Mixing
converge pointwise but not uniformly
How can I prove that
$$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}}$$
converges pointwise on $[-pi, pi]$ but not uniformly?
For the pointwise part, I tried to prove it by comparison, using
$$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}} leq sum_{n=1}^infty frac{nx}{sqrt{n}} = sum_{n=1}^infty xsqrt{n},$$
which does not converge.
I also tried
$$ sum_{n=1}^infty left| frac{sin(nx)}{sqrt{n}}right| leq sum_{n=1}^infty left| frac{1}{sqrt{n}}right|,$$
which does not converge either.
sequences-and-series convergence uniform-convergence pointwise-convergence
edited Nov 21 '18 at 21:40
rafa11111
1,116417
asked Nov 21 '18 at 21:04
kit katkit kat
111
add a comment |
How can I prove that
$$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}}$$
converges pointwise on $[-pi, pi]$ but not uniformly?
For the pointwise part, I tried to prove it by comparison, using
$$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}} leq sum_{n=1}^infty frac{nx}{sqrt{n}} = sum_{n=1}^infty xsqrt{n},$$
which does not converge.
I also tried
$$ sum_{n=1}^infty left| frac{sin(nx)}{sqrt{n}}right| leq sum_{n=1}^infty left| frac{1}{sqrt{n}}right|,$$
which does not converge either.
sequences-and-series convergence uniform-convergence pointwise-convergence
edited Nov 21 '18 at 21:40
rafa11111
1,116417
asked Nov 21 '18 at 21:04
kit katkit kat
111
add a comment |
How can I prove that
$$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}}$$
converges pointwise on $[-pi, pi]$ but not uniformly?
For the pointwise part, I tried to prove it by comparison, using
$$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}} leq sum_{n=1}^infty frac{nx}{sqrt{n}} = sum_{n=1}^infty xsqrt{n},$$
which does not converge.
I also tried
$$ sum_{n=1}^infty left| frac{sin(nx)}{sqrt{n}}right| leq sum_{n=1}^infty left| frac{1}{sqrt{n}}right|,$$
which does not converge either.
sequences-and-series convergence uniform-convergence pointwise-convergence
edited Nov 21 '18 at 21:40
rafa11111
1,116417
asked Nov 21 '18 at 21:04
kit katkit kat
111
How can I prove that
$$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}}$$
converges pointwise on $[-pi, pi]$ but not uniformly?
For the pointwise part, I tried to prove it by comparison, using
$$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}} leq sum_{n=1}^infty frac{nx}{sqrt{n}} = sum_{n=1}^infty xsqrt{n},$$
which does not converge.
I also tried
$$ sum_{n=1}^infty left| frac{sin(nx)}{sqrt{n}}right| leq sum_{n=1}^infty left| frac{1}{sqrt{n}}right|,$$
which does not converge either.
sequences-and-series convergence uniform-convergence pointwise-convergence
sequences-and-series convergence uniform-convergence pointwise-convergence
edited Nov 21 '18 at 21:40
rafa11111
1,116417
asked Nov 21 '18 at 21:04
kit katkit kat
111
edited Nov 21 '18 at 21:40
rafa11111
1,116417
asked Nov 21 '18 at 21:04
kit katkit kat
111
edited Nov 21 '18 at 21:40
rafa11111
1,116417
edited Nov 21 '18 at 21:40
rafa11111
1,116417
edited Nov 21 '18 at 21:40
rafa11111
1,116417
1,116417
asked Nov 21 '18 at 21:04
kit katkit kat
111
asked Nov 21 '18 at 21:04
kit katkit kat
111
asked Nov 21 '18 at 21:04
kit katkit kat
111
111
add a comment |
add a comment |
2 Answers
2
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oldest
votes
For the pointwise convergence, use Dirichlet's test.
