Question about radius of convergence - powerseries
$begingroup$
Let $P(z)$ be of the form $sum_{n=0}^{infty}a_nz^n$.
It is known that
Lemma
$(i)$if $P(z_o)$ converges then all $P(z)$ with $|z|<|z_o|inmathbb{C}$ absolutely converge.
$(ii)$ if $P(z_0)$ converges absolutely then all $P(z)$ with $|z|leq|z_o|inmathbb{C}$ absolutely converge
Define $R:=sup{|z|:P(z)$ converges$}$ and
$R':=sup{|z|:P(z)$ converges absolutely$}$
With the lemma one can conclude that $R=R'$
One also has the notation $D_R(0)={zinmathbb{C}:z<R}$ and $bar{D}_R(0)={zinmathbb{C}:zleq R}$
Now my questions:
Can somebody give me an example where ${zinmathbb{C},|z|=R}$ but it is not clear whether the elements converge or converge absolutely?
Second question why does the convergencebehaviour solely depends on the coefficients i.e. $(a_n)_{ninmathbb{N}_0}$
sequences-and-series power-series
asked Dec 23 '18 at 17:00
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
Let $P(z)$ be of the form $sum_{n=0}^{infty}a_nz^n$.
It is known that
Lemma
$(i)$if $P(z_o)$ converges then all $P(z)$ with $|z|<|z_o|inmathbb{C}$ absolutely converge.
$(ii)$ if $P(z_0)$ converges absolutely then all $P(z)$ with $|z|leq|z_o|inmathbb{C}$ absolutely converge
Define $R:=sup{|z|:P(z)$ converges$}$ and
$R':=sup{|z|:P(z)$ converges absolutely$}$
With the lemma one can conclude that $R=R'$
One also has the notation $D_R(0)={zinmathbb{C}:z<R}$ and $bar{D}_R(0)={zinmathbb{C}:zleq R}$
Now my questions:
Can somebody give me an example where ${zinmathbb{C},|z|=R}$ but it is not clear whether the elements converge or converge absolutely?
Second question why does the convergencebehaviour solely depends on the coefficients i.e. $(a_n)_{ninmathbb{N}_0}$
sequences-and-series power-series
asked Dec 23 '18 at 17:00
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
Let $P(z)$ be of the form $sum_{n=0}^{infty}a_nz^n$.
It is known that
Lemma
$(i)$if $P(z_o)$ converges then all $P(z)$ with $|z|<|z_o|inmathbb{C}$ absolutely converge.
$(ii)$ if $P(z_0)$ converges absolutely then all $P(z)$ with $|z|leq|z_o|inmathbb{C}$ absolutely converge
Define $R:=sup{|z|:P(z)$ converges$}$ and
$R':=sup{|z|:P(z)$ converges absolutely$}$
With the lemma one can conclude that $R=R'$
One also has the notation $D_R(0)={zinmathbb{C}:z<R}$ and $bar{D}_R(0)={zinmathbb{C}:zleq R}$
Now my questions:
Can somebody give me an example where ${zinmathbb{C},|z|=R}$ but it is not clear whether the elements converge or converge absolutely?
Second question why does the convergencebehaviour solely depends on the coefficients i.e. $(a_n)_{ninmathbb{N}_0}$
sequences-and-series power-series
asked Dec 23 '18 at 17:00
RM777RM777
38312
$endgroup$
Let $P(z)$ be of the form $sum_{n=0}^{infty}a_nz^n$.
It is known that
Lemma
$(i)$if $P(z_o)$ converges then all $P(z)$ with $|z|<|z_o|inmathbb{C}$ absolutely converge.
$(ii)$ if $P(z_0)$ converges absolutely then all $P(z)$ with $|z|leq|z_o|inmathbb{C}$ absolutely converge
Define $R:=sup{|z|:P(z)$ converges$}$ and
$R':=sup{|z|:P(z)$ converges absolutely$}$
With the lemma one can conclude that $R=R'$
One also has the notation $D_R(0)={zinmathbb{C}:z<R}$ and $bar{D}_R(0)={zinmathbb{C}:zleq R}$
Now my questions:
Can somebody give me an example where ${zinmathbb{C},|z|=R}$ but it is not clear whether the elements converge or converge absolutely?
Second question why does the convergencebehaviour solely depends on the coefficients i.e. $(a_n)_{ninmathbb{N}_0}$
sequences-and-series power-series
sequences-and-series power-series
asked Dec 23 '18 at 17:00
RM777RM777
38312
asked Dec 23 '18 at 17:00
RM777RM777
38312
asked Dec 23 '18 at 17:00
RM777RM777
38312
asked Dec 23 '18 at 17:00
RM777RM777
38312
asked Dec 23 '18 at 17:00
RM777RM777
38312
38312
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First Question:
In general, as you noted, the power series on ${zinmathbb{C},|z|=R}$ could converge for some values of $z$ and diverge for others. An example would be:
$sum_{n=1}^{infty }frac{z^n}{n}$ which converges for all $|z|=1, zneq 1$ and diverges for $z=1$.
