Exponential Function and Expand
- To: mathgroup at smc.vnet.net
- Subject: [mg18114] Exponential Function and Expand
- From: Blimbaum Jerry DLPC <BlimbaumJE at ncsc.navy.mil>
- Date: Thu, 17 Jun 1999 12:26:45 -0400
- Delivery-date: Thu Jun 17 15:53:31 1999
- Sender: owner-wri-mathgroup at wolfram.com
In mg: 18029 Drago Ganic wrote: "Why Mathematica automatically expand E^u E^v to E^(u+v) but never the inverse situation, i.e. E^(u+v) to 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 Complex numbers) except a = 0..." One response stated the importance of transforming a^b*a^c into a^(b+c) because of its importance for many transformations. I wanted to add the example that suppose u = a and v = -a, hence E^a E^-a transforms to E^(a+(-a)) = E^0 = 1. If you did it the other way you end up with E^a*E^-a which really doesnt do you any good. However, in order to show that Mathematica, in fact, does that I typed in the following: u = a; v = -a; Expand[E^u*E^v]//TracePrint with the expectation that at some point I would see the transformation E^(a-a) by which Mathematica would get E^0 , hence the output 1. But to my surprise the last few outputs were: E^a*E^-a E^-a*E^a 1 Expand[1] 1 So where does the transformation a^b*a^c into a^(b+c) occur?? Why did Mathematica simply change the order of the exponentials and then come up with a value of 1 rather then showing the explicit addition of a + (-a) = 0?? Also, why would it try to Expand a number? THANKS. Jerry Blimbaum NSWC Panama City, Fl