Has the Riemann Rearrangement Theorem ever helped in computation rather than just being a warning?
A decent course in elementary analysis will eventually discuss series, absolute convergence, conditional convergence, and the Riemann Rearrangement Theorem. However, in any presentation I've seen, in person or within a text, the discussion sort of ends after Riemann's theorem is given---quite content in proving to the student(s) or the reader that one shouldn't extend finite intuition to infinite settings without providing a proof.
I agree that this is an important moral to impart; however, I'm interested in something else:
Has the Riemann Rearrangement theorem been used as a computational aid to explicitly calculate a sum?
By this vague question, I specifically have in mind that piece of Riemann's Theorem that states an absolutely convergent series is commutatively convergent. So, to be slightly more narrow in scope:
Has there been a series $sum a_k$ which is fairly easy to show absolutely converges; however, the sum itself was computed by a clever choice of bijection $sigma:mathbb{N}rightarrowmathbb{N}$ and by working with the partial sums of $sum a_{sigma(k)}$?
This is a rather vague question, and I don't expect it to have much of an absolute answer. But I'm interested in any variety of answers, and I'm sure they'd be demonstrative and helpful to future readers.
real-analysis sequences-and-series big-list riemann-sum absolute-convergence
asked Dec 2 '18 at 6:20
Robert Wolfe
5,68722463
add a comment |
A decent course in elementary analysis will eventually discuss series, absolute convergence, conditional convergence, and the Riemann Rearrangement Theorem. However, in any presentation I've seen, in person or within a text, the discussion sort of ends after Riemann's theorem is given---quite content in proving to the student(s) or the reader that one shouldn't extend finite intuition to infinite settings without providing a proof.
I agree that this is an important moral to impart; however, I'm interested in something else:
Has the Riemann Rearrangement theorem been used as a computational aid to explicitly calculate a sum?
By this vague question, I specifically have in mind that piece of Riemann's Theorem that states an absolutely convergent series is commutatively convergent. So, to be slightly more narrow in scope:
Has there been a series $sum a_k$ which is fairly easy to show absolutely converges; however, the sum itself was computed by a clever choice of bijection $sigma:mathbb{N}rightarrowmathbb{N}$ and by working with the partial sums of $sum a_{sigma(k)}$?
This is a rather vague question, and I don't expect it to have much of an absolute answer. But I'm interested in any variety of answers, and I'm sure they'd be demonstrative and helpful to future readers.
real-analysis sequences-and-series big-list riemann-sum absolute-convergence
asked Dec 2 '18 at 6:20
Robert Wolfe
5,68722463
13
Riemann rearrangement illuminates the idea that, for absolutely convergent sum we can rearrange its order of summation. Although this sounds not very practical for sums over $mathbb{N}$, it is indeed very useful for sums over $mathbb{N}^k$, i.e., multiple sums. For instance, for $exp(x)=sum_{ngeq0}frac{x^n}{n!}$ we can rearrange sums to prove $$exp(x)exp(y)=sum_{m,ngeq0}frac{x^my^n}{m!n!}=sum_{lgeq0}frac{1}{l!}left(sum_{m+n=l}frac{l!}{m!n!}x^my^nright)=sum_{lgeq0}frac{(x+y)^l}{l!}=exp(x+y).$$ And this extends to integrals, yielding the famous Fubini's theorem.
– Sangchul Lee
Dec 2 '18 at 6:28
2
In comment to the previous comment, you need the rearrangement theorem to prove already the first equality, as to show that the infinite summation over infinite summations can be compressed into one series at all, and because the summation over the grid does not imply any specific linear order.
– LutzL
Dec 2 '18 at 10:36
You don't need the rearrangement theorem. Mertens' theorems states that you need only the absolute convergence of one of the series in order to get that the Cauchy product converges and the limes is the product of both series. See also here for a reference.
– p4sch
Dec 2 '18 at 11:10
I would guess that there are doubly infinite sums that would be taxing to evaluate in some natural order of summation, but much easier in some other order. As Sangchul Lee pointed out, analogously with Fubini's theorem! My answer is relatively weak in the sense that the amount of rearranging is very minor when done in the order of increasing $|m|+|n|$. It was just the first thing that came to my mind :-/
– Jyrki Lahtonen
Dec 4 '18 at 18:44
1
Robert, well the second bounty came out empty. A bit disappointing as I was sure somebody would have a nice example. Shrug. I guess there are too many calculus and series questions to choose from. I know that my post was not quite what you were looking for. May be somebody shows up later. Ping me, if something great comes up :-)
– Jyrki Lahtonen
Dec 18 '18 at 19:28
add a comment |
A decent course in elementary analysis will eventually discuss series, absolute convergence, conditional convergence, and the Riemann Rearrangement Theorem. However, in any presentation I've seen, in person or within a text, the discussion sort of ends after Riemann's theorem is given---quite content in proving to the student(s) or the reader that one shouldn't extend finite intuition to infinite settings without providing a proof.
