Analytic properties of a series
$begingroup$
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
asked Dec 3 '18 at 18:29
FraFra
24615
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add a comment |
$begingroup$
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
asked Dec 3 '18 at 18:29
FraFra
24615
$endgroup$
add a comment |
$begingroup$
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
asked Dec 3 '18 at 18:29
FraFra
24615
$endgroup$
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 '18 at 18:29
FraFra
24615
asked Dec 3 '18 at 18:29
FraFra
24615
asked Dec 3 '18 at 18:29
FraFra
24615
asked Dec 3 '18 at 18:29
FraFra
24615
asked Dec 3 '18 at 18:29
FraFra
24615
24615
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Another is
$sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
=dfrac{pi}{sin{pi z}}
$.
answered Dec 3 '18 at 22:12
marty cohenmarty cohen
73.2k549128
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:50
FraFra
24615
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Another is
$sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
=dfrac{pi}{sin{pi z}}
$.
answered Dec 3 '18 at 22:12
marty cohenmarty cohen
73.2k549128
$endgroup$
add a comment |
$begingroup$
Another is
$sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
=dfrac{pi}{sin{pi z}}
$.
answered Dec 3 '18 at 22:12
marty cohenmarty cohen
73.2k549128
$endgroup$
add a comment |
$begingroup$
Another is
$sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
=dfrac{pi}{sin{pi z}}
$.
answered Dec 3 '18 at 22:12
marty cohenmarty cohen
73.2k549128
$endgroup$
Another is
$sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
=dfrac{pi}{sin{pi z}}
$.
answered Dec 3 '18 at 22:12
marty cohenmarty cohen
73.2k549128
answered Dec 3 '18 at 22:12
marty cohenmarty cohen
73.2k549128
answered Dec 3 '18 at 22:12
marty cohenmarty cohen
73.2k549128
answered Dec 3 '18 at 22:12
marty cohenmarty cohen
73.2k549128
73.2k549128
add a comment |
add a comment |
$begingroup$
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 '18 at 20:50
FraFra
24615
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:50
FraFra
24615
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:50
FraFra
24615
$endgroup$
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 '18 at 20:50
FraFra
24615
answered Dec 3 '18 at 20:50
FraFra
24615
answered Dec 3 '18 at 20:50
FraFra
24615
answered Dec 3 '18 at 20:50
FraFra
24615
24615
add a comment |
add a comment |
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