Deep understanding on exponential generating function
$begingroup$
In spite of having done some exercises, I still find it harder to understand exponential generating function deeply than ordinary generating function. Could someone explain it "deeply"? Or are there any articles on that?
combinatorics generating-functions
$endgroup$
|
show 1 more comment
$begingroup$
In spite of having done some exercises, I still find it harder to understand exponential generating function deeply than ordinary generating function. Could someone explain it "deeply"? Or are there any articles on that?
combinatorics generating-functions
$endgroup$
$begingroup$
I suggest that you either read the book by Bergeron, Labelle and Leroux or that you ask a more precise question. The book might help to convince you that egfs are simpler than ogfs. books.google.com/…
$endgroup$
– Phira
Nov 6 '11 at 0:02
1
$begingroup$
I like "Generatingfunctionology" by Wilf. The 2nd edition can be downloaded from math.upenn.edu/~wilf/DownldGF.html. The 3rd edition is available from the publisher.
$endgroup$
– marty cohen
Nov 6 '11 at 0:17
1
$begingroup$
Chapter $3$ of Miklós Bóna’s Introduction to Enumerative Combinatorics gives a nice introductory sense of the respective contexts in which egf’s and ogf’s are useful.
$endgroup$
– Brian M. Scott
Nov 6 '11 at 0:23
5
$begingroup$
Think of it this way: there are situations where the EGF is simple/easy to manipulate when the OGF is unwieldy or has no closed form, and vice-versa...
$endgroup$
– J. M. is not a mathematician
Nov 6 '11 at 1:08
3
$begingroup$
anon guessed correctly, here's an old blog post by Qiaochu somewhat related to this.
$endgroup$
– Ragib Zaman
Nov 6 '11 at 2:19
|
show 1 more comment
$begingroup$
In spite of having done some exercises, I still find it harder to understand exponential generating function deeply than ordinary generating function. Could someone explain it "deeply"? Or are there any articles on that?
combinatorics generating-functions
$endgroup$
In spite of having done some exercises, I still find it harder to understand exponential generating function deeply than ordinary generating function. Could someone explain it "deeply"? Or are there any articles on that?
combinatorics generating-functions
combinatorics generating-functions
asked Nov 5 '11 at 23:40
VladimirVladimir
1,11311231
1,11311231
$begingroup$
I suggest that you either read the book by Bergeron, Labelle and Leroux or that you ask a more precise question. The book might help to convince you that egfs are simpler than ogfs. books.google.com/…
$endgroup$
– Phira
Nov 6 '11 at 0:02
1
$begingroup$
I like "Generatingfunctionology" by Wilf. The 2nd edition can be downloaded from math.upenn.edu/~wilf/DownldGF.html. The 3rd edition is available from the publisher.
$endgroup$
– marty cohen
Nov 6 '11 at 0:17
1
$begingroup$
Chapter $3$ of Miklós Bóna’s Introduction to Enumerative Combinatorics gives a nice introductory sense of the respective contexts in which egf’s and ogf’s are useful.
$endgroup$
– Brian M. Scott
Nov 6 '11 at 0:23
5
$begingroup$
Think of it this way: there are situations where the EGF is simple/easy to manipulate when the OGF is unwieldy or has no closed form, and vice-versa...
$endgroup$
– J. M. is not a mathematician
Nov 6 '11 at 1:08
3
$begingroup$
anon guessed correctly, here's an old blog post by Qiaochu somewhat related to this.
$endgroup$
– Ragib Zaman
Nov 6 '11 at 2:19
|
show 1 more comment
$begingroup$
I suggest that you either read the book by Bergeron, Labelle and Leroux or that you ask a more precise question. The book might help to convince you that egfs are simpler than ogfs. books.google.com/…
$endgroup$
– Phira
Nov 6 '11 at 0:02
1
$begingroup$
I like "Generatingfunctionology" by Wilf. The 2nd edition can be downloaded from math.upenn.edu/~wilf/DownldGF.html. The 3rd edition is available from the publisher.
$endgroup$
– marty cohen
Nov 6 '11 at 0:17
1
$begingroup$
Chapter $3$ of Miklós Bóna’s Introduction to Enumerative Combinatorics gives a nice introductory sense of the respective contexts in which egf’s and ogf’s are useful.
$endgroup$
– Brian M. Scott
Nov 6 '11 at 0:23
5
$begingroup$
Think of it this way: there are situations where the EGF is simple/easy to manipulate when the OGF is unwieldy or has no closed form, and vice-versa...
$endgroup$
– J. M. is not a mathematician
Nov 6 '11 at 1:08
3
$begingroup$
anon guessed correctly, here's an old blog post by Qiaochu somewhat related to this.
$endgroup$
– Ragib Zaman
Nov 6 '11 at 2:19
$begingroup$
I suggest that you either read the book by Bergeron, Labelle and Leroux or that you ask a more precise question. The book might help to convince you that egfs are simpler than ogfs. books.google.com/…
$endgroup$
– Phira
Nov 6 '11 at 0:02
$begingroup$
I suggest that you either read the book by Bergeron, Labelle and Leroux or that you ask a more precise question. The book might help to convince you that egfs are simpler than ogfs. books.google.com/…
$endgroup$
– Phira
Nov 6 '11 at 0:02
1
1
$begingroup$
I like "Generatingfunctionology" by Wilf. The 2nd edition can be downloaded from math.upenn.edu/~wilf/DownldGF.html. The 3rd edition is available from the publisher.
