       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

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
```

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