Mixing
Continuity of an exponential series [duplicate]
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
This question already has an answer here:
continuity of power series
1 answer
Let $(a_n)_n in mathbbN$ be an infinite positive sequence of integers.
Assume that $f(lambda) = sumlimits_n=0^infty a_n lambda^n$ is finite for any $lambda in [0, lambda_0)$. Does this necessarily imply that $f(lambda)$ is also continuous in $(0, lambda_0)$, or might something strange happen?
real-analysis complex-analysis functional-analysis analysis continuity
asked Aug 3 at 19:04
QuantumLogarithm
468313
marked as duplicate by mathcounterexamples.net, Hans Lundmark, Math1000, max_zorn, Xander Henderson Aug 4 at 2:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
3
down vote
favorite
This question already has an answer here:
continuity of power series
1 answer
Let $(a_n)_n in mathbbN$ be an infinite positive sequence of integers.
Assume that $f(lambda) = sumlimits_n=0^infty a_n lambda^n$ is finite for any $lambda in [0, lambda_0)$. Does this necessarily imply that $f(lambda)$ is also continuous in $(0, lambda_0)$, or might something strange happen?
real-analysis complex-analysis functional-analysis analysis continuity
asked Aug 3 at 19:04
QuantumLogarithm
468313
marked as duplicate by mathcounterexamples.net, Hans Lundmark, Math1000, max_zorn, Xander Henderson Aug 4 at 2:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Yes, it is continuous. It is in fact analytic.
– Andrés E. Caicedo
Aug 3 at 19:07
1
(For positive $a_n$ this is easy, because you very easily establish have uniform bounds, so you can apply Weierstrass M-test.)
– Andrés E. Caicedo
Aug 3 at 19:09
Thank you. Is there a theorem I can cite without a proof being required?
– QuantumLogarithm
Aug 3 at 19:37
1
Two results :(1). From the details of the proof of the Cauchy-Hadamard Radius Formula ( sometimes called the Hadamard Radius Formula) we can easily show that the sequence $(sum_n=0^ja_n x^n)_jin Bbb N$ converges uniformly on any closed subset of $D=z$....(2). A uniform limit of a sequence of continuous real or complex functions on $D$ is continuous.
– DanielWainfleet
Aug 3 at 19:49
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This question already has an answer here:
continuity of power series
1 answer
Let $(a_n)_n in mathbbN$ be an infinite positive sequence of integers.
Assume that $f(lambda) = sumlimits_n=0^infty a_n lambda^n$ is finite for any $lambda in [0, lambda_0)$. Does this necessarily imply that $f(lambda)$ is also continuous in $(0, lambda_0)$, or might something strange happen?
real-analysis complex-analysis functional-analysis analysis continuity
asked Aug 3 at 19:04
QuantumLogarithm
468313
This question already has an answer here:
continuity of power series
1 answer
Let $(a_n)_n in mathbbN$ be an infinite positive sequence of integers.
Assume that $f(lambda) = sumlimits_n=0^infty a_n lambda^n$ is finite for any $lambda in [0, lambda_0)$. Does this necessarily imply that $f(lambda)$ is also continuous in $(0, lambda_0)$, or might something strange happen?
This question already has an answer here:
continuity of power series
1 answer
real-analysis complex-analysis functional-analysis analysis continuity
asked Aug 3 at 19:04
QuantumLogarithm
468313
edited Aug 3 at 19:08
asked Aug 3 at 19:04
QuantumLogarithm
468313
asked Aug 3 at 19:04
QuantumLogarithm
468313
asked Aug 3 at 19:04
QuantumLogarithm
468313
468313
marked as duplicate by mathcounterexamples.net, Hans Lundmark, Math1000, max_zorn, Xander Henderson Aug 4 at 2:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by mathcounterexamples.net, Hans Lundmark, Math1000, max_zorn, Xander Henderson Aug 4 at 2:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Yes, it is continuous. It is in fact analytic.
– Andrés E. Caicedo
Aug 3 at 19:07
1
(For positive $a_n$ this is easy, because you very easily establish have uniform bounds, so you can apply Weierstrass M-test.)
– Andrés E. Caicedo
Aug 3 at 19:09
Thank you. Is there a theorem I can cite without a proof being required?
– QuantumLogarithm
Aug 3 at 19:37
1
Two results :(1). From the details of the proof of the Cauchy-Hadamard Radius Formula ( sometimes called the Hadamard Radius Formula) we can easily show that the sequence $(sum_n=0^ja_n x^n)_jin Bbb N$ converges uniformly on any closed subset of $D=z$....(2). A uniform limit of a sequence of continuous real or complex functions on $D$ is continuous.
– DanielWainfleet
Aug 3 at 19:49
add a comment |Â
1
Yes, it is continuous. It is in fact analytic.
– Andrés E. Caicedo
Aug 3 at 19:07
1
(For positive $a_n$ this is easy, because you very easily establish have uniform bounds, so you can apply Weierstrass M-test.)
– Andrés E. Caicedo
Aug 3 at 19:09
Thank you. Is there a theorem I can cite without a proof being required?
– QuantumLogarithm
Aug 3 at 19:37
1
Two results :(1). From the details of the proof of the Cauchy-Hadamard Radius Formula ( sometimes called the Hadamard Radius Formula) we can easily show that the sequence $(sum_n=0^ja_n x^n)_jin Bbb N$ converges uniformly on any closed subset of $D=z$....(2). A uniform limit of a sequence of continuous real or complex functions on $D$ is continuous.
