       RE: Expand Exp[a(b-c)]

• To: mathgroup at smc.vnet.net
• Subject: [mg16988] RE: [mg16940] Expand Exp[a(b-c)]
• From: "Ersek, Ted R" <ErsekTR at navair.navy.mil>
• Date: Sat, 10 Apr 1999 02:13:28 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```Cyril Fisher wrote:
-----------------
How can I expand Exp[a(b+c)] to E^(a b) E^(-a c).
There should be active a rule, which re-evaluate expanded result
back to E^(a b-a c).
PowerExpand does not work in this case.

-----------------

Yes a lot of us would like to have a way to prevent the kernel from
completing the normal evaluation process.  In the notebook below I give you
other things I would like to do along the same lines.  For now things like
this are a little difficult to accomplish.

You can copy the Notebook stuff below into Mathematica 3.0.

Regards,
Ted Ersek

(-----------------)

Notebook[{

Cell[CellGroupData[{
Cell[BoxData[
\(Simplify[2/\((1 - w^2*x^2*\((y*z)\)^2)\)]\)], "Input"],

Cell[BoxData[
\(2\/\(1 - \((w\ x\ y\ z)\)\^2\)\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
\(FullSimplify[\((18446744073709551616 - n)\)/5]\)], "Input"],

Cell[BoxData[
\(\(2\^64 - n\)\/5\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
\(poly =
2\ e\^2 - 4\ e\ h + 2\ h\^2 + 3\ e\ s - 3\ h\ s + 5\ k \((e - h)\)\^3;

\nnewpoly = Series[poly, {e, h, 5}] // Normal\)], "Input"],

Cell[BoxData[
\(3\ s\ \((e - h)\) + 2\ \((e - h)\)\^2 + 5\ k\ \((e - h)\)\^3\)],
"Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
\(ser =
\((Series[BesselI[0, x], {x, 0, 8}, FullEvaluation -> False] //
Normal)
\)\)], "Input"],

Cell[BoxData[
\(1 + x\^2\/2\^2 + x\^4\/\(\(2\^4\) \((\(2!\))\)\^2\) +
x\^6\/\(\(2\^6\) \((\(3!\))\)\^2\) +
x\^8\/\(\(2\^8\) \((\(4!\))\)\^2\)\)], "Output"]
}, Open  ]]
},
FrontEndVersion->"Microsoft Windows 3.0",
ScreenRectangle->{{0, 800}, {0, 544}},
WindowSize->{664, 420},
WindowMargins->{{40, Automatic}, {Automatic, 5}}
]

```

• Prev by Date: Re: Help Formatting Outputs?
• Next by Date: ParametricPlot help please
• Previous by thread: Expand Exp[a(b-c)]
• Next by thread: Re: Expand Exp[a(b-c)]