Mixing
Does this also imply uniform convergence? and hence preserves continuity on the interval $[0, infty)$
$begingroup$
I am asked to show pointwise convergence on $[0, infty)$ of the series:
$$sum_{n=1}^infty frac{1}{n^2 +n^3 x} $$
But I could just remark that since $x geq 0$ we have that $n^2 +n^3 x> n^2$
$$left|frac{1}{n^2 +n^3 x}right|=frac{1}{n^2 +n^3 x} leq frac{1}{n^2} =M_n$$
Since $|f_n| leq M_n $ and $sum M_n$ convergent p-series, by the Weierstrass $M$ test, the original series converges uniformly on the domain.
Now I simply conclude that uniform convergence implies pointwise convergence.
The reason I want to show this is that in the next question we are asked to show that:
Prove that the function $s: [0, infty) rightarrow mathbb R$ defined by the series, is continuous.
We know that uniform convergence preserves continuity. We only need to make the case that the individual functions are continuous. We know that quotients of polynomials are continuous, as long as we do not divide by zero, so $frac{1}{n^2+n^3 x}$ runs into a problem if $x=-frac{1}{n}$, but $n>0$ and $xgeq0$ so this can never happen. Therefore our polynomial quotient is continuous and so is the original function by uniform continuity.
real-analysis proof-verification uniform-convergence
asked Jan 27 at 9:22
Wesley StrikWesley Strik
2,209424
$endgroup$
add a comment |
$begingroup$
I am asked to show pointwise convergence on $[0, infty)$ of the series:
$$sum_{n=1}^infty frac{1}{n^2 +n^3 x} $$
But I could just remark that since $x geq 0$ we have that $n^2 +n^3 x> n^2$
$$left|frac{1}{n^2 +n^3 x}right|=frac{1}{n^2 +n^3 x} leq frac{1}{n^2} =M_n$$
Since $|f_n| leq M_n $ and $sum M_n$ convergent p-series, by the Weierstrass $M$ test, the original series converges uniformly on the domain.
Now I simply conclude that uniform convergence implies pointwise convergence.
The reason I want to show this is that in the next question we are asked to show that:
Prove that the function $s: [0, infty) rightarrow mathbb R$ defined by the series, is continuous.
We know that uniform convergence preserves continuity. We only need to make the case that the individual functions are continuous. We know that quotients of polynomials are continuous, as long as we do not divide by zero, so $frac{1}{n^2+n^3 x}$ runs into a problem if $x=-frac{1}{n}$, but $n>0$ and $xgeq0$ so this can never happen. Therefore our polynomial quotient is continuous and so is the original function by uniform continuity.
real-analysis proof-verification uniform-convergence
asked Jan 27 at 9:22
Wesley StrikWesley Strik
2,209424
$endgroup$
$begingroup$
So yes, even though the question only asked for pointwise, if I prove uniform convergence, I can get continuity of the function series.
$endgroup$
– Wesley Strik
Jan 27 at 9:35
add a comment |
$begingroup$
I am asked to show pointwise convergence on $[0, infty)$ of the series:
$$sum_{n=1}^infty frac{1}{n^2 +n^3 x} $$
But I could just remark that since $x geq 0$ we have that $n^2 +n^3 x> n^2$
$$left|frac{1}{n^2 +n^3 x}right|=frac{1}{n^2 +n^3 x} leq frac{1}{n^2} =M_n$$
Since $|f_n| leq M_n $ and $sum M_n$ convergent p-series, by the Weierstrass $M$ test, the original series converges uniformly on the domain.
Now I simply conclude that uniform convergence implies pointwise convergence.
The reason I want to show this is that in the next question we are asked to show that:
Prove that the function $s: [0, infty) rightarrow mathbb R$ defined by the series, is continuous.
We know that uniform convergence preserves continuity. We only need to make the case that the individual functions are continuous. We know that quotients of polynomials are continuous, as long as we do not divide by zero, so $frac{1}{n^2+n^3 x}$ runs into a problem if $x=-frac{1}{n}$, but $n>0$ and $xgeq0$ so this can never happen. Therefore our polynomial quotient is continuous and so is the original function by uniform continuity.
real-analysis proof-verification uniform-convergence
asked Jan 27 at 9:22
Wesley StrikWesley Strik
2,209424
$endgroup$
I am asked to show pointwise convergence on $[0, infty)$ of the series:
$$sum_{n=1}^infty frac{1}{n^2 +n^3 x} $$
But I could just remark that since $x geq 0$ we have that $n^2 +n^3 x> n^2$
$$left|frac{1}{n^2 +n^3 x}right|=frac{1}{n^2 +n^3 x} leq frac{1}{n^2} =M_n$$
Since $|f_n| leq M_n $ and $sum M_n$ convergent p-series, by the Weierstrass $M$ test, the original series converges uniformly on the domain.
