Expressions without absolute value signs

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Homework Help Overview

The discussion revolves around rewriting expressions without absolute value signs, specifically focusing on the expression a - |a - |x||. Participants are exploring the implications of absolute values in the context of piecewise functions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the number of cases that need to be considered when rewriting the expression, with some suggesting there are two cases while others argue for four based on different conditions of x and a.

Discussion Status

The discussion is active, with participants presenting different interpretations of the number of cases involved. Some have offered reasoning for their positions, but there is no explicit consensus on the correct number of cases to consider.

Contextual Notes

Participants are questioning the necessity of certain absolute functions and the implications of different ranges for x and a. There is an acknowledgment of the complexity introduced by the absolute values in the expression.

garyljc
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Homework Statement


Rewrite the following expressions without absolute value signs, treating various cases separately where neccesary


Homework Equations


a-Abs[(a-(abs)a)]


the question is do i have 2 answers to this ?
 
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garyljc said:
the question is do i have 2 answers to this ?
Yes there will be two cases. At first inspection one might conclude that there would be four cases, but you should note that one of the absolute functions is not necessary.
 
There is only one[/b\] function for this, which may have different parts.

If f(x)= a- |a- |x||, then, of course, you need to look at x< 0 and x> 0, then at the cases |x|> a and |x|< a.

You could also write this as a single "formula" using the Heaviside step function which is defined as H(x)= 0 if x< 0, H(x)= 1 if x\ge 0.
 
Hootenanny said:
Yes there will be two cases. At first inspection one might conclude that there would be four cases, but you should note that one of the absolute functions is not necessary.

why would there be 4 cases ?
 
Hmmm, the function is a-|a- |x||. If x> 0 that is a- |a- x| so if x< a, that is if a-x> 0, we have a- (a- x)= x. If x> a, that is if a- x< 0, |a-x|= -(a-x) and we have a+(a-x)= 2a- x.

If x< 0, so we have a- |a+ x|. Now if a> -x, so x+ a> 0, that is a- (a+ x)= -x. Finally if a< -x so x+ a< 0, that is a+ (a+x)= 2a+ x. Looks like 4 cases to me.
 

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