Formula for composition of formal power series with binomial coefficient
$begingroup$
Let $f=sumlimits_{ngeq 0}{f_n x^n}$ and $g=sumlimits_{ngeq 1}{g_n x^n}$ be formal power series. The $x^n$ coefficient of $f(g)$ is
$$
sumlimits_{mathbb{i} in mathcal{C}_{n}}
{f_k ,g_{i_1}cdots g_{i_k}}
,
$$
where $mathcal{C}_{n} = { (i_1, ldots, i_k) : 1 leq k leq n, i_1 + cdots + i_k = n }$
is the set of compositions of $n$ into $k$ parts.
Consider the power series $h=sumlimits_{ngeq 0}{h_n x^n}$ with coefficient
$$
h_n =
sumlimits_{mathbb{i} in mathcal{C}_{n}}
{binom{n}{k} f_k ,g_{i_1}cdots g_{i_k}}
.
$$
Is there a "nice" way to express $h$ in terms of $f$ and $g$ or related generating series (eg. their e.g.f.s), where by "nice", I mean a formula that can be evaluated if $f$ and $g$ are at least rational functions with $g(0)=0$?
UPDATE
If we let $F=sumlimits_{ngeq 0}{f_n frac{x^n}{n!}}$ and $H=sumlimits_{ngeq 0}{h_n frac{x^n}{n!}}$ then, using the inverse Laplace transform $mathcal{L}^{-1}$, we can write
$$H = mathcal{L}^{-1}{ F(s ,g(x/s))/s}(1).$$
This is because $$mathcal{L}^{-1}{ s^{k-n}/s}(1)=frac{1}{(n-k)!}.$$
However this isn't what I want because
I want $h$, not $H$, and;
this formula is not "nice", as if $f=frac{1}{1-x}$ (so $F=e^x$) and $g=frac{x}{1-x}$, the expression for $H$ is an intractable integral, whereas $h=frac{1}{2} + frac{1}{2sqrt{1-4x}}$.
combinatorics power-series binomial-coefficients generating-functions formal-power-series
asked Jun 13 '15 at 14:17
Nasos EvangelouNasos Evangelou
7713
$endgroup$
add a comment |
$begingroup$
Let $f=sumlimits_{ngeq 0}{f_n x^n}$ and $g=sumlimits_{ngeq 1}{g_n x^n}$ be formal power series. The $x^n$ coefficient of $f(g)$ is
$$
sumlimits_{mathbb{i} in mathcal{C}_{n}}
{f_k ,g_{i_1}cdots g_{i_k}}
,
$$
where $mathcal{C}_{n} = { (i_1, ldots, i_k) : 1 leq k leq n, i_1 + cdots + i_k = n }$
is the set of compositions of $n$ into $k$ parts.
Consider the power series $h=sumlimits_{ngeq 0}{h_n x^n}$ with coefficient
$$
h_n =
sumlimits_{mathbb{i} in mathcal{C}_{n}}
{binom{n}{k} f_k ,g_{i_1}cdots g_{i_k}}
.
$$
Is there a "nice" way to express $h$ in terms of $f$ and $g$ or related generating series (eg. their e.g.f.s), where by "nice", I mean a formula that can be evaluated if $f$ and $g$ are at least rational functions with $g(0)=0$?
UPDATE
If we let $F=sumlimits_{ngeq 0}{f_n frac{x^n}{n!}}$ and $H=sumlimits_{ngeq 0}{h_n frac{x^n}{n!}}$ then, using the inverse Laplace transform $mathcal{L}^{-1}$, we can write
$$H = mathcal{L}^{-1}{ F(s ,g(x/s))/s}(1).$$
This is because $$mathcal{L}^{-1}{ s^{k-n}/s}(1)=frac{1}{(n-k)!}.$$
However this isn't what I want because
I want $h$, not $H$, and;
this formula is not "nice", as if $f=frac{1}{1-x}$ (so $F=e^x$) and $g=frac{x}{1-x}$, the expression for $H$ is an intractable integral, whereas $h=frac{1}{2} + frac{1}{2sqrt{1-4x}}$.
combinatorics power-series binomial-coefficients generating-functions formal-power-series
asked Jun 13 '15 at 14:17
Nasos EvangelouNasos Evangelou
7713
$endgroup$
1
$begingroup$
It's not exactly what you want, but if you let $G = sum b_n frac{x^n}{n!}$ and $D(x) = f(G) = sum d_n frac{x^n}{n!}$, then you have $$ d_n = sumlimits_{i in mathcal{C}_n}binom{n}{i_1,i_2,ldots,i_k}a_k b_{i_1}cdots b_{i_k}. $$
$endgroup$
– Marcus M
Jun 19 '15 at 2:28
$begingroup$
Thanks for the idea, but unfortunately this doesn't help me! Also I'm sorry that I changed notation in my question to make things clearer.
