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