$endgroup$
– marty cohen
Nov 6 '11 at 0:17
$begingroup$
I like "Generatingfunctionology" by Wilf. The 2nd edition can be downloaded from math.upenn.edu/~wilf/DownldGF.html. The 3rd edition is available from the publisher.
$endgroup$
– marty cohen
Nov 6 '11 at 0:17
1
1
$begingroup$
Chapter $3$ of Miklós Bóna’s Introduction to Enumerative Combinatorics gives a nice introductory sense of the respective contexts in which egf’s and ogf’s are useful.
$endgroup$
– Brian M. Scott
Nov 6 '11 at 0:23
$begingroup$
Chapter $3$ of Miklós Bóna’s Introduction to Enumerative Combinatorics gives a nice introductory sense of the respective contexts in which egf’s and ogf’s are useful.
$endgroup$
– Brian M. Scott
Nov 6 '11 at 0:23
5
5
$begingroup$
Think of it this way: there are situations where the EGF is simple/easy to manipulate when the OGF is unwieldy or has no closed form, and vice-versa...
$endgroup$
– J. M. is not a mathematician
Nov 6 '11 at 1:08
$begingroup$
Think of it this way: there are situations where the EGF is simple/easy to manipulate when the OGF is unwieldy or has no closed form, and vice-versa...
$endgroup$
– J. M. is not a mathematician
Nov 6 '11 at 1:08
3
3
$begingroup$
anon guessed correctly, here's an old blog post by Qiaochu somewhat related to this.
$endgroup$
– Ragib Zaman
Nov 6 '11 at 2:19
$begingroup$
anon guessed correctly, here's an old blog post by Qiaochu somewhat related to this.
$endgroup$
– Ragib Zaman
Nov 6 '11 at 2:19
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
We typically use ordinary generating functions when counting unlabelled objects and exponentially generating functions when counting labelled objects. For instance combinatorial enumeration of permutations is most convenient analysed by exponential generating functions.
Here I'd like to give an indication which structures do we count when applying the binomial convolution
begin{align*}
sum_{k=0}^nbinom{n}{k}b_kc_{n-k}
end{align*}
of two exponential generating functions $B(z)=sum_{j=0}^infty b_j frac{z^j}{j!}$ and $C(z)=sum_{k=0}^infty c_kfrac{z^k}{k!}$.
Permutations:
We introduce a universe $mathcal{P}$ of (labelled) permutations. We label the atoms of the permutations of size $n$ as usual with $1,2,ldots,n$. We also add an empty permutation $varepsilon$ of length $0$ and consider
begin{align*}
mathcal{P}={varepsilon,1,12,21,123,132,213,231,312,321,ldots}tag{1}
end{align*}
The coefficients $a_n$ of the corresponding exponential generating function $A(z)=sum_{n=0}^infty a_n frac{z^n}{n!}$ give the number $n!$ of permutations of size $n$. We obtain
begin{align*}
color{blue}{A(z)}=sum_{n=0}^infty n!frac{z^n}{n!}=sum_{n=0}^infty z^ncolor{blue}{=frac{1}{1-z}}
end{align*}
We observe that generating functions like $frac{1}{1-z}$ may serve as both, ordinary generating functions
begin{align*}
frac{1}{1-z}=sum_{n=0}^infty z^n=color{blue}{1}+color{blue}{1}cdot z+color{blue}{1}cdot z^2+color{blue}{1}cdot z^3+cdots
end{align*}
to count for instance a sequence of unary words
begin{align*}
mathcal{S}={varepsilon,bullet,bulletbullet,bulletbulletbullet,ldots}tag{2}
end{align*}
as well as exponential generating functions
begin{align*}
frac{1}{1-z}=sum_{n=0}^infty n!frac{z^n}{n!}=color{blue}{0!}+color{blue}{1!}cdot frac{z}{1!}+color{blue}{2!}cdot frac{z^2}{2!}+color{blue}{3!}cdot frac{z^3}{3!}+cdots
end{align*}
to count permutations according to (1).
Note, the simplicity of the geometric series $frac{1}{1-z}=1+z+z^2+cdots$ strongly indicates the universe of permutations is the prototypical structure approachable by exponential generating functions.
Warmup: Multiplicative structure of ordinary generating functions.
We consider $mathcal{B}={bullet,bulletbulletbullet},mathcal{C}={bullet,bulletbullet}$ with corresponding generating functions $B(z)=z+z^3, C(z)=z+z^2$ and obtain due to the multiplicative rule
begin{align*}
color{blue}{mathrm{card}left(mathcal{B}times mathcal{C}right)=mathrm{card}left(mathcal{B}right)cdot mathrm{card}left(mathcal{C}right)}
end{align*}
the set $mathcal{A}={(bullet,bullet),(bullet,bulletbullet),(bulletbulletbullet,bullet),(bulletbulletbullet,bulletbullet)}$ and the following relationship
begin{align*}
A(z)&=sum_{ainmathcal{A}}z^{|a|}=sum_{(beta,gamma)in(mathcal{B}timesmathcal{C})}z^{|beta|+|gamma|}
=left(sum_{betain mathcal{B}}z^{|beta|}right)timesleft(sum_{gammain mathcal{C}}z^{|C|}right)=B(z)cdot C(z)\
A(z)&=z^2+z^3+z^4+z^5\
&=z^{1+1}+z^{1+2}+z^{3+1}+z^{3+2}=left(z^1+z^3right)timesleft(z^1+z^2right)=(z+z^3)(z+z^2)
end{align*}
which follows from distributing products over sums.
