The derivatives of the integrals with Leibniz Rule?
$begingroup$
Can we obtain the following result for ... $f(x)={x-lfloor x rfloor}$ ... ?
It is very simple but important question. Here ${lfloor x rfloor}$ is floor function and $a in mathbb{R}$ . Thank you for your kind comment.
$$
frac{{frac {d} {du}}left[int_1^u f(x) cdot x^{-a-1} dxright]}
{{frac {d} {du}}left[int_1^u f(x) cdot x^{a-2}dxright]}
=u^{1-2a}
$$
real-analysis calculus integration derivatives
asked Dec 6 '18 at 14:13
Testform Testform
185
$endgroup$
add a comment |
$begingroup$
Can we obtain the following result for ... $f(x)={x-lfloor x rfloor}$ ... ?
It is very simple but important question. Here ${lfloor x rfloor}$ is floor function and $a in mathbb{R}$ . Thank you for your kind comment.
$$
frac{{frac {d} {du}}left[int_1^u f(x) cdot x^{-a-1} dxright]}
{{frac {d} {du}}left[int_1^u f(x) cdot x^{a-2}dxright]}
=u^{1-2a}
$$
real-analysis calculus integration derivatives
asked Dec 6 '18 at 14:13
Testform Testform
185
$endgroup$
1
$begingroup$
Why is this important? And what were your attempts solving this problem?
$endgroup$
– James
Dec 6 '18 at 14:14
$begingroup$
We are discussing about it very seriously with some mathematicians .
$endgroup$
– Testform
Dec 6 '18 at 14:18
$begingroup$
Can I get a more solid answer ? I mean we can obtain or not ?
$endgroup$
– Testform
Dec 6 '18 at 16:20
add a comment |
$begingroup$
Can we obtain the following result for ... $f(x)={x-lfloor x rfloor}$ ... ?
It is very simple but important question. Here ${lfloor x rfloor}$ is floor function and $a in mathbb{R}$ . Thank you for your kind comment.
$$
frac{{frac {d} {du}}left[int_1^u f(x) cdot x^{-a-1} dxright]}
{{frac {d} {du}}left[int_1^u f(x) cdot x^{a-2}dxright]}
=u^{1-2a}
$$
real-analysis calculus integration derivatives
asked Dec 6 '18 at 14:13
Testform Testform
185
$endgroup$
Can we obtain the following result for ... $f(x)={x-lfloor x rfloor}$ ... ?
It is very simple but important question. Here ${lfloor x rfloor}$ is floor function and $a in mathbb{R}$ . Thank you for your kind comment.
$$
frac{{frac {d} {du}}left[int_1^u f(x) cdot x^{-a-1} dxright]}
{{frac {d} {du}}left[int_1^u f(x) cdot x^{a-2}dxright]}
=u^{1-2a}
$$
real-analysis calculus integration derivatives
real-analysis calculus integration derivatives
asked Dec 6 '18 at 14:13
Testform Testform
185
asked Dec 6 '18 at 14:13
Testform Testform
185
edited Dec 6 '18 at 15:04
Testform
asked Dec 6 '18 at 14:13
Testform Testform
185
asked Dec 6 '18 at 14:13
Testform Testform
185
asked Dec 6 '18 at 14:13
Testform Testform
185
185
1
$begingroup$
Why is this important? And what were your attempts solving this problem?
$endgroup$
– James
Dec 6 '18 at 14:14
$begingroup$
We are discussing about it very seriously with some mathematicians .
$endgroup$
– Testform
Dec 6 '18 at 14:18
$begingroup$
Can I get a more solid answer ? I mean we can obtain or not ?
$endgroup$
– Testform
Dec 6 '18 at 16:20
add a comment |
1
$begingroup$
Why is this important? And what were your attempts solving this problem?
$endgroup$
– James
Dec 6 '18 at 14:14
$begingroup$
We are discussing about it very seriously with some mathematicians .
$endgroup$
– Testform
Dec 6 '18 at 14:18
$begingroup$
Can I get a more solid answer ? I mean we can obtain or not ?
