- #1

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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

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- Thread starter maximus101
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- #1

- 22

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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

- #2

- 256

- 2

Did you try actually integrating f(x) ?

- #3

- 76

- 0

- #4

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Doesn't Riemann integrationassumecontinuity?

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

Science Advisor

Homework Helper

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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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derivativeof 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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