Understanding infinite products convergence The 2019 Stack Overflow Developer Survey Results Are Insufficiency and necessity of convergence of $sum a_n$ wrt convergence of $prod (1 + a_n)$Does order matter for the convergence of infinite productsQuestion concerning the conditional convergence of some infinite productsIf $prodlimits_n(1+a_n)$ converges, does $sumlimits_nfraca_n1+a_n$ converge?A question about the convergence of an infinite product in Complex AnalysisSimple questions about infinite productsEquivalence of convergence of a series and convergence of an infinite productDoes infinite product $ prod ( 1 - frac12^n ) $ diverge to 0 or convergeParticular infinite product convergenceUnderstanding entire functions
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Understanding infinite products convergence
The 2019 Stack Overflow Developer Survey Results Are Insufficiency and necessity of convergence of $sum a_n$ wrt convergence of $prod (1 + a_n)$Does order matter for the convergence of infinite productsQuestion concerning the conditional convergence of some infinite productsIf $prodlimits_n(1+a_n)$ converges, does $sumlimits_nfraca_n1+a_n$ converge?A question about the convergence of an infinite product in Complex AnalysisSimple questions about infinite productsEquivalence of convergence of a series and convergence of an infinite productDoes infinite product $ prod ( 1 - frac12^n ) $ diverge to 0 or convergeParticular infinite product convergenceUnderstanding entire functions
$begingroup$
I am reading about a convergence problem with infinite products and I am told:
Any finite sequence $c_n$ in the complex plane has an associated polynomial p(z) that has zeroes precisely at the points of that sequence, $p(z)=,prod _n(z-c_n)$.
Also, any polynomial function p(z) in the complex plane has a factorization $,p(z)=aprod _n(z-c_n)$, where a is a non-zero constant and $c_n$ are the zeroes of p.
It then says that the infinite product does not converge:
When one considers the product $,prod _n(z-c_n)$ when the sequence $c_n$ is not finite then it can never define an entire function, because the infinite product does not converge.
It is somewhat unclear to me why it does not converge, but in any case, it then goes on to state that:
A necessary condition for convergence of the infinite product in question is that each factor $(z-c_n)$ must approach 1 as $nto infty$
This last part flew right over my head. Why is it necessary for $(z-c_n)$ to approach 1? There's no explanation provided. I must be lacking some fundamental knowledge so can someone explain to me in Layman's Terms why this rule holds true, or perhaps show a proof that might help me understand?
convergence complex-numbers infinite-product entire-functions
asked Mar 24 at 6:44
BolboaBolboa
408616
$endgroup$
add a comment |
$begingroup$
I am reading about a convergence problem with infinite products and I am told:
Any finite sequence $c_n$ in the complex plane has an associated polynomial p(z) that has zeroes precisely at the points of that sequence, $p(z)=,prod _n(z-c_n)$.
Also, any polynomial function p(z) in the complex plane has a factorization $,p(z)=aprod _n(z-c_n)$, where a is a non-zero constant and $c_n$ are the zeroes of p.
It then says that the infinite product does not converge:
When one considers the product $,prod _n(z-c_n)$ when the sequence $c_n$ is not finite then it can never define an entire function, because the infinite product does not converge.
It is somewhat unclear to me why it does not converge, but in any case, it then goes on to state that:
A necessary condition for convergence of the infinite product in question is that each factor $(z-c_n)$ must approach 1 as $nto infty$
This last part flew right over my head. Why is it necessary for $(z-c_n)$ to approach 1? There's no explanation provided. I must be lacking some fundamental knowledge so can someone explain to me in Layman's Terms why this rule holds true, or perhaps show a proof that might help me understand?
convergence complex-numbers infinite-product entire-functions
asked Mar 24 at 6:44
BolboaBolboa
408616
$endgroup$
$begingroup$
This is similar to the term test for an infinite series. If the series $sum_n geqslant 1a_n$ converges then it is necessary for $lim_n to infty a_n = 0$ since the general term is the difference between two partial sums converging to the same value. An infinite series is a sequence of partial sums, an infinite product is a sequence of partial products. So if the infinite product converges then the general term, as a ratio of partial products converging to the same value, must converge to $1$.
$endgroup$
– RRL
Mar 24 at 8:44
add a comment |
$begingroup$
I am reading about a convergence problem with infinite products and I am told:
Any finite sequence $c_n$ in the complex plane has an associated polynomial p(z) that has zeroes precisely at the points of that sequence, $p(z)=,prod _n(z-c_n)$.
Also, any polynomial function p(z) in the complex plane has a factorization $,p(z)=aprod _n(z-c_n)$, where a is a non-zero constant and $c_n$ are the zeroes of p.
It then says that the infinite product does not converge:
When one considers the product $,prod _n(z-c_n)$ when the sequence $c_n$ is not finite then it can never define an entire function, because the infinite product does not converge.
It is somewhat unclear to me why it does not converge, but in any case, it then goes on to state that:
A necessary condition for convergence of the infinite product in question is that each factor $(z-c_n)$ must approach 1 as $nto infty$
This last part flew right over my head. Why is it necessary for $(z-c_n)$ to approach 1? There's no explanation provided. I must be lacking some fundamental knowledge so can someone explain to me in Layman's Terms why this rule holds true, or perhaps show a proof that might help me understand?
convergence complex-numbers infinite-product entire-functions
asked Mar 24 at 6:44
BolboaBolboa
408616
$endgroup$
I am reading about a convergence problem with infinite products and I am told:
Any finite sequence $c_n$ in the complex plane has an associated polynomial p(z) that has zeroes precisely at the points of that sequence, $p(z)=,prod _n(z-c_n)$.
Also, any polynomial function p(z) in the complex plane has a factorization $,p(z)=aprod _n(z-c_n)$, where a is a non-zero constant and $c_n$ are the zeroes of p.
It then says that the infinite product does not converge:
When one considers the product $,prod _n(z-c_n)$ when the sequence $c_n$ is not finite then it can never define an entire function, because the infinite product does not converge.
It is somewhat unclear to me why it does not converge, but in any case, it then goes on to state that:
A necessary condition for convergence of the infinite product in question is that each factor $(z-c_n)$ must approach 1 as $nto infty$
This last part flew right over my head. Why is it necessary for $(z-c_n)$ to approach 1? There's no explanation provided. I must be lacking some fundamental knowledge so can someone explain to me in Layman's Terms why this rule holds true, or perhaps show a proof that might help me understand?
convergence complex-numbers infinite-product entire-functions
convergence complex-numbers infinite-product entire-functions
asked Mar 24 at 6:44
BolboaBolboa
408616
asked Mar 24 at 6:44
BolboaBolboa
408616
asked Mar 24 at 6:44
BolboaBolboa
408616
asked Mar 24 at 6:44
BolboaBolboa
408616
asked Mar 24 at 6:44
BolboaBolboa
408616
408616
$begingroup$
This is similar to the term test for an infinite series. If the series $sum_n geqslant 1a_n$ converges then it is necessary for $lim_n to infty a_n = 0$ since the general term is the difference between two partial sums converging to the same value. An infinite series is a sequence of partial sums, an infinite product is a sequence of partial products. So if the infinite product converges then the general term, as a ratio of partial products converging to the same value, must converge to $1$.
$endgroup$
– RRL
Mar 24 at 8:44
add a comment |
$begingroup$
This is similar to the term test for an infinite series. If the series $sum_n geqslant 1a_n$ converges then it is necessary for $lim_n to infty a_n = 0$ since the general term is the difference between two partial sums converging to the same value. An infinite series is a sequence of partial sums, an infinite product is a sequence of partial products. So if the infinite product converges then the general term, as a ratio of partial products converging to the same value, must converge to $1$.
$endgroup$
– RRL
Mar 24 at 8:44
$begingroup$
This is similar to the term test for an infinite series. If the series $sum_n geqslant 1a_n$ converges then it is necessary for $lim_n to infty a_n = 0$ since the general term is the difference between two partial sums converging to the same value. An infinite series is a sequence of partial sums, an infinite product is a sequence of partial products. So if the infinite product converges then the general term, as a ratio of partial products converging to the same value, must converge to $1$.
$endgroup$
– RRL
Mar 24 at 8:44
$begingroup$
This is similar to the term test for an infinite series. If the series $sum_n geqslant 1a_n$ converges then it is necessary for $lim_n to infty a_n = 0$ since the general term is the difference between two partial sums converging to the same value. An infinite series is a sequence of partial sums, an infinite product is a sequence of partial products. So if the infinite product converges then the general term, as a ratio of partial products converging to the same value, must converge to $1$.
$endgroup$
– RRL
Mar 24 at 8:44
add a comment |
1 Answer
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$begingroup$
If the infinite product is convergent, i.e., $P_n = prod_k=1^n(z - c_k) to P$ as $n to infty$, then
$$lim_n to infty (z- c_n) = lim_n to inftyfracP_nP_n-1 = fraclim_n to inftyP_nlim_n to inftyP_n-1 =1$$
answered Mar 24 at 7:23
RRLRRL
53.6k52574
$endgroup$
add a comment |
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$begingroup$
If the infinite product is convergent, i.e., $P_n = prod_k=1^n(z - c_k) to P$ as $n to infty$, then
$$lim_n to infty (z- c_n) = lim_n to inftyfracP_nP_n-1 = fraclim_n to inftyP_nlim_n to inftyP_n-1 =1$$
answered Mar 24 at 7:23
RRLRRL
53.6k52574
$endgroup$
add a comment |
$begingroup$
If the infinite product is convergent, i.e., $P_n = prod_k=1^n(z - c_k) to P$ as $n to infty$, then
$$lim_n to infty (z- c_n) = lim_n to inftyfracP_nP_n-1 = fraclim_n to inftyP_nlim_n to inftyP_n-1 =1$$
answered Mar 24 at 7:23
RRLRRL
53.6k52574
$endgroup$
add a comment |
$begingroup$
If the infinite product is convergent, i.e., $P_n = prod_k=1^n(z - c_k) to P$ as $n to infty$, then
$$lim_n to infty (z- c_n) = lim_n to inftyfracP_nP_n-1 = fraclim_n to inftyP_nlim_n to inftyP_n-1 =1$$
answered Mar 24 at 7:23
RRLRRL
53.6k52574
$endgroup$
If the infinite product is convergent, i.e., $P_n = prod_k=1^n(z - c_k) to P$ as $n to infty$, then
$$lim_n to infty (z- c_n) = lim_n to inftyfracP_nP_n-1 = fraclim_n to inftyP_nlim_n to inftyP_n-1 =1$$
answered Mar 24 at 7:23
RRLRRL
53.6k52574
answered Mar 24 at 7:23
RRLRRL
53.6k52574
answered Mar 24 at 7:23
RRLRRL
53.6k52574
answered Mar 24 at 7:23
RRLRRL
53.6k52574
53.6k52574
add a comment |
add a comment |
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$begingroup$
This is similar to the term test for an infinite series. If the series $sum_n geqslant 1a_n$ converges then it is necessary for $lim_n to infty a_n = 0$ since the general term is the difference between two partial sums converging to the same value. An infinite series is a sequence of partial sums, an infinite product is a sequence of partial products. So if the infinite product converges then the general term, as a ratio of partial products converging to the same value, must converge to $1$.
$endgroup$
– RRL
Mar 24 at 8:44