Mixing
Show uniform convergence of $f_n (x)=sqrt {sin left(frac {x}{n}right)+cosleft(frac {x}{n}right) }$
$begingroup$
How can I prove uniform convergence of
$$
f_n (x)
=sqrt {sin left(frac {x}{n}right)+cosleft(frac {x}{n}right)}
$$
for $x in left[0,frac{pi}{2}right]$
I have been able to show its pointwise limit is $1$.
real-analysis analysis convergence uniform-convergence
edited Jan 26 at 14:58
Viktor Glombik
1,2032528
$endgroup$
add a comment |
$begingroup$
How can I prove uniform convergence of
$$
f_n (x)
=sqrt {sin left(frac {x}{n}right)+cosleft(frac {x}{n}right)}
$$
for $x in left[0,frac{pi}{2}right]$
I have been able to show its pointwise limit is $1$.
real-analysis analysis convergence uniform-convergence
edited Jan 26 at 14:58
Viktor Glombik
1,2032528
$endgroup$
$begingroup$
Have have you tried? How did you show the pointwise limit?
$endgroup$
– Viktor Glombik
Jan 26 at 14:16
$begingroup$
@ViktorGlombik, for any fixed $x $, by the continuity of the square root, sine and cosine functions, as n approaches infinity, we have $sqrt (1)=1$
$endgroup$
– user397197
Jan 26 at 14:19
add a comment |
$begingroup$
How can I prove uniform convergence of
$$
f_n (x)
=sqrt {sin left(frac {x}{n}right)+cosleft(frac {x}{n}right)}
$$
for $x in left[0,frac{pi}{2}right]$
I have been able to show its pointwise limit is $1$.
real-analysis analysis convergence uniform-convergence
edited Jan 26 at 14:58
Viktor Glombik
1,2032528
$endgroup$
How can I prove uniform convergence of
$$
f_n (x)
=sqrt {sin left(frac {x}{n}right)+cosleft(frac {x}{n}right)}
$$
for $x in left[0,frac{pi}{2}right]$
I have been able to show its pointwise limit is $1$.
real-analysis analysis convergence uniform-convergence
real-analysis analysis convergence uniform-convergence
edited Jan 26 at 14:58
Viktor Glombik
1,2032528
edited Jan 26 at 14:58
Viktor Glombik
1,2032528
edited Jan 26 at 14:58
Viktor Glombik
1,2032528
edited Jan 26 at 14:58
Viktor Glombik
1,2032528
edited Jan 26 at 14:58
Viktor Glombik
1,2032528
1,2032528
asked Jan 26 at 14:10
user397197
$begingroup$
Have have you tried? How did you show the pointwise limit?
$endgroup$
– Viktor Glombik
Jan 26 at 14:16
$begingroup$
@ViktorGlombik, for any fixed $x $, by the continuity of the square root, sine and cosine functions, as n approaches infinity, we have $sqrt (1)=1$
$endgroup$
– user397197
Jan 26 at 14:19
add a comment |
$begingroup$
Have have you tried? How did you show the pointwise limit?
$endgroup$
– Viktor Glombik
Jan 26 at 14:16
$begingroup$
@ViktorGlombik, for any fixed $x $, by the continuity of the square root, sine and cosine functions, as n approaches infinity, we have $sqrt (1)=1$
$endgroup$
– user397197
Jan 26 at 14:19
$begingroup$
Have have you tried? How did you show the pointwise limit?
$endgroup$
– Viktor Glombik
Jan 26 at 14:16
$begingroup$
Have have you tried? How did you show the pointwise limit?
$endgroup$
– Viktor Glombik
Jan 26 at 14:16
$begingroup$
@ViktorGlombik, for any fixed $x $, by the continuity of the square root, sine and cosine functions, as n approaches infinity, we have $sqrt (1)=1$
$endgroup$
– user397197
Jan 26 at 14:19
$begingroup$
@ViktorGlombik, for any fixed $x $, by the continuity of the square root, sine and cosine functions, as n approaches infinity, we have $sqrt (1)=1$
$endgroup$
– user397197
Jan 26 at 14:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Define
$$f(x)= sqrt{sin(x) + cos(x)}$$
This function is continuous and thus uniformly continuous on $[0; frac{pi}{2}]$.
