Proving the uniform convergence of the average sequence of $f_n(x)=sin(nx)$
2
$begingroup$
I was asked to prove that:
1) $f_n(x)=sin(nx)$ does not converge pointwise.
2) The average sequence of $f_n(x)=sin(nx)$ is uniformly convergent.
I secceed to prove the first part but I cannot prove the other one. In addition it is not allowed to use the M-test.
Thanks. (I do not know how to use the function signs.)
sequences-and-series uniform-convergence
edited Dec 29 '18 at 12:50
Saad
20.1k92352
asked Dec 29 '18 at 12:37
DANIEL SHALAMDANIEL SHALAM
115
$endgroup$
add a comment |
2
$begingroup$
I was asked to prove that:
1) $f_n(x)=sin(nx)$ does not converge pointwise.
2) The average sequence of $f_n(x)=sin(nx)$ is uniformly convergent.
I secceed to prove the first part but I cannot prove the other one. In addition it is not allowed to use the M-test.
Thanks. (I do not know how to use the function signs.)
sequences-and-series uniform-convergence
edited Dec 29 '18 at 12:50
Saad
20.1k92352
asked Dec 29 '18 at 12:37
DANIEL SHALAMDANIEL SHALAM
115
$endgroup$
$begingroup$
I expect the only property of $sin(x)$ you will need is that it has an irrational period. The average will converge to the integral over the period.
$endgroup$
– SmileyCraft
Dec 29 '18 at 12:59
$begingroup$
I forgot to mention pi/3=>x>=2pi/3 . can you explain how it helps?
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:19
add a comment |
2
2
2
$begingroup$
I was asked to prove that:
1) $f_n(x)=sin(nx)$ does not converge pointwise.
2) The average sequence of $f_n(x)=sin(nx)$ is uniformly convergent.
I secceed to prove the first part but I cannot prove the other one. In addition it is not allowed to use the M-test.
Thanks. (I do not know how to use the function signs.)
sequences-and-series uniform-convergence
edited Dec 29 '18 at 12:50
Saad
20.1k92352
asked Dec 29 '18 at 12:37
DANIEL SHALAMDANIEL SHALAM
115
$endgroup$
I was asked to prove that:
1) $f_n(x)=sin(nx)$ does not converge pointwise.
2) The average sequence of $f_n(x)=sin(nx)$ is uniformly convergent.
I secceed to prove the first part but I cannot prove the other one. In addition it is not allowed to use the M-test.
Thanks. (I do not know how to use the function signs.)
sequences-and-series uniform-convergence
sequences-and-series uniform-convergence
edited Dec 29 '18 at 12:50
Saad
20.1k92352
asked Dec 29 '18 at 12:37
DANIEL SHALAMDANIEL SHALAM
115
edited Dec 29 '18 at 12:50
Saad
20.1k92352
asked Dec 29 '18 at 12:37
DANIEL SHALAMDANIEL SHALAM
115
edited Dec 29 '18 at 12:50
Saad
20.1k92352
edited Dec 29 '18 at 12:50
Saad
20.1k92352
edited Dec 29 '18 at 12:50
Saad
20.1k92352
20.1k92352
asked Dec 29 '18 at 12:37
DANIEL SHALAMDANIEL SHALAM
115
asked Dec 29 '18 at 12:37
DANIEL SHALAMDANIEL SHALAM
115
asked Dec 29 '18 at 12:37
DANIEL SHALAMDANIEL SHALAM
115
115
$begingroup$
I expect the only property of $sin(x)$ you will need is that it has an irrational period. The average will converge to the integral over the period.
$endgroup$
– SmileyCraft
Dec 29 '18 at 12:59
$begingroup$
I forgot to mention pi/3=>x>=2pi/3 . can you explain how it helps?
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:19
add a comment |
$begingroup$
I expect the only property of $sin(x)$ you will need is that it has an irrational period. The average will converge to the integral over the period.
$endgroup$
– SmileyCraft
Dec 29 '18 at 12:59
$begingroup$
I forgot to mention pi/3=>x>=2pi/3 . can you explain how it helps?
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:19
$begingroup$
I expect the only property of $sin(x)$ you will need is that it has an irrational period. The average will converge to the integral over the period.
$endgroup$
– SmileyCraft
Dec 29 '18 at 12:59
$begingroup$
I expect the only property of $sin(x)$ you will need is that it has an irrational period. The average will converge to the integral over the period.
$endgroup$
– SmileyCraft
Dec 29 '18 at 12:59
$begingroup$
I forgot to mention pi/3=>x>=2pi/3 . can you explain how it helps?
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:19
$begingroup$
I forgot to mention pi/3=>x>=2pi/3 . can you explain how it helps?
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:19
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
Hint:
$$frac1nsum_{k=0}^{n-1} e^{ikx}=frac{e^{inx}-1}{n(e^{ix}-1)}.$$
answered Dec 29 '18 at 12:57
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
I think you mean $sum_{i=0}^{n-1}e^{inx}$.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:00
$begingroup$
We do not use complex numbers at our solutions..
