Prove that $xmapsto sum^infty_n=0f_n(x)=sum^infty_n=0fracx^2(1+x^2)^n$ does not converge uniformly on $[-1,1]$
Clash Royale CLAN TAG#URR8PPP
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For each $n,$ we define beginalign f_n:BbbRto BbbR endalign
beginalign xmapsto f_n(x)=fracx^2(1+x^2)^nendalign
We consider the function
beginalign f:BbbRto BbbR endalign
beginalign xmapsto f(x)=sum^infty_n=0f_n(x)=sum^infty_n=0fracx^2(1+x^2)^nendalign
I want to show that the series does not converge uniformly on $[-1,1]$ but I'm finding it difficult to do that.
First of all, I considered the Weierstrass M test.
MY TRIAL
beginalignleft|f_n(x)right|=left|fracx^2(1+x^2)^nright|leq frac1(1+x^2)^n ,;;forall ;xin[-1,1],;forall;ninBbbNendalign
I'm also thinking that the $beta_n=suplimits_xin[-1,1]|sum^n_i=0f_i(x)-sum^infty_i=0f_i(x)|$ approach could be very helpful too!
sequences-and-series convergence power-series uniform-convergence
asked Sep 5 at 8:59
Micheal
25510
add a comment |Â
up vote
3
down vote
favorite
For each $n,$ we define beginalign f_n:BbbRto BbbR endalign
beginalign xmapsto f_n(x)=fracx^2(1+x^2)^nendalign
We consider the function
beginalign f:BbbRto BbbR endalign
beginalign xmapsto f(x)=sum^infty_n=0f_n(x)=sum^infty_n=0fracx^2(1+x^2)^nendalign
I want to show that the series does not converge uniformly on $[-1,1]$ but I'm finding it difficult to do that.
First of all, I considered the Weierstrass M test.
MY TRIAL
beginalignleft|f_n(x)right|=left|fracx^2(1+x^2)^nright|leq frac1(1+x^2)^n ,;;forall ;xin[-1,1],;forall;ninBbbNendalign
I'm also thinking that the $beta_n=suplimits_xin[-1,1]|sum^n_i=0f_i(x)-sum^infty_i=0f_i(x)|$ approach could be very helpful too!
sequences-and-series convergence power-series uniform-convergence
asked Sep 5 at 8:59
Micheal
25510
1
Do you have to prove that the sequence $f_n$ converges (this is what you say in the title) or that the series $sum_n=0^infty f_n$ converges?
– 5xum
Sep 5 at 9:02
1
@5xum: The series!
– Micheal
Sep 5 at 9:04
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For each $n,$ we define beginalign f_n:BbbRto BbbR endalign
beginalign xmapsto f_n(x)=fracx^2(1+x^2)^nendalign
We consider the function
beginalign f:BbbRto BbbR endalign
beginalign xmapsto f(x)=sum^infty_n=0f_n(x)=sum^infty_n=0fracx^2(1+x^2)^nendalign
I want to show that the series does not converge uniformly on $[-1,1]$ but I'm finding it difficult to do that.
First of all, I considered the Weierstrass M test.
MY TRIAL
beginalignleft|f_n(x)right|=left|fracx^2(1+x^2)^nright|leq frac1(1+x^2)^n ,;;forall ;xin[-1,1],;forall;ninBbbNendalign
I'm also thinking that the $beta_n=suplimits_xin[-1,1]|sum^n_i=0f_i(x)-sum^infty_i=0f_i(x)|$ approach could be very helpful too!
sequences-and-series convergence power-series uniform-convergence
asked Sep 5 at 8:59
Micheal
25510
For each $n,$ we define beginalign f_n:BbbRto BbbR endalign
beginalign xmapsto f_n(x)=fracx^2(1+x^2)^nendalign
We consider the function
beginalign f:BbbRto BbbR endalign
beginalign xmapsto f(x)=sum^infty_n=0f_n(x)=sum^infty_n=0fracx^2(1+x^2)^nendalign
I want to show that the series does not converge uniformly on $[-1,1]$ but I'm finding it difficult to do that.
First of all, I considered the Weierstrass M test.
MY TRIAL
beginalignleft|f_n(x)right|=left|fracx^2(1+x^2)^nright|leq frac1(1+x^2)^n ,;;forall ;xin[-1,1],;forall;ninBbbNendalign
I'm also thinking that the $beta_n=suplimits_xin[-1,1]|sum^n_i=0f_i(x)-sum^infty_i=0f_i(x)|$ approach could be very helpful too!
sequences-and-series convergence power-series uniform-convergence
sequences-and-series convergence power-series uniform-convergence
asked Sep 5 at 8:59
Micheal
25510
asked Sep 5 at 8:59
Micheal
25510
edited Sep 5 at 9:32
asked Sep 5 at 8:59
Micheal
25510
asked Sep 5 at 8:59
Micheal
25510
asked Sep 5 at 8:59
Micheal
25510
25510
1
Do you have to prove that the sequence $f_n$ converges (this is what you say in the title) or that the series $sum_n=0^infty f_n$ converges?
