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Re: Asymptotic Expansions of acosh(x) for large x (was Unidentified s ub
From: |
Miroslaw Kwasniak |
Subject: |
Re: Asymptotic Expansions of acosh(x) for large x (was Unidentified s ubject!) |
Date: |
Thu, 5 May 2005 11:06:13 +0200 |
User-agent: |
Mutt/1.5.6+20040907i |
On Wed, May 04, 2005 at 05:09:36PM -0400, Hall, Benjamin wrote:
> >I've got a problem calculating "acosh". I have to calculate cosh^(-1) of
> >a very large complex number (-7.2718e+15 + 1.3382e+15i). If I try to
>
>
> The wolfram website indicates some asymtotic expansions for acosh(x) for
> large x
>
> http://functions.wolfram.com/ElementaryFunctions/ArcCosh/06/01/04/
>
> For your case it looks like 2 terms may be enough:
>
>
> octave:20> z = -7.28e+15 + 1.34e+15i
> z = -7.2800e+15 + 1.3400e+15i
>
> ### Asymtotic expansion for acosh(z) for large z
> octave:21> y = log( -4*z^2 ) / 2 - pi*sqrt( -z^2 ) / (2*z)
> y = 37.2337 + 2.9596i
>
> octave:22> z2 = cosh(y)
> z2 = -7.2800e+15 + 1.3400e+15i
>
> octave:25> abs(z - z2)/abs(z)
> ans = 1.2275e-15
Even whithout doing asymptotics a generic formula from
http://mathworld.wolfram.com/InverseHyperbolicCosine.html
gives proper result:
octave2.9:98> function y=acosh1(x); y=log(x+sqrt(x+1).*sqrt(x-1)); endfunction
octave2.9:99> z = -7.28e+15 + 1.34e+15i
z = -7.2800e+15 + 1.3400e+15i
octave2.9:100> y1=acosh1(z)
y1 = 37.2337 + 2.9596i
octave2.9:101> y = log( -4*z^2 ) / 2 - pi*sqrt( -z^2 ) / (2*z)
y = 37.2337 + 2.9596i
octave2.9:102> y1-y
ans = 0
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