Multiplicative structure of exponential generating functions:
In order to cover the binomial convolution $sum_{k=0}^nbinom{n}{k}b_kc_{n-k}$ of exponential generating functions we will encounter a more sophisticated multiplicative rule for labelled structures. Nevertheless we will do it similarly to the warmup above.
We consider $mathcal{B}={123,213},mathcal{C}={12}$ with generating functions $B(z)=2frac{z^3}{3!}, C(z)=frac{z^2}{2!}$ and consider the multiplicative rule
begin{align*}
color{blue}{mathcal{B}starmathcal{C}=bigcup_{betainmathcal{B},gammainmathcal{C}}left(betastargammaright)}tag{3}
end{align*}
which is defined as:
- The labelled product (3) of $mathcal{B}$ and $mathcal{C}$, denoted $mathcal{B}starmathcal{C}$, is obtained by forming ordered pairs from $Bstar C$ and performing all possible order-consistent relabellings.
The last phrase all possible order-consistent relabellings needs to be explained.
With the example $mathcal{B}={123,213}$ and $mathcal{C}={12}$ we perform a relabeling.
At first we start with a relabelling and select wlog $mathcal{C}={45}$ by respecting the order ($1<2$, i.e. $4<5$). We now obtain all order-consistent relabellings as follows
begin{align*}
mathcal{B}starmathcal{C}&=bigcup_{betainmathcal{B},gammainmathcal{C}}left(betastargammaright)\
&=left({123}star{45}right)cupleft({213}star{45}right)\
&={12345,12435,12534,13425,13524,14523,23415,23514,24513,34512}tag{4}\
&qquadcup{21345,21435,21534,31425,31524,41523,32415,32514,42513,43512}tag{5}\
end{align*}
When looking at the set ${123}star{45}$ we can find all order-consistent relabelings as follows: We have to generate permutations of length $5$. We create all $binom{5}{3}=10$ strings of length $3$ from ${1,2,3,4,5}$ which follow the order $1<2<3$. We obtain
begin{align*}
{123,124,125,134,135,145,234,235,245,345}.
end{align*}
Note, that each element in the set above follows the ordering $(mathrm{smallest} < mathrm{middle} < mathrm {largest})$. We complete (4) by concatenating each $3$-element string with a string consisting of the remaining two elements from ${1,2,3,4,5}$ following the order $4<5$, i.e. $(mathrm{smallest} < mathrm{largest})$. Similarly we build (5) by respecting the order $2<1<3$, i.e. $(mathrm{middle} < mathrm{smallest} < mathrm{largest})$.
We observe from (4), (5):
begin{align*}
|mathcal{B}starmathcal{C}|=2binom{5}{3}=2cdot 10=20
end{align*}
and the relationship with the generating functions $B(z)$ and $C(z)$ is given as
begin{align*}
color{blue}{A(z)}&=B(z)cdot C(z)=left(2cdot frac{z^3}{3!}right)cdotleft(frac{z^2}{2!}right)\
&,,color{blue}{=2binom{5}{3}frac{z^5}{5!}}
end{align*}
Hint: This answer is based upon the great presentation of admissible structures in chapter 1 and 2 of Analytic Combinatorics by P. Flajolet and R. Sedgewick.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f79355%2fdeep-understanding-on-exponential-generating-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We typically use ordinary generating functions when counting unlabelled objects and exponentially generating functions when counting labelled objects. For instance combinatorial enumeration of permutations is most convenient analysed by exponential generating functions.
Here I'd like to give an indication which structures do we count when applying the binomial convolution
begin{align*}
sum_{k=0}^nbinom{n}{k}b_kc_{n-k}
end{align*}
of two exponential generating functions $B(z)=sum_{j=0}^infty b_j frac{z^j}{j!}$ and $C(z)=sum_{k=0}^infty c_kfrac{z^k}{k!}$.
Permutations:
We introduce a universe $mathcal{P}$ of (labelled) permutations. We label the atoms of the permutations of size $n$ as usual with $1,2,ldots,n$. We also add an empty permutation $varepsilon$ of length $0$ and consider
begin{align*}
mathcal{P}={varepsilon,1,12,21,123,132,213,231,312,321,ldots}tag{1}
end{align*}
The coefficients $a_n$ of the corresponding exponential generating function $A(z)=sum_{n=0}^infty a_n frac{z^n}{n!}$ give the number $n!$ of permutations of size $n$. We obtain
begin{align*}
color{blue}{A(z)}=sum_{n=0}^infty n!frac{z^n}{n!}=sum_{n=0}^infty z^ncolor{blue}{=frac{1}{1-z}}
end{align*}
We observe that generating functions like $frac{1}{1-z}$ may serve as both, ordinary generating functions
begin{align*}
frac{1}{1-z}=sum_{n=0}^infty z^n=color{blue}{1}+color{blue}{1}cdot z+color{blue}{1}cdot z^2+color{blue}{1}cdot z^3+cdots
end{align*}
to count for instance a sequence of unary words
begin{align*}
mathcal{S}={varepsilon,bullet,bulletbullet,bulletbulletbullet,ldots}tag{2}
end{align*}
as well as exponential generating functions
begin{align*}
frac{1}{1-z}=sum_{n=0}^infty n!frac{z^n}{n!}=color{blue}{0!}+color{blue}{1!}cdot frac{z}{1!}+color{blue}{2!}cdot frac{z^2}{2!}+color{blue}{3!}cdot frac{z^3}{3!}+cdots
end{align*}
to count permutations according to (1).
