prove that $f_n(x) = frac{n^2x}{1 + n^2x^2log n}$ converge but not uniform
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I have to prove that the sequence of functions
$$
f_n(x) = frac{n^2x}{1 + n^2x^2log n},
$$ converges.
So I think that I can just show that $lim_{n to infty} (f_n) = 0$ pointwise. But to show that the sequence does not converge uniformly, I need to show that $f_n$ has a point of maximum $x = frac{1}{n sqrt{log n}}$ and at this point, the limit is equal to infinity. Is that correct?
And finally I need to show that $forall alpha > 0$ in the interval $mathbb{R} setminus (-alpha,+alpha)$, the sequence $f_n$ converges uniformly, but I don't have ideas how to do it.
real-analysis sequences-and-series analysis
asked Dec 15 '18 at 21:24
Matheus FachiniMatheus Fachini
859
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add a comment |
$begingroup$
I have to prove that the sequence of functions
$$
f_n(x) = frac{n^2x}{1 + n^2x^2log n},
$$ converges.
So I think that I can just show that $lim_{n to infty} (f_n) = 0$ pointwise. But to show that the sequence does not converge uniformly, I need to show that $f_n$ has a point of maximum $x = frac{1}{n sqrt{log n}}$ and at this point, the limit is equal to infinity. Is that correct?
And finally I need to show that $forall alpha > 0$ in the interval $mathbb{R} setminus (-alpha,+alpha)$, the sequence $f_n$ converges uniformly, but I don't have ideas how to do it.
real-analysis sequences-and-series analysis
asked Dec 15 '18 at 21:24
Matheus FachiniMatheus Fachini
859
$endgroup$
$begingroup$
You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 21:39
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Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
$endgroup$
– Matheus Fachini
Dec 15 '18 at 21:45
1
$begingroup$
Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 22:23
add a comment |
$begingroup$
I have to prove that the sequence of functions
$$
f_n(x) = frac{n^2x}{1 + n^2x^2log n},
$$ converges.
So I think that I can just show that $lim_{n to infty} (f_n) = 0$ pointwise. But to show that the sequence does not converge uniformly, I need to show that $f_n$ has a point of maximum $x = frac{1}{n sqrt{log n}}$ and at this point, the limit is equal to infinity. Is that correct?
And finally I need to show that $forall alpha > 0$ in the interval $mathbb{R} setminus (-alpha,+alpha)$, the sequence $f_n$ converges uniformly, but I don't have ideas how to do it.
real-analysis sequences-and-series analysis
asked Dec 15 '18 at 21:24
Matheus FachiniMatheus Fachini
859
$endgroup$
I have to prove that the sequence of functions
$$
f_n(x) = frac{n^2x}{1 + n^2x^2log n},
$$ converges.
So I think that I can just show that $lim_{n to infty} (f_n) = 0$ pointwise. But to show that the sequence does not converge uniformly, I need to show that $f_n$ has a point of maximum $x = frac{1}{n sqrt{log n}}$ and at this point, the limit is equal to infinity. Is that correct?
And finally I need to show that $forall alpha > 0$ in the interval $mathbb{R} setminus (-alpha,+alpha)$, the sequence $f_n$ converges uniformly, but I don't have ideas how to do it.
real-analysis sequences-and-series analysis
real-analysis sequences-and-series analysis
asked Dec 15 '18 at 21:24
Matheus FachiniMatheus Fachini
859
asked Dec 15 '18 at 21:24
Matheus FachiniMatheus Fachini
859
edited Dec 15 '18 at 21:33
user587192
asked Dec 15 '18 at 21:24
Matheus FachiniMatheus Fachini
859
asked Dec 15 '18 at 21:24
Matheus FachiniMatheus Fachini
859
asked Dec 15 '18 at 21:24
Matheus FachiniMatheus Fachini
859
859
$begingroup$
You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 21:39
$begingroup$
Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
$endgroup$
– Matheus Fachini
Dec 15 '18 at 21:45
1
$begingroup$
Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 22:23
add a comment |
$begingroup$
You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 21:39
$begingroup$
Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
$endgroup$
– Matheus Fachini
Dec 15 '18 at 21:45
1
$begingroup$
Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 22:23
$begingroup$
You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 21:39
$begingroup$
You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 21:39
$begingroup$
Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
$endgroup$
– Matheus Fachini
Dec 15 '18 at 21:45
$begingroup$
Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
$endgroup$
– Matheus Fachini
Dec 15 '18 at 21:45
1
1
$begingroup$
Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 22:23
$begingroup$
Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 22:23
add a comment |
1 Answer
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$1). $ If $x=0$ the sequence clearly converges to $0$. If not, then $|f_n(x)|le frac{1}{xlog n}to 0$ as $nto infty$ so $(f_n)$ is pointwise convergent.
$2). $ The trouble will be near zero, so take $x_n=1/n$. Then $x_nto 0$ whereas $f_n(x_n)to infty$, and so $(f_n)$ is not uniformly convergent.
$3). $ $f_n$ is odd for each $nge 1$ so it suffices to check $[alpha,infty).$ In this case,$ |f_n(x)|le frac{1}{alphalog n}$ so $(f_n)$ is uniformly bounded, and so converges uniformly.
answered Dec 15 '18 at 22:24
MatematletaMatematleta
11.5k2920
$endgroup$
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Because $f_n(1/n)=n/(1+ln n)$
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– Matematleta
Dec 16 '18 at 2:39
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I had misread the numerator as $n^2x^2$.