I agree that this is an important moral to impart; however, I'm interested in something else:
Has the Riemann Rearrangement theorem been used as a computational aid to explicitly calculate a sum?
By this vague question, I specifically have in mind that piece of Riemann's Theorem that states an absolutely convergent series is commutatively convergent. So, to be slightly more narrow in scope:
Has there been a series $sum a_k$ which is fairly easy to show absolutely converges; however, the sum itself was computed by a clever choice of bijection $sigma:mathbb{N}rightarrowmathbb{N}$ and by working with the partial sums of $sum a_{sigma(k)}$?
This is a rather vague question, and I don't expect it to have much of an absolute answer. But I'm interested in any variety of answers, and I'm sure they'd be demonstrative and helpful to future readers.
real-analysis sequences-and-series big-list riemann-sum absolute-convergence
asked Dec 2 '18 at 6:20
Robert Wolfe
5,68722463
A decent course in elementary analysis will eventually discuss series, absolute convergence, conditional convergence, and the Riemann Rearrangement Theorem. However, in any presentation I've seen, in person or within a text, the discussion sort of ends after Riemann's theorem is given---quite content in proving to the student(s) or the reader that one shouldn't extend finite intuition to infinite settings without providing a proof.
I agree that this is an important moral to impart; however, I'm interested in something else:
Has the Riemann Rearrangement theorem been used as a computational aid to explicitly calculate a sum?
By this vague question, I specifically have in mind that piece of Riemann's Theorem that states an absolutely convergent series is commutatively convergent. So, to be slightly more narrow in scope:
Has there been a series $sum a_k$ which is fairly easy to show absolutely converges; however, the sum itself was computed by a clever choice of bijection $sigma:mathbb{N}rightarrowmathbb{N}$ and by working with the partial sums of $sum a_{sigma(k)}$?
This is a rather vague question, and I don't expect it to have much of an absolute answer. But I'm interested in any variety of answers, and I'm sure they'd be demonstrative and helpful to future readers.
real-analysis sequences-and-series big-list riemann-sum absolute-convergence
real-analysis sequences-and-series big-list riemann-sum absolute-convergence
asked Dec 2 '18 at 6:20
Robert Wolfe
5,68722463
asked Dec 2 '18 at 6:20
Robert Wolfe
5,68722463
asked Dec 2 '18 at 6:20
Robert Wolfe
5,68722463
asked Dec 2 '18 at 6:20
Robert Wolfe
5,68722463
asked Dec 2 '18 at 6:20
Robert Wolfe
5,68722463
5,68722463
13
Riemann rearrangement illuminates the idea that, for absolutely convergent sum we can rearrange its order of summation. Although this sounds not very practical for sums over $mathbb{N}$, it is indeed very useful for sums over $mathbb{N}^k$, i.e., multiple sums. For instance, for $exp(x)=sum_{ngeq0}frac{x^n}{n!}$ we can rearrange sums to prove $$exp(x)exp(y)=sum_{m,ngeq0}frac{x^my^n}{m!n!}=sum_{lgeq0}frac{1}{l!}left(sum_{m+n=l}frac{l!}{m!n!}x^my^nright)=sum_{lgeq0}frac{(x+y)^l}{l!}=exp(x+y).$$ And this extends to integrals, yielding the famous Fubini's theorem.
– Sangchul Lee
Dec 2 '18 at 6:28
2
In comment to the previous comment, you need the rearrangement theorem to prove already the first equality, as to show that the infinite summation over infinite summations can be compressed into one series at all, and because the summation over the grid does not imply any specific linear order.
– LutzL
Dec 2 '18 at 10:36
You don't need the rearrangement theorem. Mertens' theorems states that you need only the absolute convergence of one of the series in order to get that the Cauchy product converges and the limes is the product of both series. See also here for a reference.
