Uniform convergence of $sin(frac{x}{n}) , x>0$ [closed]
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I am having trouble in proving the uniform converge of $sin(frac{x}{n})$ for $x > 0$.
convergence
edited Nov 26 at 19:42
hamam_Abdallah
37.7k21634
asked Nov 26 at 18:33
Mikhail Terentyev
32
closed as unclear what you're asking by gammatester, Bungo, T. Bongers, Jean-Claude Arbaut, Shailesh Nov 27 at 0:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-3
down vote
favorite
I am having trouble in proving the uniform converge of $sin(frac{x}{n})$ for $x > 0$.
convergence
edited Nov 26 at 19:42
hamam_Abdallah
37.7k21634
asked Nov 26 at 18:33
Mikhail Terentyev
32
closed as unclear what you're asking by gammatester, Bungo, T. Bongers, Jean-Claude Arbaut, Shailesh Nov 27 at 0:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
@hamam_Abdallah I have to prove it
– Mikhail Terentyev
Nov 26 at 18:37
1
Title and text are very different.
– gammatester
Nov 26 at 18:41
Welcome to Math StackExchange. Please explain your attempt at this and where you are stuck so that others can give an answer relevant to you.
– Swapnil
Nov 26 at 18:46
@gammatester It's my fault. I corrected question.
– Mikhail Terentyev
Nov 26 at 18:51
1
Sorry, but do you have any clue what you should do? How can you accept an answer, that does not handle your problem? This is your fault, because you edited the question in a way that it is now completely different. I voted to close the question.
– gammatester
Nov 26 at 19:25
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show 3 more comments
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
I am having trouble in proving the uniform converge of $sin(frac{x}{n})$ for $x > 0$.
convergence
edited Nov 26 at 19:42
hamam_Abdallah
37.7k21634
asked Nov 26 at 18:33
Mikhail Terentyev
32
I am having trouble in proving the uniform converge of $sin(frac{x}{n})$ for $x > 0$.
convergence
convergence
edited Nov 26 at 19:42
hamam_Abdallah
37.7k21634
asked Nov 26 at 18:33
Mikhail Terentyev
32
edited Nov 26 at 19:42
hamam_Abdallah
37.7k21634
asked Nov 26 at 18:33
Mikhail Terentyev
32
edited Nov 26 at 19:42
hamam_Abdallah
37.7k21634
edited Nov 26 at 19:42
hamam_Abdallah
37.7k21634
edited Nov 26 at 19:42
hamam_Abdallah
37.7k21634
37.7k21634
asked Nov 26 at 18:33
Mikhail Terentyev
32
asked Nov 26 at 18:33
Mikhail Terentyev
32
asked Nov 26 at 18:33
Mikhail Terentyev
32
32
closed as unclear what you're asking by gammatester, Bungo, T. Bongers, Jean-Claude Arbaut, Shailesh Nov 27 at 0:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by gammatester, Bungo, T. Bongers, Jean-Claude Arbaut, Shailesh Nov 27 at 0:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
@hamam_Abdallah I have to prove it
– Mikhail Terentyev
Nov 26 at 18:37
1
Title and text are very different.
– gammatester
Nov 26 at 18:41
Welcome to Math StackExchange. Please explain your attempt at this and where you are stuck so that others can give an answer relevant to you.
– Swapnil
Nov 26 at 18:46
@gammatester It's my fault. I corrected question.
– Mikhail Terentyev
Nov 26 at 18:51
1
Sorry, but do you have any clue what you should do? How can you accept an answer, that does not handle your problem? This is your fault, because you edited the question in a way that it is now completely different. I voted to close the question.
– gammatester
Nov 26 at 19:25
|
show 3 more comments
@hamam_Abdallah I have to prove it
– Mikhail Terentyev
Nov 26 at 18:37
1
Title and text are very different.
– gammatester
Nov 26 at 18:41
Welcome to Math StackExchange. Please explain your attempt at this and where you are stuck so that others can give an answer relevant to you.
– Swapnil
Nov 26 at 18:46
@gammatester It's my fault. I corrected question.
– Mikhail Terentyev
Nov 26 at 18:51
1
Sorry, but do you have any clue what you should do? How can you accept an answer, that does not handle your problem? This is your fault, because you edited the question in a way that it is now completely different. I voted to close the question.
– gammatester
Nov 26 at 19:25
@hamam_Abdallah I have to prove it
– Mikhail Terentyev
Nov 26 at 18:37
@hamam_Abdallah I have to prove it
– Mikhail Terentyev
Nov 26 at 18:37
1
1
Title and text are very different.