– 5xum
Sep 5 at 9:02
1
@5xum: The series!
– Micheal
Sep 5 at 9:04
add a comment |Â
1
Do you have to prove that the sequence $f_n$ converges (this is what you say in the title) or that the series $sum_n=0^infty f_n$ converges?
– 5xum
Sep 5 at 9:02
1
@5xum: The series!
– Micheal
Sep 5 at 9:04
1
1
Do you have to prove that the sequence $f_n$ converges (this is what you say in the title) or that the series $sum_n=0^infty f_n$ converges?
– 5xum
Sep 5 at 9:02
Do you have to prove that the sequence $f_n$ converges (this is what you say in the title) or that the series $sum_n=0^infty f_n$ converges?
– 5xum
Sep 5 at 9:02
1
1
@5xum: The series!
– Micheal
Sep 5 at 9:04
@5xum: The series!
– Micheal
Sep 5 at 9:04
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
The convergence is not uniform.
For $x ne 0$ this is a geometric series:
$$sum^infty_n=0fracx^2(1+x^2)^n = x^2cdot frac11-frac11+x^2= x^2+1$$
and for $x = 0$ the sum is $0$.
Therefore
$$f(x) = begincases
x^2+1, &text if x ne 0\
0, &text if x = 0
endcases$$
Since $f$ is not continuous, the convergence cannot be uniform.
answered Sep 5 at 9:16
mechanodroid
24.4k62245
Thank you very much but I have shown this before now. I want to show that the series converges uniformly on $[-1,1]$ and not on $BbbR$
– Micheal
Sep 5 at 9:18
1
@Micheal The convergence is not uniform on $[-1,1]$ so it can't be shown. The sum is the same on $[-1,1]$.
– mechanodroid
Sep 5 at 9:19
I was thinking so. Is it because of the discontinuity at $0$?
– Micheal
Sep 5 at 9:20
1
@Micheal Yes, uniform limit of continuous functions has to be continuous.
– mechanodroid
Sep 5 at 9:21
2
If convergence is uniform on $[-1,1]$, sum function $f$ must be continuous!
– Riemann
Sep 5 at 9:21
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The convergence is not uniform.
For $x ne 0$ this is a geometric series:
$$sum^infty_n=0fracx^2(1+x^2)^n = x^2cdot frac11-frac11+x^2= x^2+1$$
and for $x = 0$ the sum is $0$.
Therefore
$$f(x) = begincases
x^2+1, &text if x ne 0\
0, &text if x = 0
endcases$$
Since $f$ is not continuous, the convergence cannot be uniform.
answered Sep 5 at 9:16
mechanodroid
24.4k62245
Thank you very much but I have shown this before now. I want to show that the series converges uniformly on $[-1,1]$ and not on $BbbR$
– Micheal
Sep 5 at 9:18
1
@Micheal The convergence is not uniform on $[-1,1]$ so it can't be shown. The sum is the same on $[-1,1]$.
– mechanodroid
Sep 5 at 9:19
I was thinking so. Is it because of the discontinuity at $0$?
– Micheal
Sep 5 at 9:20
1
@Micheal Yes, uniform limit of continuous functions has to be continuous.
– mechanodroid
Sep 5 at 9:21
2
If convergence is uniform on $[-1,1]$, sum function $f$ must be continuous!
– Riemann
Sep 5 at 9:21
 |Â
show 3 more comments
up vote
4
down vote
accepted
The convergence is not uniform.
For $x ne 0$ this is a geometric series:
$$sum^infty_n=0fracx^2(1+x^2)^n = x^2cdot frac11-frac11+x^2= x^2+1$$
and for $x = 0$ the sum is $0$.
Therefore
$$f(x) = begincases
x^2+1, &text if x ne 0\
0, &text if x = 0
endcases$$
Since $f$ is not continuous, the convergence cannot be uniform.
answered Sep 5 at 9:16
mechanodroid
24.4k62245
Thank you very much but I have shown this before now. I want to show that the series converges uniformly on $[-1,1]$ and not on $BbbR$
– Micheal
Sep 5 at 9:18
1
@Micheal The convergence is not uniform on $[-1,1]$ so it can't be shown. The sum is the same on $[-1,1]$.
– mechanodroid
Sep 5 at 9:19
I was thinking so. Is it because of the discontinuity at $0$?
