Analytic properties of a series
up vote
0
down vote
favorite
I was wondering about whether it is possible that the analytic properties of a series are different from the ones of the partial sums. For instance suppose we have the series
$$S=sum_{n=0}^{infty}frac{1}{z-c_n}$$
where $c_n$ are positive real numbers.
Clearly the partial sums
$$S_N=sum_{n=0}^{N}frac{1}{z-c_n}$$
are analytic functions with a collection of simple poles on the real line.
Is it possible that the series develops branch cuts or something it isn't possible to infer from the partial sum? An anser would be welcome even if it isn't related to the particular example I produced.
sequences-and-series analyticity
add a comment |
up vote
0
down vote
favorite
I was wondering about whether it is possible that the analytic properties of a series are different from the ones of the partial sums. For instance suppose we have the series
$$S=sum_{n=0}^{infty}frac{1}{z-c_n}$$
where $c_n$ are positive real numbers.
Clearly the partial sums
$$S_N=sum_{n=0}^{N}frac{1}{z-c_n}$$
are analytic functions with a collection of simple poles on the real line.
Is it possible that the series develops branch cuts or something it isn't possible to infer from the partial sum? An anser would be welcome even if it isn't related to the particular example I produced.
sequences-and-series analyticity
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was wondering about whether it is possible that the analytic properties of a series are different from the ones of the partial sums. For instance suppose we have the series
$$S=sum_{n=0}^{infty}frac{1}{z-c_n}$$
where $c_n$ are positive real numbers.
Clearly the partial sums
$$S_N=sum_{n=0}^{N}frac{1}{z-c_n}$$
are analytic functions with a collection of simple poles on the real line.
Is it possible that the series develops branch cuts or something it isn't possible to infer from the partial sum? An anser would be welcome even if it isn't related to the particular example I produced.
sequences-and-series analyticity
I was wondering about whether it is possible that the analytic properties of a series are different from the ones of the partial sums. For instance suppose we have the series
$$S=sum_{n=0}^{infty}frac{1}{z-c_n}$$
where $c_n$ are positive real numbers.
Clearly the partial sums
$$S_N=sum_{n=0}^{N}frac{1}{z-c_n}$$
are analytic functions with a collection of simple poles on the real line.
Is it possible that the series develops branch cuts or something it isn't possible to infer from the partial sum? An anser would be welcome even if it isn't related to the particular example I produced.
sequences-and-series analyticity
sequences-and-series analyticity
asked Dec 3 at 18:29
Fra
24615
24615
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
Another is
$sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
=dfrac{pi}{sin{pi z}}
$.
add a comment |
up vote
0
down vote
The doubt came and went rather fast. The answer seems to be yes, the series can have analytic properties different from the partial sum. Since I don't have the means to prove it in general, I thought an example might just do the trick to show that it is possible.
Consider the case in which $c_n= a n^2 +b$ where $a,bin mathbb{R}^+$, then
$$S(z)=sum_{n=0}^{infty}frac{1}{a n^2+b-z}=frac{pi sqrt{z-b} cot left(frac{pi sqrt{z-b}}{sqrt{a}}right)-sqrt{a}}{2 sqrt{a} (b-z)}$$
which clearly shows that $S(z)$ has a branch cut, while the partial sum does not.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Another is
$sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
=dfrac{pi}{sin{pi z}}
$.
add a comment |
up vote
1
down vote
Another is
$sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
=dfrac{pi}{sin{pi z}}
$.
add a comment |
up vote
1
down vote
up vote
1
down vote
Another is
$sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
=dfrac{pi}{sin{pi z}}
$.
Another is
$sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
=dfrac{pi}{sin{pi z}}
$.
answered Dec 3 at 22:12
marty cohen
71.7k546124
71.7k546124
add a comment |
add a comment |
up vote
0
down vote
The doubt came and went rather fast. The answer seems to be yes, the series can have analytic properties different from the partial sum. Since I don't have the means to prove it in general, I thought an example might just do the trick to show that it is possible.
Consider the case in which $c_n= a n^2 +b$ where $a,bin mathbb{R}^+$, then
$$S(z)=sum_{n=0}^{infty}frac{1}{a n^2+b-z}=frac{pi sqrt{z-b} cot left(frac{pi sqrt{z-b}}{sqrt{a}}right)-sqrt{a}}{2 sqrt{a} (b-z)}$$
which clearly shows that $S(z)$ has a branch cut, while the partial sum does not.
add a comment |
up vote
0
down vote
The doubt came and went rather fast. The answer seems to be yes, the series can have analytic properties different from the partial sum. Since I don't have the means to prove it in general, I thought an example might just do the trick to show that it is possible.
Consider the case in which $c_n= a n^2 +b$ where $a,bin mathbb{R}^+$, then
$$S(z)=sum_{n=0}^{infty}frac{1}{a n^2+b-z}=frac{pi sqrt{z-b} cot left(frac{pi sqrt{z-b}}{sqrt{a}}right)-sqrt{a}}{2 sqrt{a} (b-z)}$$
which clearly shows that $S(z)$ has a branch cut, while the partial sum does not.
add a comment |
up vote
0
down vote
up vote
0
down vote
The doubt came and went rather fast. The answer seems to be yes, the series can have analytic properties different from the partial sum. Since I don't have the means to prove it in general, I thought an example might just do the trick to show that it is possible.
Consider the case in which $c_n= a n^2 +b$ where $a,bin mathbb{R}^+$, then
$$S(z)=sum_{n=0}^{infty}frac{1}{a n^2+b-z}=frac{pi sqrt{z-b} cot left(frac{pi sqrt{z-b}}{sqrt{a}}right)-sqrt{a}}{2 sqrt{a} (b-z)}$$
which clearly shows that $S(z)$ has a branch cut, while the partial sum does not.
The doubt came and went rather fast. The answer seems to be yes, the series can have analytic properties different from the partial sum. Since I don't have the means to prove it in general, I thought an example might just do the trick to show that it is possible.
Consider the case in which $c_n= a n^2 +b$ where $a,bin mathbb{R}^+$, then
$$S(z)=sum_{n=0}^{infty}frac{1}{a n^2+b-z}=frac{pi sqrt{z-b} cot left(frac{pi sqrt{z-b}}{sqrt{a}}right)-sqrt{a}}{2 sqrt{a} (b-z)}$$
which clearly shows that $S(z)$ has a branch cut, while the partial sum does not.
answered Dec 3 at 20:50
Fra
24615
24615
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024480%2fanalytic-properties-of-a-series%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown