Finding where a series converges pointwise and uniformly
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Consider
$$sum_{n=0}^{infty} x(1 - x)^{2n} $$
Where does the sum converge pointwise and uniformly?
I think $[0, 2)$ pointwise and $(0, 2)$ uniformly because it becomes a geometric series. Also the function to which it converges to needs to be continuous
real-analysis sequences-and-series
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add a comment |
$begingroup$
Consider
$$sum_{n=0}^{infty} x(1 - x)^{2n} $$
Where does the sum converge pointwise and uniformly?
I think $[0, 2)$ pointwise and $(0, 2)$ uniformly because it becomes a geometric series. Also the function to which it converges to needs to be continuous
real-analysis sequences-and-series
$endgroup$
5
$begingroup$
What has the title to do with the question?
$endgroup$
– egreg
Dec 14 '18 at 21:08
add a comment |
$begingroup$
Consider
$$sum_{n=0}^{infty} x(1 - x)^{2n} $$
Where does the sum converge pointwise and uniformly?
I think $[0, 2)$ pointwise and $(0, 2)$ uniformly because it becomes a geometric series. Also the function to which it converges to needs to be continuous
real-analysis sequences-and-series
$endgroup$
Consider
$$sum_{n=0}^{infty} x(1 - x)^{2n} $$
Where does the sum converge pointwise and uniformly?
I think $[0, 2)$ pointwise and $(0, 2)$ uniformly because it becomes a geometric series. Also the function to which it converges to needs to be continuous
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Dec 27 '18 at 7:06
RRL
52.3k42573
52.3k42573
asked Dec 14 '18 at 21:05
josephjoseph
496111
496111
5
$begingroup$
What has the title to do with the question?
$endgroup$
– egreg
Dec 14 '18 at 21:08
add a comment |
5
$begingroup$
What has the title to do with the question?
$endgroup$
– egreg
Dec 14 '18 at 21:08
5
5
$begingroup$
What has the title to do with the question?
$endgroup$
– egreg
Dec 14 '18 at 21:08
$begingroup$
What has the title to do with the question?
$endgroup$
– egreg
Dec 14 '18 at 21:08
add a comment |
1 Answer
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$begingroup$
You are correct about recognizing the geometric series converging pointwise on $[0,2)$.
For this series to converge uniformly on $(0,2)$ it is necessary that
$$tag{*}sup_{x in (0,2)}left|sum_{n=N}^infty x(1-x)^{2n}right| xrightarrow[N to infty]{} 0$$
However,
$$left|sum_{n=N}^infty x(1-x)^{2n}right| = x(1-x)^{2N}sum_{n=0}^infty (1-x)^{2n} = frac{x(1-x)^{2N}}{1 - (1-x)^2} = frac{(1-x)^{2N}}{2 - x},$$
and
$$sup_{x in (0,2)}frac{(1-x)^{2N}}{2 - x} = +infty$$
So the convergence is not uniform on $(0,2)$. Noting that a problem arises due to behavior near $x = 2$ (and $x=0$), try on your own to determine and justify where convergence is uniform.
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$begingroup$
So it's uniform only at $0$?
$endgroup$
– joseph
Dec 14 '18 at 23:56
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@joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
$endgroup$
– RRL
Dec 15 '18 at 0:10
add a comment |
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1 Answer
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$begingroup$
You are correct about recognizing the geometric series converging pointwise on $[0,2)$.
For this series to converge uniformly on $(0,2)$ it is necessary that
$$tag{*}sup_{x in (0,2)}left|sum_{n=N}^infty x(1-x)^{2n}right| xrightarrow[N to infty]{} 0$$
However,
$$left|sum_{n=N}^infty x(1-x)^{2n}right| = x(1-x)^{2N}sum_{n=0}^infty (1-x)^{2n} = frac{x(1-x)^{2N}}{1 - (1-x)^2} = frac{(1-x)^{2N}}{2 - x},$$
and
$$sup_{x in (0,2)}frac{(1-x)^{2N}}{2 - x} = +infty$$
So the convergence is not uniform on $(0,2)$. Noting that a problem arises due to behavior near $x = 2$ (and $x=0$), try on your own to determine and justify where convergence is uniform.
$endgroup$
$begingroup$
So it's uniform only at $0$?
