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MathGroup Archive 2004

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Re: Expansion of an exponential expression

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47228] Re: Expansion of an exponential expression
  • From: adam.smith at hillsdale.edu (Adam Smith)
  • Date: Wed, 31 Mar 2004 02:58:29 -0500 (EST)
  • References: <c4bdrj$6uu$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

carlos at colorado.edu (Carlos Felippa) wrote in message 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?

Would the following do what you want?

In[1]:=
anlog= Log[ ((2 + Sqrt[1 + 2*h])^(-1))^(t/h) ] 

Out[1]=
Log[(1/(2 + Sqrt[1 + 2*h]))^(t/h)]

In[2]:=
simpanlog = Simplify[ anlog,{h >0,t>0}]

Out[2]=
-((t*Log[2 + Sqrt[1 + 2*h]])/h)

Though not exactly the form you specified above, it is very close and
can be used in the Series command

In[3]:=
thing = Normal[ Series[simpanlog,{h,0,3}]]

Out[3]=
-(t/3) + (2*h*t)/9 - (19*h^2*t)/81 + (97*h^3*t)/324 - 
  (t*Log[3])/h

Adam Smith


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