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Evaluate $limlimits_xtoinftyfrac1xint_0^xmaxsin t,sin(tsqrt2)dt$
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I want to evaluate
$$L=lim_xtoinftyfrac1xint_0^xmaxsin t,sin(tsqrt2)dt$$
My attempt
$$L=lim_xtoinftyfrac12xint_0^xBig(sin t+sin(tsqrt2)+big|sin t-sin(tsqrt2)big|Big)dt\
=lim_xtoinftyfrac12xint_0^xbig|sin t-sin(tsqrt2)big|dt\
=lim_xtoinftyfrac1xint_0^xbigg|cosfracsqrt2+12tcdotsinfracsqrt2-12tbigg|dt$$
Denote $s_n$ the $n$th zero point of $cosfracsqrt2+12tcdotsinfracsqrt2-12t (tge0)$. Since $1$, $sqrt2$ and $pi$ are linear independent in $mathbb Q$, the order of the zero points should be $1$. According to the squeeze theorem, we have
$$L=lim_ntoinftyfrac1s_n+1sum_k=0^n(-1)^kint_s_k^s_k+1big(sin t-sin(tsqrt2)big)dt\
=lim_ntoinftyfrac1s_n+1sum_k=0^n(-1)^kbigg(cos s_k-cos s_k+1+fraccossqrt2s_k-cossqrt2s_k+1sqrt2bigg)dt$$
I can't go further. I think the zero points of that function is the key point.
calculus integration definite-integrals
asked Jul 22 at 9:12
LittleCuteKemono
46812
 |Â
show 2 more comments
up vote
19
down vote
favorite
I want to evaluate
$$L=lim_xtoinftyfrac1xint_0^xmaxsin t,sin(tsqrt2)dt$$
My attempt
$$L=lim_xtoinftyfrac12xint_0^xBig(sin t+sin(tsqrt2)+big|sin t-sin(tsqrt2)big|Big)dt\
=lim_xtoinftyfrac12xint_0^xbig|sin t-sin(tsqrt2)big|dt\
=lim_xtoinftyfrac1xint_0^xbigg|cosfracsqrt2+12tcdotsinfracsqrt2-12tbigg|dt$$
Denote $s_n$ the $n$th zero point of $cosfracsqrt2+12tcdotsinfracsqrt2-12t (tge0)$. Since $1$, $sqrt2$ and $pi$ are linear independent in $mathbb Q$, the order of the zero points should be $1$. According to the squeeze theorem, we have
$$L=lim_ntoinftyfrac1s_n+1sum_k=0^n(-1)^kint_s_k^s_k+1big(sin t-sin(tsqrt2)big)dt\
=lim_ntoinftyfrac1s_n+1sum_k=0^n(-1)^kbigg(cos s_k-cos s_k+1+fraccossqrt2s_k-cossqrt2s_k+1sqrt2bigg)dt$$
I can't go further. I think the zero points of that function is the key point.
calculus integration definite-integrals
asked Jul 22 at 9:12
LittleCuteKemono
46812
I'm a bit puzzled by the first row in your attempt. Why is this true? And I suppose in the second row you split into three integrals and dumped zeros?
– dEmigOd
Jul 22 at 10:15
@dEmigOd I guess the integral of first two terms are bounded and would vanish in the limit.
– Szeto
Jul 22 at 10:18
1
@dEmigOd I used the identity $maxa,b=(a+b+|a-b|)/2$. The first and the second integral I splitted divided by $x$ is $0$ as $xtoinfty$.
– LittleCuteKemono
Jul 22 at 10:19
Does the function $|sin t-sin(tsqrt2)|$ have a period?
– Szeto
Jul 22 at 10:35
1
@Szeto: of course not, $sqrt2$ is irrational.
– Yves Daoust
Jul 22 at 10:36
 |Â
show 2 more comments
up vote
19
down vote
favorite
up vote
19
down vote
favorite
I want to evaluate
$$L=lim_xtoinftyfrac1xint_0^xmaxsin t,sin(tsqrt2)dt$$
My attempt
$$L=lim_xtoinftyfrac12xint_0^xBig(sin t+sin(tsqrt2)+big|sin t-sin(tsqrt2)big|Big)dt\
=lim_xtoinftyfrac12xint_0^xbig|sin t-sin(tsqrt2)big|dt\
=lim_xtoinftyfrac1xint_0^xbigg|cosfracsqrt2+12tcdotsinfracsqrt2-12tbigg|dt$$
Denote $s_n$ the $n$th zero point of $cosfracsqrt2+12tcdotsinfracsqrt2-12t (tge0)$. Since $1$, $sqrt2$ and $pi$ are linear independent in $mathbb Q$, the order of the zero points should be $1$. According to the squeeze theorem, we have
$$L=lim_ntoinftyfrac1s_n+1sum_k=0^n(-1)^kint_s_k^s_k+1big(sin t-sin(tsqrt2)big)dt\
=lim_ntoinftyfrac1s_n+1sum_k=0^n(-1)^kbigg(cos s_k-cos s_k+1+fraccossqrt2s_k-cossqrt2s_k+1sqrt2bigg)dt$$
I can't go further. I think the zero points of that function is the key point.
