What is known about the transformation of a power series in which $z^n$ is replaced with $z^{n^2}$?
up vote
4
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Say we have the function $$G(z) = sum_{n geq 0} g_n z^n.$$
Is there a name for the transform T defined so that $$(T(G))(z) = sum_{n geq 0} g_n z^{n^2}?$$
Is there anything known about this kind of transformation?
transformation power-series
asked Nov 12 '10 at 23:22
Andrew MacFie
10213
|
show 5 more comments
up vote
4
down vote
favorite
Say we have the function $$G(z) = sum_{n geq 0} g_n z^n.$$
Is there a name for the transform T defined so that $$(T(G))(z) = sum_{n geq 0} g_n z^{n^2}?$$
Is there anything known about this kind of transformation?
transformation power-series
asked Nov 12 '10 at 23:22
Andrew MacFie
10213
2
Well, they become lacunary most of the time, and z usually becomes restricted to be within the unit disk after such a change.
– J. M. is not a mathematician
Nov 12 '10 at 23:25
2
The original series does not really bear a strong relation to the new series. I don't think this is a useful definition.
– Qiaochu Yuan
Nov 12 '10 at 23:38
@J.M. I don't think it actually becomes lacunary in this case. I think that it pretty much has the same properties as the original series: en.wikipedia.org/wiki/Lacunary_function
– Eric Haengel
Nov 12 '10 at 23:49
5
@Eric: it really doesn't. For example, the "transform" of the innocent rational function 1/(1-z) is a much more complicated beast: en.wikipedia.org/wiki/Theta_function
– Qiaochu Yuan
Nov 13 '10 at 0:15
3
@Qiaochu Yuan: Sure, you may be right. However, to me it is not obvious that there is no application. :)
– AD.
Nov 13 '10 at 11:55
|
show 5 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Say we have the function $$G(z) = sum_{n geq 0} g_n z^n.$$
Is there a name for the transform T defined so that $$(T(G))(z) = sum_{n geq 0} g_n z^{n^2}?$$
Is there anything known about this kind of transformation?
transformation power-series
asked Nov 12 '10 at 23:22
Andrew MacFie
10213
Say we have the function $$G(z) = sum_{n geq 0} g_n z^n.$$
Is there a name for the transform T defined so that $$(T(G))(z) = sum_{n geq 0} g_n z^{n^2}?$$
Is there anything known about this kind of transformation?
transformation power-series
transformation power-series
asked Nov 12 '10 at 23:22
Andrew MacFie
10213
asked Nov 12 '10 at 23:22
Andrew MacFie
10213
asked Nov 12 '10 at 23:22
Andrew MacFie
10213
asked Nov 12 '10 at 23:22
Andrew MacFie
10213
asked Nov 12 '10 at 23:22
Andrew MacFie
10213
10213
2
Well, they become lacunary most of the time, and z usually becomes restricted to be within the unit disk after such a change.
– J. M. is not a mathematician
Nov 12 '10 at 23:25
2
The original series does not really bear a strong relation to the new series. I don't think this is a useful definition.
– Qiaochu Yuan
Nov 12 '10 at 23:38
@J.M. I don't think it actually becomes lacunary in this case. I think that it pretty much has the same properties as the original series: en.wikipedia.org/wiki/Lacunary_function
– Eric Haengel
Nov 12 '10 at 23:49
5
@Eric: it really doesn't. For example, the "transform" of the innocent rational function 1/(1-z) is a much more complicated beast: en.wikipedia.org/wiki/Theta_function
– Qiaochu Yuan
Nov 13 '10 at 0:15
3
@Qiaochu Yuan: Sure, you may be right. However, to me it is not obvious that there is no application. :)
– AD.
Nov 13 '10 at 11:55
|
show 5 more comments
2
Well, they become lacunary most of the time, and z usually becomes restricted to be within the unit disk after such a change.
– J. M. is not a mathematician
Nov 12 '10 at 23:25
2
The original series does not really bear a strong relation to the new series. I don't think this is a useful definition.
– Qiaochu Yuan
Nov 12 '10 at 23:38
@J.M. I don't think it actually becomes lacunary in this case. I think that it pretty much has the same properties as the original series: en.wikipedia.org/wiki/Lacunary_function
– Eric Haengel
Nov 12 '10 at 23:49
5
@Eric: it really doesn't. For example, the "transform" of the innocent rational function 1/(1-z) is a much more complicated beast: en.wikipedia.org/wiki/Theta_function
– Qiaochu Yuan
Nov 13 '10 at 0:15
3
@Qiaochu Yuan: Sure, you may be right. However, to me it is not obvious that there is no application. :)
– AD.
Nov 13 '10 at 11:55
2
2
Well, they become lacunary most of the time, and z usually becomes restricted to be within the unit disk after such a change.
– J. M. is not a mathematician
Nov 12 '10 at 23:25
Well, they become lacunary most of the time, and z usually becomes restricted to be within the unit disk after such a change.