$endgroup$
– Testform
Dec 6 '18 at 16:20
1
1
$begingroup$
Why is this important? And what were your attempts solving this problem?
$endgroup$
– James
Dec 6 '18 at 14:14
$begingroup$
Why is this important? And what were your attempts solving this problem?
$endgroup$
– James
Dec 6 '18 at 14:14
$begingroup$
We are discussing about it very seriously with some mathematicians .
$endgroup$
– Testform
Dec 6 '18 at 14:18
$begingroup$
We are discussing about it very seriously with some mathematicians .
$endgroup$
– Testform
Dec 6 '18 at 14:18
$begingroup$
Can I get a more solid answer ? I mean we can obtain or not ?
$endgroup$
– Testform
Dec 6 '18 at 16:20
$begingroup$
Can I get a more solid answer ? I mean we can obtain or not ?
$endgroup$
– Testform
Dec 6 '18 at 16:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
HINT
Use the Fundamental Theorem of Calculus. For any (constant) $a in mathbb{R}$:
$$
frac{d}{dz} int_a^z f(u)du = f(z).
$$
That should simplify your problem greatly.
UPDATE
In your case $f(x) = x - lfloor x rfloor$, so it has a countable number of discontinuities, which does not prevent the Riemann integral from existing. You have to define, for example,
$$
int_1^x g(f(u)) du
:= int_{lfloor x rfloor}^x g(f(u))du
+ sum_{k=2}^{lfloor x rfloor} int_1^k g(f(u)) du.
$$
This way, all the integrals on the RHS are well-defined and the LHS can be defined in a meaningful way.
answered Dec 6 '18 at 14:19
gt6989bgt6989b
34.7k22456
$endgroup$
$begingroup$
@Testform why not? discontinuities in $f$ are discrete, it is still integrable.
$endgroup$
– gt6989b
Dec 6 '18 at 14:26
$begingroup$
I have thought exactly like you. Even, I have tested it only with f(x) on Wolfram/Alpha. And I got the result exactly what you said. But I can not verify it for $f(x)={x-lfloor x rfloor}$ , the program gives nothing.
$endgroup$
– Testform
Dec 6 '18 at 14:32
$begingroup$
@Testform see update
$endgroup$
– gt6989b
Dec 6 '18 at 14:38
$begingroup$
thus, it means can we obtain the result in my question?
$endgroup$
– Testform
Dec 6 '18 at 14:41
$begingroup$
Thank you my freind. But someone say that you can not obtain it for $f(x)={x-lfloor x rfloor}$ . He say : Because, the floor function is not differentiable
$endgroup$
– Testform
Dec 6 '18 at 14:46
|
show 2 more comments
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT
Use the Fundamental Theorem of Calculus. For any (constant) $a in mathbb{R}$:
$$
frac{d}{dz} int_a^z f(u)du = f(z).
$$
That should simplify your problem greatly.
UPDATE
In your case $f(x) = x - lfloor x rfloor$, so it has a countable number of discontinuities, which does not prevent the Riemann integral from existing. You have to define, for example,
$$
int_1^x g(f(u)) du
:= int_{lfloor x rfloor}^x g(f(u))du
+ sum_{k=2}^{lfloor x rfloor} int_1^k g(f(u)) du.
$$
This way, all the integrals on the RHS are well-defined and the LHS can be defined in a meaningful way.
answered Dec 6 '18 at 14:19
gt6989bgt6989b
34.7k22456
$endgroup$
$begingroup$
@Testform why not? discontinuities in $f$ are discrete, it is still integrable.
$endgroup$
– gt6989b
Dec 6 '18 at 14:26
$begingroup$
I have thought exactly like you. Even, I have tested it only with f(x) on Wolfram/Alpha. And I got the result exactly what you said. But I can not verify it for $f(x)={x-lfloor x rfloor}$ , the program gives nothing.
$endgroup$
– Testform
Dec 6 '18 at 14:32
$begingroup$
@Testform see update
$endgroup$
– gt6989b
Dec 6 '18 at 14:38
$begingroup$
thus, it means can we obtain the result in my question?
