Mixing
Product of Stirling numbers of second kind
$begingroup$
What does the following expression evaluate to:
begin{equation}
k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
We know that $k! begin{Bmatrix} n \ k end{Bmatrix} = n![x^n]:(e^x-1)^k$, where $[x^k]:f(x)$ represents the coefficient of $x^k$ in the power series for $f(x)$. I was wondering if squaring takes us to a different power series or just to a different coefficient in the same power series?
combinatorics power-series stirling-numbers
asked Jan 20 at 4:43
SinghSingh
454
$endgroup$
add a comment |
$begingroup$
What does the following expression evaluate to:
begin{equation}
k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
We know that $k! begin{Bmatrix} n \ k end{Bmatrix} = n![x^n]:(e^x-1)^k$, where $[x^k]:f(x)$ represents the coefficient of $x^k$ in the power series for $f(x)$. I was wondering if squaring takes us to a different power series or just to a different coefficient in the same power series?
combinatorics power-series stirling-numbers
asked Jan 20 at 4:43
SinghSingh
454
$endgroup$
$begingroup$
You are squaring $k! begin{Bmatrix} n \ k end{Bmatrix} $?
$endgroup$
– Lord Shark the Unknown
Jan 20 at 5:12
1
$begingroup$
@LordSharktheUnknown yes, I am wondering whether squaring takes us to a different power series or to a different coefficient in the same power series?
$endgroup$
– Singh
Jan 20 at 12:15
add a comment |
$begingroup$
What does the following expression evaluate to:
begin{equation}
k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
We know that $k! begin{Bmatrix} n \ k end{Bmatrix} = n![x^n]:(e^x-1)^k$, where $[x^k]:f(x)$ represents the coefficient of $x^k$ in the power series for $f(x)$. I was wondering if squaring takes us to a different power series or just to a different coefficient in the same power series?
combinatorics power-series stirling-numbers
asked Jan 20 at 4:43
SinghSingh
454
$endgroup$
What does the following expression evaluate to:
begin{equation}
k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
We know that $k! begin{Bmatrix} n \ k end{Bmatrix} = n![x^n]:(e^x-1)^k$, where $[x^k]:f(x)$ represents the coefficient of $x^k$ in the power series for $f(x)$. I was wondering if squaring takes us to a different power series or just to a different coefficient in the same power series?
combinatorics power-series stirling-numbers
combinatorics power-series stirling-numbers
asked Jan 20 at 4:43
SinghSingh
454
asked Jan 20 at 4:43
SinghSingh
454
edited Jan 20 at 16:18
Singh
asked Jan 20 at 4:43
SinghSingh
454
asked Jan 20 at 4:43
SinghSingh
454
asked Jan 20 at 4:43
SinghSingh
454
454
$begingroup$
You are squaring $k! begin{Bmatrix} n \ k end{Bmatrix} $?
$endgroup$
– Lord Shark the Unknown
Jan 20 at 5:12
1
$begingroup$
@LordSharktheUnknown yes, I am wondering whether squaring takes us to a different power series or to a different coefficient in the same power series?
$endgroup$
– Singh
Jan 20 at 12:15
add a comment |
$begingroup$
You are squaring $k! begin{Bmatrix} n \ k end{Bmatrix} $?
$endgroup$
– Lord Shark the Unknown
Jan 20 at 5:12
1
$begingroup$
@LordSharktheUnknown yes, I am wondering whether squaring takes us to a different power series or to a different coefficient in the same power series?
$endgroup$
– Singh
Jan 20 at 12:15
$begingroup$
You are squaring $k! begin{Bmatrix} n \ k end{Bmatrix} $?
$endgroup$
– Lord Shark the Unknown
Jan 20 at 5:12
$begingroup$
You are squaring $k! begin{Bmatrix} n \ k end{Bmatrix} $?
$endgroup$
– Lord Shark the Unknown
Jan 20 at 5:12
1
1
$begingroup$
@LordSharktheUnknown yes, I am wondering whether squaring takes us to a different power series or to a different coefficient in the same power series?
$endgroup$
– Singh
Jan 20 at 12:15
$begingroup$
@LordSharktheUnknown yes, I am wondering whether squaring takes us to a different power series or to a different coefficient in the same power series?
$endgroup$
– Singh
Jan 20 at 12:15
add a comment |
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$begingroup$
You are squaring $k! begin{Bmatrix} n \ k end{Bmatrix} $?
$endgroup$
– Lord Shark the Unknown
Jan 20 at 5:12
1
$begingroup$
@LordSharktheUnknown yes, I am wondering whether squaring takes us to a different power series or to a different coefficient in the same power series?
$endgroup$
– Singh
Jan 20 at 12:15