– J. M. is not a mathematician
Nov 12 '10 at 23:25
2
2
The original series does not really bear a strong relation to the new series. I don't think this is a useful definition.
– Qiaochu Yuan
Nov 12 '10 at 23:38
The original series does not really bear a strong relation to the new series. I don't think this is a useful definition.
– Qiaochu Yuan
Nov 12 '10 at 23:38
@J.M. I don't think it actually becomes lacunary in this case. I think that it pretty much has the same properties as the original series: en.wikipedia.org/wiki/Lacunary_function
– Eric Haengel
Nov 12 '10 at 23:49
@J.M. I don't think it actually becomes lacunary in this case. I think that it pretty much has the same properties as the original series: en.wikipedia.org/wiki/Lacunary_function
– Eric Haengel
Nov 12 '10 at 23:49
5
5
@Eric: it really doesn't. For example, the "transform" of the innocent rational function 1/(1-z) is a much more complicated beast: en.wikipedia.org/wiki/Theta_function
– Qiaochu Yuan
Nov 13 '10 at 0:15
@Eric: it really doesn't. For example, the "transform" of the innocent rational function 1/(1-z) is a much more complicated beast: en.wikipedia.org/wiki/Theta_function
– Qiaochu Yuan
Nov 13 '10 at 0:15
3
3
@Qiaochu Yuan: Sure, you may be right. However, to me it is not obvious that there is no application. :)
– AD.
Nov 13 '10 at 11:55
@Qiaochu Yuan: Sure, you may be right. However, to me it is not obvious that there is no application. :)
– AD.
Nov 13 '10 at 11:55
|
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If you know a formula for the ordinary generating function of the sequence and its $j^{th}$ derivatives, which must exist for all $j geq 0$, then this article (2017) provides you with an integral representation of the transformed series in question. In particular, if $G(z)$ is the ordinary generating function of the sequence ${g_n}_{n geq 0}$ and $q in mathbb{C}$ is such that $0 < |q| < 1$, then we have proved in the article that
$$sum_{n geq 0} g_n q^{n^2} z^n = frac{1}{sqrt{2pi}} int_0^{infty}
left[sum_{b = pm 1} Gleft(e^{bt sqrt{2log(q)} z}right)right] e^{-t^2 / 2} dt. $$ The article terms this general procedure for modifying the original sequence generating function a square series transformation integral, but more generally, some of the most interesting applications of this method include new integral representations for theta functions and classical identities such as the series expansion for Jacobi's triple product.
answered May 24 at 14:51
mds
44218
links seems to be broken now
– Andrew MacFie
Nov 7 at 0:50
1
Fixed the link to the arxiv version. Thanks.
– mds
Nov 7 at 1:02
Inside $G$ it should be $e^{bt sqrt{2log(q)}}z$, right?
– Andrew MacFie
Nov 26 at 17:12
Yes, fixing it now.
– mds
Nov 27 at 0:04
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If you know a formula for the ordinary generating function of the sequence and its $j^{th}$ derivatives, which must exist for all $j geq 0$, then this article (2017) provides you with an integral representation of the transformed series in question. In particular, if $G(z)$ is the ordinary generating function of the sequence ${g_n}_{n geq 0}$ and $q in mathbb{C}$ is such that $0 < |q| < 1$, then we have proved in the article that
$$sum_{n geq 0} g_n q^{n^2} z^n = frac{1}{sqrt{2pi}} int_0^{infty}
left[sum_{b = pm 1} Gleft(e^{bt sqrt{2log(q)} z}right)right] e^{-t^2 / 2} dt. $$ The article terms this general procedure for modifying the original sequence generating function a square series transformation integral, but more generally, some of the most interesting applications of this method include new integral representations for theta functions and classical identities such as the series expansion for Jacobi's triple product.
answered May 24 at 14:51
mds
44218
links seems to be broken now
– Andrew MacFie
Nov 7 at 0:50
1
Fixed the link to the arxiv version. Thanks.
– mds
Nov 7 at 1:02
Inside $G$ it should be $e^{bt sqrt{2log(q)}}z$, right?
– Andrew MacFie
Nov 26 at 17:12
Yes, fixing it now.
– mds
Nov 27 at 0:04
add a comment |
up vote
2
down vote
accepted
If you know a formula for the ordinary generating function of the sequence and its $j^{th}$ derivatives, which must exist for all $j geq 0$, then this article (2017) provides you with an integral representation of the transformed series in question. In particular, if $G(z)$ is the ordinary generating function of the sequence ${g_n}_{n geq 0}$ and $q in mathbb{C}$ is such that $0 < |q| < 1$, then we have proved in the article that
$$sum_{n geq 0} g_n q^{n^2} z^n = frac{1}{sqrt{2pi}} int_0^{infty}
left[sum_{b = pm 1} Gleft(e^{bt sqrt{2log(q)} z}right)right] e^{-t^2 / 2} dt. $$ The article terms this general procedure for modifying the original sequence generating function a square series transformation integral, but more generally, some of the most interesting applications of this method include new integral representations for theta functions and classical identities such as the series expansion for Jacobi's triple product.
answered May 24 at 14:51
mds
44218
links seems to be broken now
– Andrew MacFie
Nov 7 at 0:50
1
Fixed the link to the arxiv version. Thanks.
