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Evaluating $lim_ntoinftyfrac1nint_0^pi /2fracsin^2nxsin^2 xf(x);dx$, for continuous $f$
Clash Royale CLAN TAG#URR8PPP
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Let $f:left[ 0,pi /2; right]to mathbbR$ be a continuous function. What is the value of the following limit?
$$undersetnto infty mathoplim ,frac1nint_0^pi /2;fracsin ^2left( nx right)sin ^2left( x right)fleft( x right)dx$$
I strongly believe that the solution depends on this integral
$$int_0^pi /2;fracsin ^2left( nx right)sin ^2left( x right)dx=fracnpi 2$$
and using Weierstrass Approximation Theorem.
real-analysis integration limits trigonometry
edited Aug 1 at 2:51
Blue
43.6k868141
asked Jul 31 at 17:53
Vincent Law
1,287110
add a comment |Â
up vote
4
down vote
favorite
Let $f:left[ 0,pi /2; right]to mathbbR$ be a continuous function. What is the value of the following limit?
$$undersetnto infty mathoplim ,frac1nint_0^pi /2;fracsin ^2left( nx right)sin ^2left( x right)fleft( x right)dx$$
I strongly believe that the solution depends on this integral
$$int_0^pi /2;fracsin ^2left( nx right)sin ^2left( x right)dx=fracnpi 2$$
and using Weierstrass Approximation Theorem.
real-analysis integration limits trigonometry
edited Aug 1 at 2:51
Blue
43.6k868141
asked Jul 31 at 17:53
Vincent Law
1,287110
I didn't exactly work out but I think Stolz cesaro theorem might help
– Manthanein
Jul 31 at 18:17
1
Are you sure no other information about f is given like if $f$ is an even function or odd function
– Manthanein
Jul 31 at 18:26
As much as I could work out (If I have correctly applied Stolz Cesaro and used Dirichlet kernel appropriately) I reached up-to $$frac 12int_0^pi f(t)dt +sum_k=1^inftyleft(int_0^pi left(cos (kt) f(t)right) dtright)$$
– Manthanein
Jul 31 at 18:31
1
This converges to $pi f(0)$ (possibly with a constant I've lost) by Fejér's theorem.
– Chappers
Jul 31 at 18:48
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $f:left[ 0,pi /2; right]to mathbbR$ be a continuous function. What is the value of the following limit?
$$undersetnto infty mathoplim ,frac1nint_0^pi /2;fracsin ^2left( nx right)sin ^2left( x right)fleft( x right)dx$$
I strongly believe that the solution depends on this integral
$$int_0^pi /2;fracsin ^2left( nx right)sin ^2left( x right)dx=fracnpi 2$$
and using Weierstrass Approximation Theorem.
real-analysis integration limits trigonometry
edited Aug 1 at 2:51
Blue
43.6k868141
asked Jul 31 at 17:53
Vincent Law
1,287110
Let $f:left[ 0,pi /2; right]to mathbbR$ be a continuous function. What is the value of the following limit?
$$undersetnto infty mathoplim ,frac1nint_0^pi /2;fracsin ^2left( nx right)sin ^2left( x right)fleft( x right)dx$$
I strongly believe that the solution depends on this integral
$$int_0^pi /2;fracsin ^2left( nx right)sin ^2left( x right)dx=fracnpi 2$$
and using Weierstrass Approximation Theorem.
real-analysis integration limits trigonometry
edited Aug 1 at 2:51
Blue
43.6k868141
asked Jul 31 at 17:53
Vincent Law
1,287110
edited Aug 1 at 2:51
Blue
43.6k868141
edited Aug 1 at 2:51
Blue
43.6k868141
edited Aug 1 at 2:51
Blue
43.6k868141
43.6k868141
asked Jul 31 at 17:53
Vincent Law
1,287110
asked Jul 31 at 17:53
Vincent Law
1,287110
asked Jul 31 at 17:53
Vincent Law
1,287110
1,287110
I didn't exactly work out but I think Stolz cesaro theorem might help
– Manthanein
Jul 31 at 18:17
1
Are you sure no other information about f is given like if $f$ is an even function or odd function
– Manthanein
Jul 31 at 18:26
As much as I could work out (If I have correctly applied Stolz Cesaro and used Dirichlet kernel appropriately) I reached up-to $$frac 12int_0^pi f(t)dt +sum_k=1^inftyleft(int_0^pi left(cos (kt) f(t)right) dtright)$$
– Manthanein
Jul 31 at 18:31
1
This converges to $pi f(0)$ (possibly with a constant I've lost) by Fejér's theorem.