$endgroup$
– Nasos Evangelou
Jun 21 '15 at 9:32
$begingroup$
The $x^n$ coefficient of $fcirc g$ should be $sum_{k=1}^nf_ksum_{iin C_{n,k}}g_{i_1}...g_{i_k}$, shouldn't it? If not, what does the $k$ refer to in your first formula ? I would also ask the same question about the $h_n$ you want to know, is there a sum over $k$ ?
$endgroup$
– Clément Guérin
Jun 24 '15 at 7:53
$begingroup$
@ClémentGuérin: Yes the sum is over $k$ in both cases--I think we can omit the summation sign as long as we write $mathbb{i} in mathcal{C}_{n}$ instead of $mathbb{i} in mathcal{C}_{n,k}$.
$endgroup$
– Nasos Evangelou
Jun 24 '15 at 10:14
add a comment |
$begingroup$
Let $f=sumlimits_{ngeq 0}{f_n x^n}$ and $g=sumlimits_{ngeq 1}{g_n x^n}$ be formal power series. The $x^n$ coefficient of $f(g)$ is
$$
sumlimits_{mathbb{i} in mathcal{C}_{n}}
{f_k ,g_{i_1}cdots g_{i_k}}
,
$$
where $mathcal{C}_{n} = { (i_1, ldots, i_k) : 1 leq k leq n, i_1 + cdots + i_k = n }$
is the set of compositions of $n$ into $k$ parts.
Consider the power series $h=sumlimits_{ngeq 0}{h_n x^n}$ with coefficient
$$
h_n =
sumlimits_{mathbb{i} in mathcal{C}_{n}}
{binom{n}{k} f_k ,g_{i_1}cdots g_{i_k}}
.
$$
Is there a "nice" way to express $h$ in terms of $f$ and $g$ or related generating series (eg. their e.g.f.s), where by "nice", I mean a formula that can be evaluated if $f$ and $g$ are at least rational functions with $g(0)=0$?
UPDATE
If we let $F=sumlimits_{ngeq 0}{f_n frac{x^n}{n!}}$ and $H=sumlimits_{ngeq 0}{h_n frac{x^n}{n!}}$ then, using the inverse Laplace transform $mathcal{L}^{-1}$, we can write
$$H = mathcal{L}^{-1}{ F(s ,g(x/s))/s}(1).$$
This is because $$mathcal{L}^{-1}{ s^{k-n}/s}(1)=frac{1}{(n-k)!}.$$
However this isn't what I want because
I want $h$, not $H$, and;
this formula is not "nice", as if $f=frac{1}{1-x}$ (so $F=e^x$) and $g=frac{x}{1-x}$, the expression for $H$ is an intractable integral, whereas $h=frac{1}{2} + frac{1}{2sqrt{1-4x}}$.
combinatorics power-series binomial-coefficients generating-functions formal-power-series
asked Jun 13 '15 at 14:17
Nasos EvangelouNasos Evangelou
7713
$endgroup$
Let $f=sumlimits_{ngeq 0}{f_n x^n}$ and $g=sumlimits_{ngeq 1}{g_n x^n}$ be formal power series. The $x^n$ coefficient of $f(g)$ is
$$
sumlimits_{mathbb{i} in mathcal{C}_{n}}
{f_k ,g_{i_1}cdots g_{i_k}}
,
$$
where $mathcal{C}_{n} = { (i_1, ldots, i_k) : 1 leq k leq n, i_1 + cdots + i_k = n }$
is the set of compositions of $n$ into $k$ parts.
Consider the power series $h=sumlimits_{ngeq 0}{h_n x^n}$ with coefficient
$$
h_n =
sumlimits_{mathbb{i} in mathcal{C}_{n}}
{binom{n}{k} f_k ,g_{i_1}cdots g_{i_k}}
.
$$
Is there a "nice" way to express $h$ in terms of $f$ and $g$ or related generating series (eg. their e.g.f.s), where by "nice", I mean a formula that can be evaluated if $f$ and $g$ are at least rational functions with $g(0)=0$?