– mds
Nov 7 at 1:02
Inside $G$ it should be $e^{bt sqrt{2log(q)}}z$, right?
– Andrew MacFie
Nov 26 at 17:12
Yes, fixing it now.
– mds
Nov 27 at 0:04
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If you know a formula for the ordinary generating function of the sequence and its $j^{th}$ derivatives, which must exist for all $j geq 0$, then this article (2017) provides you with an integral representation of the transformed series in question. In particular, if $G(z)$ is the ordinary generating function of the sequence ${g_n}_{n geq 0}$ and $q in mathbb{C}$ is such that $0 < |q| < 1$, then we have proved in the article that
$$sum_{n geq 0} g_n q^{n^2} z^n = frac{1}{sqrt{2pi}} int_0^{infty}
left[sum_{b = pm 1} Gleft(e^{bt sqrt{2log(q)} z}right)right] e^{-t^2 / 2} dt. $$ The article terms this general procedure for modifying the original sequence generating function a square series transformation integral, but more generally, some of the most interesting applications of this method include new integral representations for theta functions and classical identities such as the series expansion for Jacobi's triple product.
answered May 24 at 14:51
mds
44218
If you know a formula for the ordinary generating function of the sequence and its $j^{th}$ derivatives, which must exist for all $j geq 0$, then this article (2017) provides you with an integral representation of the transformed series in question. In particular, if $G(z)$ is the ordinary generating function of the sequence ${g_n}_{n geq 0}$ and $q in mathbb{C}$ is such that $0 < |q| < 1$, then we have proved in the article that
$$sum_{n geq 0} g_n q^{n^2} z^n = frac{1}{sqrt{2pi}} int_0^{infty}
left[sum_{b = pm 1} Gleft(e^{bt sqrt{2log(q)} z}right)right] e^{-t^2 / 2} dt. $$ The article terms this general procedure for modifying the original sequence generating function a square series transformation integral, but more generally, some of the most interesting applications of this method include new integral representations for theta functions and classical identities such as the series expansion for Jacobi's triple product.
answered May 24 at 14:51
mds
44218
edited Nov 27 at 0:05
answered May 24 at 14:51
mds
44218
answered May 24 at 14:51
mds
44218
answered May 24 at 14:51
mds
44218
44218
links seems to be broken now
– Andrew MacFie
Nov 7 at 0:50
1
Fixed the link to the arxiv version. Thanks.
– mds
Nov 7 at 1:02
Inside $G$ it should be $e^{bt sqrt{2log(q)}}z$, right?
– Andrew MacFie
Nov 26 at 17:12
Yes, fixing it now.
– mds
Nov 27 at 0:04
add a comment |
links seems to be broken now
– Andrew MacFie
Nov 7 at 0:50
1
Fixed the link to the arxiv version. Thanks.
– mds
Nov 7 at 1:02
Inside $G$ it should be $e^{bt sqrt{2log(q)}}z$, right?
– Andrew MacFie
Nov 26 at 17:12
Yes, fixing it now.
– mds
Nov 27 at 0:04
links seems to be broken now
– Andrew MacFie
Nov 7 at 0:50
links seems to be broken now
– Andrew MacFie
Nov 7 at 0:50
1
1
Fixed the link to the arxiv version. Thanks.
– mds
Nov 7 at 1:02
Fixed the link to the arxiv version. Thanks.
– mds
Nov 7 at 1:02
Inside $G$ it should be $e^{bt sqrt{2log(q)}}z$, right?
– Andrew MacFie
Nov 26 at 17:12
Inside $G$ it should be $e^{bt sqrt{2log(q)}}z$, right?
– Andrew MacFie
Nov 26 at 17:12
Yes, fixing it now.
– mds
Nov 27 at 0:04
Yes, fixing it now.
– mds
Nov 27 at 0:04
add a comment |
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2
Well, they become lacunary most of the time, and z usually becomes restricted to be within the unit disk after such a change.
– J. M. is not a mathematician
Nov 12 '10 at 23:25
2
The original series does not really bear a strong relation to the new series. I don't think this is a useful definition.
– Qiaochu Yuan
Nov 12 '10 at 23:38
@J.M. I don't think it actually becomes lacunary in this case. I think that it pretty much has the same properties as the original series: en.wikipedia.org/wiki/Lacunary_function
– Eric Haengel
Nov 12 '10 at 23:49
5
@Eric: it really doesn't. For example, the "transform" of the innocent rational function 1/(1-z) is a much more complicated beast: en.wikipedia.org/wiki/Theta_function
– Qiaochu Yuan
Nov 13 '10 at 0:15
3
@Qiaochu Yuan: Sure, you may be right. However, to me it is not obvious that there is no application. :)
– AD.
Nov 13 '10 at 11:55