calculus integration definite-integrals
asked Jul 22 at 9:12
LittleCuteKemono
46812
I want to evaluate
$$L=lim_xtoinftyfrac1xint_0^xmaxsin t,sin(tsqrt2)dt$$
My attempt
$$L=lim_xtoinftyfrac12xint_0^xBig(sin t+sin(tsqrt2)+big|sin t-sin(tsqrt2)big|Big)dt\
=lim_xtoinftyfrac12xint_0^xbig|sin t-sin(tsqrt2)big|dt\
=lim_xtoinftyfrac1xint_0^xbigg|cosfracsqrt2+12tcdotsinfracsqrt2-12tbigg|dt$$
Denote $s_n$ the $n$th zero point of $cosfracsqrt2+12tcdotsinfracsqrt2-12t (tge0)$. Since $1$, $sqrt2$ and $pi$ are linear independent in $mathbb Q$, the order of the zero points should be $1$. According to the squeeze theorem, we have
$$L=lim_ntoinftyfrac1s_n+1sum_k=0^n(-1)^kint_s_k^s_k+1big(sin t-sin(tsqrt2)big)dt\
=lim_ntoinftyfrac1s_n+1sum_k=0^n(-1)^kbigg(cos s_k-cos s_k+1+fraccossqrt2s_k-cossqrt2s_k+1sqrt2bigg)dt$$
I can't go further. I think the zero points of that function is the key point.
calculus integration definite-integrals
asked Jul 22 at 9:12
LittleCuteKemono
46812
edited Jul 22 at 10:26
asked Jul 22 at 9:12
LittleCuteKemono
46812
asked Jul 22 at 9:12
LittleCuteKemono
46812
asked Jul 22 at 9:12
LittleCuteKemono
46812
46812
I'm a bit puzzled by the first row in your attempt. Why is this true? And I suppose in the second row you split into three integrals and dumped zeros?
– dEmigOd
Jul 22 at 10:15
@dEmigOd I guess the integral of first two terms are bounded and would vanish in the limit.
– Szeto
Jul 22 at 10:18
1
@dEmigOd I used the identity $maxa,b=(a+b+|a-b|)/2$. The first and the second integral I splitted divided by $x$ is $0$ as $xtoinfty$.
– LittleCuteKemono
Jul 22 at 10:19
Does the function $|sin t-sin(tsqrt2)|$ have a period?
– Szeto
Jul 22 at 10:35
1
@Szeto: of course not, $sqrt2$ is irrational.
– Yves Daoust
Jul 22 at 10:36
 |Â
show 2 more comments
I'm a bit puzzled by the first row in your attempt. Why is this true? And I suppose in the second row you split into three integrals and dumped zeros?
– dEmigOd
Jul 22 at 10:15
@dEmigOd I guess the integral of first two terms are bounded and would vanish in the limit.
– Szeto
Jul 22 at 10:18
1
@dEmigOd I used the identity $maxa,b=(a+b+|a-b|)/2$. The first and the second integral I splitted divided by $x$ is $0$ as $xtoinfty$.
– LittleCuteKemono
Jul 22 at 10:19
Does the function $|sin t-sin(tsqrt2)|$ have a period?
– Szeto
Jul 22 at 10:35
1
@Szeto: of course not, $sqrt2$ is irrational.
– Yves Daoust
Jul 22 at 10:36
I'm a bit puzzled by the first row in your attempt. Why is this true? And I suppose in the second row you split into three integrals and dumped zeros?
– dEmigOd
Jul 22 at 10:15
I'm a bit puzzled by the first row in your attempt. Why is this true? And I suppose in the second row you split into three integrals and dumped zeros?
– dEmigOd
Jul 22 at 10:15
@dEmigOd I guess the integral of first two terms are bounded and would vanish in the limit.
– Szeto
Jul 22 at 10:18
@dEmigOd I guess the integral of first two terms are bounded and would vanish in the limit.
– Szeto
Jul 22 at 10:18
1
1
@dEmigOd I used the identity $maxa,b=(a+b+|a-b|)/2$. The first and the second integral I splitted divided by $x$ is $0$ as $xtoinfty$.
– LittleCuteKemono
Jul 22 at 10:19
@dEmigOd I used the identity $maxa,b=(a+b+|a-b|)/2$. The first and the second integral I splitted divided by $x$ is $0$ as $xtoinfty$.
– LittleCuteKemono
Jul 22 at 10:19
Does the function $|sin t-sin(tsqrt2)|$ have a period?
– Szeto
Jul 22 at 10:35
Does the function $|sin t-sin(tsqrt2)|$ have a period?
– Szeto
Jul 22 at 10:35
1
1
@Szeto: of course not, $sqrt2$ is irrational.
– Yves Daoust
Jul 22 at 10:36
@Szeto: of course not, $sqrt2$ is irrational.