Added: We have that for every $xin [0; frac{pi}{2}]$ that
$$ leftvert frac{x}{n} - frac{x}{m} rightvert
= leftvert frac{1}{n} - frac{1}{m} rightvert cdot vert x vert
leq frac{pi}{2} leftvert frac{1}{n} - frac{1}{m} rightvert $$
Furthermore, we have
$$ vert f_n(x) - f_m(x) vert = vert f(frac{x}{n}) - f(frac{x}{m}) vert. $$
Of course, we don't need the uniform continuity (as shown in the comments by p4sch), but I consider it usefull to think this way. It allows one to also tackle different problems (like for example the uniform convergence of $g_n(x)=sqrt{x^2+frac{1}{n}}$ to the absolute value on the whole real line).
answered Jan 26 at 14:18
Severin SchravenSeverin Schraven
6,4501935
$endgroup$
$begingroup$
I'm not sure how this should help to show that $f_n$ congerves uniformly. I think of $f_n(x)=x^n$, which are uniformly continuous on $[0,1]$ as well while it doesn't converge uniformly on $[0,1]$.
$endgroup$
– Mundron Schmidt
Jan 26 at 14:21
$begingroup$
I know this but how does how does uniform continuity and uniform convergence relate in this case. Is there any theoren I am missing?
$endgroup$
– user397197
Jan 26 at 14:24
1
$begingroup$
You have $f_n(x) = f(x/n)$. Since $|f(0) -f(x)|< varepsilon$ for all $|x|< delta$, you get for all $1/n < delta$ and $x in [0,pi/2]$ that $|f_n(x)| < varepsilon$.
$endgroup$
– p4sch
Jan 26 at 14:30
$begingroup$
@MundronSchmidt Honestly I do not see how your example is related to what I wrote. I consider a single function which reproduces the family and is uniformly continuous, but in your example only every individual function is uniformly continuous. Maybe you could elaborate on your comment?
$endgroup$
– Severin Schraven
Jan 26 at 14:51
$begingroup$
@stackuser I did not add the second part so that you could try on your own. But it seems it was not clear enough. I guess now you should be able to fill in the gaps. Or do you want me to do it as well?
$endgroup$
– Severin Schraven
Jan 26 at 14:53
|
show 2 more comments
$begingroup$
Hint: use the inequalities $|sin(t)| leq |t|$ and $|cos(t)-1|leq frac{t^2}{2}$.
answered Jan 26 at 14:17
Ewan DelanoyEwan Delanoy
41.9k443104
$endgroup$
$begingroup$
I am yet to know how to apply the hint. please guide me
$endgroup$
– user397197
Jan 26 at 15:39
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
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oldest
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$begingroup$
Hint: Define
$$f(x)= sqrt{sin(x) + cos(x)}$$
This function is continuous and thus uniformly continuous on $[0; frac{pi}{2}]$.
Added: We have that for every $xin [0; frac{pi}{2}]$ that
$$ leftvert frac{x}{n} - frac{x}{m} rightvert
= leftvert frac{1}{n} - frac{1}{m} rightvert cdot vert x vert
leq frac{pi}{2} leftvert frac{1}{n} - frac{1}{m} rightvert $$
Furthermore, we have
$$ vert f_n(x) - f_m(x) vert = vert f(frac{x}{n}) - f(frac{x}{m}) vert. $$
Of course, we don't need the uniform continuity (as shown in the comments by p4sch), but I consider it usefull to think this way. It allows one to also tackle different problems (like for example the uniform convergence of $g_n(x)=sqrt{x^2+frac{1}{n}}$ to the absolute value on the whole real line).
answered Jan 26 at 14:18
Severin SchravenSeverin Schraven
6,4501935
$endgroup$
$begingroup$
I'm not sure how this should help to show that $f_n$ congerves uniformly. I think of $f_n(x)=x^n$, which are uniformly continuous on $[0,1]$ as well while it doesn't converge uniformly on $[0,1]$.
$endgroup$
– Mundron Schmidt
Jan 26 at 14:21
$begingroup$
I know this but how does how does uniform continuity and uniform convergence relate in this case. Is there any theoren I am missing?
$endgroup$
– user397197
Jan 26 at 14:24
1
$begingroup$
You have $f_n(x) = f(x/n)$. Since $|f(0) -f(x)|< varepsilon$ for all $|x|< delta$, you get for all $1/n < delta$ and $x in [0,pi/2]$ that $|f_n(x)| < varepsilon$.
$endgroup$
– p4sch
Jan 26 at 14:30
$begingroup$
@MundronSchmidt Honestly I do not see how your example is related to what I wrote. I consider a single function which reproduces the family and is uniformly continuous, but in your example only every individual function is uniformly continuous. Maybe you could elaborate on your comment?