To see that it's not uniform, note that $sin(nx) > 2 n x/pi$ if $0 < n x < pi/2$. Take $x = 1/N$ and consider the $N$'th partial sum
$$ eqalign{sum_{n=1}^N frac{sin(n/N)}{
sqrt{n}} &ge frac{2}{pi N} sum_{n=1}^N sqrt{n} cr&ge frac{2}{pi N} int_0^N
sqrt{t}; dt = frac{4sqrt{N}}{3pi}}$$
answered Nov 21 '18 at 21:20
Robert IsraelRobert Israel
319k23208457
add a comment |
Pointwise convergence is granted by Dirichlet's test, but a series of continuous functions cannot converge uniformly to an unbounded function. We may compute $lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}$ through a convolution with an approximate identity:
$$begin{eqnarray*}lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}int_{0}^{+infty}msin(nx) e^{-mx},dx\&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}cdotfrac{mn}{m^2+n^2}.end{eqnarray*}$$
The RHS is $+infty$ since by Riemann sums
$$ lim_{mto +infty}sum_{ngeq 1}frac{sqrt{mn}}{m^2+n^2}=int_{0}^{+infty}frac{sqrt{x}}{1+x^2},dx = frac{pi}{sqrt{2}}.$$
Much simpler, once $f(x)$ is defined as $sum_{ngeq 1}frac{sin(nx)}{sqrt{n}}$ we have
$$ int_{-pi}^{pi}f(x)^2,dx = pisum_{ngeq 1}frac{1}{n} = +infty $$
by Parseval's theorem, hence $f(x)$ cannot be bounded on $[-pi,pi]$.
answered Nov 21 '18 at 21:34
Jack D'AurizioJack D'Aurizio
287k33280658
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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votes
For the pointwise convergence, use Dirichlet's test.
To see that it's not uniform, note that $sin(nx) > 2 n x/pi$ if $0 < n x < pi/2$. Take $x = 1/N$ and consider the $N$'th partial sum
$$ eqalign{sum_{n=1}^N frac{sin(n/N)}{
sqrt{n}} &ge frac{2}{pi N} sum_{n=1}^N sqrt{n} cr&ge frac{2}{pi N} int_0^N
sqrt{t}; dt = frac{4sqrt{N}}{3pi}}$$
answered Nov 21 '18 at 21:20
Robert IsraelRobert Israel
319k23208457
add a comment |
For the pointwise convergence, use Dirichlet's test.
To see that it's not uniform, note that $sin(nx) > 2 n x/pi$ if $0 < n x < pi/2$. Take $x = 1/N$ and consider the $N$'th partial sum
$$ eqalign{sum_{n=1}^N frac{sin(n/N)}{
sqrt{n}} &ge frac{2}{pi N} sum_{n=1}^N sqrt{n} cr&ge frac{2}{pi N} int_0^N
sqrt{t}; dt = frac{4sqrt{N}}{3pi}}$$
answered Nov 21 '18 at 21:20
Robert IsraelRobert Israel
319k23208457
add a comment |
For the pointwise convergence, use Dirichlet's test.
To see that it's not uniform, note that $sin(nx) > 2 n x/pi$ if $0 < n x < pi/2$. Take $x = 1/N$ and consider the $N$'th partial sum
$$ eqalign{sum_{n=1}^N frac{sin(n/N)}{
sqrt{n}} &ge frac{2}{pi N} sum_{n=1}^N sqrt{n} cr&ge frac{2}{pi N} int_0^N
sqrt{t}; dt = frac{4sqrt{N}}{3pi}}$$
answered Nov 21 '18 at 21:20
Robert IsraelRobert Israel
319k23208457
For the pointwise convergence, use Dirichlet's test.
To see that it's not uniform, note that $sin(nx) > 2 n x/pi$ if $0 < n x < pi/2$. Take $x = 1/N$ and consider the $N$'th partial sum
$$ eqalign{sum_{n=1}^N frac{sin(n/N)}{
sqrt{n}} &ge frac{2}{pi N} sum_{n=1}^N sqrt{n} cr&ge frac{2}{pi N} int_0^N
sqrt{t}; dt = frac{4sqrt{N}}{3pi}}$$
answered Nov 21 '18 at 21:20
Robert IsraelRobert Israel
319k23208457
answered Nov 21 '18 at 21:20
Robert IsraelRobert Israel
319k23208457
answered Nov 21 '18 at 21:20
Robert IsraelRobert Israel
319k23208457
answered Nov 21 '18 at 21:20
Robert IsraelRobert Israel
319k23208457
319k23208457
add a comment |
add a comment |
Pointwise convergence is granted by Dirichlet's test, but a series of continuous functions cannot converge uniformly to an unbounded function. We may compute $lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}$ through a convolution with an approximate identity:
$$begin{eqnarray*}lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}int_{0}^{+infty}msin(nx) e^{-mx},dx\&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}cdotfrac{mn}{m^2+n^2}.end{eqnarray*}$$
The RHS is $+infty$ since by Riemann sums
$$ lim_{mto +infty}sum_{ngeq 1}frac{sqrt{mn}}{m^2+n^2}=int_{0}^{+infty}frac{sqrt{x}}{1+x^2},dx = frac{pi}{sqrt{2}}.$$
Much simpler, once $f(x)$ is defined as $sum_{ngeq 1}frac{sin(nx)}{sqrt{n}}$ we have
$$ int_{-pi}^{pi}f(x)^2,dx = pisum_{ngeq 1}frac{1}{n} = +infty $$
by Parseval's theorem, hence $f(x)$ cannot be bounded on $[-pi,pi]$.