– Yves Daoust
Jul 22 at 10:36
 |Â
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
9
down vote
accepted
As pointed out by OP, we can use $maxa,b = fraca+b2 + frac2$ to discover that
$$ L = lim_xtoinfty frac1x int_0^x left|cosleft(fracsqrt2+12tright)sinleft(fracsqrt2-12tright)right| , dt. $$
Applying the substitution $fracsqrt2+12t = pi u$ and writing $alpha=(sqrt2-1)^2$ followed by the substitution $y = fracsqrt2+12pix$, we see that
beginalign*
L
&= lim_ytoinfty frac1y int_0^y left| cos(pi u)sin(pi alpha u) right| , du \
&= lim_Ntoinfty frac1N int_0^N left| cos(pi u)sin(pi alpha u) right| , du \
&= lim_Ntoinfty int_0^1 left| cos(pi u) right| left( frac1N sum_n=0^N-1 left| sin(pi alpha k + pi alpha u) right| right) , du
endalign*
Since $alpha$ is irrational, the equidistribution theorem applied to $v mapsto left| sin(pi v + pi alpha u) right|$ for each fixed $u$ tells that
$$ forall uinmathbbR : lim_Ntoinfty frac1N sum_n=0^N-1 left| sin(pi alpha k + pi alpha u) right| = int_0^1 left| sin(pi v) right| , dv. $$
Therefore by the dominated convergence theorem,
$$ L = left( int_0^1 left| cos(pi u) right| , du right)left( int_0^1 left| sin(pi v) right| , dv right) = frac4pi^2. $$
answered Jul 22 at 13:12
Sangchul Lee
85.5k12155253
I think you have a typo: $maxa,b=fraca+b2+|fraca-b2|$.
– LittleCuteKemono
Jul 23 at 5:43
@LittleCuteKemono, Thank you for pointing out the typo. It is now fixed :)
– Sangchul Lee
Jul 23 at 10:48
add a comment |Â
up vote
7
down vote
I strongly conjecture that the limit is $4overpi^2=0.405285$. Integrating numerically over the interval $[0,1000]$ Mathematica obtained $0.406966$, but warned that the error might be larger than Mathematica's standard.
Consider the torus $T:=bigl(mathbb R/(2pimathbb Z)bigr)^2$and on $T$ the function
$$f(x,y):=maxsin x,sin y .$$
Draw a figure of the fundamental domain $[-pi,pi]^2$ in order to identify the parts of $T$ where $sin x$, resp. $sin y$, is larger. Then compute the required double integrals and obtain
$$int_T f(x,y)>rm d(x,y)=16 .$$
This means that the "space average" $E$ of $f$ is given by
$$E(f)=16over rm area(T)=4overpi^2 .$$
Now the orbit $$tmapsto bigl(x(t),y(t)bigr):=(t,>sqrt2,t)$$
projects to a line with irrational slope on $T$. In such a situation an "ergodic principle" is at work. According to this principle the "time average" of $f$ coincides with the "space average" $E(f)$. Theoretical basis for this to happen is the fact that the multiples $k/sqrt2$ $(kinmathbb N)$ are uniformly distributed mod $1$.
answered Jul 22 at 12:39
Christian Blatter
163k7107306
add a comment |Â
up vote
3
down vote
For simplicity I'll notate $alpha = fracsqrt2+12$ and $beta = fracsqrt2-12$. Now note that the integral basically only depends on the asymptotic behaviour of the integrand, meaning that for all $y$:
$$
beginalign
L =lim_xtoinfty frac1x int_0^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt
&= lim_xtoinfty frac1x int_y^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt\
&= lim_xtoinfty frac1x+y int_0^x bigg| cos alpha (t+y) cdot sin beta (t+y)bigg| ,mathrmdt\
&= lim_xtoinfty frac1x int_0^x bigg| cos alpha (t+y) cdot sin beta (t+y)bigg| ,mathrmdt
endalign
$$
Letting $y = frac2pi nalpha$ for $n in mathbbN$ this implies:
$$
L = lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt
$$
now since $alpha$ and $beta$ are irrational we get (using the ergodic theorem):
$$
lim_Ntoinfty frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| = frac12piint_0^2pi bigg| sin(s) bigg| ,mathrmds = frac2pi
$$
for almost all $t$. With this we can now solve the original integral as follows:
$$
beginalign
L =lim_xtoinfty frac1x int_0^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt
&= lim_Ntoinfty frac1N sum_n=0^N lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_Ntoinfty lim_xtoinfty frac1x frac1N sum_n=0^Nint_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_Ntoinfty lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdot frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdot frac2pi ,mathrmdt\
&= frac4pi^2
endalign
$$
this second to last step still requires some justification however. This can be done by noting that ergodic theorem guarantees that $frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg|$ converges in the $L^2$ sense on $[0, 2pi / beta]$ to the constant function $frac2pi$. This is enough since, if we have $g$ bounded and $f_n$ periodic with period $2pi / beta$ and $f_n to f$ in the $L^2$ sense over that period then:
$$
beginalign
lim_Ntoinfty left| lim_xtoinfty frac1x int_0^x g cdot f_n ,mathrmdt - lim_xtoinfty frac1x int_0^x g cdot f ,mathrmdt right|
&le lim_Ntoinfty lim_xtoinfty frac1x int_0^x | g cdot f_n - g cdot f |^2 ,mathrmdt\
&le lim_Ntoinfty lim_xtoinfty frac1x int_0^x C | f_n - f |^2 ,mathrmdt\
&le lim_Ntoinfty lim_xtoinfty fracCx leftlceilfracx2pi / betarightrceil | f_n - f |_2^2\
&le lim_Ntoinfty fracC2pi / beta | f_n - f |_2^2 to 0
endalign
$$
It's possible there's a less fiddly proof to show convergence, but combining asymptotic density and ergodic theory does make things a bit tricky.
