeriesData[x, 0, {1, 0, 1/6, 0, 3/40, 0, 5/112, 0,
35/1152, 0, 63/2816}, 1, 12, 1]},
{ArcSinh[x], SeriesData[x, 0, {1, 0, -1/6, 0, 3/40, 0, -5/112, 0,
35/1152, 0, -63/2816}, 1, 12, 1]},
{ArcTan[x], SeriesData[x, 0, {1, 0, -1/3, 0, 1/5, 0, -1/7, 0, 1/9,
0, -1/11}, 1, 12, 1]},
{ArcTanh[x], SeriesData[x, 0, {1, 0, 1/3, 0, 1/5, 0, 1/7, 0, 1/9, 0,
1/11}, 1, 12, 1]}}

Best Regards
Dimitris

> Try:
>
> Series[ArcCosh[x], {x, 0, 11}]
>
> and now try
>
> ArcCosh[x] + O[x]^12
>
> At least with my version of Mathematica:
>
> \$Version
> 5.2 for Mac OS X (February 24, 2006)
>
>
> I do not get the same answer (in fact in the latter case the input is
> returned unevaluated). With ArcSinh and any other function that I
> have tried in place of ArcCosh  the outputs are always the same.
>
> Andrzej Kozlowski

```

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