Dirichlet kernel inequality
$begingroup$
Let's define
$$A := frac{1}{pi} int_{-pi}^pi big| D_{n}(t) big| , dt,$$
where $D_{n}(t)$ is Dirichlet Kernel
$$D_n(t):=frac{1}{2} + sum_{k=1}^{n} cos(kt)= frac{sin left(tleft(n+frac{1}{2}right)right)}{2sin left(frac{t}{2}right)}.$$
I need to prove that $$frac{4}{pi^2} ln(n) leq A leq 3 + ln(n).$$
Any ideas, clues or hints on how to prove it? Something to start from?
real-analysis analysis fourier-series
edited Jan 16 at 20:09
AEngineer
1,6521318
asked Jan 16 at 20:02
BreadBread
104
$endgroup$
add a comment |
$begingroup$
Let's define
$$A := frac{1}{pi} int_{-pi}^pi big| D_{n}(t) big| , dt,$$
where $D_{n}(t)$ is Dirichlet Kernel
$$D_n(t):=frac{1}{2} + sum_{k=1}^{n} cos(kt)= frac{sin left(tleft(n+frac{1}{2}right)right)}{2sin left(frac{t}{2}right)}.$$
I need to prove that $$frac{4}{pi^2} ln(n) leq A leq 3 + ln(n).$$
Any ideas, clues or hints on how to prove it? Something to start from?
real-analysis analysis fourier-series
edited Jan 16 at 20:09
AEngineer
1,6521318
asked Jan 16 at 20:02
BreadBread
104
$endgroup$
add a comment |
$begingroup$
Let's define
$$A := frac{1}{pi} int_{-pi}^pi big| D_{n}(t) big| , dt,$$
where $D_{n}(t)$ is Dirichlet Kernel
$$D_n(t):=frac{1}{2} + sum_{k=1}^{n} cos(kt)= frac{sin left(tleft(n+frac{1}{2}right)right)}{2sin left(frac{t}{2}right)}.$$
I need to prove that $$frac{4}{pi^2} ln(n) leq A leq 3 + ln(n).$$
Any ideas, clues or hints on how to prove it? Something to start from?
real-analysis analysis fourier-series
edited Jan 16 at 20:09
AEngineer
1,6521318
asked Jan 16 at 20:02
BreadBread
104
$endgroup$
Let's define
$$A := frac{1}{pi} int_{-pi}^pi big| D_{n}(t) big| , dt,$$
where $D_{n}(t)$ is Dirichlet Kernel
$$D_n(t):=frac{1}{2} + sum_{k=1}^{n} cos(kt)= frac{sin left(tleft(n+frac{1}{2}right)right)}{2sin left(frac{t}{2}right)}.$$
I need to prove that $$frac{4}{pi^2} ln(n) leq A leq 3 + ln(n).$$
Any ideas, clues or hints on how to prove it? Something to start from?
real-analysis analysis fourier-series
real-analysis analysis fourier-series
edited Jan 16 at 20:09
AEngineer
1,6521318
asked Jan 16 at 20:02
BreadBread
104
edited Jan 16 at 20:09
AEngineer
1,6521318
asked Jan 16 at 20:02
BreadBread
104
edited Jan 16 at 20:09
AEngineer
1,6521318
edited Jan 16 at 20:09
AEngineer
1,6521318
edited Jan 16 at 20:09
AEngineer
1,6521318
1,6521318
asked Jan 16 at 20:02
BreadBread
104
asked Jan 16 at 20:02
BreadBread
104
asked Jan 16 at 20:02
BreadBread
104
104
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
All you need to know is that $3 < pi < 4$ and $e > 2$.
As $n ge 1$ you have $1 + dfrac 1{2n} le dfrac 32$ so that $dfrac 2pi left( 1 + dfrac 1{2n} le dfrac 32 right) le dfrac 3 pi < 1$,
and $ln pi < log_2 4 = 2$.
answered Jan 24 at 20:42
Umberto P.Umberto P.
40.4k13370
$endgroup$
$begingroup$
Thank you kindly. :D
$endgroup$
– Bread
Jan 25 at 1:04
add a comment |
$begingroup$
What I managed to find out is: $$A := frac{1}{pi} int_{-pi}^pi big| D_{n}(t) big| , dt leq frac{2}{pi} int_{0}^{frac{1}{n}} (n+frac{1}{2}) dt + frac{2}{pi} int_{frac{1}{n}}^{pi} frac{pi}{2t}dt.$$
After integration I get: $$frac{2}{pi}(nt+frac{t}{2}) + frac{2}{pi} frac{pi}{2} ln|t| = frac{2}{pi}(1+frac{1}{2n}) + ln pi + ln n.$$
Any idea how to get 3 in this?
answered Jan 24 at 17:07
BreadBread
104
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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$begingroup$
All you need to know is that $3 < pi < 4$ and $e > 2$.
As $n ge 1$ you have $1 + dfrac 1{2n} le dfrac 32$ so that $dfrac 2pi left( 1 + dfrac 1{2n} le dfrac 32 right) le dfrac 3 pi < 1$,
and $ln pi < log_2 4 = 2$.
answered Jan 24 at 20:42
Umberto P.Umberto P.