– p4sch
Dec 2 '18 at 11:10
I would guess that there are doubly infinite sums that would be taxing to evaluate in some natural order of summation, but much easier in some other order. As Sangchul Lee pointed out, analogously with Fubini's theorem! My answer is relatively weak in the sense that the amount of rearranging is very minor when done in the order of increasing $|m|+|n|$. It was just the first thing that came to my mind :-/
– Jyrki Lahtonen
Dec 4 '18 at 18:44
1
Robert, well the second bounty came out empty. A bit disappointing as I was sure somebody would have a nice example. Shrug. I guess there are too many calculus and series questions to choose from. I know that my post was not quite what you were looking for. May be somebody shows up later. Ping me, if something great comes up :-)
– Jyrki Lahtonen
Dec 18 '18 at 19:28
add a comment |
13
Riemann rearrangement illuminates the idea that, for absolutely convergent sum we can rearrange its order of summation. Although this sounds not very practical for sums over $mathbb{N}$, it is indeed very useful for sums over $mathbb{N}^k$, i.e., multiple sums. For instance, for $exp(x)=sum_{ngeq0}frac{x^n}{n!}$ we can rearrange sums to prove $$exp(x)exp(y)=sum_{m,ngeq0}frac{x^my^n}{m!n!}=sum_{lgeq0}frac{1}{l!}left(sum_{m+n=l}frac{l!}{m!n!}x^my^nright)=sum_{lgeq0}frac{(x+y)^l}{l!}=exp(x+y).$$ And this extends to integrals, yielding the famous Fubini's theorem.
– Sangchul Lee
Dec 2 '18 at 6:28
2
In comment to the previous comment, you need the rearrangement theorem to prove already the first equality, as to show that the infinite summation over infinite summations can be compressed into one series at all, and because the summation over the grid does not imply any specific linear order.
– LutzL
Dec 2 '18 at 10:36
You don't need the rearrangement theorem. Mertens' theorems states that you need only the absolute convergence of one of the series in order to get that the Cauchy product converges and the limes is the product of both series. See also here for a reference.
– p4sch
Dec 2 '18 at 11:10
I would guess that there are doubly infinite sums that would be taxing to evaluate in some natural order of summation, but much easier in some other order. As Sangchul Lee pointed out, analogously with Fubini's theorem! My answer is relatively weak in the sense that the amount of rearranging is very minor when done in the order of increasing $|m|+|n|$. It was just the first thing that came to my mind :-/
– Jyrki Lahtonen
Dec 4 '18 at 18:44
1
Robert, well the second bounty came out empty. A bit disappointing as I was sure somebody would have a nice example. Shrug. I guess there are too many calculus and series questions to choose from. I know that my post was not quite what you were looking for. May be somebody shows up later. Ping me, if something great comes up :-)
– Jyrki Lahtonen
Dec 18 '18 at 19:28
13
13
Riemann rearrangement illuminates the idea that, for absolutely convergent sum we can rearrange its order of summation. Although this sounds not very practical for sums over $mathbb{N}$, it is indeed very useful for sums over $mathbb{N}^k$, i.e., multiple sums. For instance, for $exp(x)=sum_{ngeq0}frac{x^n}{n!}$ we can rearrange sums to prove $$exp(x)exp(y)=sum_{m,ngeq0}frac{x^my^n}{m!n!}=sum_{lgeq0}frac{1}{l!}left(sum_{m+n=l}frac{l!}{m!n!}x^my^nright)=sum_{lgeq0}frac{(x+y)^l}{l!}=exp(x+y).$$ And this extends to integrals, yielding the famous Fubini's theorem.
– Sangchul Lee
Dec 2 '18 at 6:28
Riemann rearrangement illuminates the idea that, for absolutely convergent sum we can rearrange its order of summation. Although this sounds not very practical for sums over $mathbb{N}$, it is indeed very useful for sums over $mathbb{N}^k$, i.e., multiple sums. For instance, for $exp(x)=sum_{ngeq0}frac{x^n}{n!}$ we can rearrange sums to prove $$exp(x)exp(y)=sum_{m,ngeq0}frac{x^my^n}{m!n!}=sum_{lgeq0}frac{1}{l!}left(sum_{m+n=l}frac{l!}{m!n!}x^my^nright)=sum_{lgeq0}frac{(x+y)^l}{l!}=exp(x+y).$$ And this extends to integrals, yielding the famous Fubini's theorem.
– Sangchul Lee
Dec 2 '18 at 6:28
2
2
In comment to the previous comment, you need the rearrangement theorem to prove already the first equality, as to show that the infinite summation over infinite summations can be compressed into one series at all, and because the summation over the grid does not imply any specific linear order.
– LutzL
Dec 2 '18 at 10:36
In comment to the previous comment, you need the rearrangement theorem to prove already the first equality, as to show that the infinite summation over infinite summations can be compressed into one series at all, and because the summation over the grid does not imply any specific linear order.
– LutzL
Dec 2 '18 at 10:36
You don't need the rearrangement theorem. Mertens' theorems states that you need only the absolute convergence of one of the series in order to get that the Cauchy product converges and the limes is the product of both series. See also here for a reference.