answered Jul 22 at 13:34
Contravariant
1,24416
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
As pointed out by OP, we can use $maxa,b = fraca+b2 + frac2$ to discover that
$$ L = lim_xtoinfty frac1x int_0^x left|cosleft(fracsqrt2+12tright)sinleft(fracsqrt2-12tright)right| , dt. $$
Applying the substitution $fracsqrt2+12t = pi u$ and writing $alpha=(sqrt2-1)^2$ followed by the substitution $y = fracsqrt2+12pix$, we see that
beginalign*
L
&= lim_ytoinfty frac1y int_0^y left| cos(pi u)sin(pi alpha u) right| , du \
&= lim_Ntoinfty frac1N int_0^N left| cos(pi u)sin(pi alpha u) right| , du \
&= lim_Ntoinfty int_0^1 left| cos(pi u) right| left( frac1N sum_n=0^N-1 left| sin(pi alpha k + pi alpha u) right| right) , du
endalign*
Since $alpha$ is irrational, the equidistribution theorem applied to $v mapsto left| sin(pi v + pi alpha u) right|$ for each fixed $u$ tells that
$$ forall uinmathbbR : lim_Ntoinfty frac1N sum_n=0^N-1 left| sin(pi alpha k + pi alpha u) right| = int_0^1 left| sin(pi v) right| , dv. $$
Therefore by the dominated convergence theorem,
$$ L = left( int_0^1 left| cos(pi u) right| , du right)left( int_0^1 left| sin(pi v) right| , dv right) = frac4pi^2. $$
answered Jul 22 at 13:12
Sangchul Lee
85.5k12155253
I think you have a typo: $maxa,b=fraca+b2+|fraca-b2|$.
– LittleCuteKemono
Jul 23 at 5:43
@LittleCuteKemono, Thank you for pointing out the typo. It is now fixed :)
– Sangchul Lee
Jul 23 at 10:48
add a comment |Â
up vote
9
down vote
accepted
As pointed out by OP, we can use $maxa,b = fraca+b2 + frac2$ to discover that
$$ L = lim_xtoinfty frac1x int_0^x left|cosleft(fracsqrt2+12tright)sinleft(fracsqrt2-12tright)right| , dt. $$
Applying the substitution $fracsqrt2+12t = pi u$ and writing $alpha=(sqrt2-1)^2$ followed by the substitution $y = fracsqrt2+12pix$, we see that
beginalign*
L
&= lim_ytoinfty frac1y int_0^y left| cos(pi u)sin(pi alpha u) right| , du \
&= lim_Ntoinfty frac1N int_0^N left| cos(pi u)sin(pi alpha u) right| , du \
&= lim_Ntoinfty int_0^1 left| cos(pi u) right| left( frac1N sum_n=0^N-1 left| sin(pi alpha k + pi alpha u) right| right) , du
endalign*
Since $alpha$ is irrational, the equidistribution theorem applied to $v mapsto left| sin(pi v + pi alpha u) right|$ for each fixed $u$ tells that
$$ forall uinmathbbR : lim_Ntoinfty frac1N sum_n=0^N-1 left| sin(pi alpha k + pi alpha u) right| = int_0^1 left| sin(pi v) right| , dv. $$
Therefore by the dominated convergence theorem,
$$ L = left( int_0^1 left| cos(pi u) right| , du right)left( int_0^1 left| sin(pi v) right| , dv right) = frac4pi^2. $$
answered Jul 22 at 13:12
Sangchul Lee
85.5k12155253
I think you have a typo: $maxa,b=fraca+b2+|fraca-b2|$.
