# Math Idea

Discussion in 'Math' started by adrian.dmc, Jan 22, 2008.

Feb 22, 2007
52
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I as studying for math and I as studying the equation |x| and I as thinking in a way to draw that function in my calculator but I didn't know how to represent |x| so I got the idea of plotting sqrt(x^2) and it worked. Later I as confronted with a function like: -1 for -1 < x < 0 and 1 for 0 < x < 1 and because |x|' its that function I derivate the function sqrt(x^2) and obtained x/sqrt(x^2) and it worked its exactly the same except that the function is not defined in x = 0.

Is this "property" used, in my opinion I think it would be very useful for example in "heaviside" functions? Am I being stupid?

Example:
5 + 3 * heaveside(x-2) = 5 + 3 * ((x/sqrt(x^2) + 1)/2)

2. ### Dave Retired Moderator

Nov 17, 2003
6,960
146
Yes, it is known as the modulus of the number and is very widely used. None more so than in complex numbers that are widely used in many branches of electrical and electronics engineering.

See Volume 2 - Chapter 2 for examples of how it is extensively used in electronics.

Dave

Feb 22, 2007
52
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I was not referring to |x| I as referring to the equivalence between |x| and sqrt(x^2) and to the "special" function provided by the derivate of that.

4. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
|x| = sqrt(x^2) for all real numbers x.
d/dx (|x|) = -1 for x < 0
= 1 for x > 0

The absolute value function is not differentiable for x = 0, which is easy to understand because of the sharp corner at (0,0).

The Heaviside step function is defined in a similar way, but isn't the same. It's defined in this way:
H(x) = 0 for x < 0
H(x) = 1 for x > 0
To make it defined everywhere on its domain, you can set H(0) = 0.5, which is something of an arbitrary value. Doing this makes it piecewise continuous. There's no possible value of H(0) that will make the function continuous.
Mark

Feb 22, 2007
52
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We are right I'm wrong... (LOL) The last part of the post (the Example) is completely wrong... Sorry.

6. ### Dave Retired Moderator

Nov 17, 2003
6,960
146
Apologies, I misinterpreted your initial question. I concur with the answer provided by Mark44.

Dave