Mixing
Estimate of Fourier Coefficient
$begingroup$
Knowing $f in C^r(mathbb{R})$
$$c_n(hat{f}) = frac{1}{2pi} int_{0}^{2pi} f(x)e^{-inx}dx, n in mathbb{Z}$$
I am trying to show that $$mid c_n mid leqslant C_r (1 + {mid n mid}^{-r}) $$
I have been struggling the whole evening and I still can't see how the estimate depends on the smoothness $r$ and $n$. Since I just replaced $e^{-inx}$ by $1$ and I really don't know any other ways to deal with complex numbers. Thank you very much for your hint.
*Edit, I just found out that I am asked to show that $$mid c_n mid leqslant C_r (1 + {mid n mid})^{-r} $$
real-analysis fourier-analysis
asked Jan 28 at 3:40
Alvis NordkovichAlvis Nordkovich
256110
$endgroup$
add a comment |
$begingroup$
Knowing $f in C^r(mathbb{R})$
$$c_n(hat{f}) = frac{1}{2pi} int_{0}^{2pi} f(x)e^{-inx}dx, n in mathbb{Z}$$
I am trying to show that $$mid c_n mid leqslant C_r (1 + {mid n mid}^{-r}) $$
I have been struggling the whole evening and I still can't see how the estimate depends on the smoothness $r$ and $n$. Since I just replaced $e^{-inx}$ by $1$ and I really don't know any other ways to deal with complex numbers. Thank you very much for your hint.
*Edit, I just found out that I am asked to show that $$mid c_n mid leqslant C_r (1 + {mid n mid})^{-r} $$
real-analysis fourier-analysis
asked Jan 28 at 3:40
Alvis NordkovichAlvis Nordkovich
256110
$endgroup$
add a comment |
$begingroup$
Knowing $f in C^r(mathbb{R})$
$$c_n(hat{f}) = frac{1}{2pi} int_{0}^{2pi} f(x)e^{-inx}dx, n in mathbb{Z}$$
I am trying to show that $$mid c_n mid leqslant C_r (1 + {mid n mid}^{-r}) $$
I have been struggling the whole evening and I still can't see how the estimate depends on the smoothness $r$ and $n$. Since I just replaced $e^{-inx}$ by $1$ and I really don't know any other ways to deal with complex numbers. Thank you very much for your hint.
*Edit, I just found out that I am asked to show that $$mid c_n mid leqslant C_r (1 + {mid n mid})^{-r} $$
real-analysis fourier-analysis
asked Jan 28 at 3:40
Alvis NordkovichAlvis Nordkovich
256110
$endgroup$
Knowing $f in C^r(mathbb{R})$
$$c_n(hat{f}) = frac{1}{2pi} int_{0}^{2pi} f(x)e^{-inx}dx, n in mathbb{Z}$$
I am trying to show that $$mid c_n mid leqslant C_r (1 + {mid n mid}^{-r}) $$
I have been struggling the whole evening and I still can't see how the estimate depends on the smoothness $r$ and $n$. Since I just replaced $e^{-inx}$ by $1$ and I really don't know any other ways to deal with complex numbers. Thank you very much for your hint.
*Edit, I just found out that I am asked to show that $$mid c_n mid leqslant C_r (1 + {mid n mid})^{-r} $$
real-analysis fourier-analysis
real-analysis fourier-analysis
asked Jan 28 at 3:40
Alvis NordkovichAlvis Nordkovich
256110
asked Jan 28 at 3:40
Alvis NordkovichAlvis Nordkovich
256110
edited Feb 3 at 3:53
Alvis Nordkovich
asked Jan 28 at 3:40
Alvis NordkovichAlvis Nordkovich
256110
asked Jan 28 at 3:40
Alvis NordkovichAlvis Nordkovich
256110
asked Jan 28 at 3:40
Alvis NordkovichAlvis Nordkovich
256110
256110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: In general $c_n = o(|n|^{-r})$ for $2pi$-periodic $f in C^r(mathbb{R})$ and getting the exact form for your bound requires a little manipulation.