Note, the simplicity of the geometric series $frac{1}{1-z}=1+z+z^2+cdots$ strongly indicates the universe of permutations is the prototypical structure approachable by exponential generating functions.
Warmup: Multiplicative structure of ordinary generating functions.
We consider $mathcal{B}={bullet,bulletbulletbullet},mathcal{C}={bullet,bulletbullet}$ with corresponding generating functions $B(z)=z+z^3, C(z)=z+z^2$ and obtain due to the multiplicative rule
begin{align*}
color{blue}{mathrm{card}left(mathcal{B}times mathcal{C}right)=mathrm{card}left(mathcal{B}right)cdot mathrm{card}left(mathcal{C}right)}
end{align*}
the set $mathcal{A}={(bullet,bullet),(bullet,bulletbullet),(bulletbulletbullet,bullet),(bulletbulletbullet,bulletbullet)}$ and the following relationship
begin{align*}
A(z)&=sum_{ainmathcal{A}}z^{|a|}=sum_{(beta,gamma)in(mathcal{B}timesmathcal{C})}z^{|beta|+|gamma|}
=left(sum_{betain mathcal{B}}z^{|beta|}right)timesleft(sum_{gammain mathcal{C}}z^{|C|}right)=B(z)cdot C(z)\
A(z)&=z^2+z^3+z^4+z^5\
&=z^{1+1}+z^{1+2}+z^{3+1}+z^{3+2}=left(z^1+z^3right)timesleft(z^1+z^2right)=(z+z^3)(z+z^2)
end{align*}
which follows from distributing products over sums.
Multiplicative structure of exponential generating functions:
In order to cover the binomial convolution $sum_{k=0}^nbinom{n}{k}b_kc_{n-k}$ of exponential generating functions we will encounter a more sophisticated multiplicative rule for labelled structures. Nevertheless we will do it similarly to the warmup above.
We consider $mathcal{B}={123,213},mathcal{C}={12}$ with generating functions $B(z)=2frac{z^3}{3!}, C(z)=frac{z^2}{2!}$ and consider the multiplicative rule
begin{align*}
color{blue}{mathcal{B}starmathcal{C}=bigcup_{betainmathcal{B},gammainmathcal{C}}left(betastargammaright)}tag{3}
end{align*}
which is defined as:
- The labelled product (3) of $mathcal{B}$ and $mathcal{C}$, denoted $mathcal{B}starmathcal{C}$, is obtained by forming ordered pairs from $Bstar C$ and performing all possible order-consistent relabellings.
The last phrase all possible order-consistent relabellings needs to be explained.
With the example $mathcal{B}={123,213}$ and $mathcal{C}={12}$ we perform a relabeling.
At first we start with a relabelling and select wlog $mathcal{C}={45}$ by respecting the order ($1<2$, i.e. $4<5$). We now obtain all order-consistent relabellings as follows
begin{align*}
mathcal{B}starmathcal{C}&=bigcup_{betainmathcal{B},gammainmathcal{C}}left(betastargammaright)\
&=left({123}star{45}right)cupleft({213}star{45}right)\
&={12345,12435,12534,13425,13524,14523,23415,23514,24513,34512}tag{4}\
&qquadcup{21345,21435,21534,31425,31524,41523,32415,32514,42513,43512}tag{5}\
end{align*}
When looking at the set ${123}star{45}$ we can find all order-consistent relabelings as follows: We have to generate permutations of length $5$. We create all $binom{5}{3}=10$ strings of length $3$ from ${1,2,3,4,5}$ which follow the order $1<2<3$. We obtain
begin{align*}
{123,124,125,134,135,145,234,235,245,345}.
end{align*}
Note, that each element in the set above follows the ordering $(mathrm{smallest} < mathrm{middle} < mathrm {largest})$. We complete (4) by concatenating each $3$-element string with a string consisting of the remaining two elements from ${1,2,3,4,5}$ following the order $4<5$, i.e. $(mathrm{smallest} < mathrm{largest})$. Similarly we build (5) by respecting the order $2<1<3$, i.e. $(mathrm{middle} < mathrm{smallest} < mathrm{largest})$.
We observe from (4), (5):
begin{align*}
|mathcal{B}starmathcal{C}|=2binom{5}{3}=2cdot 10=20
end{align*}
and the relationship with the generating functions $B(z)$ and $C(z)$ is given as
begin{align*}
color{blue}{A(z)}&=B(z)cdot C(z)=left(2cdot frac{z^3}{3!}right)cdotleft(frac{z^2}{2!}right)\
&,,color{blue}{=2binom{5}{3}frac{z^5}{5!}}
end{align*}
Hint: This answer is based upon the great presentation of admissible structures in chapter 1 and 2 of Analytic Combinatorics by P. Flajolet and R. Sedgewick.
$endgroup$
add a comment |
$begingroup$
We typically use ordinary generating functions when counting unlabelled objects and exponentially generating functions when counting labelled objects. For instance combinatorial enumeration of permutations is most convenient analysed by exponential generating functions.
Here I'd like to give an indication which structures do we count when applying the binomial convolution
begin{align*}
sum_{k=0}^nbinom{n}{k}b_kc_{n-k}
end{align*}
of two exponential generating functions $B(z)=sum_{j=0}^infty b_j frac{z^j}{j!}$ and $C(z)=sum_{k=0}^infty c_kfrac{z^k}{k!}$.