– p4sch
Dec 2 '18 at 11:10
You don't need the rearrangement theorem. Mertens' theorems states that you need only the absolute convergence of one of the series in order to get that the Cauchy product converges and the limes is the product of both series. See also here for a reference.
– p4sch
Dec 2 '18 at 11:10
I would guess that there are doubly infinite sums that would be taxing to evaluate in some natural order of summation, but much easier in some other order. As Sangchul Lee pointed out, analogously with Fubini's theorem! My answer is relatively weak in the sense that the amount of rearranging is very minor when done in the order of increasing $|m|+|n|$. It was just the first thing that came to my mind :-/
– Jyrki Lahtonen
Dec 4 '18 at 18:44
I would guess that there are doubly infinite sums that would be taxing to evaluate in some natural order of summation, but much easier in some other order. As Sangchul Lee pointed out, analogously with Fubini's theorem! My answer is relatively weak in the sense that the amount of rearranging is very minor when done in the order of increasing $|m|+|n|$. It was just the first thing that came to my mind :-/
– Jyrki Lahtonen
Dec 4 '18 at 18:44
1
1
Robert, well the second bounty came out empty. A bit disappointing as I was sure somebody would have a nice example. Shrug. I guess there are too many calculus and series questions to choose from. I know that my post was not quite what you were looking for. May be somebody shows up later. Ping me, if something great comes up :-)
– Jyrki Lahtonen
Dec 18 '18 at 19:28
Robert, well the second bounty came out empty. A bit disappointing as I was sure somebody would have a nice example. Shrug. I guess there are too many calculus and series questions to choose from. I know that my post was not quite what you were looking for. May be somebody shows up later. Ping me, if something great comes up :-)
– Jyrki Lahtonen
Dec 18 '18 at 19:28
add a comment |
1 Answer
1
active
oldest
votes
Proffering an application of the rearrangement theorem. I'm not sure that this is exactly what you want, because there is no actual calculation of the sum in a closed form.
The existence of doubly periodic functions. That is, functions $f(z)$ that are meromorphic on the entire complex plane, and have two periods, say $omega_1,omega_2inBbb{C}$ that are linearly independent over $Bbb{R}$. In other words, we require that the identity
$$
f(z+momega_1+nomega_2)=f(z)
$$
holds for all non-poles $z$ and all integers $m,n$.
Let us write
$$Lambda={momega_1+nomega_2mid m,ninBbb{Z}}.$$
Under the assumption that $omega_1/omega_2notinBbb{R}$ we have that $Lambda$ is a discrete subset of $Bbb{C}$ that is also an additive free abelian group of rank two.
A construction idea is to start with a doubly infinite sum
$$
f_{Lambda}(z)=sum_{lambdainLambda}frac1{(z-lambda)^3}.
$$
If we fix an $epsilon>0$ it is not difficult to show that this sum converges absolutely and uniformly in the set $U(epsilon)=Bbb{C}setminusbigcup_{lambdainLambda}B(lambda,epsilon),$ where $B(lambda,epsilon)$ is the ball of radius $epsilon$ around the point $lambda$. A key ingredient in the proof is that, when grouping the terms according to $max{|m|,|n|}$ we get a series majorized by $sum_k 1/k^2$.
Consequently we are allowed rearrange the (countably infinitely many) terms of the series $f_Lambda(z)$ as we see fit.
As the point $omega_1+lambda$ ranges over the additive group $Lambda$ while $lambda$ does, the rearrangement theorem precisely tells us that $$f_{Lambda}(z+omega_1)=f(z)$$ for all $zin U(epsilon)$. Repeating the same argument shows that $omega_2$ is also a period.
It follows from standard results (Weierstrass M-test and the like) that the function $f_Lambda(z)$ is then holomorphic in $Bbb{C}setminusLambda$, and has a triple pole at all the points of the lattice $Lambda$. The famous Weierstrass $wp$-function can be constructed from here. Essentially the function $f_Lambda(z)$ is the derivative $wp_{Lambda}'(z)$.
These doubly periodic functions come in handy when working on (complex) elliptic curves and such. I recommend the first chapter of Apostol's Modular Functions and Dirichlet Series for more on the details, and books dedicated to elliptic curves for more.
answered Dec 4 '18 at 14:41
Jyrki Lahtonen
108k12166367
This shares with the Cauchy product (see Sangchul Lee's comment under main) the feature of making sense of a doubly infinite sum like $$sum_{minBbb{Z}}sum_{ninBbb{Z}}a_{m,n}.$$ As long as the number of terms is countably infinite, it is a series, and the rearrangement theorem makes sure that we can do certain types of manipulations.
– Jyrki Lahtonen
Dec 4 '18 at 18:39
I took an introductory course to modular forms, and this was either week two or three. Unrelated to this question, but I remember wondering whether meromorphic functions whose poles were a triangular or hexagonal lattice were any much as special.
