Convergence of $h_n(x) = x^ 1 +frac12n-1 $ defined on [-1,1]
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1.
$$ h_n(x) = x^ 1 + frac12n-1 = x^ frac2n-1+12n-1 = x^frac2n2n-1 = (x^2)^fracn2n-1 = (x^2 )^ frac12-frac1n $$
then $$ lim_nrightarrow infty h_n(x) = (x^2)^frac12 = |x| $$
- $ $
I want to know if the convergence is pointwise or uniform.
$ $ Let $x in [-1,1]$
$$ | h_n(x)-|x| | = | x^ 1 + frac12n-1 - |x|| = | x^1 x^frac12n-1 - |x| | = | xbig( x^frac12n-1-(pm1) big) | leq |x^frac12n-1-(pm1) big)| $$ since $|x|leq 1 Big(xin [-1,1]Big)$
but I'm stuck. How would you do that?
real-analysis uniform-convergence pointwise-convergence
edited Aug 17 at 5:16
Holo
4,2972629
asked Aug 17 at 4:37
tnt235711
447
add a comment |Â
up vote
0
down vote
favorite
1.
$$ h_n(x) = x^ 1 + frac12n-1 = x^ frac2n-1+12n-1 = x^frac2n2n-1 = (x^2)^fracn2n-1 = (x^2 )^ frac12-frac1n $$
then $$ lim_nrightarrow infty h_n(x) = (x^2)^frac12 = |x| $$
- $ $
I want to know if the convergence is pointwise or uniform.
$ $ Let $x in [-1,1]$
$$ | h_n(x)-|x| | = | x^ 1 + frac12n-1 - |x|| = | x^1 x^frac12n-1 - |x| | = | xbig( x^frac12n-1-(pm1) big) | leq |x^frac12n-1-(pm1) big)| $$ since $|x|leq 1 Big(xin [-1,1]Big)$
but I'm stuck. How would you do that?
real-analysis uniform-convergence pointwise-convergence
edited Aug 17 at 5:16
Holo
4,2972629
asked Aug 17 at 4:37
tnt235711
447
1
You could say $$lim h_n(x) = lim_infty x^ 1 +frac12n-1 =x^1+frac1infty=x$$
– Nosrati
Aug 17 at 4:41
@Nosrati I don't get it...Because in part 1. I showed that $lim h_n(x) = |x|$
– tnt235711
Aug 17 at 4:57
You need to be careful about 'changing the order' of powers: $$x=x^1=x^2/2ne (x^2)^1/2=|x|$$You can change it as you like assuming $xge 0$
– Holo
Aug 17 at 5:06
(note that you have $2n-1$[odd] so you can always take a real root of $x$, even when $xin[-1,0)$)
– Holo
Aug 17 at 5:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
1.
$$ h_n(x) = x^ 1 + frac12n-1 = x^ frac2n-1+12n-1 = x^frac2n2n-1 = (x^2)^fracn2n-1 = (x^2 )^ frac12-frac1n $$
then $$ lim_nrightarrow infty h_n(x) = (x^2)^frac12 = |x| $$
- $ $
I want to know if the convergence is pointwise or uniform.
$ $ Let $x in [-1,1]$
$$ | h_n(x)-|x| | = | x^ 1 + frac12n-1 - |x|| = | x^1 x^frac12n-1 - |x| | = | xbig( x^frac12n-1-(pm1) big) | leq |x^frac12n-1-(pm1) big)| $$ since $|x|leq 1 Big(xin [-1,1]Big)$
but I'm stuck. How would you do that?
real-analysis uniform-convergence pointwise-convergence
edited Aug 17 at 5:16
Holo
4,2972629
asked Aug 17 at 4:37
tnt235711
447
1.
$$ h_n(x) = x^ 1 + frac12n-1 = x^ frac2n-1+12n-1 = x^frac2n2n-1 = (x^2)^fracn2n-1 = (x^2 )^ frac12-frac1n $$
then $$ lim_nrightarrow infty h_n(x) = (x^2)^frac12 = |x| $$
- $ $
I want to know if the convergence is pointwise or uniform.