Permutations:
We introduce a universe $mathcal{P}$ of (labelled) permutations. We label the atoms of the permutations of size $n$ as usual with $1,2,ldots,n$. We also add an empty permutation $varepsilon$ of length $0$ and consider
begin{align*}
mathcal{P}={varepsilon,1,12,21,123,132,213,231,312,321,ldots}tag{1}
end{align*}
The coefficients $a_n$ of the corresponding exponential generating function $A(z)=sum_{n=0}^infty a_n frac{z^n}{n!}$ give the number $n!$ of permutations of size $n$. We obtain
begin{align*}
color{blue}{A(z)}=sum_{n=0}^infty n!frac{z^n}{n!}=sum_{n=0}^infty z^ncolor{blue}{=frac{1}{1-z}}
end{align*}
We observe that generating functions like $frac{1}{1-z}$ may serve as both, ordinary generating functions
begin{align*}
frac{1}{1-z}=sum_{n=0}^infty z^n=color{blue}{1}+color{blue}{1}cdot z+color{blue}{1}cdot z^2+color{blue}{1}cdot z^3+cdots
end{align*}
to count for instance a sequence of unary words
begin{align*}
mathcal{S}={varepsilon,bullet,bulletbullet,bulletbulletbullet,ldots}tag{2}
end{align*}
as well as exponential generating functions
begin{align*}
frac{1}{1-z}=sum_{n=0}^infty n!frac{z^n}{n!}=color{blue}{0!}+color{blue}{1!}cdot frac{z}{1!}+color{blue}{2!}cdot frac{z^2}{2!}+color{blue}{3!}cdot frac{z^3}{3!}+cdots
end{align*}
to count permutations according to (1).
Note, the simplicity of the geometric series $frac{1}{1-z}=1+z+z^2+cdots$ strongly indicates the universe of permutations is the prototypical structure approachable by exponential generating functions.
Warmup: Multiplicative structure of ordinary generating functions.
We consider $mathcal{B}={bullet,bulletbulletbullet},mathcal{C}={bullet,bulletbullet}$ with corresponding generating functions $B(z)=z+z^3, C(z)=z+z^2$ and obtain due to the multiplicative rule
begin{align*}
color{blue}{mathrm{card}left(mathcal{B}times mathcal{C}right)=mathrm{card}left(mathcal{B}right)cdot mathrm{card}left(mathcal{C}right)}
end{align*}
the set $mathcal{A}={(bullet,bullet),(bullet,bulletbullet),(bulletbulletbullet,bullet),(bulletbulletbullet,bulletbullet)}$ and the following relationship
begin{align*}
A(z)&=sum_{ainmathcal{A}}z^{|a|}=sum_{(beta,gamma)in(mathcal{B}timesmathcal{C})}z^{|beta|+|gamma|}
=left(sum_{betain mathcal{B}}z^{|beta|}right)timesleft(sum_{gammain mathcal{C}}z^{|C|}right)=B(z)cdot C(z)\
A(z)&=z^2+z^3+z^4+z^5\
&=z^{1+1}+z^{1+2}+z^{3+1}+z^{3+2}=left(z^1+z^3right)timesleft(z^1+z^2right)=(z+z^3)(z+z^2)
end{align*}
which follows from distributing products over sums.
Multiplicative structure of exponential generating functions:
In order to cover the binomial convolution $sum_{k=0}^nbinom{n}{k}b_kc_{n-k}$ of exponential generating functions we will encounter a more sophisticated multiplicative rule for labelled structures. Nevertheless we will do it similarly to the warmup above.
We consider $mathcal{B}={123,213},mathcal{C}={12}$ with generating functions $B(z)=2frac{z^3}{3!}, C(z)=frac{z^2}{2!}$ and consider the multiplicative rule
begin{align*}
color{blue}{mathcal{B}starmathcal{C}=bigcup_{betainmathcal{B},gammainmathcal{C}}left(betastargammaright)}tag{3}
end{align*}
which is defined as:
- The labelled product (3) of $mathcal{B}$ and $mathcal{C}$, denoted $mathcal{B}starmathcal{C}$, is obtained by forming ordered pairs from $Bstar C$ and performing all possible order-consistent relabellings.
The last phrase all possible order-consistent relabellings needs to be explained.
With the example $mathcal{B}={123,213}$ and $mathcal{C}={12}$ we perform a relabeling.
At first we start with a relabelling and select wlog $mathcal{C}={45}$ by respecting the order ($1<2$, i.e. $4<5$). We now obtain all order-consistent relabellings as follows
begin{align*}
mathcal{B}starmathcal{C}&=bigcup_{betainmathcal{B},gammainmathcal{C}}left(betastargammaright)\
&=left({123}star{45}right)cupleft({213}star{45}right)\
&={12345,12435,12534,13425,13524,14523,23415,23514,24513,34512}tag{4}\
&qquadcup{21345,21435,21534,31425,31524,41523,32415,32514,42513,43512}tag{5}\
end{align*}
When looking at the set ${123}star{45}$ we can find all order-consistent relabelings as follows: We have to generate permutations of length $5$. We create all $binom{5}{3}=10$ strings of length $3$ from ${1,2,3,4,5}$ which follow the order $1<2<3$. We obtain
begin{align*}
{123,124,125,134,135,145,234,235,245,345}.
end{align*}
Note, that each element in the set above follows the ordering $(mathrm{smallest} < mathrm{middle} < mathrm {largest})$. We complete (4) by concatenating each $3$-element string with a string consisting of the remaining two elements from ${1,2,3,4,5}$ following the order $4<5$, i.e. $(mathrm{smallest} < mathrm{largest})$. Similarly we build (5) by respecting the order $2<1<3$, i.e. $(mathrm{middle} < mathrm{smallest} < mathrm{largest})$.