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:21
$begingroup$
You can use Euler's formula to get back $sin$ and $cos$. The solution Yves is hinting at is a really neat use of complex numbers.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:22
$begingroup$
I think you do need an edge case for $x=2kpi$ by the way.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:29
$begingroup$
I can not see how it helps yet... another hint can help.
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:32
|
show 5 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Hint:
$$frac1nsum_{k=0}^{n-1} e^{ikx}=frac{e^{inx}-1}{n(e^{ix}-1)}.$$
answered Dec 29 '18 at 12:57
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
I think you mean $sum_{i=0}^{n-1}e^{inx}$.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:00
$begingroup$
We do not use complex numbers at our solutions..
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:21
$begingroup$
You can use Euler's formula to get back $sin$ and $cos$. The solution Yves is hinting at is a really neat use of complex numbers.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:22
$begingroup$
I think you do need an edge case for $x=2kpi$ by the way.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:29
$begingroup$
I can not see how it helps yet... another hint can help.
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:32
|
show 5 more comments
1
$begingroup$
Hint:
$$frac1nsum_{k=0}^{n-1} e^{ikx}=frac{e^{inx}-1}{n(e^{ix}-1)}.$$
answered Dec 29 '18 at 12:57
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
I think you mean $sum_{i=0}^{n-1}e^{inx}$.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:00
$begingroup$
We do not use complex numbers at our solutions..
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:21
$begingroup$
You can use Euler's formula to get back $sin$ and $cos$. The solution Yves is hinting at is a really neat use of complex numbers.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:22
$begingroup$
I think you do need an edge case for $x=2kpi$ by the way.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:29
$begingroup$
I can not see how it helps yet... another hint can help.
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:32
|
show 5 more comments
1
1
1
$begingroup$
Hint:
$$frac1nsum_{k=0}^{n-1} e^{ikx}=frac{e^{inx}-1}{n(e^{ix}-1)}.$$
answered Dec 29 '18 at 12:57
Yves DaoustYves Daoust
131k676229
$endgroup$
Hint:
$$frac1nsum_{k=0}^{n-1} e^{ikx}=frac{e^{inx}-1}{n(e^{ix}-1)}.$$
answered Dec 29 '18 at 12:57
Yves DaoustYves Daoust
131k676229
edited Dec 29 '18 at 13:45
answered Dec 29 '18 at 12:57
Yves DaoustYves Daoust
131k676229
answered Dec 29 '18 at 12:57
Yves DaoustYves Daoust
131k676229
answered Dec 29 '18 at 12:57
Yves DaoustYves Daoust
131k676229
131k676229
$begingroup$
I think you mean $sum_{i=0}^{n-1}e^{inx}$.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:00
$begingroup$
We do not use complex numbers at our solutions..
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:21
$begingroup$
You can use Euler's formula to get back $sin$ and $cos$. The solution Yves is hinting at is a really neat use of complex numbers.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:22
$begingroup$
I think you do need an edge case for $x=2kpi$ by the way.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:29
$begingroup$
I can not see how it helps yet... another hint can help.
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:32
|
show 5 more comments
$begingroup$
I think you mean $sum_{i=0}^{n-1}e^{inx}$.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:00
$begingroup$
We do not use complex numbers at our solutions..
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:21
$begingroup$
You can use Euler's formula to get back $sin$ and $cos$. The solution Yves is hinting at is a really neat use of complex numbers.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:22
$begingroup$
I think you do need an edge case for $x=2kpi$ by the way.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:29
$begingroup$
I can not see how it helps yet... another hint can help.
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:32
$begingroup$
I think you mean $sum_{i=0}^{n-1}e^{inx}$.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:00
$begingroup$
I think you mean $sum_{i=0}^{n-1}e^{inx}$.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:00
$begingroup$
We do not use complex numbers at our solutions..
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:21
$begingroup$
We do not use complex numbers at our solutions..
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:21
$begingroup$
You can use Euler's formula to get back $sin$ and $cos$. The solution Yves is hinting at is a really neat use of complex numbers.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:22
$begingroup$
You can use Euler's formula to get back $sin$ and $cos$. The solution Yves is hinting at is a really neat use of complex numbers.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:22
$begingroup$
I think you do need an edge case for $x=2kpi$ by the way.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:29
$begingroup$
I think you do need an edge case for $x=2kpi$ by the way.
$endgroup$
– SmileyCraft
Dec 29 '18 at 13:29
$begingroup$
I can not see how it helps yet... another hint can help.
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:32
$begingroup$
I can not see how it helps yet... another hint can help.
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:32
|
show 5 more comments
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$begingroup$
I expect the only property of $sin(x)$ you will need is that it has an irrational period. The average will converge to the integral over the period.
$endgroup$
– SmileyCraft
Dec 29 '18 at 12:59
$begingroup$
I forgot to mention pi/3=>x>=2pi/3 . can you explain how it helps?
$endgroup$
– DANIEL SHALAM
Dec 29 '18 at 13:19