– Chappers
Jul 31 at 18:48
add a comment |Â
I didn't exactly work out but I think Stolz cesaro theorem might help
– Manthanein
Jul 31 at 18:17
1
Are you sure no other information about f is given like if $f$ is an even function or odd function
– Manthanein
Jul 31 at 18:26
As much as I could work out (If I have correctly applied Stolz Cesaro and used Dirichlet kernel appropriately) I reached up-to $$frac 12int_0^pi f(t)dt +sum_k=1^inftyleft(int_0^pi left(cos (kt) f(t)right) dtright)$$
– Manthanein
Jul 31 at 18:31
1
This converges to $pi f(0)$ (possibly with a constant I've lost) by Fejér's theorem.
– Chappers
Jul 31 at 18:48
I didn't exactly work out but I think Stolz cesaro theorem might help
– Manthanein
Jul 31 at 18:17
I didn't exactly work out but I think Stolz cesaro theorem might help
– Manthanein
Jul 31 at 18:17
1
1
Are you sure no other information about f is given like if $f$ is an even function or odd function
– Manthanein
Jul 31 at 18:26
Are you sure no other information about f is given like if $f$ is an even function or odd function
– Manthanein
Jul 31 at 18:26
As much as I could work out (If I have correctly applied Stolz Cesaro and used Dirichlet kernel appropriately) I reached up-to $$frac 12int_0^pi f(t)dt +sum_k=1^inftyleft(int_0^pi left(cos (kt) f(t)right) dtright)$$
– Manthanein
Jul 31 at 18:31
As much as I could work out (If I have correctly applied Stolz Cesaro and used Dirichlet kernel appropriately) I reached up-to $$frac 12int_0^pi f(t)dt +sum_k=1^inftyleft(int_0^pi left(cos (kt) f(t)right) dtright)$$
– Manthanein
Jul 31 at 18:31
1
1
This converges to $pi f(0)$ (possibly with a constant I've lost) by Fejér's theorem.
– Chappers
Jul 31 at 18:48
This converges to $pi f(0)$ (possibly with a constant I've lost) by Fejér's theorem.
– Chappers
Jul 31 at 18:48
add a comment |Â
1 Answer
1
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oldest
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up vote
4
down vote
accepted
The limit is $displaystyle fracpi2f(0)$.
Note that
$$tag* left|int_0^pi/2 fracsin^2(nx)nsin^2 xf(x) , dx - fracpi2f(0)right| \leqslant left|int_0^delta fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright|+ left|int_delta^pi/2 fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright|$$
Since $f$ is continuous on $[0,pi/2]$ and, hence, bounded we have $|f(x)| leqslant M$ and $|f(x) - f(0)| leqslant 2M$ and an estimate for the second integral on the RHS of (*) is
$$left|int_delta^pi/2 fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright| leqslant frac2Mnint_delta^pi/2 frac1sin^2 x , dx to_n to infty 0$$
Since the second integral converges to $0$ for any $delta$ , it remains to prove that the first integral on the RHS of (*) can be made arbitrarily close to $0$ with a suitable choice for $delta$.
For any $epsilon >0$ we can chose $delta$ such that $|f(x) - f(0)| < epsilon$ for $0 leqslant x leqslant delta$.
Using $sin x > 2x/pi$ we get
$$left|int_0^delta fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright| leqslant fracpi^2 epsilon4n int_0^delta fracsin^2 (nx)x^2 , dx \ = fracpi^2 epsilon4 int_0^ndelta fracsin^2 uu^2 , du \ leqslant fracpi^2 epsilon4 int_0^infty fracsin^2 uu^2 , du = fracpi^3 epsilon8$$
Since, $epsilon$ can be arbitrarily close to $0$ we are done.
answered Jul 31 at 19:16
RRL
43.4k42260
Who said anything about "$f$" being bounded?
