Is there such a theorem about uniform convergence? The 2019 Stack Overflow Developer Survey Results Are InProve that $sumlimits_n=1(-1)^nfracx^2+nn^2$ uniformly convergent on $[a,b]$A problem for the uniform convergence of seriesSeries and uniform convergenceUniform convergence on singletonUniform convergence of a sequence of functions?Does $f_nrightarrow f$ uniformly suggest $sum f_nrightarrowsum f$ uniformly?Uniform convergence of $sqrtx^2+n^2^-1-n^-1$Examine uniform convergence of the series of functions with $f_n: [1,2] to mathbbR: x mapsto xe^-nx$prove uniform convergence?Check my proof of uniform convergenceShowing $sum x_nsin(nx)$ converges uniformly
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Is there such a theorem about uniform convergence?
The 2019 Stack Overflow Developer Survey Results Are InProve that $sumlimits_n=1(-1)^nfracx^2+nn^2$ uniformly convergent on $[a,b]$A problem for the uniform convergence of seriesSeries and uniform convergenceUniform convergence on singletonUniform convergence of a sequence of functions?Does $f_nrightarrow f$ uniformly suggest $sum f_nrightarrowsum f$ uniformly?Uniform convergence of $sqrtx^2+n^2^-1-n^-1$Examine uniform convergence of the series of functions with $f_n: [1,2] to mathbbR: x mapsto xe^-nx$prove uniform convergence?Check my proof of uniform convergenceShowing $sum x_nsin(nx)$ converges uniformly
$begingroup$
If we have $sum (-1)^nx_n$ and if $x_n>0$, and $lim_nrightarrow +infty x_n=0$ and $x_n$ is a decreasing sequence then
$left | sum_n=q+1^p (-1)^nx_nright |leq x_q+1$
And if $undersetxsup(x_q+1)rightarrow0 $ then the series converges uniformly
real-analysis sequences-and-series uniform-convergence
asked Mar 30 at 17:20
Pedro AlvarèsPedro Alvarès
946
$endgroup$
|
show 1 more comment
$begingroup$
If we have $sum (-1)^nx_n$ and if $x_n>0$, and $lim_nrightarrow +infty x_n=0$ and $x_n$ is a decreasing sequence then
$left | sum_n=q+1^p (-1)^nx_nright |leq x_q+1$
And if $undersetxsup(x_q+1)rightarrow0 $ then the series converges uniformly
real-analysis sequences-and-series uniform-convergence
asked Mar 30 at 17:20
Pedro AlvarèsPedro Alvarès
946
$endgroup$
1
$begingroup$
the first statement is true. The second makes no sense, what do you mean by supremum over all $x$'s?
$endgroup$
– Yanko
Mar 30 at 17:21
$begingroup$
To prove uniform convergence, I took supremum of both sides, if the R.H.S tends to zero ,then we're finshed.
$endgroup$
– Pedro Alvarès
Mar 30 at 17:52
$begingroup$
Maybe I'm wrong, it should be $x_q+1$ that tends to zero ?
$endgroup$
– Pedro Alvarès
Mar 30 at 17:53
$begingroup$
It is given that $x_n$ goes to zero.
$endgroup$
– Yanko
Mar 30 at 18:01
$begingroup$
Oh okay, but the thing I didn't understad is why $left | sum_n=q+1^p (-1)^nx_nright |leq x_q+1$
$endgroup$
– Pedro Alvarès
Mar 30 at 18:33
|
show 1 more comment
$begingroup$
If we have $sum (-1)^nx_n$ and if $x_n>0$, and $lim_nrightarrow +infty x_n=0$ and $x_n$ is a decreasing sequence then
$left | sum_n=q+1^p (-1)^nx_nright |leq x_q+1$
And if $undersetxsup(x_q+1)rightarrow0 $ then the series converges uniformly
real-analysis sequences-and-series uniform-convergence
asked Mar 30 at 17:20
Pedro AlvarèsPedro Alvarès
946
$endgroup$
If we have $sum (-1)^nx_n$ and if $x_n>0$, and $lim_nrightarrow +infty x_n=0$ and $x_n$ is a decreasing sequence then
$left | sum_n=q+1^p (-1)^nx_nright |leq x_q+1$
And if $undersetxsup(x_q+1)rightarrow0 $ then the series converges uniformly
real-analysis sequences-and-series uniform-convergence
real-analysis sequences-and-series uniform-convergence
asked Mar 30 at 17:20
Pedro AlvarèsPedro Alvarès
946
asked Mar 30 at 17:20
Pedro AlvarèsPedro Alvarès
946
asked Mar 30 at 17:20
Pedro AlvarèsPedro Alvarès
946
asked Mar 30 at 17:20
Pedro AlvarèsPedro Alvarès
946
asked Mar 30 at 17:20
Pedro AlvarèsPedro Alvarès
946
946
1
$begingroup$
the first statement is true. The second makes no sense, what do you mean by supremum over all $x$'s?
