Can we find the $a$ value?
$begingroup$
We have a function with ${lfloor x rfloor}$ floor function and $a
in mathbb{R}$ and $
u in mathbb{R}$ .
$$lim_{u rightarrow infty}
frac{f(a)-int_1^u ( {x-lfloor x rfloor}) cdot x^{-a-1} dx}
{g(a)-int_1^u ({x-lfloor x rfloor}) cdot x^{a-2}dx}
=1
$$
We know also the following derivatives results. It means $f(a)$ and $g(a)$ are functions of only $a$ ( they are not any related with $u$).
$$
frac{d}
{du}f(a)
=0....and ....frac{d}
{du}g(a)
=0.
$$
Thus, if the limit appearance of the left side (under $u→∞$) is
$0/0$ , can we find the $a$ value by applying Hospital rule ?
NOTE: Please notice also the following result and the comment when you will try to solve it:
$$
frac{{frac {d} {du}}left[int_1^u {(x-lfloor x rfloor}) cdot x^{-a-1} dxright]}
{{frac {d} {du}}left[int_1^u {(x-lfloor x rfloor}) cdot x^{a-2}dxright]}
=u^{1-2a}
$$
The floor function does not create a fatal problem on above equation.
It is true that both the numerator and denominator are discontinuous at integers, but the ratio of them is continuous in punctured neighborhoods of integers, and so can be extended by continuity to get the result $u^{1-2a}$.
real-analysis calculus integration limits derivatives
asked Dec 9 '18 at 13:27
Testform Testform
64
$endgroup$
add a comment |
$begingroup$
We have a function with ${lfloor x rfloor}$ floor function and $a
in mathbb{R}$ and $
u in mathbb{R}$ .
$$lim_{u rightarrow infty}
frac{f(a)-int_1^u ( {x-lfloor x rfloor}) cdot x^{-a-1} dx}
{g(a)-int_1^u ({x-lfloor x rfloor}) cdot x^{a-2}dx}
=1
$$
We know also the following derivatives results. It means $f(a)$ and $g(a)$ are functions of only $a$ ( they are not any related with $u$).
$$
frac{d}
{du}f(a)
=0....and ....frac{d}
{du}g(a)
=0.
$$
Thus, if the limit appearance of the left side (under $u→∞$) is
$0/0$ , can we find the $a$ value by applying Hospital rule ?
NOTE: Please notice also the following result and the comment when you will try to solve it:
$$
frac{{frac {d} {du}}left[int_1^u {(x-lfloor x rfloor}) cdot x^{-a-1} dxright]}
{{frac {d} {du}}left[int_1^u {(x-lfloor x rfloor}) cdot x^{a-2}dxright]}
=u^{1-2a}
$$
The floor function does not create a fatal problem on above equation.
It is true that both the numerator and denominator are discontinuous at integers, but the ratio of them is continuous in punctured neighborhoods of integers, and so can be extended by continuity to get the result $u^{1-2a}$.
real-analysis calculus integration limits derivatives
asked Dec 9 '18 at 13:27
Testform Testform
64
$endgroup$
$begingroup$
I guess you can try to calculate the integral by breaking the interval $[1,u]$ into $[1,2]cup[2,3] cup cdots [lfloor u rfloor -1,lfloor urfloor ] cup [lfloor u rfloor,u]$. If $xin [i,i+1]$ then $x-lfloor x rfloor = x-i$.
$endgroup$
– Tengu
Dec 10 '18 at 23:38
add a comment |
$begingroup$
We have a function with ${lfloor x rfloor}$ floor function and $a
in mathbb{R}$ and $
u in mathbb{R}$ .
$$lim_{u rightarrow infty}
frac{f(a)-int_1^u ( {x-lfloor x rfloor}) cdot x^{-a-1} dx}
{g(a)-int_1^u ({x-lfloor x rfloor}) cdot x^{a-2}dx}
=1
$$
We know also the following derivatives results. It means $f(a)$ and $g(a)$ are functions of only $a$ ( they are not any related with $u$).
$$
frac{d}
{du}f(a)
=0....and ....frac{d}
{du}g(a)
=0.
$$
Thus, if the limit appearance of the left side (under $u→∞$) is
$0/0$ , can we find the $a$ value by applying Hospital rule ?