For $f in C^1$ and $n neq 0$, integration by parts yields
$$begin{align}|c_n| &= frac{1}{2pi}left|ifrac{f(2pi)e^{-i2pi n} - f(0)}{n} - frac{i}{n}int_0^{2pi} f'(x)e^{-inx} , dxright| \ &leqslant frac{|f(2pi) - f(0)|}{2pi}+ frac{1}{2pi|n|}int_0^{2pi} |f'(x)| , dx \ &= frac{1}{2pi}left|int_0^{2pi}f'(x) , dx right|+ frac{1}{2pi|n|}int_0^{2pi} |f'(x)| , dx \ & leqslant frac{1}{2pi}int_0^{2pi}|f'(x)| , dx left(1 + |n|^{-1} right) end{align}$$
For $f in C^r$ use repeated integration by parts.
answered Jan 28 at 6:16
RRLRRL
53k42573
$endgroup$
$begingroup$
Thank you for the answer. Can you also explain the case when $n = 0$? I struggled to find the estimate since integration by part will not work.
$endgroup$
– Alvis Nordkovich
Feb 3 at 3:52
$begingroup$
@AlvisNordkovich: You're welcome. The constant I gave you for the $C^1$ case is $C = frac{1}{2pi}int_0^{2pi}|f'(x)| , dx$ and since $f'$ is bounded we can get a larger bound $C leqslant frac{1}{2pi}int_0^{2pi}|f'|_{infty} , dx = |f'|_{infty}$. Why worry about the case $n=0$ since the expression $(1 +|n|^{-1})$ is not even defined? Also $c_0 = frac{1}{2pi}int_0^{2pi} f(x) , dx$ is just a number.
$endgroup$
– RRL
Feb 3 at 6:02
$begingroup$
Maybe try to bound $c_0$ in terms of $|f'|_{infty}$ using $int_0^{2pi} f(x) , dx = int_0^{2pi} left( f(0) + int_0^x f'(t) , dt right) , dx$ to try to get some common constant. However, at this point I don't see the context of these extensions to the original question. If it gets too much more complicated you should ask another question.
$endgroup$
– RRL
Feb 3 at 6:11
add a comment |
$begingroup$
I am assuming that your function $f$ has period $2pi$. If $g=f^{(r)}$ then $c_n (hat {g}) =(in)^{r} c_n(hat {f})$. [This is proved by repeated integration by parts]. From this the inequality follow easily.
answered Jan 28 at 6:12
Kavi Rama MurthyKavi Rama Murthy
70.4k53170
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint: In general $c_n = o(|n|^{-r})$ for $2pi$-periodic $f in C^r(mathbb{R})$ and getting the exact form for your bound requires a little manipulation.
For $f in C^1$ and $n neq 0$, integration by parts yields
$$begin{align}|c_n| &= frac{1}{2pi}left|ifrac{f(2pi)e^{-i2pi n} - f(0)}{n} - frac{i}{n}int_0^{2pi} f'(x)e^{-inx} , dxright| \ &leqslant frac{|f(2pi) - f(0)|}{2pi}+ frac{1}{2pi|n|}int_0^{2pi} |f'(x)| , dx \ &= frac{1}{2pi}left|int_0^{2pi}f'(x) , dx right|+ frac{1}{2pi|n|}int_0^{2pi} |f'(x)| , dx \ & leqslant frac{1}{2pi}int_0^{2pi}|f'(x)| , dx left(1 + |n|^{-1} right) end{align}$$
For $f in C^r$ use repeated integration by parts.
answered Jan 28 at 6:16
RRLRRL
53k42573
$endgroup$
$begingroup$
Thank you for the answer. Can you also explain the case when $n = 0$? I struggled to find the estimate since integration by part will not work.
$endgroup$
– Alvis Nordkovich
Feb 3 at 3:52
$begingroup$
@AlvisNordkovich: You're welcome. The constant I gave you for the $C^1$ case is $C = frac{1}{2pi}int_0^{2pi}|f'(x)| , dx$ and since $f'$ is bounded we can get a larger bound $C leqslant frac{1}{2pi}int_0^{2pi}|f'|_{infty} , dx = |f'|_{infty}$. Why worry about the case $n=0$ since the expression $(1 +|n|^{-1})$ is not even defined? Also $c_0 = frac{1}{2pi}int_0^{2pi} f(x) , dx$ is just a number.