Now I simply conclude that uniform convergence implies pointwise convergence.
The reason I want to show this is that in the next question we are asked to show that:
Prove that the function $s: [0, infty) rightarrow mathbb R$ defined by the series, is continuous.
We know that uniform convergence preserves continuity. We only need to make the case that the individual functions are continuous. We know that quotients of polynomials are continuous, as long as we do not divide by zero, so $frac{1}{n^2+n^3 x}$ runs into a problem if $x=-frac{1}{n}$, but $n>0$ and $xgeq0$ so this can never happen. Therefore our polynomial quotient is continuous and so is the original function by uniform continuity.
real-analysis proof-verification uniform-convergence
real-analysis proof-verification uniform-convergence
asked Jan 27 at 9:22
Wesley StrikWesley Strik
2,209424
asked Jan 27 at 9:22
Wesley StrikWesley Strik
2,209424
edited Jan 27 at 22:27
Wesley Strik
asked Jan 27 at 9:22
Wesley StrikWesley Strik
2,209424
asked Jan 27 at 9:22
Wesley StrikWesley Strik
2,209424
asked Jan 27 at 9:22
Wesley StrikWesley Strik
2,209424
2,209424
$begingroup$
So yes, even though the question only asked for pointwise, if I prove uniform convergence, I can get continuity of the function series.
$endgroup$
– Wesley Strik
Jan 27 at 9:35
add a comment |
$begingroup$
So yes, even though the question only asked for pointwise, if I prove uniform convergence, I can get continuity of the function series.
$endgroup$
– Wesley Strik
Jan 27 at 9:35
$begingroup$
So yes, even though the question only asked for pointwise, if I prove uniform convergence, I can get continuity of the function series.
$endgroup$
– Wesley Strik
Jan 27 at 9:35
$begingroup$
So yes, even though the question only asked for pointwise, if I prove uniform convergence, I can get continuity of the function series.
$endgroup$
– Wesley Strik
Jan 27 at 9:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, you are correct the function
$$f(x)=sum_{n=1}^infty frac{1}{n^2 +n^3 x}$$
is continuous in $[0,+infty)$ because it is the uniform limit of the sequence of continuous functions,
$$f_N(x)=sum_{n=1}^N frac{1}{n^2 +n^3 x}.$$
Indeed, as $N$ goes to infinity,
$$sup_{[0,+infty)}|f(x)-f_N(x)|leq sum_{n=N+1}^infty frac{1}{n^2}to 0$$
because $sum_{n=1}^infty frac{1}{n^2}$ is a convergent series.
answered Jan 27 at 9:35
Robert ZRobert Z
101k1070143
$endgroup$
add a comment |
$begingroup$
You've used the Weierstrass M test incorrectly. A series $sum f_n$ of positive functions that satisfies
$$sum_{n=1}^{infty}f_n(x) le sum_{n=1}^{infty}frac{1}{n^2}$$
need not be uniformly convergent. However, what we have is
$$0le frac{1}{n^2+n^3x} le frac{1}{n^2}$$
for $ n=1,2,dots$ and $xge 0.$ Now Weierstrass M gives the desired result.
answered Jan 27 at 19:51
zhw.zhw.
74.7k43175
$endgroup$
$begingroup$
"the original series with only positive terms "
$endgroup$
– Wesley Strik
Jan 27 at 20:18
$begingroup$
You might want to read my answer again. I was assuming the $f_n$ are positive. W-M requires an estimate on each $f_n,$ not on $sum f_n.$
$endgroup$
– zhw.
Jan 27 at 21:31
$begingroup$
the $f_n$ are the terms
$endgroup$
– Wesley Strik
Jan 27 at 21:43
$begingroup$
What you wrote was $$sum_{n=1}^infty frac{1}{n^2 +n^3 x} leqsum_{n=1}^infty frac{1}{n^2}$$ What I'm saying is you can't apply WM based on that. i don't know how to make it clearer.
$endgroup$
– zhw.
Jan 27 at 22:09
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
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$begingroup$
Yes, you are correct the function
$$f(x)=sum_{n=1}^infty frac{1}{n^2 +n^3 x}$$
is continuous in $[0,+infty)$ because it is the uniform limit of the sequence of continuous functions,
$$f_N(x)=sum_{n=1}^N frac{1}{n^2 +n^3 x}.$$
Indeed, as $N$ goes to infinity,
$$sup_{[0,+infty)}|f(x)-f_N(x)|leq sum_{n=N+1}^infty frac{1}{n^2}to 0$$
because $sum_{n=1}^infty frac{1}{n^2}$ is a convergent series.
