Integrability of the derivative of $f(x) = x^2 sin (1/x^2)$ if $x ne 0$ and 0 otherwise.
why is the derivative of the function $f(x) = x^2 sin (1/x^2)$ if $x ne 0$ and 0 otherwise, not integrable?
real-analysis calculus integration riemann-integration
edited Dec 6 at 11:21
Larry
1,3862722
asked Dec 6 at 11:11
hopefully
129112
add a comment |
why is the derivative of the function $f(x) = x^2 sin (1/x^2)$ if $x ne 0$ and 0 otherwise, not integrable?
real-analysis calculus integration riemann-integration
edited Dec 6 at 11:21
Larry
1,3862722
asked Dec 6 at 11:11
hopefully
129112
The least you can do is compute the derivative first. Then we can think about why it is not integrable.
– астон вілла олоф мэллбэрг
Dec 6 at 11:22
Have you plotted this derivative and seen what it looks like?
– Arthur
Dec 6 at 11:22
I have computed it in my notebook @астонвіллаолофмэллбэрг
– hopefully
Dec 6 at 11:27
1
@hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
– астон вілла олоф мэллбэрг
Dec 6 at 11:29
1
I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
– Daphna Keidar
Dec 6 at 11:35
add a comment |
why is the derivative of the function $f(x) = x^2 sin (1/x^2)$ if $x ne 0$ and 0 otherwise, not integrable?
real-analysis calculus integration riemann-integration
edited Dec 6 at 11:21
Larry
1,3862722
asked Dec 6 at 11:11
hopefully
129112
why is the derivative of the function $f(x) = x^2 sin (1/x^2)$ if $x ne 0$ and 0 otherwise, not integrable?
real-analysis calculus integration riemann-integration
real-analysis calculus integration riemann-integration
edited Dec 6 at 11:21
Larry
1,3862722
asked Dec 6 at 11:11
hopefully
129112
edited Dec 6 at 11:21
Larry
1,3862722
asked Dec 6 at 11:11
hopefully
129112
edited Dec 6 at 11:21
Larry
1,3862722
edited Dec 6 at 11:21
Larry
1,3862722
edited Dec 6 at 11:21
Larry
1,3862722
1,3862722
asked Dec 6 at 11:11
hopefully
129112
asked Dec 6 at 11:11
hopefully
129112
asked Dec 6 at 11:11
hopefully
129112
129112
The least you can do is compute the derivative first. Then we can think about why it is not integrable.
– астон вілла олоф мэллбэрг
Dec 6 at 11:22
Have you plotted this derivative and seen what it looks like?
– Arthur
Dec 6 at 11:22
I have computed it in my notebook @астонвіллаолофмэллбэрг
– hopefully
Dec 6 at 11:27
1
@hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
– астон вілла олоф мэллбэрг
Dec 6 at 11:29
1
I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
– Daphna Keidar
Dec 6 at 11:35
add a comment |
The least you can do is compute the derivative first. Then we can think about why it is not integrable.
– астон вілла олоф мэллбэрг
Dec 6 at 11:22
Have you plotted this derivative and seen what it looks like?
– Arthur
Dec 6 at 11:22
I have computed it in my notebook @астонвіллаолофмэллбэрг
– hopefully
Dec 6 at 11:27
1
@hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
– астон вілла олоф мэллбэрг
Dec 6 at 11:29
1
I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
– Daphna Keidar
Dec 6 at 11:35
The least you can do is compute the derivative first. Then we can think about why it is not integrable.
– астон вілла олоф мэллбэрг
Dec 6 at 11:22
The least you can do is compute the derivative first. Then we can think about why it is not integrable.
– астон вілла олоф мэллбэрг
Dec 6 at 11:22
Have you plotted this derivative and seen what it looks like?
– Arthur
Dec 6 at 11:22
Have you plotted this derivative and seen what it looks like?