– Robert Wolfe
Dec 4 '18 at 23:32
1
The elliptic curve coming out of the triangular lattice ($omega_1=1,omega_2=e^{2pi i/3}$ is somewhat exceptional as a consequence of the extra symmetries. I know too little about it to be useful. Wikipedia article on CM-curves says a bit more.
– Jyrki Lahtonen
Dec 5 '18 at 3:57
add a comment |
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Proffering an application of the rearrangement theorem. I'm not sure that this is exactly what you want, because there is no actual calculation of the sum in a closed form.
The existence of doubly periodic functions. That is, functions $f(z)$ that are meromorphic on the entire complex plane, and have two periods, say $omega_1,omega_2inBbb{C}$ that are linearly independent over $Bbb{R}$. In other words, we require that the identity
$$
f(z+momega_1+nomega_2)=f(z)
$$
holds for all non-poles $z$ and all integers $m,n$.
Let us write
$$Lambda={momega_1+nomega_2mid m,ninBbb{Z}}.$$
Under the assumption that $omega_1/omega_2notinBbb{R}$ we have that $Lambda$ is a discrete subset of $Bbb{C}$ that is also an additive free abelian group of rank two.
A construction idea is to start with a doubly infinite sum
$$
f_{Lambda}(z)=sum_{lambdainLambda}frac1{(z-lambda)^3}.
$$
If we fix an $epsilon>0$ it is not difficult to show that this sum converges absolutely and uniformly in the set $U(epsilon)=Bbb{C}setminusbigcup_{lambdainLambda}B(lambda,epsilon),$ where $B(lambda,epsilon)$ is the ball of radius $epsilon$ around the point $lambda$. A key ingredient in the proof is that, when grouping the terms according to $max{|m|,|n|}$ we get a series majorized by $sum_k 1/k^2$.
Consequently we are allowed rearrange the (countably infinitely many) terms of the series $f_Lambda(z)$ as we see fit.
As the point $omega_1+lambda$ ranges over the additive group $Lambda$ while $lambda$ does, the rearrangement theorem precisely tells us that $$f_{Lambda}(z+omega_1)=f(z)$$ for all $zin U(epsilon)$. Repeating the same argument shows that $omega_2$ is also a period.
It follows from standard results (Weierstrass M-test and the like) that the function $f_Lambda(z)$ is then holomorphic in $Bbb{C}setminusLambda$, and has a triple pole at all the points of the lattice $Lambda$. The famous Weierstrass $wp$-function can be constructed from here. Essentially the function $f_Lambda(z)$ is the derivative $wp_{Lambda}'(z)$.
These doubly periodic functions come in handy when working on (complex) elliptic curves and such. I recommend the first chapter of Apostol's Modular Functions and Dirichlet Series for more on the details, and books dedicated to elliptic curves for more.
answered Dec 4 '18 at 14:41
Jyrki Lahtonen
108k12166367
This shares with the Cauchy product (see Sangchul Lee's comment under main) the feature of making sense of a doubly infinite sum like $$sum_{minBbb{Z}}sum_{ninBbb{Z}}a_{m,n}.$$ As long as the number of terms is countably infinite, it is a series, and the rearrangement theorem makes sure that we can do certain types of manipulations.
– Jyrki Lahtonen
Dec 4 '18 at 18:39
I took an introductory course to modular forms, and this was either week two or three. Unrelated to this question, but I remember wondering whether meromorphic functions whose poles were a triangular or hexagonal lattice were any much as special.
– Robert Wolfe
Dec 4 '18 at 23:32
1
The elliptic curve coming out of the triangular lattice ($omega_1=1,omega_2=e^{2pi i/3}$ is somewhat exceptional as a consequence of the extra symmetries. I know too little about it to be useful. Wikipedia article on CM-curves says a bit more.
– Jyrki Lahtonen
Dec 5 '18 at 3:57
add a comment |
Proffering an application of the rearrangement theorem. I'm not sure that this is exactly what you want, because there is no actual calculation of the sum in a closed form.
The existence of doubly periodic functions. That is, functions $f(z)$ that are meromorphic on the entire complex plane, and have two periods, say $omega_1,omega_2inBbb{C}$ that are linearly independent over $Bbb{R}$. In other words, we require that the identity
$$
f(z+momega_1+nomega_2)=f(z)
$$
holds for all non-poles $z$ and all integers $m,n$.
Let us write
$$Lambda={momega_1+nomega_2mid m,ninBbb{Z}}.$$
Under the assumption that $omega_1/omega_2notinBbb{R}$ we have that $Lambda$ is a discrete subset of $Bbb{C}$ that is also an additive free abelian group of rank two.
