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 |
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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