– Micheal
Sep 5 at 9:20
1
@Micheal Yes, uniform limit of continuous functions has to be continuous.
– mechanodroid
Sep 5 at 9:21
2
If convergence is uniform on $[-1,1]$, sum function $f$ must be continuous!
– Riemann
Sep 5 at 9:21
 |Â
show 3 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The convergence is not uniform.
For $x ne 0$ this is a geometric series:
$$sum^infty_n=0fracx^2(1+x^2)^n = x^2cdot frac11-frac11+x^2= x^2+1$$
and for $x = 0$ the sum is $0$.
Therefore
$$f(x) = begincases
x^2+1, &text if x ne 0\
0, &text if x = 0
endcases$$
Since $f$ is not continuous, the convergence cannot be uniform.
answered Sep 5 at 9:16
mechanodroid
24.4k62245
The convergence is not uniform.
For $x ne 0$ this is a geometric series:
$$sum^infty_n=0fracx^2(1+x^2)^n = x^2cdot frac11-frac11+x^2= x^2+1$$
and for $x = 0$ the sum is $0$.
Therefore
$$f(x) = begincases
x^2+1, &text if x ne 0\
0, &text if x = 0
endcases$$
Since $f$ is not continuous, the convergence cannot be uniform.
answered Sep 5 at 9:16
mechanodroid
24.4k62245
answered Sep 5 at 9:16
mechanodroid
24.4k62245
answered Sep 5 at 9:16
mechanodroid
24.4k62245
answered Sep 5 at 9:16
mechanodroid
24.4k62245
24.4k62245
Thank you very much but I have shown this before now. I want to show that the series converges uniformly on $[-1,1]$ and not on $BbbR$
– Micheal
Sep 5 at 9:18
1
@Micheal The convergence is not uniform on $[-1,1]$ so it can't be shown. The sum is the same on $[-1,1]$.
– mechanodroid
Sep 5 at 9:19
I was thinking so. Is it because of the discontinuity at $0$?
– Micheal
Sep 5 at 9:20
1
@Micheal Yes, uniform limit of continuous functions has to be continuous.
– mechanodroid
Sep 5 at 9:21
2
If convergence is uniform on $[-1,1]$, sum function $f$ must be continuous!
– Riemann
Sep 5 at 9:21
 |Â
show 3 more comments
Thank you very much but I have shown this before now. I want to show that the series converges uniformly on $[-1,1]$ and not on $BbbR$
– Micheal
Sep 5 at 9:18
1
@Micheal The convergence is not uniform on $[-1,1]$ so it can't be shown. The sum is the same on $[-1,1]$.
– mechanodroid
Sep 5 at 9:19
I was thinking so. Is it because of the discontinuity at $0$?
– Micheal
Sep 5 at 9:20
1
@Micheal Yes, uniform limit of continuous functions has to be continuous.
– mechanodroid
Sep 5 at 9:21
2
If convergence is uniform on $[-1,1]$, sum function $f$ must be continuous!
– Riemann
Sep 5 at 9:21
Thank you very much but I have shown this before now. I want to show that the series converges uniformly on $[-1,1]$ and not on $BbbR$
– Micheal
Sep 5 at 9:18
Thank you very much but I have shown this before now. I want to show that the series converges uniformly on $[-1,1]$ and not on $BbbR$
– Micheal
Sep 5 at 9:18
1
1
@Micheal The convergence is not uniform on $[-1,1]$ so it can't be shown. The sum is the same on $[-1,1]$.
– mechanodroid
Sep 5 at 9:19
@Micheal The convergence is not uniform on $[-1,1]$ so it can't be shown. The sum is the same on $[-1,1]$.
– mechanodroid
Sep 5 at 9:19
I was thinking so. Is it because of the discontinuity at $0$?
– Micheal
Sep 5 at 9:20
I was thinking so. Is it because of the discontinuity at $0$?
– Micheal
Sep 5 at 9:20
1
1
@Micheal Yes, uniform limit of continuous functions has to be continuous.
– mechanodroid
Sep 5 at 9:21
@Micheal Yes, uniform limit of continuous functions has to be continuous.
– mechanodroid
Sep 5 at 9:21
2
2
If convergence is uniform on $[-1,1]$, sum function $f$ must be continuous!
– Riemann
Sep 5 at 9:21
If convergence is uniform on $[-1,1]$, sum function $f$ must be continuous!
– Riemann
Sep 5 at 9:21
 |Â
show 3 more comments
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1
Do you have to prove that the sequence $f_n$ converges (this is what you say in the title) or that the series $sum_n=0^infty f_n$ converges?
– 5xum
Sep 5 at 9:02
1
@5xum: The series!
– Micheal
Sep 5 at 9:04