$endgroup$
– Testform
Dec 6 '18 at 14:41
$begingroup$
Thank you my freind. But someone say that you can not obtain it for $f(x)={x-lfloor x rfloor}$ . He say : Because, the floor function is not differentiable
$endgroup$
– Testform
Dec 6 '18 at 14:46
|
show 2 more comments
$begingroup$
HINT
Use the Fundamental Theorem of Calculus. For any (constant) $a in mathbb{R}$:
$$
frac{d}{dz} int_a^z f(u)du = f(z).
$$
That should simplify your problem greatly.
UPDATE
In your case $f(x) = x - lfloor x rfloor$, so it has a countable number of discontinuities, which does not prevent the Riemann integral from existing. You have to define, for example,
$$
int_1^x g(f(u)) du
:= int_{lfloor x rfloor}^x g(f(u))du
+ sum_{k=2}^{lfloor x rfloor} int_1^k g(f(u)) du.
$$
This way, all the integrals on the RHS are well-defined and the LHS can be defined in a meaningful way.
answered Dec 6 '18 at 14:19
gt6989bgt6989b
34.7k22456
$endgroup$
$begingroup$
@Testform why not? discontinuities in $f$ are discrete, it is still integrable.
$endgroup$
– gt6989b
Dec 6 '18 at 14:26
$begingroup$
I have thought exactly like you. Even, I have tested it only with f(x) on Wolfram/Alpha. And I got the result exactly what you said. But I can not verify it for $f(x)={x-lfloor x rfloor}$ , the program gives nothing.
$endgroup$
– Testform
Dec 6 '18 at 14:32
$begingroup$
@Testform see update
$endgroup$
– gt6989b
Dec 6 '18 at 14:38
$begingroup$
thus, it means can we obtain the result in my question?
$endgroup$
– Testform
Dec 6 '18 at 14:41
$begingroup$
Thank you my freind. But someone say that you can not obtain it for $f(x)={x-lfloor x rfloor}$ . He say : Because, the floor function is not differentiable
$endgroup$
– Testform
Dec 6 '18 at 14:46
|
show 2 more comments
$begingroup$
HINT
Use the Fundamental Theorem of Calculus. For any (constant) $a in mathbb{R}$:
$$
frac{d}{dz} int_a^z f(u)du = f(z).
$$
That should simplify your problem greatly.
UPDATE
In your case $f(x) = x - lfloor x rfloor$, so it has a countable number of discontinuities, which does not prevent the Riemann integral from existing. You have to define, for example,
$$
int_1^x g(f(u)) du
:= int_{lfloor x rfloor}^x g(f(u))du
+ sum_{k=2}^{lfloor x rfloor} int_1^k g(f(u)) du.
$$
This way, all the integrals on the RHS are well-defined and the LHS can be defined in a meaningful way.
answered Dec 6 '18 at 14:19
gt6989bgt6989b
34.7k22456
$endgroup$
HINT
Use the Fundamental Theorem of Calculus. For any (constant) $a in mathbb{R}$:
$$
frac{d}{dz} int_a^z f(u)du = f(z).
$$
That should simplify your problem greatly.
UPDATE
In your case $f(x) = x - lfloor x rfloor$, so it has a countable number of discontinuities, which does not prevent the Riemann integral from existing. You have to define, for example,
$$
int_1^x g(f(u)) du
:= int_{lfloor x rfloor}^x g(f(u))du
+ sum_{k=2}^{lfloor x rfloor} int_1^k g(f(u)) du.
$$
This way, all the integrals on the RHS are well-defined and the LHS can be defined in a meaningful way.
answered Dec 6 '18 at 14:19
gt6989bgt6989b
34.7k22456
edited Dec 6 '18 at 14:38
answered Dec 6 '18 at 14:19
gt6989bgt6989b
34.7k22456
answered Dec 6 '18 at 14:19
gt6989bgt6989b
34.7k22456
answered Dec 6 '18 at 14:19
gt6989bgt6989b
34.7k22456
34.7k22456
$begingroup$
@Testform why not? discontinuities in $f$ are discrete, it is still integrable.