– Manthanein
Aug 1 at 2:08
@Manthanein: With $u= nx$ we have $fracsin^2(nx)nx^2 , dx = fracsin^2(u)n(u/n)^2 , d(u/n) = fracsin^2(u)u^2 , du$
– RRL
Aug 1 at 2:28
@Manthanein: $f$ is continuous on the compact interval $[0,pi/2]$. Please read the very first sentence of OP.
– RRL
Aug 1 at 2:29
@Manthanein: A continuous function on a closed, bounded interval is bounded.
– RRL
Aug 1 at 2:37
Oh yes I didn't notice the closed intervals. Thanks
– Manthanein
Aug 1 at 3:24
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The limit is $displaystyle fracpi2f(0)$.
Note that
$$tag* left|int_0^pi/2 fracsin^2(nx)nsin^2 xf(x) , dx - fracpi2f(0)right| \leqslant left|int_0^delta fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright|+ left|int_delta^pi/2 fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright|$$
Since $f$ is continuous on $[0,pi/2]$ and, hence, bounded we have $|f(x)| leqslant M$ and $|f(x) - f(0)| leqslant 2M$ and an estimate for the second integral on the RHS of (*) is
$$left|int_delta^pi/2 fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright| leqslant frac2Mnint_delta^pi/2 frac1sin^2 x , dx to_n to infty 0$$
Since the second integral converges to $0$ for any $delta$ , it remains to prove that the first integral on the RHS of (*) can be made arbitrarily close to $0$ with a suitable choice for $delta$.
For any $epsilon >0$ we can chose $delta$ such that $|f(x) - f(0)| < epsilon$ for $0 leqslant x leqslant delta$.
Using $sin x > 2x/pi$ we get
$$left|int_0^delta fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright| leqslant fracpi^2 epsilon4n int_0^delta fracsin^2 (nx)x^2 , dx \ = fracpi^2 epsilon4 int_0^ndelta fracsin^2 uu^2 , du \ leqslant fracpi^2 epsilon4 int_0^infty fracsin^2 uu^2 , du = fracpi^3 epsilon8$$
Since, $epsilon$ can be arbitrarily close to $0$ we are done.
answered Jul 31 at 19:16
RRL
43.4k42260
Who said anything about "$f$" being bounded?
– Manthanein
Aug 1 at 2:08
@Manthanein: With $u= nx$ we have $fracsin^2(nx)nx^2 , dx = fracsin^2(u)n(u/n)^2 , d(u/n) = fracsin^2(u)u^2 , du$
– RRL
Aug 1 at 2:28
@Manthanein: $f$ is continuous on the compact interval $[0,pi/2]$. Please read the very first sentence of OP.
– RRL
Aug 1 at 2:29
@Manthanein: A continuous function on a closed, bounded interval is bounded.
– RRL
Aug 1 at 2:37
Oh yes I didn't notice the closed intervals. Thanks
– Manthanein
Aug 1 at 3:24
add a comment |Â
up vote
4
down vote
accepted
The limit is $displaystyle fracpi2f(0)$.
Note that
$$tag* left|int_0^pi/2 fracsin^2(nx)nsin^2 xf(x) , dx - fracpi2f(0)right| \leqslant left|int_0^delta fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright|+ left|int_delta^pi/2 fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright|$$
Since $f$ is continuous on $[0,pi/2]$ and, hence, bounded we have $|f(x)| leqslant M$ and $|f(x) - f(0)| leqslant 2M$ and an estimate for the second integral on the RHS of (*) is
$$left|int_delta^pi/2 fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright| leqslant frac2Mnint_delta^pi/2 frac1sin^2 x , dx to_n to infty 0$$
Since the second integral converges to $0$ for any $delta$ , it remains to prove that the first integral on the RHS of (*) can be made arbitrarily close to $0$ with a suitable choice for $delta$.
For any $epsilon >0$ we can chose $delta$ such that $|f(x) - f(0)| < epsilon$ for $0 leqslant x leqslant delta$.
