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Re: Absolute value

Am Mon, 28 Jun 2010 06:30:57 +0000 (UTC)
schrieb Marco Masi <marco.masi at>:

> Yes, thank you, that brough me a step forward (and yes, I forgot to
> square the Abs value in the previous example... sorry for that).
> However, there is still a step which I can't accomplish. Please try
> the following: FullSimplify[ ComplexExpand[ Abs[1/2 (A1 E^(I \[Phi]1)
> - A2 E^(I \[Phi]2) + A1 E^(I \[Phi]1) Cos[Sqrt[2] c z] + A2 E^(I
> \[Phi]2) Cos[Sqrt[2] c z])]]^2]
> I would like to have Mathematica avoiding one of the resulting Cos[2
> Sqrt[2] c z] expression, and maintain both as Cos[Sqrt[2] c z], and
> then simplify. How should I proceed?
> Regards, Mark.

Hi Mark,

try squaring the absolute value before using ComplexExpand:
   Abs[(1/2)*(A1*E^(I*\[Phi]1) - A2*E^(I*\[Phi]2) +
A1*E^(I*\[Phi]1)*Cos[Sqrt[2]*c*z] +

--> (1/4)*Abs[A2*E^(I*\[Phi]2)*(-1 + Cos[Sqrt[2]*c*z]) +
A1*E^(I*\[Phi]1)*(1 + Cos[Sqrt[2]*c*z])]^2

but I would prefer:

    Abs[(1/2)*(A1*E^(I*\[Phi]1) -
      A2*E^(I*\[Phi]2) + A1*E^(I*\[Phi]1)*Cos[Sqrt[2]*c*z] +
   TargetFunctions -> {Re,Im}],
A1 | A2, TrigFactor]

which returns
A1^2*Cos[(c*z)/Sqrt[2]]^4 + A2^2*Sin[(c*z)/Sqrt[2]]^4 -
(1/2)*A1*A2*Cos[\[Phi]1 - \[Phi]2]*Sin[Sqrt[2]*c*z]^2


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