– Arthur
Dec 6 at 11:22
I have computed it in my notebook @астонвіллаолофмэллбэрг
– hopefully
Dec 6 at 11:27
I have computed it in my notebook @астонвіллаолофмэллбэрг
– hopefully
Dec 6 at 11:27
1
1
@hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
– астон вілла олоф мэллбэрг
Dec 6 at 11:29
@hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
– астон вілла олоф мэллбэрг
Dec 6 at 11:29
1
1
I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
– Daphna Keidar
Dec 6 at 11:35
I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
– Daphna Keidar
Dec 6 at 11:35
add a comment |
2 Answers
2
active
oldest
votes
For $x ne 0$ we have $f'(x)=2x sin(1/x^2)-frac{1}{x}cos(1/x^2)$. Then:
$f'(frac{1}{sqrt{n pi}})=-sqrt{n pi}cos(n pi)=(-1)^{n+1}sqrt{n pi}$,
thus $|f'(frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$
Similar: $|f'(-frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$
Conclusion: if $[a,b]$ is an intervall with $0 in [a,b]$ , then $f'$ is not bounded on $[a,b]$, hence $f'$ is not R- integrable on $[a,b]$.
answered Dec 6 at 11:23
Fred
44.1k1645
add a comment |
What is your domain? If it is $mathbf{R}$ then $f$ is also not integrable. It is by the definition of Riemann integrability. Now if your domain is $[0,1]$ then $f$ is differentiable but its derivative $f'(x)=2xsin(frac{1}{x^2})-frac{cos(frac{1}{x^2})}{x}$ is not bounded so not differentiable.
edited Dec 6 at 12:15
Tianlalu
3,01021038
answered Dec 6 at 11:30
MANI SHANKAR PANDEY
276
I want to show that the derivative is not integrable on [-1,1].
– hopefully
Dec 6 at 12:01
Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
– user539887
Dec 6 at 12:54
In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
– MANI SHANKAR PANDEY
Dec 6 at 16:35
add a comment |
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2 Answers
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2 Answers
2
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active
oldest
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active
oldest
votes
For $x ne 0$ we have $f'(x)=2x sin(1/x^2)-frac{1}{x}cos(1/x^2)$. Then:
$f'(frac{1}{sqrt{n pi}})=-sqrt{n pi}cos(n pi)=(-1)^{n+1}sqrt{n pi}$,
thus $|f'(frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$
Similar: $|f'(-frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$
Conclusion: if $[a,b]$ is an intervall with $0 in [a,b]$ , then $f'$ is not bounded on $[a,b]$, hence $f'$ is not R- integrable on $[a,b]$.
answered Dec 6 at 11:23
Fred
44.1k1645
add a comment |
For $x ne 0$ we have $f'(x)=2x sin(1/x^2)-frac{1}{x}cos(1/x^2)$. Then:
$f'(frac{1}{sqrt{n pi}})=-sqrt{n pi}cos(n pi)=(-1)^{n+1}sqrt{n pi}$,
thus $|f'(frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$
Similar: $|f'(-frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$
Conclusion: if $[a,b]$ is an intervall with $0 in [a,b]$ , then $f'$ is not bounded on $[a,b]$, hence $f'$ is not R- integrable on $[a,b]$.
answered Dec 6 at 11:23
Fred
44.1k1645
add a comment |
For $x ne 0$ we have $f'(x)=2x sin(1/x^2)-frac{1}{x}cos(1/x^2)$. Then:
$f'(frac{1}{sqrt{n pi}})=-sqrt{n pi}cos(n pi)=(-1)^{n+1}sqrt{n pi}$,
thus $|f'(frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$
Similar: $|f'(-frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$
Conclusion: if $[a,b]$ is an intervall with $0 in [a,b]$ , then $f'$ is not bounded on $[a,b]$, hence $f'$ is not R- integrable on $[a,b]$.
answered Dec 6 at 11:23
Fred
44.1k1645
For $x ne 0$ we have $f'(x)=2x sin(1/x^2)-frac{1}{x}cos(1/x^2)$. Then:
$f'(frac{1}{sqrt{n pi}})=-sqrt{n pi}cos(n pi)=(-1)^{n+1}sqrt{n pi}$,
thus $|f'(frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$
Similar: $|f'(-frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$
Conclusion: if $[a,b]$ is an intervall with $0 in [a,b]$ , then $f'$ is not bounded on $[a,b]$, hence $f'$ is not R- integrable on $[a,b]$.
answered Dec 6 at 11:23
Fred
44.1k1645
answered Dec 6 at 11:23
Fred
44.1k1645
answered Dec 6 at 11:23
Fred
44.1k1645
answered Dec 6 at 11:23
Fred
44.1k1645
44.1k1645
add a comment |
add a comment |
What is your domain? If it is $mathbf{R}$ then $f$ is also not integrable. It is by the definition of Riemann integrability. Now if your domain is $[0,1]$ then $f$ is differentiable but its derivative $f'(x)=2xsin(frac{1}{x^2})-frac{cos(frac{1}{x^2})}{x}$ is not bounded so not differentiable.
edited Dec 6 at 12:15
Tianlalu
3,01021038
answered Dec 6 at 11:30
MANI SHANKAR PANDEY
276
I want to show that the derivative is not integrable on [-1,1].