Using $sin x > 2x/pi$ we get
$$left|int_0^delta fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright| leqslant fracpi^2 epsilon4n int_0^delta fracsin^2 (nx)x^2 , dx \ = fracpi^2 epsilon4 int_0^ndelta fracsin^2 uu^2 , du \ leqslant fracpi^2 epsilon4 int_0^infty fracsin^2 uu^2 , du = fracpi^3 epsilon8$$
Since, $epsilon$ can be arbitrarily close to $0$ we are done.
answered Jul 31 at 19:16
RRL
43.4k42260
Who said anything about "$f$" being bounded?
– Manthanein
Aug 1 at 2:08
@Manthanein: With $u= nx$ we have $fracsin^2(nx)nx^2 , dx = fracsin^2(u)n(u/n)^2 , d(u/n) = fracsin^2(u)u^2 , du$
– RRL
Aug 1 at 2:28
@Manthanein: $f$ is continuous on the compact interval $[0,pi/2]$. Please read the very first sentence of OP.
– RRL
Aug 1 at 2:29
@Manthanein: A continuous function on a closed, bounded interval is bounded.
– RRL
Aug 1 at 2:37
Oh yes I didn't notice the closed intervals. Thanks
– Manthanein
Aug 1 at 3:24
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The limit is $displaystyle fracpi2f(0)$.
Note that
$$tag* left|int_0^pi/2 fracsin^2(nx)nsin^2 xf(x) , dx - fracpi2f(0)right| \leqslant left|int_0^delta fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright|+ left|int_delta^pi/2 fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright|$$
Since $f$ is continuous on $[0,pi/2]$ and, hence, bounded we have $|f(x)| leqslant M$ and $|f(x) - f(0)| leqslant 2M$ and an estimate for the second integral on the RHS of (*) is
$$left|int_delta^pi/2 fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright| leqslant frac2Mnint_delta^pi/2 frac1sin^2 x , dx to_n to infty 0$$
Since the second integral converges to $0$ for any $delta$ , it remains to prove that the first integral on the RHS of (*) can be made arbitrarily close to $0$ with a suitable choice for $delta$.
For any $epsilon >0$ we can chose $delta$ such that $|f(x) - f(0)| < epsilon$ for $0 leqslant x leqslant delta$.
Using $sin x > 2x/pi$ we get
$$left|int_0^delta fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright| leqslant fracpi^2 epsilon4n int_0^delta fracsin^2 (nx)x^2 , dx \ = fracpi^2 epsilon4 int_0^ndelta fracsin^2 uu^2 , du \ leqslant fracpi^2 epsilon4 int_0^infty fracsin^2 uu^2 , du = fracpi^3 epsilon8$$
Since, $epsilon$ can be arbitrarily close to $0$ we are done.
answered Jul 31 at 19:16
RRL
43.4k42260
The limit is $displaystyle fracpi2f(0)$.
Note that
$$tag* left|int_0^pi/2 fracsin^2(nx)nsin^2 xf(x) , dx - fracpi2f(0)right| \leqslant left|int_0^delta fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright|+ left|int_delta^pi/2 fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright|$$
Since $f$ is continuous on $[0,pi/2]$ and, hence, bounded we have $|f(x)| leqslant M$ and $|f(x) - f(0)| leqslant 2M$ and an estimate for the second integral on the RHS of (*) is
$$left|int_delta^pi/2 fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright| leqslant frac2Mnint_delta^pi/2 frac1sin^2 x , dx to_n to infty 0$$
Since the second integral converges to $0$ for any $delta$ , it remains to prove that the first integral on the RHS of (*) can be made arbitrarily close to $0$ with a suitable choice for $delta$.
For any $epsilon >0$ we can chose $delta$ such that $|f(x) - f(0)| < epsilon$ for $0 leqslant x leqslant delta$.