$endgroup$
– joseph
Dec 14 '18 at 23:56
$begingroup$
@joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
$endgroup$
– RRL
Dec 15 '18 at 0:10
add a comment |
$begingroup$
You are correct about recognizing the geometric series converging pointwise on $[0,2)$.
For this series to converge uniformly on $(0,2)$ it is necessary that
$$tag{*}sup_{x in (0,2)}left|sum_{n=N}^infty x(1-x)^{2n}right| xrightarrow[N to infty]{} 0$$
However,
$$left|sum_{n=N}^infty x(1-x)^{2n}right| = x(1-x)^{2N}sum_{n=0}^infty (1-x)^{2n} = frac{x(1-x)^{2N}}{1 - (1-x)^2} = frac{(1-x)^{2N}}{2 - x},$$
and
$$sup_{x in (0,2)}frac{(1-x)^{2N}}{2 - x} = +infty$$
So the convergence is not uniform on $(0,2)$. Noting that a problem arises due to behavior near $x = 2$ (and $x=0$), try on your own to determine and justify where convergence is uniform.
$endgroup$
$begingroup$
So it's uniform only at $0$?
$endgroup$
– joseph
Dec 14 '18 at 23:56
$begingroup$
@joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
$endgroup$
– RRL
Dec 15 '18 at 0:10
add a comment |
$begingroup$
You are correct about recognizing the geometric series converging pointwise on $[0,2)$.
For this series to converge uniformly on $(0,2)$ it is necessary that
$$tag{*}sup_{x in (0,2)}left|sum_{n=N}^infty x(1-x)^{2n}right| xrightarrow[N to infty]{} 0$$
However,
$$left|sum_{n=N}^infty x(1-x)^{2n}right| = x(1-x)^{2N}sum_{n=0}^infty (1-x)^{2n} = frac{x(1-x)^{2N}}{1 - (1-x)^2} = frac{(1-x)^{2N}}{2 - x},$$
and
$$sup_{x in (0,2)}frac{(1-x)^{2N}}{2 - x} = +infty$$
So the convergence is not uniform on $(0,2)$. Noting that a problem arises due to behavior near $x = 2$ (and $x=0$), try on your own to determine and justify where convergence is uniform.
$endgroup$
You are correct about recognizing the geometric series converging pointwise on $[0,2)$.
For this series to converge uniformly on $(0,2)$ it is necessary that
$$tag{*}sup_{x in (0,2)}left|sum_{n=N}^infty x(1-x)^{2n}right| xrightarrow[N to infty]{} 0$$
However,
$$left|sum_{n=N}^infty x(1-x)^{2n}right| = x(1-x)^{2N}sum_{n=0}^infty (1-x)^{2n} = frac{x(1-x)^{2N}}{1 - (1-x)^2} = frac{(1-x)^{2N}}{2 - x},$$
and
$$sup_{x in (0,2)}frac{(1-x)^{2N}}{2 - x} = +infty$$
So the convergence is not uniform on $(0,2)$. Noting that a problem arises due to behavior near $x = 2$ (and $x=0$), try on your own to determine and justify where convergence is uniform.
edited Dec 27 '18 at 6:32
answered Dec 14 '18 at 21:50
RRLRRL
52.3k42573
52.3k42573
$begingroup$
So it's uniform only at $0$?
$endgroup$
– joseph
Dec 14 '18 at 23:56
$begingroup$
@joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
$endgroup$
– RRL
Dec 15 '18 at 0:10
add a comment |
$begingroup$
So it's uniform only at $0$?
$endgroup$
– joseph
Dec 14 '18 at 23:56
$begingroup$
@joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
$endgroup$
– RRL
Dec 15 '18 at 0:10
$begingroup$
So it's uniform only at $0$?
$endgroup$
– joseph
Dec 14 '18 at 23:56
$begingroup$
So it's uniform only at $0$?
$endgroup$
– joseph
Dec 14 '18 at 23:56
$begingroup$
@joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
$endgroup$
– RRL
Dec 15 '18 at 0:10
$begingroup$
@joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
$endgroup$
– RRL
Dec 15 '18 at 0:10
add a comment |
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$begingroup$
What has the title to do with the question?
$endgroup$
– egreg
Dec 14 '18 at 21:08