We observe from (4), (5):
begin{align*}
|mathcal{B}starmathcal{C}|=2binom{5}{3}=2cdot 10=20
end{align*}
and the relationship with the generating functions $B(z)$ and $C(z)$ is given as
begin{align*}
color{blue}{A(z)}&=B(z)cdot C(z)=left(2cdot frac{z^3}{3!}right)cdotleft(frac{z^2}{2!}right)\
&,,color{blue}{=2binom{5}{3}frac{z^5}{5!}}
end{align*}
Hint: This answer is based upon the great presentation of admissible structures in chapter 1 and 2 of Analytic Combinatorics by P. Flajolet and R. Sedgewick.
$endgroup$
add a comment |
$begingroup$
We typically use ordinary generating functions when counting unlabelled objects and exponentially generating functions when counting labelled objects. For instance combinatorial enumeration of permutations is most convenient analysed by exponential generating functions.
Here I'd like to give an indication which structures do we count when applying the binomial convolution
begin{align*}
sum_{k=0}^nbinom{n}{k}b_kc_{n-k}
end{align*}
of two exponential generating functions $B(z)=sum_{j=0}^infty b_j frac{z^j}{j!}$ and $C(z)=sum_{k=0}^infty c_kfrac{z^k}{k!}$.
Permutations:
We introduce a universe $mathcal{P}$ of (labelled) permutations. We label the atoms of the permutations of size $n$ as usual with $1,2,ldots,n$. We also add an empty permutation $varepsilon$ of length $0$ and consider
begin{align*}
mathcal{P}={varepsilon,1,12,21,123,132,213,231,312,321,ldots}tag{1}
end{align*}
The coefficients $a_n$ of the corresponding exponential generating function $A(z)=sum_{n=0}^infty a_n frac{z^n}{n!}$ give the number $n!$ of permutations of size $n$. We obtain
begin{align*}
color{blue}{A(z)}=sum_{n=0}^infty n!frac{z^n}{n!}=sum_{n=0}^infty z^ncolor{blue}{=frac{1}{1-z}}
end{align*}
We observe that generating functions like $frac{1}{1-z}$ may serve as both, ordinary generating functions
begin{align*}
frac{1}{1-z}=sum_{n=0}^infty z^n=color{blue}{1}+color{blue}{1}cdot z+color{blue}{1}cdot z^2+color{blue}{1}cdot z^3+cdots
end{align*}
to count for instance a sequence of unary words
begin{align*}
mathcal{S}={varepsilon,bullet,bulletbullet,bulletbulletbullet,ldots}tag{2}
end{align*}
as well as exponential generating functions
begin{align*}
frac{1}{1-z}=sum_{n=0}^infty n!frac{z^n}{n!}=color{blue}{0!}+color{blue}{1!}cdot frac{z}{1!}+color{blue}{2!}cdot frac{z^2}{2!}+color{blue}{3!}cdot frac{z^3}{3!}+cdots
end{align*}
to count permutations according to (1).
Note, the simplicity of the geometric series $frac{1}{1-z}=1+z+z^2+cdots$ strongly indicates the universe of permutations is the prototypical structure approachable by exponential generating functions.
Warmup: Multiplicative structure of ordinary generating functions.
We consider $mathcal{B}={bullet,bulletbulletbullet},mathcal{C}={bullet,bulletbullet}$ with corresponding generating functions $B(z)=z+z^3, C(z)=z+z^2$ and obtain due to the multiplicative rule
begin{align*}
color{blue}{mathrm{card}left(mathcal{B}times mathcal{C}right)=mathrm{card}left(mathcal{B}right)cdot mathrm{card}left(mathcal{C}right)}
end{align*}
the set $mathcal{A}={(bullet,bullet),(bullet,bulletbullet),(bulletbulletbullet,bullet),(bulletbulletbullet,bulletbullet)}$ and the following relationship
begin{align*}
A(z)&=sum_{ainmathcal{A}}z^{|a|}=sum_{(beta,gamma)in(mathcal{B}timesmathcal{C})}z^{|beta|+|gamma|}
=left(sum_{betain mathcal{B}}z^{|beta|}right)timesleft(sum_{gammain mathcal{C}}z^{|C|}right)=B(z)cdot C(z)\
A(z)&=z^2+z^3+z^4+z^5\
&=z^{1+1}+z^{1+2}+z^{3+1}+z^{3+2}=left(z^1+z^3right)timesleft(z^1+z^2right)=(z+z^3)(z+z^2)
end{align*}
which follows from distributing products over sums.
Multiplicative structure of exponential generating functions:
In order to cover the binomial convolution $sum_{k=0}^nbinom{n}{k}b_kc_{n-k}$ of exponential generating functions we will encounter a more sophisticated multiplicative rule for labelled structures. Nevertheless we will do it similarly to the warmup above.
We consider $mathcal{B}={123,213},mathcal{C}={12}$ with generating functions $B(z)=2frac{z^3}{3!}, C(z)=frac{z^2}{2!}$ and consider the multiplicative rule
begin{align*}
color{blue}{mathcal{B}starmathcal{C}=bigcup_{betainmathcal{B},gammainmathcal{C}}left(betastargammaright)}tag{3}
end{align*}
which is defined as:
- The labelled product (3) of $mathcal{B}$ and $mathcal{C}$, denoted $mathcal{B}starmathcal{C}$, is obtained by forming ordered pairs from $Bstar C$ and performing all possible order-consistent relabellings.