UPDATE
If we let $F=sumlimits_{ngeq 0}{f_n frac{x^n}{n!}}$ and $H=sumlimits_{ngeq 0}{h_n frac{x^n}{n!}}$ then, using the inverse Laplace transform $mathcal{L}^{-1}$, we can write
$$H = mathcal{L}^{-1}{ F(s ,g(x/s))/s}(1).$$
This is because $$mathcal{L}^{-1}{ s^{k-n}/s}(1)=frac{1}{(n-k)!}.$$
However this isn't what I want because
I want $h$, not $H$, and;
this formula is not "nice", as if $f=frac{1}{1-x}$ (so $F=e^x$) and $g=frac{x}{1-x}$, the expression for $H$ is an intractable integral, whereas $h=frac{1}{2} + frac{1}{2sqrt{1-4x}}$.
combinatorics power-series binomial-coefficients generating-functions formal-power-series
combinatorics power-series binomial-coefficients generating-functions formal-power-series
asked Jun 13 '15 at 14:17
Nasos EvangelouNasos Evangelou
7713
asked Jun 13 '15 at 14:17
Nasos EvangelouNasos Evangelou
7713
edited Jun 26 '15 at 4:19
Nasos Evangelou
asked Jun 13 '15 at 14:17
Nasos EvangelouNasos Evangelou
7713
asked Jun 13 '15 at 14:17
Nasos EvangelouNasos Evangelou
7713
asked Jun 13 '15 at 14:17
Nasos EvangelouNasos Evangelou
7713
7713
1
$begingroup$
It's not exactly what you want, but if you let $G = sum b_n frac{x^n}{n!}$ and $D(x) = f(G) = sum d_n frac{x^n}{n!}$, then you have $$ d_n = sumlimits_{i in mathcal{C}_n}binom{n}{i_1,i_2,ldots,i_k}a_k b_{i_1}cdots b_{i_k}. $$
$endgroup$
– Marcus M
Jun 19 '15 at 2:28
$begingroup$
Thanks for the idea, but unfortunately this doesn't help me! Also I'm sorry that I changed notation in my question to make things clearer.
$endgroup$
– Nasos Evangelou
Jun 21 '15 at 9:32
$begingroup$
The $x^n$ coefficient of $fcirc g$ should be $sum_{k=1}^nf_ksum_{iin C_{n,k}}g_{i_1}...g_{i_k}$, shouldn't it? If not, what does the $k$ refer to in your first formula ? I would also ask the same question about the $h_n$ you want to know, is there a sum over $k$ ?
$endgroup$
– Clément Guérin
Jun 24 '15 at 7:53
$begingroup$
@ClémentGuérin: Yes the sum is over $k$ in both cases--I think we can omit the summation sign as long as we write $mathbb{i} in mathcal{C}_{n}$ instead of $mathbb{i} in mathcal{C}_{n,k}$.
$endgroup$
– Nasos Evangelou
Jun 24 '15 at 10:14
add a comment |
1
$begingroup$
It's not exactly what you want, but if you let $G = sum b_n frac{x^n}{n!}$ and $D(x) = f(G) = sum d_n frac{x^n}{n!}$, then you have $$ d_n = sumlimits_{i in mathcal{C}_n}binom{n}{i_1,i_2,ldots,i_k}a_k b_{i_1}cdots b_{i_k}. $$
$endgroup$
– Marcus M
Jun 19 '15 at 2:28
$begingroup$
Thanks for the idea, but unfortunately this doesn't help me! Also I'm sorry that I changed notation in my question to make things clearer.
$endgroup$
– Nasos Evangelou
Jun 21 '15 at 9:32
$begingroup$
The $x^n$ coefficient of $fcirc g$ should be $sum_{k=1}^nf_ksum_{iin C_{n,k}}g_{i_1}...g_{i_k}$, shouldn't it? If not, what does the $k$ refer to in your first formula ? I would also ask the same question about the $h_n$ you want to know, is there a sum over $k$ ?
$endgroup$
– Clément Guérin
Jun 24 '15 at 7:53
$begingroup$
@ClémentGuérin: Yes the sum is over $k$ in both cases--I think we can omit the summation sign as long as we write $mathbb{i} in mathcal{C}_{n}$ instead of $mathbb{i} in mathcal{C}_{n,k}$.