$ $ Let $x in [-1,1]$
$$ | h_n(x)-|x| | = | x^ 1 + frac12n-1 - |x|| = | x^1 x^frac12n-1 - |x| | = | xbig( x^frac12n-1-(pm1) big) | leq |x^frac12n-1-(pm1) big)| $$ since $|x|leq 1 Big(xin [-1,1]Big)$
but I'm stuck. How would you do that?
real-analysis uniform-convergence pointwise-convergence
edited Aug 17 at 5:16
Holo
4,2972629
asked Aug 17 at 4:37
tnt235711
447
edited Aug 17 at 5:16
Holo
4,2972629
edited Aug 17 at 5:16
Holo
4,2972629
edited Aug 17 at 5:16
Holo
4,2972629
4,2972629
asked Aug 17 at 4:37
tnt235711
447
asked Aug 17 at 4:37
tnt235711
447
asked Aug 17 at 4:37
tnt235711
447
447
1
You could say $$lim h_n(x) = lim_infty x^ 1 +frac12n-1 =x^1+frac1infty=x$$
– Nosrati
Aug 17 at 4:41
@Nosrati I don't get it...Because in part 1. I showed that $lim h_n(x) = |x|$
– tnt235711
Aug 17 at 4:57
You need to be careful about 'changing the order' of powers: $$x=x^1=x^2/2ne (x^2)^1/2=|x|$$You can change it as you like assuming $xge 0$
– Holo
Aug 17 at 5:06
(note that you have $2n-1$[odd] so you can always take a real root of $x$, even when $xin[-1,0)$)
– Holo
Aug 17 at 5:12
add a comment |Â
1
You could say $$lim h_n(x) = lim_infty x^ 1 +frac12n-1 =x^1+frac1infty=x$$
– Nosrati
Aug 17 at 4:41
@Nosrati I don't get it...Because in part 1. I showed that $lim h_n(x) = |x|$
– tnt235711
Aug 17 at 4:57
You need to be careful about 'changing the order' of powers: $$x=x^1=x^2/2ne (x^2)^1/2=|x|$$You can change it as you like assuming $xge 0$
– Holo
Aug 17 at 5:06
(note that you have $2n-1$[odd] so you can always take a real root of $x$, even when $xin[-1,0)$)
– Holo
Aug 17 at 5:12
1
1
You could say $$lim h_n(x) = lim_infty x^ 1 +frac12n-1 =x^1+frac1infty=x$$
– Nosrati
Aug 17 at 4:41
You could say $$lim h_n(x) = lim_infty x^ 1 +frac12n-1 =x^1+frac1infty=x$$
– Nosrati
Aug 17 at 4:41
@Nosrati I don't get it...Because in part 1. I showed that $lim h_n(x) = |x|$
– tnt235711
Aug 17 at 4:57
@Nosrati I don't get it...Because in part 1. I showed that $lim h_n(x) = |x|$
– tnt235711
Aug 17 at 4:57
You need to be careful about 'changing the order' of powers: $$x=x^1=x^2/2ne (x^2)^1/2=|x|$$You can change it as you like assuming $xge 0$
– Holo
Aug 17 at 5:06
You need to be careful about 'changing the order' of powers: $$x=x^1=x^2/2ne (x^2)^1/2=|x|$$You can change it as you like assuming $xge 0$
– Holo
Aug 17 at 5:06
(note that you have $2n-1$[odd] so you can always take a real root of $x$, even when $xin[-1,0)$)
– Holo
Aug 17 at 5:12
(note that you have $2n-1$[odd] so you can always take a real root of $x$, even when $xin[-1,0)$)
– Holo
Aug 17 at 5:12
add a comment |Â
1 Answer
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up vote
1
down vote
accepted
The convergence is indeed uniform. Note that $h_n$ and $h(x)=|x|$ are both even functions so it is enough to prove uniform convergence on $[0,1]$. So consider $|x^1+frac 1 2n-1 -x|$ with $0leq x leq 1$. This is $x|1-e^frac 1 2n-1 log x|$. Use the inequality $1-e^-t leq t$ for all $t geq 0$ to see that $x|1-e^frac 1 2n-1 log x| leq frac (-xlog x) 2n-1$. Note that $xlog x$ is bounded, so the last quantity tends to $0$ uniformly.
answered Aug 17 at 5:56
Kavi Rama Murthy
22.8k2933
Nice!...that's exactly what I was looking for, thank you!