answered Nov 21 '18 at 21:34
Jack D'AurizioJack D'Aurizio
287k33280658
add a comment |
Pointwise convergence is granted by Dirichlet's test, but a series of continuous functions cannot converge uniformly to an unbounded function. We may compute $lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}$ through a convolution with an approximate identity:
$$begin{eqnarray*}lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}int_{0}^{+infty}msin(nx) e^{-mx},dx\&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}cdotfrac{mn}{m^2+n^2}.end{eqnarray*}$$
The RHS is $+infty$ since by Riemann sums
$$ lim_{mto +infty}sum_{ngeq 1}frac{sqrt{mn}}{m^2+n^2}=int_{0}^{+infty}frac{sqrt{x}}{1+x^2},dx = frac{pi}{sqrt{2}}.$$
Much simpler, once $f(x)$ is defined as $sum_{ngeq 1}frac{sin(nx)}{sqrt{n}}$ we have
$$ int_{-pi}^{pi}f(x)^2,dx = pisum_{ngeq 1}frac{1}{n} = +infty $$
by Parseval's theorem, hence $f(x)$ cannot be bounded on $[-pi,pi]$.
answered Nov 21 '18 at 21:34
Jack D'AurizioJack D'Aurizio
287k33280658
add a comment |
Pointwise convergence is granted by Dirichlet's test, but a series of continuous functions cannot converge uniformly to an unbounded function. We may compute $lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}$ through a convolution with an approximate identity:
$$begin{eqnarray*}lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}int_{0}^{+infty}msin(nx) e^{-mx},dx\&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}cdotfrac{mn}{m^2+n^2}.end{eqnarray*}$$
The RHS is $+infty$ since by Riemann sums
$$ lim_{mto +infty}sum_{ngeq 1}frac{sqrt{mn}}{m^2+n^2}=int_{0}^{+infty}frac{sqrt{x}}{1+x^2},dx = frac{pi}{sqrt{2}}.$$
Much simpler, once $f(x)$ is defined as $sum_{ngeq 1}frac{sin(nx)}{sqrt{n}}$ we have
$$ int_{-pi}^{pi}f(x)^2,dx = pisum_{ngeq 1}frac{1}{n} = +infty $$
by Parseval's theorem, hence $f(x)$ cannot be bounded on $[-pi,pi]$.
answered Nov 21 '18 at 21:34
Jack D'AurizioJack D'Aurizio
287k33280658
Pointwise convergence is granted by Dirichlet's test, but a series of continuous functions cannot converge uniformly to an unbounded function. We may compute $lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}$ through a convolution with an approximate identity:
$$begin{eqnarray*}lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}int_{0}^{+infty}msin(nx) e^{-mx},dx\&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}cdotfrac{mn}{m^2+n^2}.end{eqnarray*}$$
The RHS is $+infty$ since by Riemann sums
$$ lim_{mto +infty}sum_{ngeq 1}frac{sqrt{mn}}{m^2+n^2}=int_{0}^{+infty}frac{sqrt{x}}{1+x^2},dx = frac{pi}{sqrt{2}}.$$
Much simpler, once $f(x)$ is defined as $sum_{ngeq 1}frac{sin(nx)}{sqrt{n}}$ we have
$$ int_{-pi}^{pi}f(x)^2,dx = pisum_{ngeq 1}frac{1}{n} = +infty $$
by Parseval's theorem, hence $f(x)$ cannot be bounded on $[-pi,pi]$.
answered Nov 21 '18 at 21:34
Jack D'AurizioJack D'Aurizio
287k33280658
edited Nov 21 '18 at 21:49
answered Nov 21 '18 at 21:34
Jack D'AurizioJack D'Aurizio
287k33280658
answered Nov 21 '18 at 21:34
Jack D'AurizioJack D'Aurizio
287k33280658
answered Nov 21 '18 at 21:34
Jack D'AurizioJack D'Aurizio
287k33280658
287k33280658
add a comment |
add a comment |
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