$endgroup$
– RRL
Feb 3 at 6:02
$begingroup$
Maybe try to bound $c_0$ in terms of $|f'|_{infty}$ using $int_0^{2pi} f(x) , dx = int_0^{2pi} left( f(0) + int_0^x f'(t) , dt right) , dx$ to try to get some common constant. However, at this point I don't see the context of these extensions to the original question. If it gets too much more complicated you should ask another question.
$endgroup$
– RRL
Feb 3 at 6:11
add a comment |
$begingroup$
Hint: In general $c_n = o(|n|^{-r})$ for $2pi$-periodic $f in C^r(mathbb{R})$ and getting the exact form for your bound requires a little manipulation.
For $f in C^1$ and $n neq 0$, integration by parts yields
$$begin{align}|c_n| &= frac{1}{2pi}left|ifrac{f(2pi)e^{-i2pi n} - f(0)}{n} - frac{i}{n}int_0^{2pi} f'(x)e^{-inx} , dxright| \ &leqslant frac{|f(2pi) - f(0)|}{2pi}+ frac{1}{2pi|n|}int_0^{2pi} |f'(x)| , dx \ &= frac{1}{2pi}left|int_0^{2pi}f'(x) , dx right|+ frac{1}{2pi|n|}int_0^{2pi} |f'(x)| , dx \ & leqslant frac{1}{2pi}int_0^{2pi}|f'(x)| , dx left(1 + |n|^{-1} right) end{align}$$
For $f in C^r$ use repeated integration by parts.
answered Jan 28 at 6:16
RRLRRL
53k42573
$endgroup$
$begingroup$
Thank you for the answer. Can you also explain the case when $n = 0$? I struggled to find the estimate since integration by part will not work.
$endgroup$
– Alvis Nordkovich
Feb 3 at 3:52
$begingroup$
@AlvisNordkovich: You're welcome. The constant I gave you for the $C^1$ case is $C = frac{1}{2pi}int_0^{2pi}|f'(x)| , dx$ and since $f'$ is bounded we can get a larger bound $C leqslant frac{1}{2pi}int_0^{2pi}|f'|_{infty} , dx = |f'|_{infty}$. Why worry about the case $n=0$ since the expression $(1 +|n|^{-1})$ is not even defined? Also $c_0 = frac{1}{2pi}int_0^{2pi} f(x) , dx$ is just a number.
$endgroup$
– RRL
Feb 3 at 6:02
$begingroup$
Maybe try to bound $c_0$ in terms of $|f'|_{infty}$ using $int_0^{2pi} f(x) , dx = int_0^{2pi} left( f(0) + int_0^x f'(t) , dt right) , dx$ to try to get some common constant. However, at this point I don't see the context of these extensions to the original question. If it gets too much more complicated you should ask another question.
$endgroup$
– RRL
Feb 3 at 6:11
add a comment |
$begingroup$
Hint: In general $c_n = o(|n|^{-r})$ for $2pi$-periodic $f in C^r(mathbb{R})$ and getting the exact form for your bound requires a little manipulation.
For $f in C^1$ and $n neq 0$, integration by parts yields
$$begin{align}|c_n| &= frac{1}{2pi}left|ifrac{f(2pi)e^{-i2pi n} - f(0)}{n} - frac{i}{n}int_0^{2pi} f'(x)e^{-inx} , dxright| \ &leqslant frac{|f(2pi) - f(0)|}{2pi}+ frac{1}{2pi|n|}int_0^{2pi} |f'(x)| , dx \ &= frac{1}{2pi}left|int_0^{2pi}f'(x) , dx right|+ frac{1}{2pi|n|}int_0^{2pi} |f'(x)| , dx \ & leqslant frac{1}{2pi}int_0^{2pi}|f'(x)| , dx left(1 + |n|^{-1} right) end{align}$$
For $f in C^r$ use repeated integration by parts.
answered Jan 28 at 6:16
RRLRRL
53k42573
$endgroup$
Hint: In general $c_n = o(|n|^{-r})$ for $2pi$-periodic $f in C^r(mathbb{R})$ and getting the exact form for your bound requires a little manipulation.