NOTE: Please notice also the following result and the comment when you will try to solve it:
$$
frac{{frac {d} {du}}left[int_1^u {(x-lfloor x rfloor}) cdot x^{-a-1} dxright]}
{{frac {d} {du}}left[int_1^u {(x-lfloor x rfloor}) cdot x^{a-2}dxright]}
=u^{1-2a}
$$
The floor function does not create a fatal problem on above equation.
It is true that both the numerator and denominator are discontinuous at integers, but the ratio of them is continuous in punctured neighborhoods of integers, and so can be extended by continuity to get the result $u^{1-2a}$.
real-analysis calculus integration limits derivatives
asked Dec 9 '18 at 13:27
Testform Testform
64
$endgroup$
We have a function with ${lfloor x rfloor}$ floor function and $a
in mathbb{R}$ and $
u in mathbb{R}$ .
$$lim_{u rightarrow infty}
frac{f(a)-int_1^u ( {x-lfloor x rfloor}) cdot x^{-a-1} dx}
{g(a)-int_1^u ({x-lfloor x rfloor}) cdot x^{a-2}dx}
=1
$$
We know also the following derivatives results. It means $f(a)$ and $g(a)$ are functions of only $a$ ( they are not any related with $u$).
$$
frac{d}
{du}f(a)
=0....and ....frac{d}
{du}g(a)
=0.
$$
Thus, if the limit appearance of the left side (under $u→∞$) is
$0/0$ , can we find the $a$ value by applying Hospital rule ?
NOTE: Please notice also the following result and the comment when you will try to solve it:
$$
frac{{frac {d} {du}}left[int_1^u {(x-lfloor x rfloor}) cdot x^{-a-1} dxright]}
{{frac {d} {du}}left[int_1^u {(x-lfloor x rfloor}) cdot x^{a-2}dxright]}
=u^{1-2a}
$$
The floor function does not create a fatal problem on above equation.
It is true that both the numerator and denominator are discontinuous at integers, but the ratio of them is continuous in punctured neighborhoods of integers, and so can be extended by continuity to get the result $u^{1-2a}$.
real-analysis calculus integration limits derivatives
real-analysis calculus integration limits derivatives
asked Dec 9 '18 at 13:27
Testform Testform
64
asked Dec 9 '18 at 13:27
Testform Testform
64
edited Dec 10 '18 at 21:15
Testform
asked Dec 9 '18 at 13:27
Testform Testform
64
asked Dec 9 '18 at 13:27
Testform Testform
64
asked Dec 9 '18 at 13:27
Testform Testform
64
64
$begingroup$
I guess you can try to calculate the integral by breaking the interval $[1,u]$ into $[1,2]cup[2,3] cup cdots [lfloor u rfloor -1,lfloor urfloor ] cup [lfloor u rfloor,u]$. If $xin [i,i+1]$ then $x-lfloor x rfloor = x-i$.
$endgroup$
– Tengu
Dec 10 '18 at 23:38
add a comment |
$begingroup$
I guess you can try to calculate the integral by breaking the interval $[1,u]$ into $[1,2]cup[2,3] cup cdots [lfloor u rfloor -1,lfloor urfloor ] cup [lfloor u rfloor,u]$. If $xin [i,i+1]$ then $x-lfloor x rfloor = x-i$.
$endgroup$
– Tengu
Dec 10 '18 at 23:38
$begingroup$
I guess you can try to calculate the integral by breaking the interval $[1,u]$ into $[1,2]cup[2,3] cup cdots [lfloor u rfloor -1,lfloor urfloor ] cup [lfloor u rfloor,u]$. If $xin [i,i+1]$ then $x-lfloor x rfloor = x-i$.
$endgroup$
– Tengu
Dec 10 '18 at 23:38
$begingroup$
I guess you can try to calculate the integral by breaking the interval $[1,u]$ into $[1,2]cup[2,3] cup cdots [lfloor u rfloor -1,lfloor urfloor ] cup [lfloor u rfloor,u]$. If $xin [i,i+1]$ then $x-lfloor x rfloor = x-i$.
$endgroup$
– Tengu
Dec 10 '18 at 23:38
add a comment |
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$begingroup$
I guess you can try to calculate the integral by breaking the interval $[1,u]$ into $[1,2]cup[2,3] cup cdots [lfloor u rfloor -1,lfloor urfloor ] cup [lfloor u rfloor,u]$. If $xin [i,i+1]$ then $x-lfloor x rfloor = x-i$.
$endgroup$
– Tengu
Dec 10 '18 at 23:38