A construction idea is to start with a doubly infinite sum
$$
f_{Lambda}(z)=sum_{lambdainLambda}frac1{(z-lambda)^3}.
$$
If we fix an $epsilon>0$ it is not difficult to show that this sum converges absolutely and uniformly in the set $U(epsilon)=Bbb{C}setminusbigcup_{lambdainLambda}B(lambda,epsilon),$ where $B(lambda,epsilon)$ is the ball of radius $epsilon$ around the point $lambda$. A key ingredient in the proof is that, when grouping the terms according to $max{|m|,|n|}$ we get a series majorized by $sum_k 1/k^2$.
Consequently we are allowed rearrange the (countably infinitely many) terms of the series $f_Lambda(z)$ as we see fit.
As the point $omega_1+lambda$ ranges over the additive group $Lambda$ while $lambda$ does, the rearrangement theorem precisely tells us that $$f_{Lambda}(z+omega_1)=f(z)$$ for all $zin U(epsilon)$. Repeating the same argument shows that $omega_2$ is also a period.
It follows from standard results (Weierstrass M-test and the like) that the function $f_Lambda(z)$ is then holomorphic in $Bbb{C}setminusLambda$, and has a triple pole at all the points of the lattice $Lambda$. The famous Weierstrass $wp$-function can be constructed from here. Essentially the function $f_Lambda(z)$ is the derivative $wp_{Lambda}'(z)$.
These doubly periodic functions come in handy when working on (complex) elliptic curves and such. I recommend the first chapter of Apostol's Modular Functions and Dirichlet Series for more on the details, and books dedicated to elliptic curves for more.
answered Dec 4 '18 at 14:41
Jyrki Lahtonen
108k12166367
This shares with the Cauchy product (see Sangchul Lee's comment under main) the feature of making sense of a doubly infinite sum like $$sum_{minBbb{Z}}sum_{ninBbb{Z}}a_{m,n}.$$ As long as the number of terms is countably infinite, it is a series, and the rearrangement theorem makes sure that we can do certain types of manipulations.
– Jyrki Lahtonen
Dec 4 '18 at 18:39
I took an introductory course to modular forms, and this was either week two or three. Unrelated to this question, but I remember wondering whether meromorphic functions whose poles were a triangular or hexagonal lattice were any much as special.
– Robert Wolfe
Dec 4 '18 at 23:32
1
The elliptic curve coming out of the triangular lattice ($omega_1=1,omega_2=e^{2pi i/3}$ is somewhat exceptional as a consequence of the extra symmetries. I know too little about it to be useful. Wikipedia article on CM-curves says a bit more.
– Jyrki Lahtonen
Dec 5 '18 at 3:57
add a comment |
Proffering an application of the rearrangement theorem. I'm not sure that this is exactly what you want, because there is no actual calculation of the sum in a closed form.
The existence of doubly periodic functions. That is, functions $f(z)$ that are meromorphic on the entire complex plane, and have two periods, say $omega_1,omega_2inBbb{C}$ that are linearly independent over $Bbb{R}$. In other words, we require that the identity
$$
f(z+momega_1+nomega_2)=f(z)
$$
holds for all non-poles $z$ and all integers $m,n$.
Let us write
$$Lambda={momega_1+nomega_2mid m,ninBbb{Z}}.$$
Under the assumption that $omega_1/omega_2notinBbb{R}$ we have that $Lambda$ is a discrete subset of $Bbb{C}$ that is also an additive free abelian group of rank two.
A construction idea is to start with a doubly infinite sum
$$
f_{Lambda}(z)=sum_{lambdainLambda}frac1{(z-lambda)^3}.
$$
If we fix an $epsilon>0$ it is not difficult to show that this sum converges absolutely and uniformly in the set $U(epsilon)=Bbb{C}setminusbigcup_{lambdainLambda}B(lambda,epsilon),$ where $B(lambda,epsilon)$ is the ball of radius $epsilon$ around the point $lambda$. A key ingredient in the proof is that, when grouping the terms according to $max{|m|,|n|}$ we get a series majorized by $sum_k 1/k^2$.
Consequently we are allowed rearrange the (countably infinitely many) terms of the series $f_Lambda(z)$ as we see fit.
As the point $omega_1+lambda$ ranges over the additive group $Lambda$ while $lambda$ does, the rearrangement theorem precisely tells us that $$f_{Lambda}(z+omega_1)=f(z)$$ for all $zin U(epsilon)$. Repeating the same argument shows that $omega_2$ is also a period.