For $f in C^1$ and $n neq 0$, integration by parts yields
$$begin{align}|c_n| &= frac{1}{2pi}left|ifrac{f(2pi)e^{-i2pi n} - f(0)}{n} - frac{i}{n}int_0^{2pi} f'(x)e^{-inx} , dxright| \ &leqslant frac{|f(2pi) - f(0)|}{2pi}+ frac{1}{2pi|n|}int_0^{2pi} |f'(x)| , dx \ &= frac{1}{2pi}left|int_0^{2pi}f'(x) , dx right|+ frac{1}{2pi|n|}int_0^{2pi} |f'(x)| , dx \ & leqslant frac{1}{2pi}int_0^{2pi}|f'(x)| , dx left(1 + |n|^{-1} right) end{align}$$
For $f in C^r$ use repeated integration by parts.
answered Jan 28 at 6:16
RRLRRL
53k42573
edited Jan 28 at 6:25
answered Jan 28 at 6:16
RRLRRL
53k42573
answered Jan 28 at 6:16
RRLRRL
53k42573
answered Jan 28 at 6:16
RRLRRL
53k42573
53k42573
$begingroup$
Thank you for the answer. Can you also explain the case when $n = 0$? I struggled to find the estimate since integration by part will not work.
$endgroup$
– Alvis Nordkovich
Feb 3 at 3:52
$begingroup$
@AlvisNordkovich: You're welcome. The constant I gave you for the $C^1$ case is $C = frac{1}{2pi}int_0^{2pi}|f'(x)| , dx$ and since $f'$ is bounded we can get a larger bound $C leqslant frac{1}{2pi}int_0^{2pi}|f'|_{infty} , dx = |f'|_{infty}$. Why worry about the case $n=0$ since the expression $(1 +|n|^{-1})$ is not even defined? Also $c_0 = frac{1}{2pi}int_0^{2pi} f(x) , dx$ is just a number.
$endgroup$
– RRL
Feb 3 at 6:02
$begingroup$
Maybe try to bound $c_0$ in terms of $|f'|_{infty}$ using $int_0^{2pi} f(x) , dx = int_0^{2pi} left( f(0) + int_0^x f'(t) , dt right) , dx$ to try to get some common constant. However, at this point I don't see the context of these extensions to the original question. If it gets too much more complicated you should ask another question.
$endgroup$
– RRL
Feb 3 at 6:11
add a comment |
$begingroup$
Thank you for the answer. Can you also explain the case when $n = 0$? I struggled to find the estimate since integration by part will not work.
$endgroup$
– Alvis Nordkovich
Feb 3 at 3:52
$begingroup$
@AlvisNordkovich: You're welcome. The constant I gave you for the $C^1$ case is $C = frac{1}{2pi}int_0^{2pi}|f'(x)| , dx$ and since $f'$ is bounded we can get a larger bound $C leqslant frac{1}{2pi}int_0^{2pi}|f'|_{infty} , dx = |f'|_{infty}$. Why worry about the case $n=0$ since the expression $(1 +|n|^{-1})$ is not even defined? Also $c_0 = frac{1}{2pi}int_0^{2pi} f(x) , dx$ is just a number.
$endgroup$
– RRL
Feb 3 at 6:02
$begingroup$
Maybe try to bound $c_0$ in terms of $|f'|_{infty}$ using $int_0^{2pi} f(x) , dx = int_0^{2pi} left( f(0) + int_0^x f'(t) , dt right) , dx$ to try to get some common constant. However, at this point I don't see the context of these extensions to the original question. If it gets too much more complicated you should ask another question.
$endgroup$
– RRL
Feb 3 at 6:11
$begingroup$
Thank you for the answer. Can you also explain the case when $n = 0$? I struggled to find the estimate since integration by part will not work.
$endgroup$
– Alvis Nordkovich
Feb 3 at 3:52
$begingroup$
Thank you for the answer. Can you also explain the case when $n = 0$? I struggled to find the estimate since integration by part will not work.