– LittleCuteKemono
Jul 23 at 5:43
@LittleCuteKemono, Thank you for pointing out the typo. It is now fixed :)
– Sangchul Lee
Jul 23 at 10:48
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
As pointed out by OP, we can use $maxa,b = fraca+b2 + frac2$ to discover that
$$ L = lim_xtoinfty frac1x int_0^x left|cosleft(fracsqrt2+12tright)sinleft(fracsqrt2-12tright)right| , dt. $$
Applying the substitution $fracsqrt2+12t = pi u$ and writing $alpha=(sqrt2-1)^2$ followed by the substitution $y = fracsqrt2+12pix$, we see that
beginalign*
L
&= lim_ytoinfty frac1y int_0^y left| cos(pi u)sin(pi alpha u) right| , du \
&= lim_Ntoinfty frac1N int_0^N left| cos(pi u)sin(pi alpha u) right| , du \
&= lim_Ntoinfty int_0^1 left| cos(pi u) right| left( frac1N sum_n=0^N-1 left| sin(pi alpha k + pi alpha u) right| right) , du
endalign*
Since $alpha$ is irrational, the equidistribution theorem applied to $v mapsto left| sin(pi v + pi alpha u) right|$ for each fixed $u$ tells that
$$ forall uinmathbbR : lim_Ntoinfty frac1N sum_n=0^N-1 left| sin(pi alpha k + pi alpha u) right| = int_0^1 left| sin(pi v) right| , dv. $$
Therefore by the dominated convergence theorem,
$$ L = left( int_0^1 left| cos(pi u) right| , du right)left( int_0^1 left| sin(pi v) right| , dv right) = frac4pi^2. $$
answered Jul 22 at 13:12
Sangchul Lee
85.5k12155253
As pointed out by OP, we can use $maxa,b = fraca+b2 + frac2$ to discover that
$$ L = lim_xtoinfty frac1x int_0^x left|cosleft(fracsqrt2+12tright)sinleft(fracsqrt2-12tright)right| , dt. $$
Applying the substitution $fracsqrt2+12t = pi u$ and writing $alpha=(sqrt2-1)^2$ followed by the substitution $y = fracsqrt2+12pix$, we see that
beginalign*
L
&= lim_ytoinfty frac1y int_0^y left| cos(pi u)sin(pi alpha u) right| , du \
&= lim_Ntoinfty frac1N int_0^N left| cos(pi u)sin(pi alpha u) right| , du \
&= lim_Ntoinfty int_0^1 left| cos(pi u) right| left( frac1N sum_n=0^N-1 left| sin(pi alpha k + pi alpha u) right| right) , du
endalign*
Since $alpha$ is irrational, the equidistribution theorem applied to $v mapsto left| sin(pi v + pi alpha u) right|$ for each fixed $u$ tells that
$$ forall uinmathbbR : lim_Ntoinfty frac1N sum_n=0^N-1 left| sin(pi alpha k + pi alpha u) right| = int_0^1 left| sin(pi v) right| , dv. $$
Therefore by the dominated convergence theorem,
$$ L = left( int_0^1 left| cos(pi u) right| , du right)left( int_0^1 left| sin(pi v) right| , dv right) = frac4pi^2. $$
answered Jul 22 at 13:12
Sangchul Lee
85.5k12155253
edited Jul 23 at 10:47
answered Jul 22 at 13:12
Sangchul Lee
85.5k12155253
answered Jul 22 at 13:12
Sangchul Lee
85.5k12155253
answered Jul 22 at 13:12
Sangchul Lee
85.5k12155253
85.5k12155253
I think you have a typo: $maxa,b=fraca+b2+|fraca-b2|$.
– LittleCuteKemono
Jul 23 at 5:43
@LittleCuteKemono, Thank you for pointing out the typo. It is now fixed :)
– Sangchul Lee
Jul 23 at 10:48
add a comment |Â
I think you have a typo: $maxa,b=fraca+b2+|fraca-b2|$.
– LittleCuteKemono
Jul 23 at 5:43
@LittleCuteKemono, Thank you for pointing out the typo. It is now fixed :)
– Sangchul Lee
Jul 23 at 10:48
I think you have a typo: $maxa,b=fraca+b2+|fraca-b2|$.
– LittleCuteKemono
Jul 23 at 5:43
I think you have a typo: $maxa,b=fraca+b2+|fraca-b2|$.
– LittleCuteKemono
Jul 23 at 5:43
@LittleCuteKemono, Thank you for pointing out the typo. It is now fixed :)
– Sangchul Lee
Jul 23 at 10:48
@LittleCuteKemono, Thank you for pointing out the typo. It is now fixed :)
– Sangchul Lee
Jul 23 at 10:48
add a comment |Â
up vote
7
down vote
I strongly conjecture that the limit is $4overpi^2=0.405285$. Integrating numerically over the interval $[0,1000]$ Mathematica obtained $0.406966$, but warned that the error might be larger than Mathematica's standard.
Consider the torus $T:=bigl(mathbb R/(2pimathbb Z)bigr)^2$and on $T$ the function
$$f(x,y):=maxsin x,sin y .$$
Draw a figure of the fundamental domain $[-pi,pi]^2$ in order to identify the parts of $T$ where $sin x$, resp. $sin y$, is larger. Then compute the required double integrals and obtain
$$int_T f(x,y)>rm d(x,y)=16 .$$
This means that the "space average" $E$ of $f$ is given by
$$E(f)=16over rm area(T)=4overpi^2 .$$
Now the orbit $$tmapsto bigl(x(t),y(t)bigr):=(t,>sqrt2,t)$$
projects to a line with irrational slope on $T$. In such a situation an "ergodic principle" is at work. According to this principle the "time average" of $f$ coincides with the "space average" $E(f)$. Theoretical basis for this to happen is the fact that the multiples $k/sqrt2$ $(kinmathbb N)$ are uniformly distributed mod $1$.
answered Jul 22 at 12:39
Christian Blatter
163k7107306
add a comment |Â
up vote
7
down vote
I strongly conjecture that the limit is $4overpi^2=0.405285$. Integrating numerically over the interval $[0,1000]$ Mathematica obtained $0.406966$, but warned that the error might be larger than Mathematica's standard.
