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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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