$endgroup$
– Severin Schraven
Jan 26 at 14:51
$begingroup$
@stackuser I did not add the second part so that you could try on your own. But it seems it was not clear enough. I guess now you should be able to fill in the gaps. Or do you want me to do it as well?
$endgroup$
– Severin Schraven
Jan 26 at 14:53
|
show 2 more comments
$begingroup$
Hint: Define
$$f(x)= sqrt{sin(x) + cos(x)}$$
This function is continuous and thus uniformly continuous on $[0; frac{pi}{2}]$.
Added: We have that for every $xin [0; frac{pi}{2}]$ that
$$ leftvert frac{x}{n} - frac{x}{m} rightvert
= leftvert frac{1}{n} - frac{1}{m} rightvert cdot vert x vert
leq frac{pi}{2} leftvert frac{1}{n} - frac{1}{m} rightvert $$
Furthermore, we have
$$ vert f_n(x) - f_m(x) vert = vert f(frac{x}{n}) - f(frac{x}{m}) vert. $$
Of course, we don't need the uniform continuity (as shown in the comments by p4sch), but I consider it usefull to think this way. It allows one to also tackle different problems (like for example the uniform convergence of $g_n(x)=sqrt{x^2+frac{1}{n}}$ to the absolute value on the whole real line).
answered Jan 26 at 14:18
Severin SchravenSeverin Schraven
6,4501935
$endgroup$
$begingroup$
I'm not sure how this should help to show that $f_n$ congerves uniformly. I think of $f_n(x)=x^n$, which are uniformly continuous on $[0,1]$ as well while it doesn't converge uniformly on $[0,1]$.
$endgroup$
– Mundron Schmidt
Jan 26 at 14:21
$begingroup$
I know this but how does how does uniform continuity and uniform convergence relate in this case. Is there any theoren I am missing?
$endgroup$
– user397197
Jan 26 at 14:24
1
$begingroup$
You have $f_n(x) = f(x/n)$. Since $|f(0) -f(x)|< varepsilon$ for all $|x|< delta$, you get for all $1/n < delta$ and $x in [0,pi/2]$ that $|f_n(x)| < varepsilon$.
$endgroup$
– p4sch
Jan 26 at 14:30
$begingroup$
@MundronSchmidt Honestly I do not see how your example is related to what I wrote. I consider a single function which reproduces the family and is uniformly continuous, but in your example only every individual function is uniformly continuous. Maybe you could elaborate on your comment?
$endgroup$
– Severin Schraven
Jan 26 at 14:51
$begingroup$
@stackuser I did not add the second part so that you could try on your own. But it seems it was not clear enough. I guess now you should be able to fill in the gaps. Or do you want me to do it as well?
$endgroup$
– Severin Schraven
Jan 26 at 14:53
|
show 2 more comments
$begingroup$
Hint: Define
$$f(x)= sqrt{sin(x) + cos(x)}$$
This function is continuous and thus uniformly continuous on $[0; frac{pi}{2}]$.
Added: We have that for every $xin [0; frac{pi}{2}]$ that
$$ leftvert frac{x}{n} - frac{x}{m} rightvert
= leftvert frac{1}{n} - frac{1}{m} rightvert cdot vert x vert
leq frac{pi}{2} leftvert frac{1}{n} - frac{1}{m} rightvert $$
Furthermore, we have
$$ vert f_n(x) - f_m(x) vert = vert f(frac{x}{n}) - f(frac{x}{m}) vert. $$
Of course, we don't need the uniform continuity (as shown in the comments by p4sch), but I consider it usefull to think this way. It allows one to also tackle different problems (like for example the uniform convergence of $g_n(x)=sqrt{x^2+frac{1}{n}}$ to the absolute value on the whole real line).
answered Jan 26 at 14:18
Severin SchravenSeverin Schraven
6,4501935
$endgroup$
Hint: Define
$$f(x)= sqrt{sin(x) + cos(x)}$$
This function is continuous and thus uniformly continuous on $[0; frac{pi}{2}]$.