Consider the torus $T:=bigl(mathbb R/(2pimathbb Z)bigr)^2$and on $T$ the function
$$f(x,y):=maxsin x,sin y .$$
Draw a figure of the fundamental domain $[-pi,pi]^2$ in order to identify the parts of $T$ where $sin x$, resp. $sin y$, is larger. Then compute the required double integrals and obtain
$$int_T f(x,y)>rm d(x,y)=16 .$$
This means that the "space average" $E$ of $f$ is given by
$$E(f)=16over rm area(T)=4overpi^2 .$$
Now the orbit $$tmapsto bigl(x(t),y(t)bigr):=(t,>sqrt2,t)$$
projects to a line with irrational slope on $T$. In such a situation an "ergodic principle" is at work. According to this principle the "time average" of $f$ coincides with the "space average" $E(f)$. Theoretical basis for this to happen is the fact that the multiples $k/sqrt2$ $(kinmathbb N)$ are uniformly distributed mod $1$.
answered Jul 22 at 12:39
Christian Blatter
163k7107306
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I strongly conjecture that the limit is $4overpi^2=0.405285$. Integrating numerically over the interval $[0,1000]$ Mathematica obtained $0.406966$, but warned that the error might be larger than Mathematica's standard.
Consider the torus $T:=bigl(mathbb R/(2pimathbb Z)bigr)^2$and on $T$ the function
$$f(x,y):=maxsin x,sin y .$$
Draw a figure of the fundamental domain $[-pi,pi]^2$ in order to identify the parts of $T$ where $sin x$, resp. $sin y$, is larger. Then compute the required double integrals and obtain
$$int_T f(x,y)>rm d(x,y)=16 .$$
This means that the "space average" $E$ of $f$ is given by
$$E(f)=16over rm area(T)=4overpi^2 .$$
Now the orbit $$tmapsto bigl(x(t),y(t)bigr):=(t,>sqrt2,t)$$
projects to a line with irrational slope on $T$. In such a situation an "ergodic principle" is at work. According to this principle the "time average" of $f$ coincides with the "space average" $E(f)$. Theoretical basis for this to happen is the fact that the multiples $k/sqrt2$ $(kinmathbb N)$ are uniformly distributed mod $1$.
answered Jul 22 at 12:39
Christian Blatter
163k7107306
I strongly conjecture that the limit is $4overpi^2=0.405285$. Integrating numerically over the interval $[0,1000]$ Mathematica obtained $0.406966$, but warned that the error might be larger than Mathematica's standard.
Consider the torus $T:=bigl(mathbb R/(2pimathbb Z)bigr)^2$and on $T$ the function
$$f(x,y):=maxsin x,sin y .$$
Draw a figure of the fundamental domain $[-pi,pi]^2$ in order to identify the parts of $T$ where $sin x$, resp. $sin y$, is larger. Then compute the required double integrals and obtain
$$int_T f(x,y)>rm d(x,y)=16 .$$
This means that the "space average" $E$ of $f$ is given by
$$E(f)=16over rm area(T)=4overpi^2 .$$
Now the orbit $$tmapsto bigl(x(t),y(t)bigr):=(t,>sqrt2,t)$$
projects to a line with irrational slope on $T$. In such a situation an "ergodic principle" is at work. According to this principle the "time average" of $f$ coincides with the "space average" $E(f)$. Theoretical basis for this to happen is the fact that the multiples $k/sqrt2$ $(kinmathbb N)$ are uniformly distributed mod $1$.
answered Jul 22 at 12:39
Christian Blatter
163k7107306
answered Jul 22 at 12:39
Christian Blatter
163k7107306
answered Jul 22 at 12:39
Christian Blatter
163k7107306
answered Jul 22 at 12:39
Christian Blatter
163k7107306
163k7107306
add a comment |Â
add a comment |Â
up vote
3
down vote
For simplicity I'll notate $alpha = fracsqrt2+12$ and $beta = fracsqrt2-12$. Now note that the integral basically only depends on the asymptotic behaviour of the integrand, meaning that for all $y$:
$$
beginalign
L =lim_xtoinfty frac1x int_0^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt
&= lim_xtoinfty frac1x int_y^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt\
&= lim_xtoinfty frac1x+y int_0^x bigg| cos alpha (t+y) cdot sin beta (t+y)bigg| ,mathrmdt\
&= lim_xtoinfty frac1x int_0^x bigg| cos alpha (t+y) cdot sin beta (t+y)bigg| ,mathrmdt
endalign
$$
Letting $y = frac2pi nalpha$ for $n in mathbbN$ this implies:
$$
L = lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt
$$
now since $alpha$ and $beta$ are irrational we get (using the ergodic theorem):
$$
lim_Ntoinfty frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| = frac12piint_0^2pi bigg| sin(s) bigg| ,mathrmds = frac2pi
$$
for almost all $t$. With this we can now solve the original integral as follows:
$$
beginalign
L =lim_xtoinfty frac1x int_0^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt
&= lim_Ntoinfty frac1N sum_n=0^N lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_Ntoinfty lim_xtoinfty frac1x frac1N sum_n=0^Nint_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_Ntoinfty lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdot frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdot frac2pi ,mathrmdt\
&= frac4pi^2
endalign
$$
this second to last step still requires some justification however. This can be done by noting that ergodic theorem guarantees that $frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg|$ converges in the $L^2$ sense on $[0, 2pi / beta]$ to the constant function $frac2pi$. This is enough since, if we have $g$ bounded and $f_n$ periodic with period $2pi / beta$ and $f_n to f$ in the $L^2$ sense over that period then:
$$
beginalign
lim_Ntoinfty left| lim_xtoinfty frac1x int_0^x g cdot f_n ,mathrmdt - lim_xtoinfty frac1x int_0^x g cdot f ,mathrmdt right|
&le lim_Ntoinfty lim_xtoinfty frac1x int_0^x | g cdot f_n - g cdot f |^2 ,mathrmdt\
&le lim_Ntoinfty lim_xtoinfty frac1x int_0^x C | f_n - f |^2 ,mathrmdt\
&le lim_Ntoinfty lim_xtoinfty fracCx leftlceilfracx2pi / betarightrceil | f_n - f |_2^2\
&le lim_Ntoinfty fracC2pi / beta | f_n - f |_2^2 to 0
endalign
$$
It's possible there's a less fiddly proof to show convergence, but combining asymptotic density and ergodic theory does make things a bit tricky.
