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







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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
















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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 12:50









Saad

20.6k92452




20.6k92452










asked Dec 29 '18 at 12:37









DANIEL SHALAMDANIEL SHALAM

95




95












  • $begingroup$
    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 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










1 Answer
1






active

oldest

votes


















1












$begingroup$

Hint:



$$frac1nsum_{k=0}^{n-1} e^{ikx}=frac{e^{inx}-1}{n(e^{ix}-1)}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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












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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:



$$frac1nsum_{k=0}^{n-1} e^{ikx}=frac{e^{inx}-1}{n(e^{ix}-1)}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

Hint:



$$frac1nsum_{k=0}^{n-1} e^{ikx}=frac{e^{inx}-1}{n(e^{ix}-1)}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

Hint:



$$frac1nsum_{k=0}^{n-1} e^{ikx}=frac{e^{inx}-1}{n(e^{ix}-1)}.$$






share|cite|improve this answer











$endgroup$



Hint:



$$frac1nsum_{k=0}^{n-1} e^{ikx}=frac{e^{inx}-1}{n(e^{ix}-1)}.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 13:45

























answered Dec 29 '18 at 12:57









Yves DaoustYves Daoust

133k676232




133k676232












  • $begingroup$
    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










  • $begingroup$
    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


















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