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I tried several ways, but i could not come up with any way to have an equation as such:

|n| = ...

without using the absolute value signs on the right side of the equation.

I do not know if there is any way...

I tried using some form of n * i^(an + b) but those efforts were futile. And I couldn't think of any other way to approach this.

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  • $\begingroup$ I'm guessing you want a single expression, and not a different expression for $n<0$ and $n>0$? Maybe Fourier transform? $\endgroup$ – levitopher Oct 27 '13 at 15:39
  • $\begingroup$ $|n|=\begin{cases} n \text{ if n }>0 \\ -n \text{ if n }\le 0 \\ \end{cases}$ $\endgroup$ – Shahab Oct 27 '13 at 15:56
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One way is to note that the square root has "branches":

$$(\sqrt n)^2=n\tag 1$$

$$\sqrt{n^2}=|n|\tag 2$$

Note that $(1)$ would dip into the imaginary numbers for a moment if $n\lt 0$. $(2)$ is true because the square root of a number is one of two solutions to the equation involving the square.

$$x^2=4,x=2,-2$$

$$\sqrt 4=2,-\sqrt 4=-2$$

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  • $\begingroup$ I thought of this, but I wasn't sure if it was completely correct, as it felt dirty, since square roots technically have two solutions... Does n^(1/2) also have 2 solutions since it is equivalent to the square root? I was hoping for a very definite equation, but this works. Thanks $\endgroup$ – kukac67 Oct 27 '13 at 15:48
  • $\begingroup$ Yes, $n^{\frac 12}$ has the same two branches as the square root. $\endgroup$ – abiessu Oct 27 '13 at 15:54
  • $\begingroup$ Further, the $n$th root creates branches in the same way, consider the $\arg$ function in complex analysis. $\endgroup$ – abiessu Oct 27 '13 at 19:07

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