It follows from standard results (Weierstrass M-test and the like) that the function $f_Lambda(z)$ is then holomorphic in $Bbb{C}setminusLambda$, and has a triple pole at all the points of the lattice $Lambda$. The famous Weierstrass $wp$-function can be constructed from here. Essentially the function $f_Lambda(z)$ is the derivative $wp_{Lambda}'(z)$.
These doubly periodic functions come in handy when working on (complex) elliptic curves and such. I recommend the first chapter of Apostol's Modular Functions and Dirichlet Series for more on the details, and books dedicated to elliptic curves for more.
answered Dec 4 '18 at 14:41
Jyrki Lahtonen
108k12166367
Proffering an application of the rearrangement theorem. I'm not sure that this is exactly what you want, because there is no actual calculation of the sum in a closed form.
The existence of doubly periodic functions. That is, functions $f(z)$ that are meromorphic on the entire complex plane, and have two periods, say $omega_1,omega_2inBbb{C}$ that are linearly independent over $Bbb{R}$. In other words, we require that the identity
$$
f(z+momega_1+nomega_2)=f(z)
$$
holds for all non-poles $z$ and all integers $m,n$.
Let us write
$$Lambda={momega_1+nomega_2mid m,ninBbb{Z}}.$$
Under the assumption that $omega_1/omega_2notinBbb{R}$ we have that $Lambda$ is a discrete subset of $Bbb{C}$ that is also an additive free abelian group of rank two.
A construction idea is to start with a doubly infinite sum
$$
f_{Lambda}(z)=sum_{lambdainLambda}frac1{(z-lambda)^3}.
$$
If we fix an $epsilon>0$ it is not difficult to show that this sum converges absolutely and uniformly in the set $U(epsilon)=Bbb{C}setminusbigcup_{lambdainLambda}B(lambda,epsilon),$ where $B(lambda,epsilon)$ is the ball of radius $epsilon$ around the point $lambda$. A key ingredient in the proof is that, when grouping the terms according to $max{|m|,|n|}$ we get a series majorized by $sum_k 1/k^2$.
Consequently we are allowed rearrange the (countably infinitely many) terms of the series $f_Lambda(z)$ as we see fit.
As the point $omega_1+lambda$ ranges over the additive group $Lambda$ while $lambda$ does, the rearrangement theorem precisely tells us that $$f_{Lambda}(z+omega_1)=f(z)$$ for all $zin U(epsilon)$. Repeating the same argument shows that $omega_2$ is also a period.
It follows from standard results (Weierstrass M-test and the like) that the function $f_Lambda(z)$ is then holomorphic in $Bbb{C}setminusLambda$, and has a triple pole at all the points of the lattice $Lambda$. The famous Weierstrass $wp$-function can be constructed from here. Essentially the function $f_Lambda(z)$ is the derivative $wp_{Lambda}'(z)$.
These doubly periodic functions come in handy when working on (complex) elliptic curves and such. I recommend the first chapter of Apostol's Modular Functions and Dirichlet Series for more on the details, and books dedicated to elliptic curves for more.
answered Dec 4 '18 at 14:41
Jyrki Lahtonen
108k12166367
edited Dec 4 '18 at 14:47
answered Dec 4 '18 at 14:41
Jyrki Lahtonen
108k12166367
answered Dec 4 '18 at 14:41
Jyrki Lahtonen
108k12166367
answered Dec 4 '18 at 14:41
Jyrki Lahtonen
108k12166367
108k12166367
This shares with the Cauchy product (see Sangchul Lee's comment under main) the feature of making sense of a doubly infinite sum like $$sum_{minBbb{Z}}sum_{ninBbb{Z}}a_{m,n}.$$ As long as the number of terms is countably infinite, it is a series, and the rearrangement theorem makes sure that we can do certain types of manipulations.
– Jyrki Lahtonen
Dec 4 '18 at 18:39
I took an introductory course to modular forms, and this was either week two or three. Unrelated to this question, but I remember wondering whether meromorphic functions whose poles were a triangular or hexagonal lattice were any much as special.
– Robert Wolfe
Dec 4 '18 at 23:32
1
The elliptic curve coming out of the triangular lattice ($omega_1=1,omega_2=e^{2pi i/3}$ is somewhat exceptional as a consequence of the extra symmetries. I know too little about it to be useful. Wikipedia article on CM-curves says a bit more.
– Jyrki Lahtonen
Dec 5 '18 at 3:57
add a comment |
This shares with the Cauchy product (see Sangchul Lee's comment under main) the feature of making sense of a doubly infinite sum like $$sum_{minBbb{Z}}sum_{ninBbb{Z}}a_{m,n}.$$ As long as the number of terms is countably infinite, it is a series, and the rearrangement theorem makes sure that we can do certain types of manipulations.