40.4k13370
$endgroup$
$begingroup$
Thank you kindly. :D
$endgroup$
– Bread
Jan 25 at 1:04
add a comment |
$begingroup$
All you need to know is that $3 < pi < 4$ and $e > 2$.
As $n ge 1$ you have $1 + dfrac 1{2n} le dfrac 32$ so that $dfrac 2pi left( 1 + dfrac 1{2n} le dfrac 32 right) le dfrac 3 pi < 1$,
and $ln pi < log_2 4 = 2$.
answered Jan 24 at 20:42
Umberto P.Umberto P.
40.4k13370
$endgroup$
$begingroup$
Thank you kindly. :D
$endgroup$
– Bread
Jan 25 at 1:04
add a comment |
$begingroup$
All you need to know is that $3 < pi < 4$ and $e > 2$.
As $n ge 1$ you have $1 + dfrac 1{2n} le dfrac 32$ so that $dfrac 2pi left( 1 + dfrac 1{2n} le dfrac 32 right) le dfrac 3 pi < 1$,
and $ln pi < log_2 4 = 2$.
answered Jan 24 at 20:42
Umberto P.Umberto P.
40.4k13370
$endgroup$
All you need to know is that $3 < pi < 4$ and $e > 2$.
As $n ge 1$ you have $1 + dfrac 1{2n} le dfrac 32$ so that $dfrac 2pi left( 1 + dfrac 1{2n} le dfrac 32 right) le dfrac 3 pi < 1$,
and $ln pi < log_2 4 = 2$.
answered Jan 24 at 20:42
Umberto P.Umberto P.
40.4k13370
answered Jan 24 at 20:42
Umberto P.Umberto P.
40.4k13370
answered Jan 24 at 20:42
Umberto P.Umberto P.
40.4k13370
answered Jan 24 at 20:42
Umberto P.Umberto P.
40.4k13370
40.4k13370
$begingroup$
Thank you kindly. :D
$endgroup$
– Bread
Jan 25 at 1:04
add a comment |
$begingroup$
Thank you kindly. :D
$endgroup$
– Bread
Jan 25 at 1:04
$begingroup$
Thank you kindly. :D
$endgroup$
– Bread
Jan 25 at 1:04
$begingroup$
Thank you kindly. :D
$endgroup$
– Bread
Jan 25 at 1:04
add a comment |
$begingroup$
What I managed to find out is: $$A := frac{1}{pi} int_{-pi}^pi big| D_{n}(t) big| , dt leq frac{2}{pi} int_{0}^{frac{1}{n}} (n+frac{1}{2}) dt + frac{2}{pi} int_{frac{1}{n}}^{pi} frac{pi}{2t}dt.$$
After integration I get: $$frac{2}{pi}(nt+frac{t}{2}) + frac{2}{pi} frac{pi}{2} ln|t| = frac{2}{pi}(1+frac{1}{2n}) + ln pi + ln n.$$
Any idea how to get 3 in this?
answered Jan 24 at 17:07
BreadBread
104
$endgroup$
add a comment |
$begingroup$
What I managed to find out is: $$A := frac{1}{pi} int_{-pi}^pi big| D_{n}(t) big| , dt leq frac{2}{pi} int_{0}^{frac{1}{n}} (n+frac{1}{2}) dt + frac{2}{pi} int_{frac{1}{n}}^{pi} frac{pi}{2t}dt.$$
After integration I get: $$frac{2}{pi}(nt+frac{t}{2}) + frac{2}{pi} frac{pi}{2} ln|t| = frac{2}{pi}(1+frac{1}{2n}) + ln pi + ln n.$$
Any idea how to get 3 in this?
answered Jan 24 at 17:07
BreadBread
104
$endgroup$
add a comment |
$begingroup$
What I managed to find out is: $$A := frac{1}{pi} int_{-pi}^pi big| D_{n}(t) big| , dt leq frac{2}{pi} int_{0}^{frac{1}{n}} (n+frac{1}{2}) dt + frac{2}{pi} int_{frac{1}{n}}^{pi} frac{pi}{2t}dt.$$
After integration I get: $$frac{2}{pi}(nt+frac{t}{2}) + frac{2}{pi} frac{pi}{2} ln|t| = frac{2}{pi}(1+frac{1}{2n}) + ln pi + ln n.$$
Any idea how to get 3 in this?
answered Jan 24 at 17:07
BreadBread
104
$endgroup$
What I managed to find out is: $$A := frac{1}{pi} int_{-pi}^pi big| D_{n}(t) big| , dt leq frac{2}{pi} int_{0}^{frac{1}{n}} (n+frac{1}{2}) dt + frac{2}{pi} int_{frac{1}{n}}^{pi} frac{pi}{2t}dt.$$
After integration I get: $$frac{2}{pi}(nt+frac{t}{2}) + frac{2}{pi} frac{pi}{2} ln|t| = frac{2}{pi}(1+frac{1}{2n}) + ln pi + ln n.$$
Any idea how to get 3 in this?
answered Jan 24 at 17:07
BreadBread
104
answered Jan 24 at 17:07
BreadBread
104
answered Jan 24 at 17:07
BreadBread
104
answered Jan 24 at 17:07
BreadBread
104
104
add a comment |
add a comment |
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