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

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

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

Out=
-((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:=
thing = Normal[ Series[simpanlog,{h,0,3}]]

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