– gammatester
Nov 26 at 18:41
Title and text are very different.
– gammatester
Nov 26 at 18:41
Welcome to Math StackExchange. Please explain your attempt at this and where you are stuck so that others can give an answer relevant to you.
– Swapnil
Nov 26 at 18:46
Welcome to Math StackExchange. Please explain your attempt at this and where you are stuck so that others can give an answer relevant to you.
– Swapnil
Nov 26 at 18:46
@gammatester It's my fault. I corrected question.
– Mikhail Terentyev
Nov 26 at 18:51
@gammatester It's my fault. I corrected question.
– Mikhail Terentyev
Nov 26 at 18:51
1
1
Sorry, but do you have any clue what you should do? How can you accept an answer, that does not handle your problem? This is your fault, because you edited the question in a way that it is now completely different. I voted to close the question.
– gammatester
Nov 26 at 19:25
Sorry, but do you have any clue what you should do? How can you accept an answer, that does not handle your problem? This is your fault, because you edited the question in a way that it is now completely different. I voted to close the question.
– gammatester
Nov 26 at 19:25
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
We know that for $X>0,$ we have
$$|sin(X)|le X$$
thus for $x>0$ and $n>0$
$$|sin(frac{x}{n})|le frac xn$$
but
$$lim_{nto+infty}frac xn=0$$
therefore
$$lim_{nto+infty}sin(frac xn)=0$$
the convergence is uniform at $[0,a]$ but not at $[0,+infty)$ if you take $x_n=nfrac pi 2$.
answered Nov 26 at 18:41
hamam_Abdallah
37.7k21634
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
We know that for $X>0,$ we have
$$|sin(X)|le X$$
thus for $x>0$ and $n>0$
$$|sin(frac{x}{n})|le frac xn$$
but
$$lim_{nto+infty}frac xn=0$$
therefore
$$lim_{nto+infty}sin(frac xn)=0$$
the convergence is uniform at $[0,a]$ but not at $[0,+infty)$ if you take $x_n=nfrac pi 2$.
answered Nov 26 at 18:41
hamam_Abdallah
37.7k21634
add a comment |
up vote
0
down vote
accepted
We know that for $X>0,$ we have
$$|sin(X)|le X$$
thus for $x>0$ and $n>0$
$$|sin(frac{x}{n})|le frac xn$$
but
$$lim_{nto+infty}frac xn=0$$
therefore
$$lim_{nto+infty}sin(frac xn)=0$$
the convergence is uniform at $[0,a]$ but not at $[0,+infty)$ if you take $x_n=nfrac pi 2$.
answered Nov 26 at 18:41
hamam_Abdallah
37.7k21634
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
We know that for $X>0,$ we have
$$|sin(X)|le X$$
thus for $x>0$ and $n>0$
$$|sin(frac{x}{n})|le frac xn$$
but
$$lim_{nto+infty}frac xn=0$$
therefore
$$lim_{nto+infty}sin(frac xn)=0$$
the convergence is uniform at $[0,a]$ but not at $[0,+infty)$ if you take $x_n=nfrac pi 2$.
answered Nov 26 at 18:41
hamam_Abdallah
37.7k21634
We know that for $X>0,$ we have
$$|sin(X)|le X$$
thus for $x>0$ and $n>0$
$$|sin(frac{x}{n})|le frac xn$$
but
$$lim_{nto+infty}frac xn=0$$
therefore
$$lim_{nto+infty}sin(frac xn)=0$$
the convergence is uniform at $[0,a]$ but not at $[0,+infty)$ if you take $x_n=nfrac pi 2$.
answered Nov 26 at 18:41
hamam_Abdallah
37.7k21634
edited Nov 26 at 19:36
answered Nov 26 at 18:41
hamam_Abdallah
37.7k21634
answered Nov 26 at 18:41
hamam_Abdallah
37.7k21634
answered Nov 26 at 18:41
hamam_Abdallah
37.7k21634
37.7k21634
add a comment |
add a comment |
@hamam_Abdallah I have to prove it
– Mikhail Terentyev
Nov 26 at 18:37
1
Title and text are very different.
– gammatester
Nov 26 at 18:41
Welcome to Math StackExchange. Please explain your attempt at this and where you are stuck so that others can give an answer relevant to you.
– Swapnil
Nov 26 at 18:46
@gammatester It's my fault. I corrected question.
– Mikhail Terentyev
Nov 26 at 18:51
1
Sorry, but do you have any clue what you should do? How can you accept an answer, that does not handle your problem? This is your fault, because you edited the question in a way that it is now completely different. I voted to close the question.
– gammatester
Nov 26 at 19:25