$endgroup$
– Mark Viola
Dec 16 '18 at 14:39
add a comment |
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$begingroup$
$1). $ If $x=0$ the sequence clearly converges to $0$. If not, then $|f_n(x)|le frac{1}{xlog n}to 0$ as $nto infty$ so $(f_n)$ is pointwise convergent.
$2). $ The trouble will be near zero, so take $x_n=1/n$. Then $x_nto 0$ whereas $f_n(x_n)to infty$, and so $(f_n)$ is not uniformly convergent.
$3). $ $f_n$ is odd for each $nge 1$ so it suffices to check $[alpha,infty).$ In this case,$ |f_n(x)|le frac{1}{alphalog n}$ so $(f_n)$ is uniformly bounded, and so converges uniformly.
answered Dec 15 '18 at 22:24
MatematletaMatematleta
11.5k2920
$endgroup$
$begingroup$
Because $f_n(1/n)=n/(1+ln n)$
$endgroup$
– Matematleta
Dec 16 '18 at 2:39
$begingroup$
I had misread the numerator as $n^2x^2$.
$endgroup$
– Mark Viola
Dec 16 '18 at 14:39
add a comment |
$begingroup$
$1). $ If $x=0$ the sequence clearly converges to $0$. If not, then $|f_n(x)|le frac{1}{xlog n}to 0$ as $nto infty$ so $(f_n)$ is pointwise convergent.
$2). $ The trouble will be near zero, so take $x_n=1/n$. Then $x_nto 0$ whereas $f_n(x_n)to infty$, and so $(f_n)$ is not uniformly convergent.
$3). $ $f_n$ is odd for each $nge 1$ so it suffices to check $[alpha,infty).$ In this case,$ |f_n(x)|le frac{1}{alphalog n}$ so $(f_n)$ is uniformly bounded, and so converges uniformly.
answered Dec 15 '18 at 22:24
MatematletaMatematleta
11.5k2920
$endgroup$
$begingroup$
Because $f_n(1/n)=n/(1+ln n)$
$endgroup$
– Matematleta
Dec 16 '18 at 2:39
$begingroup$
I had misread the numerator as $n^2x^2$.
$endgroup$
– Mark Viola
Dec 16 '18 at 14:39
add a comment |
$begingroup$
$1). $ If $x=0$ the sequence clearly converges to $0$. If not, then $|f_n(x)|le frac{1}{xlog n}to 0$ as $nto infty$ so $(f_n)$ is pointwise convergent.
$2). $ The trouble will be near zero, so take $x_n=1/n$. Then $x_nto 0$ whereas $f_n(x_n)to infty$, and so $(f_n)$ is not uniformly convergent.
$3). $ $f_n$ is odd for each $nge 1$ so it suffices to check $[alpha,infty).$ In this case,$ |f_n(x)|le frac{1}{alphalog n}$ so $(f_n)$ is uniformly bounded, and so converges uniformly.
answered Dec 15 '18 at 22:24
MatematletaMatematleta
11.5k2920
$endgroup$
$1). $ If $x=0$ the sequence clearly converges to $0$. If not, then $|f_n(x)|le frac{1}{xlog n}to 0$ as $nto infty$ so $(f_n)$ is pointwise convergent.
$2). $ The trouble will be near zero, so take $x_n=1/n$. Then $x_nto 0$ whereas $f_n(x_n)to infty$, and so $(f_n)$ is not uniformly convergent.
$3). $ $f_n$ is odd for each $nge 1$ so it suffices to check $[alpha,infty).$ In this case,$ |f_n(x)|le frac{1}{alphalog n}$ so $(f_n)$ is uniformly bounded, and so converges uniformly.
answered Dec 15 '18 at 22:24
MatematletaMatematleta
11.5k2920
edited Dec 16 '18 at 4:51
answered Dec 15 '18 at 22:24
MatematletaMatematleta
11.5k2920
answered Dec 15 '18 at 22:24
MatematletaMatematleta
11.5k2920
answered Dec 15 '18 at 22:24
MatematletaMatematleta
11.5k2920
11.5k2920
$begingroup$
Because $f_n(1/n)=n/(1+ln n)$
$endgroup$
– Matematleta
Dec 16 '18 at 2:39
$begingroup$
I had misread the numerator as $n^2x^2$.
$endgroup$
– Mark Viola
Dec 16 '18 at 14:39
add a comment |
$begingroup$
Because $f_n(1/n)=n/(1+ln n)$
$endgroup$
– Matematleta
Dec 16 '18 at 2:39
$begingroup$
I had misread the numerator as $n^2x^2$.
$endgroup$
– Mark Viola
Dec 16 '18 at 14:39
$begingroup$
Because $f_n(1/n)=n/(1+ln n)$
$endgroup$
– Matematleta
Dec 16 '18 at 2:39
$begingroup$
Because $f_n(1/n)=n/(1+ln n)$
$endgroup$
– Matematleta
Dec 16 '18 at 2:39
$begingroup$
I had misread the numerator as $n^2x^2$.
$endgroup$
– Mark Viola
Dec 16 '18 at 14:39
$begingroup$
I had misread the numerator as $n^2x^2$.
$endgroup$
– Mark Viola
Dec 16 '18 at 14:39
add a comment |
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$begingroup$
You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 21:39
$begingroup$
Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
$endgroup$
– Matheus Fachini
Dec 15 '18 at 21:45
1
$begingroup$
Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 22:23