Using $sin x > 2x/pi$ we get
$$left|int_0^delta fracsin^2(nx)nsin^2 x[f(x)-f(0)] , dxright| leqslant fracpi^2 epsilon4n int_0^delta fracsin^2 (nx)x^2 , dx \ = fracpi^2 epsilon4 int_0^ndelta fracsin^2 uu^2 , du \ leqslant fracpi^2 epsilon4 int_0^infty fracsin^2 uu^2 , du = fracpi^3 epsilon8$$
Since, $epsilon$ can be arbitrarily close to $0$ we are done.
answered Jul 31 at 19:16
RRL
43.4k42260
edited Aug 1 at 2:44
answered Jul 31 at 19:16
RRL
43.4k42260
answered Jul 31 at 19:16
RRL
43.4k42260
answered Jul 31 at 19:16
RRL
43.4k42260
43.4k42260
Who said anything about "$f$" being bounded?
– Manthanein
Aug 1 at 2:08
@Manthanein: With $u= nx$ we have $fracsin^2(nx)nx^2 , dx = fracsin^2(u)n(u/n)^2 , d(u/n) = fracsin^2(u)u^2 , du$
– RRL
Aug 1 at 2:28
@Manthanein: $f$ is continuous on the compact interval $[0,pi/2]$. Please read the very first sentence of OP.
– RRL
Aug 1 at 2:29
@Manthanein: A continuous function on a closed, bounded interval is bounded.
– RRL
Aug 1 at 2:37
Oh yes I didn't notice the closed intervals. Thanks
– Manthanein
Aug 1 at 3:24
add a comment |Â
Who said anything about "$f$" being bounded?
– Manthanein
Aug 1 at 2:08
@Manthanein: With $u= nx$ we have $fracsin^2(nx)nx^2 , dx = fracsin^2(u)n(u/n)^2 , d(u/n) = fracsin^2(u)u^2 , du$
– RRL
Aug 1 at 2:28
@Manthanein: $f$ is continuous on the compact interval $[0,pi/2]$. Please read the very first sentence of OP.
– RRL
Aug 1 at 2:29
@Manthanein: A continuous function on a closed, bounded interval is bounded.
– RRL
Aug 1 at 2:37
Oh yes I didn't notice the closed intervals. Thanks
– Manthanein
Aug 1 at 3:24
Who said anything about "$f$" being bounded?
– Manthanein
Aug 1 at 2:08
Who said anything about "$f$" being bounded?
– Manthanein
Aug 1 at 2:08
@Manthanein: With $u= nx$ we have $fracsin^2(nx)nx^2 , dx = fracsin^2(u)n(u/n)^2 , d(u/n) = fracsin^2(u)u^2 , du$
– RRL
Aug 1 at 2:28
@Manthanein: With $u= nx$ we have $fracsin^2(nx)nx^2 , dx = fracsin^2(u)n(u/n)^2 , d(u/n) = fracsin^2(u)u^2 , du$
– RRL
Aug 1 at 2:28
@Manthanein: $f$ is continuous on the compact interval $[0,pi/2]$. Please read the very first sentence of OP.
– RRL
Aug 1 at 2:29
@Manthanein: $f$ is continuous on the compact interval $[0,pi/2]$. Please read the very first sentence of OP.
– RRL
Aug 1 at 2:29
@Manthanein: A continuous function on a closed, bounded interval is bounded.
– RRL
Aug 1 at 2:37
@Manthanein: A continuous function on a closed, bounded interval is bounded.
– RRL
Aug 1 at 2:37
Oh yes I didn't notice the closed intervals. Thanks
– Manthanein
Aug 1 at 3:24
Oh yes I didn't notice the closed intervals. Thanks
– Manthanein
Aug 1 at 3:24
add a comment |Â
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I didn't exactly work out but I think Stolz cesaro theorem might help
– Manthanein
Jul 31 at 18:17
1
Are you sure no other information about f is given like if $f$ is an even function or odd function
– Manthanein
Jul 31 at 18:26
As much as I could work out (If I have correctly applied Stolz Cesaro and used Dirichlet kernel appropriately) I reached up-to $$frac 12int_0^pi f(t)dt +sum_k=1^inftyleft(int_0^pi left(cos (kt) f(t)right) dtright)$$
– Manthanein
Jul 31 at 18:31
1
This converges to $pi f(0)$ (possibly with a constant I've lost) by Fejér's theorem.
– Chappers
Jul 31 at 18:48