answered Jul 22 at 13:34
Contravariant
1,24416
add a comment |Â
up vote
3
down vote
For simplicity I'll notate $alpha = fracsqrt2+12$ and $beta = fracsqrt2-12$. Now note that the integral basically only depends on the asymptotic behaviour of the integrand, meaning that for all $y$:
$$
beginalign
L =lim_xtoinfty frac1x int_0^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt
&= lim_xtoinfty frac1x int_y^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt\
&= lim_xtoinfty frac1x+y int_0^x bigg| cos alpha (t+y) cdot sin beta (t+y)bigg| ,mathrmdt\
&= lim_xtoinfty frac1x int_0^x bigg| cos alpha (t+y) cdot sin beta (t+y)bigg| ,mathrmdt
endalign
$$
Letting $y = frac2pi nalpha$ for $n in mathbbN$ this implies:
$$
L = lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt
$$
now since $alpha$ and $beta$ are irrational we get (using the ergodic theorem):
$$
lim_Ntoinfty frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| = frac12piint_0^2pi bigg| sin(s) bigg| ,mathrmds = frac2pi
$$
for almost all $t$. With this we can now solve the original integral as follows:
$$
beginalign
L =lim_xtoinfty frac1x int_0^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt
&= lim_Ntoinfty frac1N sum_n=0^N lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_Ntoinfty lim_xtoinfty frac1x frac1N sum_n=0^Nint_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_Ntoinfty lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdot frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdot frac2pi ,mathrmdt\
&= frac4pi^2
endalign
$$
this second to last step still requires some justification however. This can be done by noting that ergodic theorem guarantees that $frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg|$ converges in the $L^2$ sense on $[0, 2pi / beta]$ to the constant function $frac2pi$. This is enough since, if we have $g$ bounded and $f_n$ periodic with period $2pi / beta$ and $f_n to f$ in the $L^2$ sense over that period then:
$$
beginalign
lim_Ntoinfty left| lim_xtoinfty frac1x int_0^x g cdot f_n ,mathrmdt - lim_xtoinfty frac1x int_0^x g cdot f ,mathrmdt right|
&le lim_Ntoinfty lim_xtoinfty frac1x int_0^x | g cdot f_n - g cdot f |^2 ,mathrmdt\
&le lim_Ntoinfty lim_xtoinfty frac1x int_0^x C | f_n - f |^2 ,mathrmdt\
&le lim_Ntoinfty lim_xtoinfty fracCx leftlceilfracx2pi / betarightrceil | f_n - f |_2^2\
&le lim_Ntoinfty fracC2pi / beta | f_n - f |_2^2 to 0
endalign
$$
It's possible there's a less fiddly proof to show convergence, but combining asymptotic density and ergodic theory does make things a bit tricky.
answered Jul 22 at 13:34
Contravariant
1,24416
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For simplicity I'll notate $alpha = fracsqrt2+12$ and $beta = fracsqrt2-12$. Now note that the integral basically only depends on the asymptotic behaviour of the integrand, meaning that for all $y$:
$$
beginalign
L =lim_xtoinfty frac1x int_0^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt
&= lim_xtoinfty frac1x int_y^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt\
&= lim_xtoinfty frac1x+y int_0^x bigg| cos alpha (t+y) cdot sin beta (t+y)bigg| ,mathrmdt\
&= lim_xtoinfty frac1x int_0^x bigg| cos alpha (t+y) cdot sin beta (t+y)bigg| ,mathrmdt
endalign
$$
Letting $y = frac2pi nalpha$ for $n in mathbbN$ this implies:
$$
L = lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt
$$
now since $alpha$ and $beta$ are irrational we get (using the ergodic theorem):
$$
lim_Ntoinfty frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| = frac12piint_0^2pi bigg| sin(s) bigg| ,mathrmds = frac2pi
$$
for almost all $t$. With this we can now solve the original integral as follows:
$$
beginalign
L =lim_xtoinfty frac1x int_0^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt
&= lim_Ntoinfty frac1N sum_n=0^N lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_Ntoinfty lim_xtoinfty frac1x frac1N sum_n=0^Nint_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_Ntoinfty lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdot frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdot frac2pi ,mathrmdt\
&= frac4pi^2
endalign
$$
this second to last step still requires some justification however. This can be done by noting that ergodic theorem guarantees that $frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg|$ converges in the $L^2$ sense on $[0, 2pi / beta]$ to the constant function $frac2pi$. This is enough since, if we have $g$ bounded and $f_n$ periodic with period $2pi / beta$ and $f_n to f$ in the $L^2$ sense over that period then:
$$
beginalign
lim_Ntoinfty left| lim_xtoinfty frac1x int_0^x g cdot f_n ,mathrmdt - lim_xtoinfty frac1x int_0^x g cdot f ,mathrmdt right|
&le lim_Ntoinfty lim_xtoinfty frac1x int_0^x | g cdot f_n - g cdot f |^2 ,mathrmdt\
&le lim_Ntoinfty lim_xtoinfty frac1x int_0^x C | f_n - f |^2 ,mathrmdt\
&le lim_Ntoinfty lim_xtoinfty fracCx leftlceilfracx2pi / betarightrceil | f_n - f |_2^2\
&le lim_Ntoinfty fracC2pi / beta | f_n - f |_2^2 to 0
endalign
$$
It's possible there's a less fiddly proof to show convergence, but combining asymptotic density and ergodic theory does make things a bit tricky.