$endgroup$
– Yanko
Mar 30 at 17:21
$begingroup$
To prove uniform convergence, I took supremum of both sides, if the R.H.S tends to zero ,then we're finshed.
$endgroup$
– Pedro Alvarès
Mar 30 at 17:52
$begingroup$
Maybe I'm wrong, it should be $x_q+1$ that tends to zero ?
$endgroup$
– Pedro Alvarès
Mar 30 at 17:53
$begingroup$
It is given that $x_n$ goes to zero.
$endgroup$
– Yanko
Mar 30 at 18:01
$begingroup$
Oh okay, but the thing I didn't understad is why $left | sum_n=q+1^p (-1)^nx_nright |leq x_q+1$
$endgroup$
– Pedro Alvarès
Mar 30 at 18:33
|
show 1 more comment
1
$begingroup$
the first statement is true. The second makes no sense, what do you mean by supremum over all $x$'s?
$endgroup$
– Yanko
Mar 30 at 17:21
$begingroup$
To prove uniform convergence, I took supremum of both sides, if the R.H.S tends to zero ,then we're finshed.
$endgroup$
– Pedro Alvarès
Mar 30 at 17:52
$begingroup$
Maybe I'm wrong, it should be $x_q+1$ that tends to zero ?
$endgroup$
– Pedro Alvarès
Mar 30 at 17:53
$begingroup$
It is given that $x_n$ goes to zero.
$endgroup$
– Yanko
Mar 30 at 18:01
$begingroup$
Oh okay, but the thing I didn't understad is why $left | sum_n=q+1^p (-1)^nx_nright |leq x_q+1$
$endgroup$
– Pedro Alvarès
Mar 30 at 18:33
1
1
$begingroup$
the first statement is true. The second makes no sense, what do you mean by supremum over all $x$'s?
$endgroup$
– Yanko
Mar 30 at 17:21
$begingroup$
the first statement is true. The second makes no sense, what do you mean by supremum over all $x$'s?
$endgroup$
– Yanko
Mar 30 at 17:21
$begingroup$
To prove uniform convergence, I took supremum of both sides, if the R.H.S tends to zero ,then we're finshed.
$endgroup$
– Pedro Alvarès
Mar 30 at 17:52
$begingroup$
To prove uniform convergence, I took supremum of both sides, if the R.H.S tends to zero ,then we're finshed.
$endgroup$
– Pedro Alvarès
Mar 30 at 17:52
$begingroup$
Maybe I'm wrong, it should be $x_q+1$ that tends to zero ?
$endgroup$
– Pedro Alvarès
Mar 30 at 17:53
$begingroup$
Maybe I'm wrong, it should be $x_q+1$ that tends to zero ?
$endgroup$
– Pedro Alvarès
Mar 30 at 17:53
$begingroup$
It is given that $x_n$ goes to zero.
$endgroup$
– Yanko
Mar 30 at 18:01
$begingroup$
It is given that $x_n$ goes to zero.
$endgroup$
– Yanko
Mar 30 at 18:01
$begingroup$
Oh okay, but the thing I didn't understad is why $left | sum_n=q+1^p (-1)^nx_nright |leq x_q+1$
$endgroup$
– Pedro Alvarès
Mar 30 at 18:33
$begingroup$
Oh okay, but the thing I didn't understad is why $left | sum_n=q+1^p (-1)^nx_nright |leq x_q+1$
$endgroup$
– Pedro Alvarès
Mar 30 at 18:33
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Presumably $x_n$ is a function of $x$ in some domain $D$ as you are concerned with uniform convergence.
Since $x_n$ is a decreasing sequence, $x_n - x_n+1 > 0$ and
$$left | sum_n=q+1^p (-1)^nx_nright| = left |(-1)^q+1(x_q+1 - x_q+2 + x_q+3 - x_q+4 + ldots) right| \= x_q+1 - (, x_q+2 - x_q+3, ) - (, x_q+3 - x_q+4, ) - ldots leqslant x_q+1 leqslant sup_x in D x_q+1
$$
Thus, if the RHS tends to $0$ as $q to infty$ then uniform convergence follows as the uniform Cauchy criterion is satisfied.
answered Apr 1 at 1:51
RRLRRL
53.5k52574
$endgroup$
add a comment |
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$begingroup$
Presumably $x_n$ is a function of $x$ in some domain $D$ as you are concerned with uniform convergence.
