Continuous function on R at certain values of a

  • Thread starter maximus101
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  • #1
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If we suppose [tex]a[/tex] > 0 is some constant and f: R [tex]\rightarrow[/tex] R is


f(x) = [tex]|x|^a[/tex] sin(1/x) if x [tex]\neq[/tex] 0

and

f(x) =0 if x=0

if we let F(x) := f '(x) for x [tex]\neq[/tex] 0 and F(0) :=0. For what values of
[tex]a[/tex] is F a continuous function in R
 

Answers and Replies

  • #2
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Did you try actually integrating f(x) ?
 
  • #3
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Did you try actually integrating f(x) ?

Doesn't Riemann integration assume continuity?
 
  • #4
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Doesn't Riemann integration assume continuity?

Definitely not. For instance, the integral of a step function is certainly defined. Lebesgue also proved that a function is Riemann integrable if and only if its set of discontinuities has measure zero (http://en.wikipedia.org/wiki/Riemann_integral), so for instance a countable number of discontinuities is acceptable.
 
  • #5
HallsofIvy
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Why are you talking about integrating? The problem asked about the derivative of that function. F is clearly continuous for all x other than 0 so the only question is the derivative at x= 0.

For x> 0,
[tex]\frax{f(x)- f(0)}{x}= \frac{x^a sin(1/x)}{x}= x^{a-1}sin(1/x)[/tex]
and, for any a> 1, the limit, as x goes to 0, is 0. If [itex]a\le 1[/itex] the limit does not exist.

For x< 0 we can replace x with the positive number y= -x and, since sine is an odd function we have
[tex]\frac{f(x)- f(0)}{x}= \frac{-y^a sin(1/y)}{-y}= y^{a- 1}sin(1/y)[/tex]
and have the same result as before.
 
  • #6
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Why are you talking about integrating? The problem asked about the derivative of that function. F is clearly continuous for all x other than 0 so the only question is the derivative at x= 0.

For x> 0,
[tex]\frax{f(x)- f(0)}{x}= \frac{x^a sin(1/x)}{x}= x^{a-1}sin(1/x)[/tex]
and, for any a> 1, the limit, as x goes to 0, is 0. If [itex]a\le 1[/itex] the limit does not exist.

For x< 0 we can replace x with the positive number y= -x and, since sine is an odd function we have
[tex]\frac{f(x)- f(0)}{x}= \frac{-y^a sin(1/y)}{-y}= y^{a- 1}sin(1/y)[/tex]
and have the same result as before.

Hey thanks, could you explain why at x=0 f(x) is not differentiable for any value of a?
I think it is because the limit does not exist, for any a, but I don't know how to prove it.
 

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