answered Jan 27 at 9:35
Robert ZRobert Z
101k1070143
$endgroup$
add a comment |
$begingroup$
Yes, you are correct the function
$$f(x)=sum_{n=1}^infty frac{1}{n^2 +n^3 x}$$
is continuous in $[0,+infty)$ because it is the uniform limit of the sequence of continuous functions,
$$f_N(x)=sum_{n=1}^N frac{1}{n^2 +n^3 x}.$$
Indeed, as $N$ goes to infinity,
$$sup_{[0,+infty)}|f(x)-f_N(x)|leq sum_{n=N+1}^infty frac{1}{n^2}to 0$$
because $sum_{n=1}^infty frac{1}{n^2}$ is a convergent series.
answered Jan 27 at 9:35
Robert ZRobert Z
101k1070143
$endgroup$
add a comment |
$begingroup$
Yes, you are correct the function
$$f(x)=sum_{n=1}^infty frac{1}{n^2 +n^3 x}$$
is continuous in $[0,+infty)$ because it is the uniform limit of the sequence of continuous functions,
$$f_N(x)=sum_{n=1}^N frac{1}{n^2 +n^3 x}.$$
Indeed, as $N$ goes to infinity,
$$sup_{[0,+infty)}|f(x)-f_N(x)|leq sum_{n=N+1}^infty frac{1}{n^2}to 0$$
because $sum_{n=1}^infty frac{1}{n^2}$ is a convergent series.
answered Jan 27 at 9:35
Robert ZRobert Z
101k1070143
$endgroup$
Yes, you are correct the function
$$f(x)=sum_{n=1}^infty frac{1}{n^2 +n^3 x}$$
is continuous in $[0,+infty)$ because it is the uniform limit of the sequence of continuous functions,
$$f_N(x)=sum_{n=1}^N frac{1}{n^2 +n^3 x}.$$
Indeed, as $N$ goes to infinity,
$$sup_{[0,+infty)}|f(x)-f_N(x)|leq sum_{n=N+1}^infty frac{1}{n^2}to 0$$
because $sum_{n=1}^infty frac{1}{n^2}$ is a convergent series.
answered Jan 27 at 9:35
Robert ZRobert Z
101k1070143
answered Jan 27 at 9:35
Robert ZRobert Z
101k1070143
answered Jan 27 at 9:35
Robert ZRobert Z
101k1070143
answered Jan 27 at 9:35
Robert ZRobert Z
101k1070143
101k1070143
add a comment |
add a comment |
$begingroup$
You've used the Weierstrass M test incorrectly. A series $sum f_n$ of positive functions that satisfies
$$sum_{n=1}^{infty}f_n(x) le sum_{n=1}^{infty}frac{1}{n^2}$$
need not be uniformly convergent. However, what we have is
$$0le frac{1}{n^2+n^3x} le frac{1}{n^2}$$
for $ n=1,2,dots$ and $xge 0.$ Now Weierstrass M gives the desired result.
answered Jan 27 at 19:51
zhw.zhw.
74.7k43175
$endgroup$
$begingroup$
"the original series with only positive terms "
$endgroup$
– Wesley Strik
Jan 27 at 20:18
$begingroup$
You might want to read my answer again. I was assuming the $f_n$ are positive. W-M requires an estimate on each $f_n,$ not on $sum f_n.$
$endgroup$
– zhw.
Jan 27 at 21:31
$begingroup$
the $f_n$ are the terms
$endgroup$
– Wesley Strik
Jan 27 at 21:43
$begingroup$
What you wrote was $$sum_{n=1}^infty frac{1}{n^2 +n^3 x} leqsum_{n=1}^infty frac{1}{n^2}$$ What I'm saying is you can't apply WM based on that. i don't know how to make it clearer.
$endgroup$
– zhw.
Jan 27 at 22:09
add a comment |
$begingroup$
You've used the Weierstrass M test incorrectly. A series $sum f_n$ of positive functions that satisfies
$$sum_{n=1}^{infty}f_n(x) le sum_{n=1}^{infty}frac{1}{n^2}$$
need not be uniformly convergent. However, what we have is
$$0le frac{1}{n^2+n^3x} le frac{1}{n^2}$$
for $ n=1,2,dots$ and $xge 0.$ Now Weierstrass M gives the desired result.
answered Jan 27 at 19:51
zhw.zhw.
74.7k43175
$endgroup$
$begingroup$
"the original series with only positive terms "
$endgroup$
– Wesley Strik
Jan 27 at 20:18
$begingroup$
You might want to read my answer again. I was assuming the $f_n$ are positive. W-M requires an estimate on each $f_n,$ not on $sum f_n.$
$endgroup$
– zhw.
Jan 27 at 21:31
$begingroup$
the $f_n$ are the terms
$endgroup$
– Wesley Strik
Jan 27 at 21:43
$begingroup$
What you wrote was $$sum_{n=1}^infty frac{1}{n^2 +n^3 x} leqsum_{n=1}^infty frac{1}{n^2}$$ What I'm saying is you can't apply WM based on that. i don't know how to make it clearer.