Added: We have that for every $xin [0; frac{pi}{2}]$ that
$$ leftvert frac{x}{n} - frac{x}{m} rightvert
= leftvert frac{1}{n} - frac{1}{m} rightvert cdot vert x vert
leq frac{pi}{2} leftvert frac{1}{n} - frac{1}{m} rightvert $$
Furthermore, we have
$$ vert f_n(x) - f_m(x) vert = vert f(frac{x}{n}) - f(frac{x}{m}) vert. $$
Of course, we don't need the uniform continuity (as shown in the comments by p4sch), but I consider it usefull to think this way. It allows one to also tackle different problems (like for example the uniform convergence of $g_n(x)=sqrt{x^2+frac{1}{n}}$ to the absolute value on the whole real line).
answered Jan 26 at 14:18
Severin SchravenSeverin Schraven
6,4501935
edited Jan 26 at 14:43
answered Jan 26 at 14:18
Severin SchravenSeverin Schraven
6,4501935
answered Jan 26 at 14:18
Severin SchravenSeverin Schraven
6,4501935
answered Jan 26 at 14:18
Severin SchravenSeverin Schraven
6,4501935
6,4501935
$begingroup$
I'm not sure how this should help to show that $f_n$ congerves uniformly. I think of $f_n(x)=x^n$, which are uniformly continuous on $[0,1]$ as well while it doesn't converge uniformly on $[0,1]$.
$endgroup$
– Mundron Schmidt
Jan 26 at 14:21
$begingroup$
I know this but how does how does uniform continuity and uniform convergence relate in this case. Is there any theoren I am missing?
$endgroup$
– user397197
Jan 26 at 14:24
1
$begingroup$
You have $f_n(x) = f(x/n)$. Since $|f(0) -f(x)|< varepsilon$ for all $|x|< delta$, you get for all $1/n < delta$ and $x in [0,pi/2]$ that $|f_n(x)| < varepsilon$.
$endgroup$
– p4sch
Jan 26 at 14:30
$begingroup$
@MundronSchmidt Honestly I do not see how your example is related to what I wrote. I consider a single function which reproduces the family and is uniformly continuous, but in your example only every individual function is uniformly continuous. Maybe you could elaborate on your comment?
$endgroup$
– Severin Schraven
Jan 26 at 14:51
$begingroup$
@stackuser I did not add the second part so that you could try on your own. But it seems it was not clear enough. I guess now you should be able to fill in the gaps. Or do you want me to do it as well?
$endgroup$
– Severin Schraven
Jan 26 at 14:53
|
show 2 more comments
$begingroup$
I'm not sure how this should help to show that $f_n$ congerves uniformly. I think of $f_n(x)=x^n$, which are uniformly continuous on $[0,1]$ as well while it doesn't converge uniformly on $[0,1]$.
$endgroup$
– Mundron Schmidt
Jan 26 at 14:21
$begingroup$
I know this but how does how does uniform continuity and uniform convergence relate in this case. Is there any theoren I am missing?
$endgroup$
– user397197
Jan 26 at 14:24
1
$begingroup$
You have $f_n(x) = f(x/n)$. Since $|f(0) -f(x)|< varepsilon$ for all $|x|< delta$, you get for all $1/n < delta$ and $x in [0,pi/2]$ that $|f_n(x)| < varepsilon$.
$endgroup$
– p4sch
Jan 26 at 14:30
$begingroup$
@MundronSchmidt Honestly I do not see how your example is related to what I wrote. I consider a single function which reproduces the family and is uniformly continuous, but in your example only every individual function is uniformly continuous. Maybe you could elaborate on your comment?
$endgroup$
– Severin Schraven
Jan 26 at 14:51
$begingroup$
@stackuser I did not add the second part so that you could try on your own. But it seems it was not clear enough. I guess now you should be able to fill in the gaps. Or do you want me to do it as well?
$endgroup$
– Severin Schraven
Jan 26 at 14:53
$begingroup$
I'm not sure how this should help to show that $f_n$ congerves uniformly. I think of $f_n(x)=x^n$, which are uniformly continuous on $[0,1]$ as well while it doesn't converge uniformly on $[0,1]$.
$endgroup$
– Mundron Schmidt
Jan 26 at 14:21
$begingroup$
I'm not sure how this should help to show that $f_n$ congerves uniformly. I think of $f_n(x)=x^n$, which are uniformly continuous on $[0,1]$ as well while it doesn't converge uniformly on $[0,1]$.
$endgroup$
– Mundron Schmidt
Jan 26 at 14:21
$begingroup$
I know this but how does how does uniform continuity and uniform convergence relate in this case. Is there any theoren I am missing?