$endgroup$
– Alvis Nordkovich
Feb 3 at 3:52
$begingroup$
@AlvisNordkovich: You're welcome. The constant I gave you for the $C^1$ case is $C = frac{1}{2pi}int_0^{2pi}|f'(x)| , dx$ and since $f'$ is bounded we can get a larger bound $C leqslant frac{1}{2pi}int_0^{2pi}|f'|_{infty} , dx = |f'|_{infty}$. Why worry about the case $n=0$ since the expression $(1 +|n|^{-1})$ is not even defined? Also $c_0 = frac{1}{2pi}int_0^{2pi} f(x) , dx$ is just a number.
$endgroup$
– RRL
Feb 3 at 6:02
$begingroup$
@AlvisNordkovich: You're welcome. The constant I gave you for the $C^1$ case is $C = frac{1}{2pi}int_0^{2pi}|f'(x)| , dx$ and since $f'$ is bounded we can get a larger bound $C leqslant frac{1}{2pi}int_0^{2pi}|f'|_{infty} , dx = |f'|_{infty}$. Why worry about the case $n=0$ since the expression $(1 +|n|^{-1})$ is not even defined? Also $c_0 = frac{1}{2pi}int_0^{2pi} f(x) , dx$ is just a number.
$endgroup$
– RRL
Feb 3 at 6:02
$begingroup$
Maybe try to bound $c_0$ in terms of $|f'|_{infty}$ using $int_0^{2pi} f(x) , dx = int_0^{2pi} left( f(0) + int_0^x f'(t) , dt right) , dx$ to try to get some common constant. However, at this point I don't see the context of these extensions to the original question. If it gets too much more complicated you should ask another question.
$endgroup$
– RRL
Feb 3 at 6:11
$begingroup$
Maybe try to bound $c_0$ in terms of $|f'|_{infty}$ using $int_0^{2pi} f(x) , dx = int_0^{2pi} left( f(0) + int_0^x f'(t) , dt right) , dx$ to try to get some common constant. However, at this point I don't see the context of these extensions to the original question. If it gets too much more complicated you should ask another question.
$endgroup$
– RRL
Feb 3 at 6:11
add a comment |
$begingroup$
I am assuming that your function $f$ has period $2pi$. If $g=f^{(r)}$ then $c_n (hat {g}) =(in)^{r} c_n(hat {f})$. [This is proved by repeated integration by parts]. From this the inequality follow easily.
answered Jan 28 at 6:12
Kavi Rama MurthyKavi Rama Murthy
70.4k53170
$endgroup$
add a comment |
$begingroup$
I am assuming that your function $f$ has period $2pi$. If $g=f^{(r)}$ then $c_n (hat {g}) =(in)^{r} c_n(hat {f})$. [This is proved by repeated integration by parts]. From this the inequality follow easily.
answered Jan 28 at 6:12
Kavi Rama MurthyKavi Rama Murthy
70.4k53170
$endgroup$
add a comment |
$begingroup$
I am assuming that your function $f$ has period $2pi$. If $g=f^{(r)}$ then $c_n (hat {g}) =(in)^{r} c_n(hat {f})$. [This is proved by repeated integration by parts]. From this the inequality follow easily.
answered Jan 28 at 6:12
Kavi Rama MurthyKavi Rama Murthy
70.4k53170
$endgroup$
I am assuming that your function $f$ has period $2pi$. If $g=f^{(r)}$ then $c_n (hat {g}) =(in)^{r} c_n(hat {f})$. [This is proved by repeated integration by parts]. From this the inequality follow easily.
answered Jan 28 at 6:12
Kavi Rama MurthyKavi Rama Murthy
70.4k53170
answered Jan 28 at 6:12
Kavi Rama MurthyKavi Rama Murthy
70.4k53170
answered Jan 28 at 6:12
Kavi Rama MurthyKavi Rama Murthy
70.4k53170
answered Jan 28 at 6:12
Kavi Rama MurthyKavi Rama Murthy
70.4k53170
70.4k53170
add a comment |
add a comment |
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