$endgroup$
– zhw.
Jan 27 at 22:09
add a comment |
$begingroup$
You've used the Weierstrass M test incorrectly. A series $sum f_n$ of positive functions that satisfies
$$sum_{n=1}^{infty}f_n(x) le sum_{n=1}^{infty}frac{1}{n^2}$$
need not be uniformly convergent. However, what we have is
$$0le frac{1}{n^2+n^3x} le frac{1}{n^2}$$
for $ n=1,2,dots$ and $xge 0.$ Now Weierstrass M gives the desired result.
answered Jan 27 at 19:51
zhw.zhw.
74.7k43175
$endgroup$
You've used the Weierstrass M test incorrectly. A series $sum f_n$ of positive functions that satisfies
$$sum_{n=1}^{infty}f_n(x) le sum_{n=1}^{infty}frac{1}{n^2}$$
need not be uniformly convergent. However, what we have is
$$0le frac{1}{n^2+n^3x} le frac{1}{n^2}$$
for $ n=1,2,dots$ and $xge 0.$ Now Weierstrass M gives the desired result.
answered Jan 27 at 19:51
zhw.zhw.
74.7k43175
answered Jan 27 at 19:51
zhw.zhw.
74.7k43175
answered Jan 27 at 19:51
zhw.zhw.
74.7k43175
answered Jan 27 at 19:51
zhw.zhw.
74.7k43175
74.7k43175
$begingroup$
"the original series with only positive terms "
$endgroup$
– Wesley Strik
Jan 27 at 20:18
$begingroup$
You might want to read my answer again. I was assuming the $f_n$ are positive. W-M requires an estimate on each $f_n,$ not on $sum f_n.$
$endgroup$
– zhw.
Jan 27 at 21:31
$begingroup$
the $f_n$ are the terms
$endgroup$
– Wesley Strik
Jan 27 at 21:43
$begingroup$
What you wrote was $$sum_{n=1}^infty frac{1}{n^2 +n^3 x} leqsum_{n=1}^infty frac{1}{n^2}$$ What I'm saying is you can't apply WM based on that. i don't know how to make it clearer.
$endgroup$
– zhw.
Jan 27 at 22:09
add a comment |
$begingroup$
"the original series with only positive terms "
$endgroup$
– Wesley Strik
Jan 27 at 20:18
$begingroup$
You might want to read my answer again. I was assuming the $f_n$ are positive. W-M requires an estimate on each $f_n,$ not on $sum f_n.$
$endgroup$
– zhw.
Jan 27 at 21:31
$begingroup$
the $f_n$ are the terms
$endgroup$
– Wesley Strik
Jan 27 at 21:43
$begingroup$
What you wrote was $$sum_{n=1}^infty frac{1}{n^2 +n^3 x} leqsum_{n=1}^infty frac{1}{n^2}$$ What I'm saying is you can't apply WM based on that. i don't know how to make it clearer.
$endgroup$
– zhw.
Jan 27 at 22:09
$begingroup$
"the original series with only positive terms "
$endgroup$
– Wesley Strik
Jan 27 at 20:18
$begingroup$
"the original series with only positive terms "
$endgroup$
– Wesley Strik
Jan 27 at 20:18
$begingroup$
You might want to read my answer again. I was assuming the $f_n$ are positive. W-M requires an estimate on each $f_n,$ not on $sum f_n.$
$endgroup$
– zhw.
Jan 27 at 21:31
$begingroup$
You might want to read my answer again. I was assuming the $f_n$ are positive. W-M requires an estimate on each $f_n,$ not on $sum f_n.$
$endgroup$
– zhw.
Jan 27 at 21:31
$begingroup$
the $f_n$ are the terms
$endgroup$
– Wesley Strik
Jan 27 at 21:43
$begingroup$
the $f_n$ are the terms
$endgroup$
– Wesley Strik
Jan 27 at 21:43
$begingroup$
What you wrote was $$sum_{n=1}^infty frac{1}{n^2 +n^3 x} leqsum_{n=1}^infty frac{1}{n^2}$$ What I'm saying is you can't apply WM based on that. i don't know how to make it clearer.
$endgroup$
– zhw.
Jan 27 at 22:09
$begingroup$
What you wrote was $$sum_{n=1}^infty frac{1}{n^2 +n^3 x} leqsum_{n=1}^infty frac{1}{n^2}$$ What I'm saying is you can't apply WM based on that. i don't know how to make it clearer.
$endgroup$
– zhw.
Jan 27 at 22:09
add a comment |
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$begingroup$
So yes, even though the question only asked for pointwise, if I prove uniform convergence, I can get continuity of the function series.
$endgroup$
– Wesley Strik
Jan 27 at 9:35