Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2004
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Expansion of an exponential expression

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47240] Re: Expansion of an exponential expression
  • From: "Peter Pein" <no at spam.no>
  • Date: Wed, 31 Mar 2004 02:59:45 -0500 (EST)
  • References: <c4bdrj$6uu$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Carlos Felippa" <carlos at colorado.edu> schrieb im Newsbeitrag
news:c4bdrj$6uu$1 at smc.vnet.net...
> As a result of some calculations I have
> (a snipet of a more complex expression)
>
> anlog= Log[ ((2 + Sqrt[1 + 2*h])^(-1))^(t/h)   ]
>
> I would like to simplify this to
>
> anlog= (t/h) Log[((2 + Sqrt[1 + 2*h])^(-1))   ]
>
> so I can then take the Taylor series in h.
> But FullSimplify[anlog,h>0]  doesnt do it.
> Do i need a ComplexityFunction?
>

In[4]:= FullSimplify[anlog, h > 0 && t \[Element] Reals]
Out[4]= -((t*Log[2 + Sqrt[1 + 2*h]])/h)

You can enter <esc>elem<esc> for \[Element]
-- 
Peter Pein, Berlin
StringReplace["petsie at arcand.de",
 Rule@@(ToLowerCase@ToString@Head@#&)/@{William&&S,2b||!2b}]



  • Prev by Date: RE: sort list
  • Next by Date: Re: Constant function Integrate Assumption
  • Previous by thread: Re: Expansion of an exponential expression
  • Next by thread: Re: Expansion of an exponential expression