$endgroup$
– Nasos Evangelou
Jun 24 '15 at 10:14
1
1
$begingroup$
It's not exactly what you want, but if you let $G = sum b_n frac{x^n}{n!}$ and $D(x) = f(G) = sum d_n frac{x^n}{n!}$, then you have $$ d_n = sumlimits_{i in mathcal{C}_n}binom{n}{i_1,i_2,ldots,i_k}a_k b_{i_1}cdots b_{i_k}. $$
$endgroup$
– Marcus M
Jun 19 '15 at 2:28
$begingroup$
It's not exactly what you want, but if you let $G = sum b_n frac{x^n}{n!}$ and $D(x) = f(G) = sum d_n frac{x^n}{n!}$, then you have $$ d_n = sumlimits_{i in mathcal{C}_n}binom{n}{i_1,i_2,ldots,i_k}a_k b_{i_1}cdots b_{i_k}. $$
$endgroup$
– Marcus M
Jun 19 '15 at 2:28
$begingroup$
Thanks for the idea, but unfortunately this doesn't help me! Also I'm sorry that I changed notation in my question to make things clearer.
$endgroup$
– Nasos Evangelou
Jun 21 '15 at 9:32
$begingroup$
Thanks for the idea, but unfortunately this doesn't help me! Also I'm sorry that I changed notation in my question to make things clearer.
$endgroup$
– Nasos Evangelou
Jun 21 '15 at 9:32
$begingroup$
The $x^n$ coefficient of $fcirc g$ should be $sum_{k=1}^nf_ksum_{iin C_{n,k}}g_{i_1}...g_{i_k}$, shouldn't it? If not, what does the $k$ refer to in your first formula ? I would also ask the same question about the $h_n$ you want to know, is there a sum over $k$ ?
$endgroup$
– Clément Guérin
Jun 24 '15 at 7:53
$begingroup$
The $x^n$ coefficient of $fcirc g$ should be $sum_{k=1}^nf_ksum_{iin C_{n,k}}g_{i_1}...g_{i_k}$, shouldn't it? If not, what does the $k$ refer to in your first formula ? I would also ask the same question about the $h_n$ you want to know, is there a sum over $k$ ?
$endgroup$
– Clément Guérin
Jun 24 '15 at 7:53
$begingroup$
@ClémentGuérin: Yes the sum is over $k$ in both cases--I think we can omit the summation sign as long as we write $mathbb{i} in mathcal{C}_{n}$ instead of $mathbb{i} in mathcal{C}_{n,k}$.
$endgroup$
– Nasos Evangelou
Jun 24 '15 at 10:14
$begingroup$
@ClémentGuérin: Yes the sum is over $k$ in both cases--I think we can omit the summation sign as long as we write $mathbb{i} in mathcal{C}_{n}$ instead of $mathbb{i} in mathcal{C}_{n,k}$.
$endgroup$
– Nasos Evangelou
Jun 24 '15 at 10:14
add a comment |
0
active
oldest
votes
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%2f1323816%2fformula-for-composition-of-formal-power-series-with-binomial-coefficient%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f1323816%2fformula-for-composition-of-formal-power-series-with-binomial-coefficient%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
1
$begingroup$
It's not exactly what you want, but if you let $G = sum b_n frac{x^n}{n!}$ and $D(x) = f(G) = sum d_n frac{x^n}{n!}$, then you have $$ d_n = sumlimits_{i in mathcal{C}_n}binom{n}{i_1,i_2,ldots,i_k}a_k b_{i_1}cdots b_{i_k}. $$
$endgroup$
– Marcus M
Jun 19 '15 at 2:28
$begingroup$
Thanks for the idea, but unfortunately this doesn't help me! Also I'm sorry that I changed notation in my question to make things clearer.
$endgroup$
– Nasos Evangelou
Jun 21 '15 at 9:32
$begingroup$
The $x^n$ coefficient of $fcirc g$ should be $sum_{k=1}^nf_ksum_{iin C_{n,k}}g_{i_1}...g_{i_k}$, shouldn't it? If not, what does the $k$ refer to in your first formula ? I would also ask the same question about the $h_n$ you want to know, is there a sum over $k$ ?
$endgroup$
– Clément Guérin
Jun 24 '15 at 7:53
$begingroup$
@ClémentGuérin: Yes the sum is over $k$ in both cases--I think we can omit the summation sign as long as we write $mathbb{i} in mathcal{C}_{n}$ instead of $mathbb{i} in mathcal{C}_{n,k}$.
$endgroup$
– Nasos Evangelou
Jun 24 '15 at 10:14