Second question:
The Lemmas you've stated tell you that once you have a point $z$ at which there is convergence, you automatically have convergence for all those points with smaller modulus. The Cauchy-Hadamard theorem gives you the dependence of the supremum of the moduli of those $z$ for which you have convergence with respect to your sequence $(a_n)_{ninmathbb{N}_0}$. So then, strictly within your radius of convergence you have convergence and strictly outside you have divergence. But exactly at your radius of convergence, the behavior does depend on $z$.
answered Dec 23 '18 at 18:52
John11John11
1,0441821
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First Question:
In general, as you noted, the power series on ${zinmathbb{C},|z|=R}$ could converge for some values of $z$ and diverge for others. An example would be:
$sum_{n=1}^{infty }frac{z^n}{n}$ which converges for all $|z|=1, zneq 1$ and diverges for $z=1$.
Second question:
The Lemmas you've stated tell you that once you have a point $z$ at which there is convergence, you automatically have convergence for all those points with smaller modulus. The Cauchy-Hadamard theorem gives you the dependence of the supremum of the moduli of those $z$ for which you have convergence with respect to your sequence $(a_n)_{ninmathbb{N}_0}$. So then, strictly within your radius of convergence you have convergence and strictly outside you have divergence. But exactly at your radius of convergence, the behavior does depend on $z$.
answered Dec 23 '18 at 18:52
John11John11
1,0441821
$endgroup$
add a comment |
$begingroup$
First Question:
In general, as you noted, the power series on ${zinmathbb{C},|z|=R}$ could converge for some values of $z$ and diverge for others. An example would be:
$sum_{n=1}^{infty }frac{z^n}{n}$ which converges for all $|z|=1, zneq 1$ and diverges for $z=1$.
Second question:
The Lemmas you've stated tell you that once you have a point $z$ at which there is convergence, you automatically have convergence for all those points with smaller modulus. The Cauchy-Hadamard theorem gives you the dependence of the supremum of the moduli of those $z$ for which you have convergence with respect to your sequence $(a_n)_{ninmathbb{N}_0}$. So then, strictly within your radius of convergence you have convergence and strictly outside you have divergence. But exactly at your radius of convergence, the behavior does depend on $z$.
answered Dec 23 '18 at 18:52
John11John11
1,0441821
$endgroup$
add a comment |
$begingroup$
First Question:
In general, as you noted, the power series on ${zinmathbb{C},|z|=R}$ could converge for some values of $z$ and diverge for others. An example would be:
$sum_{n=1}^{infty }frac{z^n}{n}$ which converges for all $|z|=1, zneq 1$ and diverges for $z=1$.
Second question:
The Lemmas you've stated tell you that once you have a point $z$ at which there is convergence, you automatically have convergence for all those points with smaller modulus. The Cauchy-Hadamard theorem gives you the dependence of the supremum of the moduli of those $z$ for which you have convergence with respect to your sequence $(a_n)_{ninmathbb{N}_0}$. So then, strictly within your radius of convergence you have convergence and strictly outside you have divergence. But exactly at your radius of convergence, the behavior does depend on $z$.
answered Dec 23 '18 at 18:52
John11John11
1,0441821
$endgroup$
First Question:
In general, as you noted, the power series on ${zinmathbb{C},|z|=R}$ could converge for some values of $z$ and diverge for others. An example would be:
$sum_{n=1}^{infty }frac{z^n}{n}$ which converges for all $|z|=1, zneq 1$ and diverges for $z=1$.
Second question:
The Lemmas you've stated tell you that once you have a point $z$ at which there is convergence, you automatically have convergence for all those points with smaller modulus. The Cauchy-Hadamard theorem gives you the dependence of the supremum of the moduli of those $z$ for which you have convergence with respect to your sequence $(a_n)_{ninmathbb{N}_0}$. So then, strictly within your radius of convergence you have convergence and strictly outside you have divergence. But exactly at your radius of convergence, the behavior does depend on $z$.
answered Dec 23 '18 at 18:52
John11John11
1,0441821
answered Dec 23 '18 at 18:52
John11John11
1,0441821
answered Dec 23 '18 at 18:52
John11John11
1,0441821
answered Dec 23 '18 at 18:52
John11John11
1,0441821
1,0441821
add a comment |
add a comment |
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