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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
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