Can you define an absolute value abs(x) function as an algebraic expresion without the use of roots (or fractional exponents)? For instance, can you come up with another function other than abs(x)=(sqrt(x))^2?
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Anonymous2006-11-11 18:18
oops fuck I mean other than abs(x)=sqrt(x^2) (accepts negative x, only want real numbers)
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Anonymous2006-11-11 18:30
If x<0 then -1*x else x.
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Anonymous2006-11-11 18:40
>>1
If by "algebraic expression without the use of roots (or fractional exponents)" you mean "polynomial," the answer is no, since |x| isn't differentiable at 0 and polynomials are necessarily differentiable everywhere.
>>9
Dumbass, your link proves that piecewise functions can be algebraic if their pieces are algebraic and continuous over their domain, and the pieces are in said piecewise function.
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Anonymous2006-11-13 21:22
"An algebraic expression in variables {x_1,...,x_n} is an expression constructed with the variables and algebraic numbers using addition, multiplication, and rational powers."
Really? Show how f(x) = {0 if x>0, 1 if x <=0) meets that definition.
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Anonymous2006-11-14 12:54
sqrt(x^2)
force it to be one valued or something
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Anonymous2006-11-14 14:27
pseudocode:
$number
$numberstring = string($number)
if first character $numberstring = "-"
$numberstring = $numberstring from 2nd character on
end