– DanielWainfleet
Aug 3 at 19:49
1
1
Yes, it is continuous. It is in fact analytic.
– Andrés E. Caicedo
Aug 3 at 19:07
Yes, it is continuous. It is in fact analytic.
– Andrés E. Caicedo
Aug 3 at 19:07
1
1
(For positive $a_n$ this is easy, because you very easily establish have uniform bounds, so you can apply Weierstrass M-test.)
– Andrés E. Caicedo
Aug 3 at 19:09
(For positive $a_n$ this is easy, because you very easily establish have uniform bounds, so you can apply Weierstrass M-test.)
– Andrés E. Caicedo
Aug 3 at 19:09
Thank you. Is there a theorem I can cite without a proof being required?
– QuantumLogarithm
Aug 3 at 19:37
Thank you. Is there a theorem I can cite without a proof being required?
– QuantumLogarithm
Aug 3 at 19:37
1
1
Two results :(1). From the details of the proof of the Cauchy-Hadamard Radius Formula ( sometimes called the Hadamard Radius Formula) we can easily show that the sequence $(sum_n=0^ja_n x^n)_jin Bbb N$ converges uniformly on any closed subset of $D=z$....(2). A uniform limit of a sequence of continuous real or complex functions on $D$ is continuous.
– DanielWainfleet
Aug 3 at 19:49
Two results :(1). From the details of the proof of the Cauchy-Hadamard Radius Formula ( sometimes called the Hadamard Radius Formula) we can easily show that the sequence $(sum_n=0^ja_n x^n)_jin Bbb N$ converges uniformly on any closed subset of $D=z$....(2). A uniform limit of a sequence of continuous real or complex functions on $D$ is continuous.
– DanielWainfleet
Aug 3 at 19:49
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
If your series converges for $lambda=r$, then it converges uniformly for all $lambdain(-r,r)$. This is because for such $lambda$
$$
sum_n>ma_n|lambda|^nleqsum_n>ma_nr^nxrightarrow[mtoinfty]0.
$$
So $f(lambda)$ is a uniform limit of polynomials. And a uniform limit of continuous functions is continuous.
Your function is in fact a lot more than continuous. It is analytic.
answered Aug 3 at 22:44
Martin Argerami
115k1071164
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
If your series converges for $lambda=r$, then it converges uniformly for all $lambdain(-r,r)$. This is because for such $lambda$
$$
sum_n>ma_n|lambda|^nleqsum_n>ma_nr^nxrightarrow[mtoinfty]0.
$$
So $f(lambda)$ is a uniform limit of polynomials. And a uniform limit of continuous functions is continuous.
Your function is in fact a lot more than continuous. It is analytic.
answered Aug 3 at 22:44
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
If your series converges for $lambda=r$, then it converges uniformly for all $lambdain(-r,r)$. This is because for such $lambda$
$$
sum_n>ma_n|lambda|^nleqsum_n>ma_nr^nxrightarrow[mtoinfty]0.
$$
So $f(lambda)$ is a uniform limit of polynomials. And a uniform limit of continuous functions is continuous.
Your function is in fact a lot more than continuous. It is analytic.
answered Aug 3 at 22:44
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If your series converges for $lambda=r$, then it converges uniformly for all $lambdain(-r,r)$. This is because for such $lambda$
$$
sum_n>ma_n|lambda|^nleqsum_n>ma_nr^nxrightarrow[mtoinfty]0.
$$
So $f(lambda)$ is a uniform limit of polynomials. And a uniform limit of continuous functions is continuous.
Your function is in fact a lot more than continuous. It is analytic.
answered Aug 3 at 22:44
Martin Argerami
115k1071164
If your series converges for $lambda=r$, then it converges uniformly for all $lambdain(-r,r)$. This is because for such $lambda$
$$
sum_n>ma_n|lambda|^nleqsum_n>ma_nr^nxrightarrow[mtoinfty]0.
$$
So $f(lambda)$ is a uniform limit of polynomials. And a uniform limit of continuous functions is continuous.
Your function is in fact a lot more than continuous. It is analytic.
answered Aug 3 at 22:44
Martin Argerami
115k1071164
answered Aug 3 at 22:44
Martin Argerami
115k1071164
answered Aug 3 at 22:44
Martin Argerami
115k1071164
answered Aug 3 at 22:44
Martin Argerami
115k1071164
115k1071164
add a comment |Â
add a comment |Â
1
Yes, it is continuous. It is in fact analytic.
– Andrés E. Caicedo
Aug 3 at 19:07
1
(For positive $a_n$ this is easy, because you very easily establish have uniform bounds, so you can apply Weierstrass M-test.)
– Andrés E. Caicedo
Aug 3 at 19:09
Thank you. Is there a theorem I can cite without a proof being required?
– QuantumLogarithm
Aug 3 at 19:37
1
Two results :(1). From the details of the proof of the Cauchy-Hadamard Radius Formula ( sometimes called the Hadamard Radius Formula) we can easily show that the sequence $(sum_n=0^ja_n x^n)_jin Bbb N$ converges uniformly on any closed subset of $D=z$....(2). A uniform limit of a sequence of continuous real or complex functions on $D$ is continuous.
– DanielWainfleet
Aug 3 at 19:49