$endgroup$
– user397197
Jan 26 at 14:24
$begingroup$
I know this but how does how does uniform continuity and uniform convergence relate in this case. Is there any theoren I am missing?
$endgroup$
– user397197
Jan 26 at 14:24
1
1
$begingroup$
You have $f_n(x) = f(x/n)$. Since $|f(0) -f(x)|< varepsilon$ for all $|x|< delta$, you get for all $1/n < delta$ and $x in [0,pi/2]$ that $|f_n(x)| < varepsilon$.
$endgroup$
– p4sch
Jan 26 at 14:30
$begingroup$
You have $f_n(x) = f(x/n)$. Since $|f(0) -f(x)|< varepsilon$ for all $|x|< delta$, you get for all $1/n < delta$ and $x in [0,pi/2]$ that $|f_n(x)| < varepsilon$.
$endgroup$
– p4sch
Jan 26 at 14:30
$begingroup$
@MundronSchmidt Honestly I do not see how your example is related to what I wrote. I consider a single function which reproduces the family and is uniformly continuous, but in your example only every individual function is uniformly continuous. Maybe you could elaborate on your comment?
$endgroup$
– Severin Schraven
Jan 26 at 14:51
$begingroup$
@MundronSchmidt Honestly I do not see how your example is related to what I wrote. I consider a single function which reproduces the family and is uniformly continuous, but in your example only every individual function is uniformly continuous. Maybe you could elaborate on your comment?
$endgroup$
– Severin Schraven
Jan 26 at 14:51
$begingroup$
@stackuser I did not add the second part so that you could try on your own. But it seems it was not clear enough. I guess now you should be able to fill in the gaps. Or do you want me to do it as well?
$endgroup$
– Severin Schraven
Jan 26 at 14:53
$begingroup$
@stackuser I did not add the second part so that you could try on your own. But it seems it was not clear enough. I guess now you should be able to fill in the gaps. Or do you want me to do it as well?
$endgroup$
– Severin Schraven
Jan 26 at 14:53
|
show 2 more comments
$begingroup$
Hint: use the inequalities $|sin(t)| leq |t|$ and $|cos(t)-1|leq frac{t^2}{2}$.
answered Jan 26 at 14:17
Ewan DelanoyEwan Delanoy
41.9k443104
$endgroup$
$begingroup$
I am yet to know how to apply the hint. please guide me
$endgroup$
– user397197
Jan 26 at 15:39
add a comment |
$begingroup$
Hint: use the inequalities $|sin(t)| leq |t|$ and $|cos(t)-1|leq frac{t^2}{2}$.
answered Jan 26 at 14:17
Ewan DelanoyEwan Delanoy
41.9k443104
$endgroup$
$begingroup$
I am yet to know how to apply the hint. please guide me
$endgroup$
– user397197
Jan 26 at 15:39
add a comment |
$begingroup$
Hint: use the inequalities $|sin(t)| leq |t|$ and $|cos(t)-1|leq frac{t^2}{2}$.
answered Jan 26 at 14:17
Ewan DelanoyEwan Delanoy
41.9k443104
$endgroup$
Hint: use the inequalities $|sin(t)| leq |t|$ and $|cos(t)-1|leq frac{t^2}{2}$.
answered Jan 26 at 14:17
Ewan DelanoyEwan Delanoy
41.9k443104
answered Jan 26 at 14:17
Ewan DelanoyEwan Delanoy
41.9k443104
answered Jan 26 at 14:17
Ewan DelanoyEwan Delanoy
41.9k443104
answered Jan 26 at 14:17
Ewan DelanoyEwan Delanoy
41.9k443104
41.9k443104
$begingroup$
I am yet to know how to apply the hint. please guide me
$endgroup$
– user397197
Jan 26 at 15:39
add a comment |
$begingroup$
I am yet to know how to apply the hint. please guide me
$endgroup$
– user397197
Jan 26 at 15:39
$begingroup$
I am yet to know how to apply the hint. please guide me
$endgroup$
– user397197
Jan 26 at 15:39
$begingroup$
I am yet to know how to apply the hint. please guide me
$endgroup$
– user397197
Jan 26 at 15:39
add a comment |
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$begingroup$
Have have you tried? How did you show the pointwise limit?
$endgroup$
– Viktor Glombik
Jan 26 at 14:16
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@ViktorGlombik, for any fixed $x $, by the continuity of the square root, sine and cosine functions, as n approaches infinity, we have $sqrt (1)=1$
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– user397197
Jan 26 at 14:19