$endgroup$
– gt6989b
Dec 6 '18 at 14:26
$begingroup$
I have thought exactly like you. Even, I have tested it only with f(x) on Wolfram/Alpha. And I got the result exactly what you said. But I can not verify it for $f(x)={x-lfloor x rfloor}$ , the program gives nothing.
$endgroup$
– Testform
Dec 6 '18 at 14:32
$begingroup$
@Testform see update
$endgroup$
– gt6989b
Dec 6 '18 at 14:38
$begingroup$
thus, it means can we obtain the result in my question?
$endgroup$
– Testform
Dec 6 '18 at 14:41
$begingroup$
Thank you my freind. But someone say that you can not obtain it for $f(x)={x-lfloor x rfloor}$ . He say : Because, the floor function is not differentiable
$endgroup$
– Testform
Dec 6 '18 at 14:46
|
show 2 more comments
$begingroup$
@Testform why not? discontinuities in $f$ are discrete, it is still integrable.
$endgroup$
– gt6989b
Dec 6 '18 at 14:26
$begingroup$
I have thought exactly like you. Even, I have tested it only with f(x) on Wolfram/Alpha. And I got the result exactly what you said. But I can not verify it for $f(x)={x-lfloor x rfloor}$ , the program gives nothing.
$endgroup$
– Testform
Dec 6 '18 at 14:32
$begingroup$
@Testform see update
$endgroup$
– gt6989b
Dec 6 '18 at 14:38
$begingroup$
thus, it means can we obtain the result in my question?
$endgroup$
– Testform
Dec 6 '18 at 14:41
$begingroup$
Thank you my freind. But someone say that you can not obtain it for $f(x)={x-lfloor x rfloor}$ . He say : Because, the floor function is not differentiable
$endgroup$
– Testform
Dec 6 '18 at 14:46
$begingroup$
@Testform why not? discontinuities in $f$ are discrete, it is still integrable.
$endgroup$
– gt6989b
Dec 6 '18 at 14:26
$begingroup$
@Testform why not? discontinuities in $f$ are discrete, it is still integrable.
$endgroup$
– gt6989b
Dec 6 '18 at 14:26
$begingroup$
I have thought exactly like you. Even, I have tested it only with f(x) on Wolfram/Alpha. And I got the result exactly what you said. But I can not verify it for $f(x)={x-lfloor x rfloor}$ , the program gives nothing.
$endgroup$
– Testform
Dec 6 '18 at 14:32
$begingroup$
I have thought exactly like you. Even, I have tested it only with f(x) on Wolfram/Alpha. And I got the result exactly what you said. But I can not verify it for $f(x)={x-lfloor x rfloor}$ , the program gives nothing.
$endgroup$
– Testform
Dec 6 '18 at 14:32
$begingroup$
@Testform see update
$endgroup$
– gt6989b
Dec 6 '18 at 14:38
$begingroup$
@Testform see update
$endgroup$
– gt6989b
Dec 6 '18 at 14:38
$begingroup$
thus, it means can we obtain the result in my question?
$endgroup$
– Testform
Dec 6 '18 at 14:41
$begingroup$
thus, it means can we obtain the result in my question?
$endgroup$
– Testform
Dec 6 '18 at 14:41
$begingroup$
Thank you my freind. But someone say that you can not obtain it for $f(x)={x-lfloor x rfloor}$ . He say : Because, the floor function is not differentiable
$endgroup$
– Testform
Dec 6 '18 at 14:46
$begingroup$
Thank you my freind. But someone say that you can not obtain it for $f(x)={x-lfloor x rfloor}$ . He say : Because, the floor function is not differentiable
$endgroup$
– Testform
Dec 6 '18 at 14:46
|
show 2 more comments
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1
$begingroup$
Why is this important? And what were your attempts solving this problem?
$endgroup$
– James
Dec 6 '18 at 14:14
$begingroup$
We are discussing about it very seriously with some mathematicians .
$endgroup$
– Testform
Dec 6 '18 at 14:18
$begingroup$
Can I get a more solid answer ? I mean we can obtain or not ?
$endgroup$
– Testform
Dec 6 '18 at 16:20