Re: Simplifying the exponents
- To: mathgroup at smc.vnet.net
- Subject: [mg80053] Re: [mg79994] Simplifying the exponents
- From: "Jung-Tsung Shen" <jushen at gmail.com>
- Date: Sat, 11 Aug 2007 02:15:20 -0400 (EDT)
- References: <200708100550.BAA10857@smc.vnet.net>
Thanks, Andrzej, and Jean-Marc. That's very helpful! :-)
JT
On 8/10/07, Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
>
>
> On 10 Aug 2007, at 07:50, Jung-Tsung Shen wrote:
>
> > Hello, I would like to ask a question which I haven't been able to
> > find a solution that does not need human intervening.
> >
> > I would like to simplify the following expression
> >
> > Exp[I (q1 y1 + q2 y2 + q3 y3) - I (qp1 y1 + qp2 y2 + qp3 y3)]
> >
> > according to y1, y2, and y3 so it would look like
> >
> > Exp[I (q1-qp1) y1+ I (q2-qp2) y2 + I (q3-qp3) y3], or
> >
> > Exp[I (q1-qp1) y1] * Exp[I (q2-qp2) y2] * Exp[I (q3-qp3) y3]
> >
> > How could I achieve this in an efficient way?
> >
> > Thanks.
> >
> > JT
> >
>
>
> (Collect[#1, {y1, y2, y3}] & ) /@ (E^(I*(q1*y1 + q2*y2 + q3*y3) - I*
> (qp1*y1 + qp2*y2 + qp3*y3)))
>
> E^((I*q1 - I*qp1)*y1 + (I*q2 - I*qp2)*y2 + (I*q3 - I*qp3)*y3)
>
>
> Your second request is impossible in Mathematica (without using
> HoldForm), since Mathematica always evaluates
>
> Exp[a]*Exp[b]
> E^(a + b)
>
> so even if you execute code that returns your desired output, this
> output (unless wrapped in HoldForm) will automatically be converted
> into the form above.
>
> Andrzej Kozlowski
>
- References:
- Simplifying the exponents
- From: "Jung-Tsung Shen" <jushen@gmail.com>
- Simplifying the exponents