I am trying to normalize the following vector in Mathematica with the Normalize command.

$$(-1+2\cot[x]\csc[x]+2\csc[x]^2,\;2^\frac{1}{2} (1+\cos[x])\csc[x],\; 1)$$

I specified that the argument was real, but keep getting Abs in my answer. Is there anyway around this?

  • $\begingroup$ What's the problem with the absolute value? $\endgroup$ – Kyle Kanos Jan 24 '14 at 14:54
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    $\begingroup$ Could you post actual Mathematica code ? $\endgroup$ – b.gates.you.know.what Jan 24 '14 at 15:08
  • $\begingroup$ Real isn't enough. For example, Csc[x] is ComplexInfinity when x=Pi. $\endgroup$ – bill s Jan 24 '14 at 15:09

x being real-valued isn't enough, since some of the trig functions (like Csc[x]) can be infinite for finite x (like x=Pi). If you make the assumption that x is between 0 and Pi, then the expression can be simplified without the Abs[]

v = {-1 + 2 Cot[x] Csc[x] + 2 Csc[x] 2, 2^(1/2) (1 + Cos[x]) Csc[x], 1};
FullSimplify[Normalize[v], 0 < x < Pi]

enter image description here

It also simplifies to the same answer when Pi < x < 2 Pi so the problem really is at the singularity. It's easy to calculate what happens at the singularities by taking the limit as x->0 and x->Pi:

Limit[v, x -> 0]
{∞, ∞, 1}

Limit[v, x -> Pi]
{-∞, 0, 1}
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  • $\begingroup$ So it will not simplify then if x is between 0 and $2\pi$ $\endgroup$ – user32462 Jan 24 '14 at 15:15
  • $\begingroup$ It also simplifies if you assume Pi < x < 2 Pi, so the problem really is at the singularity. $\endgroup$ – bill s Jan 24 '14 at 15:17

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