true or false that all real function is an antiderivativeby kallazans Tags: antiderivative, doubt, function, real 

#1
Dec2003, 11:58 AM

P: 5

is it true or false that all real function is an antiderivative of some real function but neither real function have an antiderivative?
I still have the doubt! Definition(Louis Leithold,The Calculus with Analytic Geometry) Antiderivative: F is antiderivative of f in I if F'(x)=f(x) for all x in I. The question is all f have some F in some I? The question is all f is a G of some g in some I? (Real Analysis) 



#2
Dec2003, 12:03 PM

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PF Gold
P: 16,101

By "primitive" do you mean that if [itex]f(x) = \int g(x) \, dx[/itex] then [itex]f(x)[/itex] is a primitive of [itex]g(x)[/itex]? (The usual English word for this is that [itex]f(x)[/itex] is an antiderivative or an integral of [itex]g(x)[/itex])
I'm not entirely sure what you're trying to ask... though it is false that any function is an antiderivative of another function. 



#3
Dec2003, 06:43 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,896

It's not clear what you mean by "either" function. If you mean the original function in the question and its antiderivative, then obviously IF "every function had a antiderivative", then it wouldn't make sense to say that THAT function did NOT have an antiderivative.
However, as Hurkyl pointed out, it is not true that every function has a primitive (antiderivative). For example, the function, f(x)= 1 if x is rational, 0 if x is irrational, does not have an antiderivative. It IS true that every bounded function whose points of discontinuity form a set of measure 0 is integrable (has an antiderivative). In particular every continuous function has an antiderivative as well as every bounded function with only a finite number of points of discontinuity. 


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