The last phrase all possible order-consistent relabellings needs to be explained.
With the example $mathcal{B}={123,213}$ and $mathcal{C}={12}$ we perform a relabeling.
At first we start with a relabelling and select wlog $mathcal{C}={45}$ by respecting the order ($1<2$, i.e. $4<5$). We now obtain all order-consistent relabellings as follows
begin{align*}
mathcal{B}starmathcal{C}&=bigcup_{betainmathcal{B},gammainmathcal{C}}left(betastargammaright)\
&=left({123}star{45}right)cupleft({213}star{45}right)\
&={12345,12435,12534,13425,13524,14523,23415,23514,24513,34512}tag{4}\
&qquadcup{21345,21435,21534,31425,31524,41523,32415,32514,42513,43512}tag{5}\
end{align*}
When looking at the set ${123}star{45}$ we can find all order-consistent relabelings as follows: We have to generate permutations of length $5$. We create all $binom{5}{3}=10$ strings of length $3$ from ${1,2,3,4,5}$ which follow the order $1<2<3$. We obtain
begin{align*}
{123,124,125,134,135,145,234,235,245,345}.
end{align*}
Note, that each element in the set above follows the ordering $(mathrm{smallest} < mathrm{middle} < mathrm {largest})$. We complete (4) by concatenating each $3$-element string with a string consisting of the remaining two elements from ${1,2,3,4,5}$ following the order $4<5$, i.e. $(mathrm{smallest} < mathrm{largest})$. Similarly we build (5) by respecting the order $2<1<3$, i.e. $(mathrm{middle} < mathrm{smallest} < mathrm{largest})$.
We observe from (4), (5):
begin{align*}
|mathcal{B}starmathcal{C}|=2binom{5}{3}=2cdot 10=20
end{align*}
and the relationship with the generating functions $B(z)$ and $C(z)$ is given as
begin{align*}
color{blue}{A(z)}&=B(z)cdot C(z)=left(2cdot frac{z^3}{3!}right)cdotleft(frac{z^2}{2!}right)\
&,,color{blue}{=2binom{5}{3}frac{z^5}{5!}}
end{align*}
Hint: This answer is based upon the great presentation of admissible structures in chapter 1 and 2 of Analytic Combinatorics by P. Flajolet and R. Sedgewick.
$endgroup$
We typically use ordinary generating functions when counting unlabelled objects and exponentially generating functions when counting labelled objects. For instance combinatorial enumeration of permutations is most convenient analysed by exponential generating functions.
Here I'd like to give an indication which structures do we count when applying the binomial convolution
begin{align*}
sum_{k=0}^nbinom{n}{k}b_kc_{n-k}
end{align*}
of two exponential generating functions $B(z)=sum_{j=0}^infty b_j frac{z^j}{j!}$ and $C(z)=sum_{k=0}^infty c_kfrac{z^k}{k!}$.
Permutations:
We introduce a universe $mathcal{P}$ of (labelled) permutations. We label the atoms of the permutations of size $n$ as usual with $1,2,ldots,n$. We also add an empty permutation $varepsilon$ of length $0$ and consider
begin{align*}
mathcal{P}={varepsilon,1,12,21,123,132,213,231,312,321,ldots}tag{1}
end{align*}
The coefficients $a_n$ of the corresponding exponential generating function $A(z)=sum_{n=0}^infty a_n frac{z^n}{n!}$ give the number $n!$ of permutations of size $n$. We obtain
begin{align*}
color{blue}{A(z)}=sum_{n=0}^infty n!frac{z^n}{n!}=sum_{n=0}^infty z^ncolor{blue}{=frac{1}{1-z}}
end{align*}
We observe that generating functions like $frac{1}{1-z}$ may serve as both, ordinary generating functions
begin{align*}
frac{1}{1-z}=sum_{n=0}^infty z^n=color{blue}{1}+color{blue}{1}cdot z+color{blue}{1}cdot z^2+color{blue}{1}cdot z^3+cdots
end{align*}
to count for instance a sequence of unary words
begin{align*}
mathcal{S}={varepsilon,bullet,bulletbullet,bulletbulletbullet,ldots}tag{2}
end{align*}
as well as exponential generating functions
begin{align*}
frac{1}{1-z}=sum_{n=0}^infty n!frac{z^n}{n!}=color{blue}{0!}+color{blue}{1!}cdot frac{z}{1!}+color{blue}{2!}cdot frac{z^2}{2!}+color{blue}{3!}cdot frac{z^3}{3!}+cdots
end{align*}
to count permutations according to (1).
Note, the simplicity of the geometric series $frac{1}{1-z}=1+z+z^2+cdots$ strongly indicates the universe of permutations is the prototypical structure approachable by exponential generating functions.
Warmup: Multiplicative structure of ordinary generating functions.