– Jyrki Lahtonen
Dec 4 '18 at 18:39
I took an introductory course to modular forms, and this was either week two or three. Unrelated to this question, but I remember wondering whether meromorphic functions whose poles were a triangular or hexagonal lattice were any much as special.
– Robert Wolfe
Dec 4 '18 at 23:32
1
The elliptic curve coming out of the triangular lattice ($omega_1=1,omega_2=e^{2pi i/3}$ is somewhat exceptional as a consequence of the extra symmetries. I know too little about it to be useful. Wikipedia article on CM-curves says a bit more.
– Jyrki Lahtonen
Dec 5 '18 at 3:57
This shares with the Cauchy product (see Sangchul Lee's comment under main) the feature of making sense of a doubly infinite sum like $$sum_{minBbb{Z}}sum_{ninBbb{Z}}a_{m,n}.$$ As long as the number of terms is countably infinite, it is a series, and the rearrangement theorem makes sure that we can do certain types of manipulations.
– Jyrki Lahtonen
Dec 4 '18 at 18:39
This shares with the Cauchy product (see Sangchul Lee's comment under main) the feature of making sense of a doubly infinite sum like $$sum_{minBbb{Z}}sum_{ninBbb{Z}}a_{m,n}.$$ As long as the number of terms is countably infinite, it is a series, and the rearrangement theorem makes sure that we can do certain types of manipulations.
– Jyrki Lahtonen
Dec 4 '18 at 18:39
I took an introductory course to modular forms, and this was either week two or three. Unrelated to this question, but I remember wondering whether meromorphic functions whose poles were a triangular or hexagonal lattice were any much as special.
– Robert Wolfe
Dec 4 '18 at 23:32
I took an introductory course to modular forms, and this was either week two or three. Unrelated to this question, but I remember wondering whether meromorphic functions whose poles were a triangular or hexagonal lattice were any much as special.
– Robert Wolfe
Dec 4 '18 at 23:32
1
1
The elliptic curve coming out of the triangular lattice ($omega_1=1,omega_2=e^{2pi i/3}$ is somewhat exceptional as a consequence of the extra symmetries. I know too little about it to be useful. Wikipedia article on CM-curves says a bit more.
– Jyrki Lahtonen
Dec 5 '18 at 3:57
The elliptic curve coming out of the triangular lattice ($omega_1=1,omega_2=e^{2pi i/3}$ is somewhat exceptional as a consequence of the extra symmetries. I know too little about it to be useful. Wikipedia article on CM-curves says a bit more.
– Jyrki Lahtonen
Dec 5 '18 at 3:57
add a comment |
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13
Riemann rearrangement illuminates the idea that, for absolutely convergent sum we can rearrange its order of summation. Although this sounds not very practical for sums over $mathbb{N}$, it is indeed very useful for sums over $mathbb{N}^k$, i.e., multiple sums. For instance, for $exp(x)=sum_{ngeq0}frac{x^n}{n!}$ we can rearrange sums to prove $$exp(x)exp(y)=sum_{m,ngeq0}frac{x^my^n}{m!n!}=sum_{lgeq0}frac{1}{l!}left(sum_{m+n=l}frac{l!}{m!n!}x^my^nright)=sum_{lgeq0}frac{(x+y)^l}{l!}=exp(x+y).$$ And this extends to integrals, yielding the famous Fubini's theorem.
– Sangchul Lee
Dec 2 '18 at 6:28
2
In comment to the previous comment, you need the rearrangement theorem to prove already the first equality, as to show that the infinite summation over infinite summations can be compressed into one series at all, and because the summation over the grid does not imply any specific linear order.
– LutzL
Dec 2 '18 at 10:36
You don't need the rearrangement theorem. Mertens' theorems states that you need only the absolute convergence of one of the series in order to get that the Cauchy product converges and the limes is the product of both series. See also here for a reference.
– p4sch
Dec 2 '18 at 11:10
I would guess that there are doubly infinite sums that would be taxing to evaluate in some natural order of summation, but much easier in some other order. As Sangchul Lee pointed out, analogously with Fubini's theorem! My answer is relatively weak in the sense that the amount of rearranging is very minor when done in the order of increasing $|m|+|n|$. It was just the first thing that came to my mind :-/
– Jyrki Lahtonen
Dec 4 '18 at 18:44
1
Robert, well the second bounty came out empty. A bit disappointing as I was sure somebody would have a nice example. Shrug. I guess there are too many calculus and series questions to choose from. I know that my post was not quite what you were looking for. May be somebody shows up later. Ping me, if something great comes up :-)
– Jyrki Lahtonen
Dec 18 '18 at 19:28