– hopefully
Dec 6 at 12:01
Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
– user539887
Dec 6 at 12:54
In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
– MANI SHANKAR PANDEY
Dec 6 at 16:35
add a comment |
What is your domain? If it is $mathbf{R}$ then $f$ is also not integrable. It is by the definition of Riemann integrability. Now if your domain is $[0,1]$ then $f$ is differentiable but its derivative $f'(x)=2xsin(frac{1}{x^2})-frac{cos(frac{1}{x^2})}{x}$ is not bounded so not differentiable.
edited Dec 6 at 12:15
Tianlalu
3,01021038
answered Dec 6 at 11:30
MANI SHANKAR PANDEY
276
I want to show that the derivative is not integrable on [-1,1].
– hopefully
Dec 6 at 12:01
Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
– user539887
Dec 6 at 12:54
In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
– MANI SHANKAR PANDEY
Dec 6 at 16:35
add a comment |
What is your domain? If it is $mathbf{R}$ then $f$ is also not integrable. It is by the definition of Riemann integrability. Now if your domain is $[0,1]$ then $f$ is differentiable but its derivative $f'(x)=2xsin(frac{1}{x^2})-frac{cos(frac{1}{x^2})}{x}$ is not bounded so not differentiable.
edited Dec 6 at 12:15
Tianlalu
3,01021038
answered Dec 6 at 11:30
MANI SHANKAR PANDEY
276
What is your domain? If it is $mathbf{R}$ then $f$ is also not integrable. It is by the definition of Riemann integrability. Now if your domain is $[0,1]$ then $f$ is differentiable but its derivative $f'(x)=2xsin(frac{1}{x^2})-frac{cos(frac{1}{x^2})}{x}$ is not bounded so not differentiable.
edited Dec 6 at 12:15
Tianlalu
3,01021038
answered Dec 6 at 11:30
MANI SHANKAR PANDEY
276
edited Dec 6 at 12:15
Tianlalu
3,01021038
edited Dec 6 at 12:15
Tianlalu
3,01021038
edited Dec 6 at 12:15
Tianlalu
3,01021038
3,01021038
answered Dec 6 at 11:30
MANI SHANKAR PANDEY
276
answered Dec 6 at 11:30
MANI SHANKAR PANDEY
276
answered Dec 6 at 11:30
MANI SHANKAR PANDEY
276
276
I want to show that the derivative is not integrable on [-1,1].
– hopefully
Dec 6 at 12:01
Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
– user539887
Dec 6 at 12:54
In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
– MANI SHANKAR PANDEY
Dec 6 at 16:35
add a comment |
I want to show that the derivative is not integrable on [-1,1].
– hopefully
Dec 6 at 12:01
Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
– user539887
Dec 6 at 12:54
In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
– MANI SHANKAR PANDEY
Dec 6 at 16:35
I want to show that the derivative is not integrable on [-1,1].
– hopefully
Dec 6 at 12:01
I want to show that the derivative is not integrable on [-1,1].
– hopefully
Dec 6 at 12:01
Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
– user539887
Dec 6 at 12:54
Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
– user539887
Dec 6 at 12:54
In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
– MANI SHANKAR PANDEY
Dec 6 at 16:35
In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
– MANI SHANKAR PANDEY
Dec 6 at 16:35
add a comment |
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The least you can do is compute the derivative first. Then we can think about why it is not integrable.
– астон вілла олоф мэллбэрг
Dec 6 at 11:22
Have you plotted this derivative and seen what it looks like?
– Arthur
Dec 6 at 11:22
I have computed it in my notebook @астонвіллаолофмэллбэрг
– hopefully
Dec 6 at 11:27
1
@hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
– астон вілла олоф мэллбэрг
Dec 6 at 11:29
1
I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
– Daphna Keidar
Dec 6 at 11:35