We consider $mathcal{B}={bullet,bulletbulletbullet},mathcal{C}={bullet,bulletbullet}$ with corresponding generating functions $B(z)=z+z^3, C(z)=z+z^2$ and obtain due to the multiplicative rule
begin{align*}
color{blue}{mathrm{card}left(mathcal{B}times mathcal{C}right)=mathrm{card}left(mathcal{B}right)cdot mathrm{card}left(mathcal{C}right)}
end{align*}
the set $mathcal{A}={(bullet,bullet),(bullet,bulletbullet),(bulletbulletbullet,bullet),(bulletbulletbullet,bulletbullet)}$ and the following relationship
begin{align*}
A(z)&=sum_{ainmathcal{A}}z^{|a|}=sum_{(beta,gamma)in(mathcal{B}timesmathcal{C})}z^{|beta|+|gamma|}
=left(sum_{betain mathcal{B}}z^{|beta|}right)timesleft(sum_{gammain mathcal{C}}z^{|C|}right)=B(z)cdot C(z)\
A(z)&=z^2+z^3+z^4+z^5\
&=z^{1+1}+z^{1+2}+z^{3+1}+z^{3+2}=left(z^1+z^3right)timesleft(z^1+z^2right)=(z+z^3)(z+z^2)
end{align*}
which follows from distributing products over sums.
Multiplicative structure of exponential generating functions:
In order to cover the binomial convolution $sum_{k=0}^nbinom{n}{k}b_kc_{n-k}$ of exponential generating functions we will encounter a more sophisticated multiplicative rule for labelled structures. Nevertheless we will do it similarly to the warmup above.
We consider $mathcal{B}={123,213},mathcal{C}={12}$ with generating functions $B(z)=2frac{z^3}{3!}, C(z)=frac{z^2}{2!}$ and consider the multiplicative rule
begin{align*}
color{blue}{mathcal{B}starmathcal{C}=bigcup_{betainmathcal{B},gammainmathcal{C}}left(betastargammaright)}tag{3}
end{align*}
which is defined as:
- The labelled product (3) of $mathcal{B}$ and $mathcal{C}$, denoted $mathcal{B}starmathcal{C}$, is obtained by forming ordered pairs from $Bstar C$ and performing all possible order-consistent relabellings.
The last phrase all possible order-consistent relabellings needs to be explained.
With the example $mathcal{B}={123,213}$ and $mathcal{C}={12}$ we perform a relabeling.
At first we start with a relabelling and select wlog $mathcal{C}={45}$ by respecting the order ($1<2$, i.e. $4<5$). We now obtain all order-consistent relabellings as follows
begin{align*}
mathcal{B}starmathcal{C}&=bigcup_{betainmathcal{B},gammainmathcal{C}}left(betastargammaright)\
&=left({123}star{45}right)cupleft({213}star{45}right)\
&={12345,12435,12534,13425,13524,14523,23415,23514,24513,34512}tag{4}\
&qquadcup{21345,21435,21534,31425,31524,41523,32415,32514,42513,43512}tag{5}\
end{align*}
When looking at the set ${123}star{45}$ we can find all order-consistent relabelings as follows: We have to generate permutations of length $5$. We create all $binom{5}{3}=10$ strings of length $3$ from ${1,2,3,4,5}$ which follow the order $1<2<3$. We obtain
begin{align*}
{123,124,125,134,135,145,234,235,245,345}.
end{align*}
Note, that each element in the set above follows the ordering $(mathrm{smallest} < mathrm{middle} < mathrm {largest})$. We complete (4) by concatenating each $3$-element string with a string consisting of the remaining two elements from ${1,2,3,4,5}$ following the order $4<5$, i.e. $(mathrm{smallest} < mathrm{largest})$. Similarly we build (5) by respecting the order $2<1<3$, i.e. $(mathrm{middle} < mathrm{smallest} < mathrm{largest})$.
We observe from (4), (5):
begin{align*}
|mathcal{B}starmathcal{C}|=2binom{5}{3}=2cdot 10=20
end{align*}
and the relationship with the generating functions $B(z)$ and $C(z)$ is given as
begin{align*}
color{blue}{A(z)}&=B(z)cdot C(z)=left(2cdot frac{z^3}{3!}right)cdotleft(frac{z^2}{2!}right)\
&,,color{blue}{=2binom{5}{3}frac{z^5}{5!}}
end{align*}
Hint: This answer is based upon the great presentation of admissible structures in chapter 1 and 2 of Analytic Combinatorics by P. Flajolet and R. Sedgewick.
edited Jan 16 at 10:56
answered Jan 16 at 7:06
Markus ScheuerMarkus Scheuer
63.6k460152
63.6k460152
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.
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%2f79355%2fdeep-understanding-on-exponential-generating-function%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
$begingroup$
I suggest that you either read the book by Bergeron, Labelle and Leroux or that you ask a more precise question. The book might help to convince you that egfs are simpler than ogfs. books.google.com/…
$endgroup$
– Phira
Nov 6 '11 at 0:02
1
$begingroup$
I like "Generatingfunctionology" by Wilf. The 2nd edition can be downloaded from math.upenn.edu/~wilf/DownldGF.html. The 3rd edition is available from the publisher.
$endgroup$
– marty cohen
Nov 6 '11 at 0:17
1
$begingroup$
Chapter $3$ of Miklós Bóna’s Introduction to Enumerative Combinatorics gives a nice introductory sense of the respective contexts in which egf’s and ogf’s are useful.
$endgroup$
– Brian M. Scott
Nov 6 '11 at 0:23
5
$begingroup$
Think of it this way: there are situations where the EGF is simple/easy to manipulate when the OGF is unwieldy or has no closed form, and vice-versa...
$endgroup$
– J. M. is not a mathematician
Nov 6 '11 at 1:08
3
$begingroup$
anon guessed correctly, here's an old blog post by Qiaochu somewhat related to this.
$endgroup$
– Ragib Zaman
Nov 6 '11 at 2:19