Since $x_n$ is a decreasing sequence, $x_n - x_n+1 > 0$ and
$$left | sum_n=q+1^p (-1)^nx_nright| = left |(-1)^q+1(x_q+1 - x_q+2 + x_q+3 - x_q+4 + ldots) right| \= x_q+1 - (, x_q+2 - x_q+3, ) - (, x_q+3 - x_q+4, ) - ldots leqslant x_q+1 leqslant sup_x in D x_q+1
$$
Thus, if the RHS tends to $0$ as $q to infty$ then uniform convergence follows as the uniform Cauchy criterion is satisfied.
answered Apr 1 at 1:51
RRLRRL
53.5k52574
$endgroup$
add a comment |
$begingroup$
Presumably $x_n$ is a function of $x$ in some domain $D$ as you are concerned with uniform convergence.
Since $x_n$ is a decreasing sequence, $x_n - x_n+1 > 0$ and
$$left | sum_n=q+1^p (-1)^nx_nright| = left |(-1)^q+1(x_q+1 - x_q+2 + x_q+3 - x_q+4 + ldots) right| \= x_q+1 - (, x_q+2 - x_q+3, ) - (, x_q+3 - x_q+4, ) - ldots leqslant x_q+1 leqslant sup_x in D x_q+1
$$
Thus, if the RHS tends to $0$ as $q to infty$ then uniform convergence follows as the uniform Cauchy criterion is satisfied.
answered Apr 1 at 1:51
RRLRRL
53.5k52574
$endgroup$
add a comment |
$begingroup$
Presumably $x_n$ is a function of $x$ in some domain $D$ as you are concerned with uniform convergence.
Since $x_n$ is a decreasing sequence, $x_n - x_n+1 > 0$ and
$$left | sum_n=q+1^p (-1)^nx_nright| = left |(-1)^q+1(x_q+1 - x_q+2 + x_q+3 - x_q+4 + ldots) right| \= x_q+1 - (, x_q+2 - x_q+3, ) - (, x_q+3 - x_q+4, ) - ldots leqslant x_q+1 leqslant sup_x in D x_q+1
$$
Thus, if the RHS tends to $0$ as $q to infty$ then uniform convergence follows as the uniform Cauchy criterion is satisfied.
answered Apr 1 at 1:51
RRLRRL
53.5k52574
$endgroup$
Presumably $x_n$ is a function of $x$ in some domain $D$ as you are concerned with uniform convergence.
Since $x_n$ is a decreasing sequence, $x_n - x_n+1 > 0$ and
$$left | sum_n=q+1^p (-1)^nx_nright| = left |(-1)^q+1(x_q+1 - x_q+2 + x_q+3 - x_q+4 + ldots) right| \= x_q+1 - (, x_q+2 - x_q+3, ) - (, x_q+3 - x_q+4, ) - ldots leqslant x_q+1 leqslant sup_x in D x_q+1
$$
Thus, if the RHS tends to $0$ as $q to infty$ then uniform convergence follows as the uniform Cauchy criterion is satisfied.
answered Apr 1 at 1:51
RRLRRL
53.5k52574
edited Apr 1 at 1:57
answered Apr 1 at 1:51
RRLRRL
53.5k52574
answered Apr 1 at 1:51
RRLRRL
53.5k52574
answered Apr 1 at 1:51
RRLRRL
53.5k52574
53.5k52574
add a comment |
add a comment |
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1
$begingroup$
the first statement is true. The second makes no sense, what do you mean by supremum over all $x$'s?
$endgroup$
– Yanko
Mar 30 at 17:21
$begingroup$
To prove uniform convergence, I took supremum of both sides, if the R.H.S tends to zero ,then we're finshed.
$endgroup$
– Pedro Alvarès
Mar 30 at 17:52
$begingroup$
Maybe I'm wrong, it should be $x_q+1$ that tends to zero ?
$endgroup$
– Pedro Alvarès
Mar 30 at 17:53
$begingroup$
It is given that $x_n$ goes to zero.
$endgroup$
– Yanko
Mar 30 at 18:01
$begingroup$
Oh okay, but the thing I didn't understad is why $left | sum_n=q+1^p (-1)^nx_nright |leq x_q+1$
$endgroup$
– Pedro Alvarès
Mar 30 at 18:33