– tnt235711
Aug 17 at 15:48
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The convergence is indeed uniform. Note that $h_n$ and $h(x)=|x|$ are both even functions so it is enough to prove uniform convergence on $[0,1]$. So consider $|x^1+frac 1 2n-1 -x|$ with $0leq x leq 1$. This is $x|1-e^frac 1 2n-1 log x|$. Use the inequality $1-e^-t leq t$ for all $t geq 0$ to see that $x|1-e^frac 1 2n-1 log x| leq frac (-xlog x) 2n-1$. Note that $xlog x$ is bounded, so the last quantity tends to $0$ uniformly.
answered Aug 17 at 5:56
Kavi Rama Murthy
22.8k2933
Nice!...that's exactly what I was looking for, thank you!
– tnt235711
Aug 17 at 15:48
add a comment |Â
up vote
1
down vote
accepted
The convergence is indeed uniform. Note that $h_n$ and $h(x)=|x|$ are both even functions so it is enough to prove uniform convergence on $[0,1]$. So consider $|x^1+frac 1 2n-1 -x|$ with $0leq x leq 1$. This is $x|1-e^frac 1 2n-1 log x|$. Use the inequality $1-e^-t leq t$ for all $t geq 0$ to see that $x|1-e^frac 1 2n-1 log x| leq frac (-xlog x) 2n-1$. Note that $xlog x$ is bounded, so the last quantity tends to $0$ uniformly.
answered Aug 17 at 5:56
Kavi Rama Murthy
22.8k2933
Nice!...that's exactly what I was looking for, thank you!
– tnt235711
Aug 17 at 15:48
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The convergence is indeed uniform. Note that $h_n$ and $h(x)=|x|$ are both even functions so it is enough to prove uniform convergence on $[0,1]$. So consider $|x^1+frac 1 2n-1 -x|$ with $0leq x leq 1$. This is $x|1-e^frac 1 2n-1 log x|$. Use the inequality $1-e^-t leq t$ for all $t geq 0$ to see that $x|1-e^frac 1 2n-1 log x| leq frac (-xlog x) 2n-1$. Note that $xlog x$ is bounded, so the last quantity tends to $0$ uniformly.
answered Aug 17 at 5:56
Kavi Rama Murthy
22.8k2933
The convergence is indeed uniform. Note that $h_n$ and $h(x)=|x|$ are both even functions so it is enough to prove uniform convergence on $[0,1]$. So consider $|x^1+frac 1 2n-1 -x|$ with $0leq x leq 1$. This is $x|1-e^frac 1 2n-1 log x|$. Use the inequality $1-e^-t leq t$ for all $t geq 0$ to see that $x|1-e^frac 1 2n-1 log x| leq frac (-xlog x) 2n-1$. Note that $xlog x$ is bounded, so the last quantity tends to $0$ uniformly.
answered Aug 17 at 5:56
Kavi Rama Murthy
22.8k2933
answered Aug 17 at 5:56
Kavi Rama Murthy
22.8k2933
answered Aug 17 at 5:56
Kavi Rama Murthy
22.8k2933
answered Aug 17 at 5:56
Kavi Rama Murthy
22.8k2933
22.8k2933
Nice!...that's exactly what I was looking for, thank you!
– tnt235711
Aug 17 at 15:48
add a comment |Â
Nice!...that's exactly what I was looking for, thank you!
– tnt235711
Aug 17 at 15:48
Nice!...that's exactly what I was looking for, thank you!
– tnt235711
Aug 17 at 15:48
Nice!...that's exactly what I was looking for, thank you!
– tnt235711
Aug 17 at 15:48
add a comment |Â
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1
You could say $$lim h_n(x) = lim_infty x^ 1 +frac12n-1 =x^1+frac1infty=x$$
– Nosrati
Aug 17 at 4:41
@Nosrati I don't get it...Because in part 1. I showed that $lim h_n(x) = |x|$
– tnt235711
Aug 17 at 4:57
You need to be careful about 'changing the order' of powers: $$x=x^1=x^2/2ne (x^2)^1/2=|x|$$You can change it as you like assuming $xge 0$
– Holo
Aug 17 at 5:06
(note that you have $2n-1$[odd] so you can always take a real root of $x$, even when $xin[-1,0)$)
– Holo
Aug 17 at 5:12