How to prove that $left|sum_{k=0}^{n-1}sin(2k+1)xright|$ is bounded
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I'm currently studying Fourier series using a textbook which unfortunately provides only partial solutions to its explanations. To show the uniform convergence of a Fourier series I need to prove the assumption from the title above. Any help is very much appreciated.
calculus functional-analysis fourier-analysis
edited Jan 2 at 12:33
mechanodroid
28.9k62748
asked Jan 2 at 11:25
GuitarmanGuitarman
152
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add a comment |
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I'm currently studying Fourier series using a textbook which unfortunately provides only partial solutions to its explanations. To show the uniform convergence of a Fourier series I need to prove the assumption from the title above. Any help is very much appreciated.
calculus functional-analysis fourier-analysis
edited Jan 2 at 12:33
mechanodroid
28.9k62748
asked Jan 2 at 11:25
GuitarmanGuitarman
152
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2
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$|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
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– N74
Jan 2 at 11:50
add a comment |
$begingroup$
I'm currently studying Fourier series using a textbook which unfortunately provides only partial solutions to its explanations. To show the uniform convergence of a Fourier series I need to prove the assumption from the title above. Any help is very much appreciated.
calculus functional-analysis fourier-analysis
edited Jan 2 at 12:33
mechanodroid
28.9k62748
asked Jan 2 at 11:25
GuitarmanGuitarman
152
$endgroup$
I'm currently studying Fourier series using a textbook which unfortunately provides only partial solutions to its explanations. To show the uniform convergence of a Fourier series I need to prove the assumption from the title above. Any help is very much appreciated.
calculus functional-analysis fourier-analysis
calculus functional-analysis fourier-analysis
edited Jan 2 at 12:33
mechanodroid
28.9k62748
asked Jan 2 at 11:25
GuitarmanGuitarman
152
edited Jan 2 at 12:33
mechanodroid
28.9k62748
asked Jan 2 at 11:25
GuitarmanGuitarman
152
edited Jan 2 at 12:33
mechanodroid
28.9k62748
edited Jan 2 at 12:33
mechanodroid
28.9k62748
edited Jan 2 at 12:33
mechanodroid
28.9k62748
28.9k62748
asked Jan 2 at 11:25
GuitarmanGuitarman
152
asked Jan 2 at 11:25
GuitarmanGuitarman
152
asked Jan 2 at 11:25
GuitarmanGuitarman
152
152
2
$begingroup$
$|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
$endgroup$
– N74
Jan 2 at 11:50
add a comment |
2
$begingroup$
$|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
$endgroup$
– N74
Jan 2 at 11:50
2
2
$begingroup$
$|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
$endgroup$
– N74
Jan 2 at 11:50
$begingroup$
$|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
$endgroup$
– N74
Jan 2 at 11:50
add a comment |
1 Answer
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$begingroup$
You can calculate this sum explicitly:
begin{align}
sum_{k=0}^{n-1}sin(2k+1)x &= operatorname{Im}left(sum_{k=0}^{n-1}e^{(2k+1)ix}right)\
&= operatorname{Im}left(e^{ix}sum_{k=0}^{n-1}left(e^{2ix}right)^{k}right)\
&= operatorname{Im}left(e^{ix}frac{e^{2nix}-1}{e^{2ix}-1}right)\
&= operatorname{Im}left(frac{e^{2nix}-1}{e^{ix}-e^{-ix}}right)\
&= operatorname{Im}left(frac{cos(2nix)-1+isin(2pi ix)}{2isin x}right)\
&= frac{sin(2nx)}{2sin x}
end{align}
Now it should be clear that it is bounded.
answered Jan 2 at 12:33
mechanodroidmechanodroid
28.9k62748
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
You can calculate this sum explicitly:
begin{align}
sum_{k=0}^{n-1}sin(2k+1)x &= operatorname{Im}left(sum_{k=0}^{n-1}e^{(2k+1)ix}right)\
&= operatorname{Im}left(e^{ix}sum_{k=0}^{n-1}left(e^{2ix}right)^{k}right)\
&= operatorname{Im}left(e^{ix}frac{e^{2nix}-1}{e^{2ix}-1}right)\
&= operatorname{Im}left(frac{e^{2nix}-1}{e^{ix}-e^{-ix}}right)\
&= operatorname{Im}left(frac{cos(2nix)-1+isin(2pi ix)}{2isin x}right)\
&= frac{sin(2nx)}{2sin x}
end{align}
Now it should be clear that it is bounded.
answered Jan 2 at 12:33
mechanodroidmechanodroid
28.9k62748
$endgroup$
add a comment |
$begingroup$
You can calculate this sum explicitly:
begin{align}
sum_{k=0}^{n-1}sin(2k+1)x &= operatorname{Im}left(sum_{k=0}^{n-1}e^{(2k+1)ix}right)\
&= operatorname{Im}left(e^{ix}sum_{k=0}^{n-1}left(e^{2ix}right)^{k}right)\
&= operatorname{Im}left(e^{ix}frac{e^{2nix}-1}{e^{2ix}-1}right)\
&= operatorname{Im}left(frac{e^{2nix}-1}{e^{ix}-e^{-ix}}right)\
&= operatorname{Im}left(frac{cos(2nix)-1+isin(2pi ix)}{2isin x}right)\
&= frac{sin(2nx)}{2sin x}
end{align}
Now it should be clear that it is bounded.
answered Jan 2 at 12:33
mechanodroidmechanodroid
28.9k62748
$endgroup$
add a comment |
$begingroup$
You can calculate this sum explicitly:
begin{align}
sum_{k=0}^{n-1}sin(2k+1)x &= operatorname{Im}left(sum_{k=0}^{n-1}e^{(2k+1)ix}right)\
&= operatorname{Im}left(e^{ix}sum_{k=0}^{n-1}left(e^{2ix}right)^{k}right)\
&= operatorname{Im}left(e^{ix}frac{e^{2nix}-1}{e^{2ix}-1}right)\
&= operatorname{Im}left(frac{e^{2nix}-1}{e^{ix}-e^{-ix}}right)\
&= operatorname{Im}left(frac{cos(2nix)-1+isin(2pi ix)}{2isin x}right)\
&= frac{sin(2nx)}{2sin x}
end{align}
Now it should be clear that it is bounded.
answered Jan 2 at 12:33
mechanodroidmechanodroid
28.9k62748
$endgroup$
You can calculate this sum explicitly:
begin{align}
sum_{k=0}^{n-1}sin(2k+1)x &= operatorname{Im}left(sum_{k=0}^{n-1}e^{(2k+1)ix}right)\
&= operatorname{Im}left(e^{ix}sum_{k=0}^{n-1}left(e^{2ix}right)^{k}right)\
&= operatorname{Im}left(e^{ix}frac{e^{2nix}-1}{e^{2ix}-1}right)\
&= operatorname{Im}left(frac{e^{2nix}-1}{e^{ix}-e^{-ix}}right)\
&= operatorname{Im}left(frac{cos(2nix)-1+isin(2pi ix)}{2isin x}right)\
&= frac{sin(2nx)}{2sin x}
end{align}
Now it should be clear that it is bounded.
answered Jan 2 at 12:33
mechanodroidmechanodroid
28.9k62748
answered Jan 2 at 12:33
mechanodroidmechanodroid
28.9k62748
answered Jan 2 at 12:33
mechanodroidmechanodroid
28.9k62748
answered Jan 2 at 12:33
mechanodroidmechanodroid
28.9k62748
28.9k62748
add a comment |
add a comment |
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$begingroup$
$|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
$endgroup$
– N74
Jan 2 at 11:50