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Simplifying expressions with Abs

Posted 10 years ago

I have a long expression which seems really complicated due to some Abs[] terms that should be simplified. However, even with the right assumptions, the expression is not simplified.

For example

Assuming[-1 <= a <= 1 && 0 <= b <= \[Pi]/4 && L \[Element] Integers && 1 - (-1)^L Sqrt[1 - a^2] Cos[2 b] > 0, Abs[1 - (-1)^L Sqrt[1 - a^2] Cos[2 b]] == 1 - (-1)^L Sqrt[1 - a^2] Cos[2 b] // FullSimplify]

should return True, but instead it returns (-1)^L Sqrt[1 - a^2] Cos[2 b] <= 1. On the contrary

Assuming[-1 <= a <= 1 && 0 <= b <= \[Pi]/4 && L \[Element] Integers && 1 - E^(I \[Pi] L) Sqrt[1 - a^2] Cos[2 b] >= 0, Abs[1 - E^(I \[Pi] L) Sqrt[1 - a^2] Cos[2 b]] == 1 - E^(I \[Pi] L) Sqrt[1 - a^2] Cos[2 b] // FullSimplify]

returns True.

Why is that?

Note that the trick to replace (-1)^L with E^(I \[Pi] L) do not always works, as in the case of

Assuming[-1 <= a <= 1 && 0 <= b <= \[Pi]/4 && 0 <= Q <= 1 && L \[Element] Integers && 1 - E^(I \[Pi] L) Sqrt[1 - a^2] Cos[2 b] >= 0 && 1 - 2 Q + E^(I \[Pi] L) Sqrt[1 - a^2] Cos[2 b] > 0, Sqrt[Abs[(1 - E^(I \[Pi] L)  Sqrt[1 - a^2] Cos[2 b])/(1 - 2 Q + E^(I \[Pi] L)  Sqrt[1 - a^2] Cos[2 b])]] == Sqrt[(1 - E^(I \[Pi] L)  Sqrt[1 - a^2] Cos[2 b])/(1 - 2 Q + E^(I \[Pi] L) Sqrt[1 - a^2] Cos[2 b])] // FullSimplify]
POSTED BY: N DP
Posted 10 years ago

A Mathematica user with Version 9.0.1.0 (Windows 8 x64) says all of them return True. I am using Mathematica 10.0 on a Windows 7 (Home Premium x64). Anyone aware of any issues?

POSTED BY: N DP
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