Re: Mathematical question: Exponetial function

*To*: mathgroup at smc.vnet.net*Subject*: [mg18071] Re: [mg18029] Mathematical question: Exponetial function*From*: "Andrzej Kozlowski" <andrzej at tuins.ac.jp>*Date*: Tue, 15 Jun 1999 01:43:30 -0400*Delivery-date*: Tue Jun 15 04:39:07 1999*Sender*: owner-wri-mathgroup at wolfram.com

Mathematica automatically transforms a^b*a^c into a^(b+c) because this is exactly what is needed in the vast majority of transformations that one wants Mathematica to do. If Mathematica did not do this automatically but required some special command to be entered it's interface would become very cumbersome. However, because Mathematica automatically transforms a^b*a^c into a^(b+c) it is difficult even to define a function that would do the opposite without using something like Hold (note also that the "opposite" only has a meaning for certain symbolic expressions, i.e. although there is a unique way to transform a^(2+3) there isn't one for a^5). You can see exactly what happens if we define a new function power and format it to look like the built in Power: In[1]:= Format[power[a_,b_]]:=a^b In[2]:= General::"spell1": "Possible spelling error: new symbol name \"\!\(power\)\" is similar to \ existing symbol \"\!\(Power\)\"." Now power[a,b] Out[2]= b a With power it is easy to do what you were asking about: In[3]:= Distribute[power[a,b+c],Plus,power,Times] Out[3]= b c a a However, if we now replace power by the built in Power the exprerssion will be transformed back into a^(b+c): In[4]:= %/.power->Power Out[4]= b + c a -- Andrzej Kozlowski Toyama International University JAPAN http://sigma.tuins.ac.jp http://eri2.tuins.ac.jp ---------- >From: "Drago Ganic" <drago.ganic at in2.hr> To: mathgroup at smc.vnet.net >To: mathgroup at smc.vnet.net >Subject: [mg18071] [mg18029] Mathematical question: Exponetial function >Date: Sun, Jun 13, 1999, 4:52 AM > > Why Mathematica automaticaly expand > E^u E^v > to > E^(u+v). > > but never the inverse situation, ie. E^(u+v) -> E^u E^v. > > In my textbooks I find that the relation a^(u+v) = a^u a^v is valid for any > a, u, v (also Comlex numbers) except a=0. > > Thanks in advance. > > Drago, > Croatia > >