answered Jul 22 at 13:34
Contravariant
1,24416
For simplicity I'll notate $alpha = fracsqrt2+12$ and $beta = fracsqrt2-12$. Now note that the integral basically only depends on the asymptotic behaviour of the integrand, meaning that for all $y$:
$$
beginalign
L =lim_xtoinfty frac1x int_0^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt
&= lim_xtoinfty frac1x int_y^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt\
&= lim_xtoinfty frac1x+y int_0^x bigg| cos alpha (t+y) cdot sin beta (t+y)bigg| ,mathrmdt\
&= lim_xtoinfty frac1x int_0^x bigg| cos alpha (t+y) cdot sin beta (t+y)bigg| ,mathrmdt
endalign
$$
Letting $y = frac2pi nalpha$ for $n in mathbbN$ this implies:
$$
L = lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt
$$
now since $alpha$ and $beta$ are irrational we get (using the ergodic theorem):
$$
lim_Ntoinfty frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| = frac12piint_0^2pi bigg| sin(s) bigg| ,mathrmds = frac2pi
$$
for almost all $t$. With this we can now solve the original integral as follows:
$$
beginalign
L =lim_xtoinfty frac1x int_0^x bigg| cos alpha t cdot sin beta tbigg| ,mathrmdt
&= lim_Ntoinfty frac1N sum_n=0^N lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_Ntoinfty lim_xtoinfty frac1x frac1N sum_n=0^Nint_0^x bigg| cos alpha t bigg| cdotbigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_Ntoinfty lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdot frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg| ,mathrmdt\
&= lim_xtoinfty frac1x int_0^x bigg| cos alpha t bigg| cdot frac2pi ,mathrmdt\
&= frac4pi^2
endalign
$$
this second to last step still requires some justification however. This can be done by noting that ergodic theorem guarantees that $frac1N sum_n=0^N bigg| sin beta Bigl(t+frac2pi nalphaBigr)bigg|$ converges in the $L^2$ sense on $[0, 2pi / beta]$ to the constant function $frac2pi$. This is enough since, if we have $g$ bounded and $f_n$ periodic with period $2pi / beta$ and $f_n to f$ in the $L^2$ sense over that period then:
$$
beginalign
lim_Ntoinfty left| lim_xtoinfty frac1x int_0^x g cdot f_n ,mathrmdt - lim_xtoinfty frac1x int_0^x g cdot f ,mathrmdt right|
&le lim_Ntoinfty lim_xtoinfty frac1x int_0^x | g cdot f_n - g cdot f |^2 ,mathrmdt\
&le lim_Ntoinfty lim_xtoinfty frac1x int_0^x C | f_n - f |^2 ,mathrmdt\
&le lim_Ntoinfty lim_xtoinfty fracCx leftlceilfracx2pi / betarightrceil | f_n - f |_2^2\
&le lim_Ntoinfty fracC2pi / beta | f_n - f |_2^2 to 0
endalign
$$
It's possible there's a less fiddly proof to show convergence, but combining asymptotic density and ergodic theory does make things a bit tricky.
answered Jul 22 at 13:34
Contravariant
1,24416
edited Jul 22 at 13:48
answered Jul 22 at 13:34
Contravariant
1,24416
answered Jul 22 at 13:34
Contravariant
1,24416
answered Jul 22 at 13:34
Contravariant
1,24416
1,24416
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I'm a bit puzzled by the first row in your attempt. Why is this true? And I suppose in the second row you split into three integrals and dumped zeros?
– dEmigOd
Jul 22 at 10:15
@dEmigOd I guess the integral of first two terms are bounded and would vanish in the limit.
– Szeto
Jul 22 at 10:18
1
@dEmigOd I used the identity $maxa,b=(a+b+|a-b|)/2$. The first and the second integral I splitted divided by $x$ is $0$ as $xtoinfty$.
– LittleCuteKemono
Jul 22 at 10:19
Does the function $|sin t-sin(tsqrt2)|$ have a period?
– Szeto
Jul 22 at 10:35
1
@Szeto: of course not, $sqrt2$ is irrational.
– Yves Daoust
Jul 22 at 10:36