Polynomial defined by products of binomial coefficients
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Consider the polynomial in $x$,
$$sum_{k=1}^n {n choose k} {m+k-1 choose m} x^k , $$
where $m$ and $n$ are positive non-zero integers.
Question: can it be expressed in terms of known function(s) of $x$?
complex-analysis polynomials
asked Nov 26 at 21:35
sirron
111
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up vote
2
down vote
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Consider the polynomial in $x$,
$$sum_{k=1}^n {n choose k} {m+k-1 choose m} x^k , $$
where $m$ and $n$ are positive non-zero integers.
Question: can it be expressed in terms of known function(s) of $x$?
complex-analysis polynomials
asked Nov 26 at 21:35
sirron
111
It would be interesting (and possibly helpful) to know what motivates your Question. As you define it, each fixed positive integer $m$ will give rise to a family of polynomials with degrees $n=1,2,3,ldots$. Checking a few examples might help by ruling out some of the well-known polynomial families.
– hardmath
Nov 26 at 21:45
2
The expression is the result of an integral based on a conformal transformation, $x$ is a parameter in the mapping, $m$ and $n$ are Fourier coefficients in the original and mapped domains.
– sirron
Nov 26 at 21:59
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the polynomial in $x$,
$$sum_{k=1}^n {n choose k} {m+k-1 choose m} x^k , $$
where $m$ and $n$ are positive non-zero integers.
Question: can it be expressed in terms of known function(s) of $x$?
complex-analysis polynomials
asked Nov 26 at 21:35
sirron
111
Consider the polynomial in $x$,
$$sum_{k=1}^n {n choose k} {m+k-1 choose m} x^k , $$
where $m$ and $n$ are positive non-zero integers.
Question: can it be expressed in terms of known function(s) of $x$?
complex-analysis polynomials
complex-analysis polynomials
asked Nov 26 at 21:35
sirron
111
asked Nov 26 at 21:35
sirron
111
asked Nov 26 at 21:35
sirron
111
asked Nov 26 at 21:35
sirron
111
asked Nov 26 at 21:35
sirron
111
111
It would be interesting (and possibly helpful) to know what motivates your Question. As you define it, each fixed positive integer $m$ will give rise to a family of polynomials with degrees $n=1,2,3,ldots$. Checking a few examples might help by ruling out some of the well-known polynomial families.
– hardmath
Nov 26 at 21:45
2
The expression is the result of an integral based on a conformal transformation, $x$ is a parameter in the mapping, $m$ and $n$ are Fourier coefficients in the original and mapped domains.
– sirron
Nov 26 at 21:59
add a comment |
It would be interesting (and possibly helpful) to know what motivates your Question. As you define it, each fixed positive integer $m$ will give rise to a family of polynomials with degrees $n=1,2,3,ldots$. Checking a few examples might help by ruling out some of the well-known polynomial families.
– hardmath
Nov 26 at 21:45
2
The expression is the result of an integral based on a conformal transformation, $x$ is a parameter in the mapping, $m$ and $n$ are Fourier coefficients in the original and mapped domains.
– sirron
Nov 26 at 21:59
It would be interesting (and possibly helpful) to know what motivates your Question. As you define it, each fixed positive integer $m$ will give rise to a family of polynomials with degrees $n=1,2,3,ldots$. Checking a few examples might help by ruling out some of the well-known polynomial families.
– hardmath
Nov 26 at 21:45
It would be interesting (and possibly helpful) to know what motivates your Question. As you define it, each fixed positive integer $m$ will give rise to a family of polynomials with degrees $n=1,2,3,ldots$. Checking a few examples might help by ruling out some of the well-known polynomial families.
– hardmath
Nov 26 at 21:45
2
2
The expression is the result of an integral based on a conformal transformation, $x$ is a parameter in the mapping, $m$ and $n$ are Fourier coefficients in the original and mapped domains.
– sirron
Nov 26 at 21:59
The expression is the result of an integral based on a conformal transformation, $x$ is a parameter in the mapping, $m$ and $n$ are Fourier coefficients in the original and mapped domains.
– sirron
Nov 26 at 21:59
add a comment |
1 Answer
1
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oldest
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up vote
1
down vote
If you want the generic function$$f_{m,n}(x)=sum_{k=1}^n {n choose k} {m+k-1 choose m} x^k$$ $$f_{m,n}(x)=n, x ,, _2F_1(m+1,1-n;2;-x)$$ where appears the Gaussian or ordinary hypergeometric function (see here).
A few expressions
$$left(
begin{array}{cc}
m & f_{m,n}(x) \
0 & (x+1)^n-1 \
1 & n x (x+1)^{n-1} \
2 & frac{1}{2} n x (x+1)^{n-2} ((n+1) x+2) \
3 & frac{1}{6} n x (x+1)^{n-3} left((n^2 +3 n +2) x^2+6 (n+1) x+6right)
end{array}
right)$$ which are just polynomials of degree $n$.
If you define
$$g_{m,n}(x)=frac{m!} {n ,x, (x+1)^{n-m}}, f_{m,n}(x)=P_{m-1}(x)$$
answered Nov 27 at 6:29
Claude Leibovici
117k1156131
The polynomial $P_{m-1}$ can be identified using the identity (see <a href="functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/…) $$, _2F_{1}(a,b;c;z) = (1-z)^{c-a-b} , _2F_{1}(c-a,c-b;c;z)$$ as $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
– sirron
Nov 29 at 15:51
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If you want the generic function$$f_{m,n}(x)=sum_{k=1}^n {n choose k} {m+k-1 choose m} x^k$$ $$f_{m,n}(x)=n, x ,, _2F_1(m+1,1-n;2;-x)$$ where appears the Gaussian or ordinary hypergeometric function (see here).
A few expressions
$$left(
begin{array}{cc}
m & f_{m,n}(x) \
0 & (x+1)^n-1 \
1 & n x (x+1)^{n-1} \
2 & frac{1}{2} n x (x+1)^{n-2} ((n+1) x+2) \
3 & frac{1}{6} n x (x+1)^{n-3} left((n^2 +3 n +2) x^2+6 (n+1) x+6right)
end{array}
right)$$ which are just polynomials of degree $n$.
If you define
$$g_{m,n}(x)=frac{m!} {n ,x, (x+1)^{n-m}}, f_{m,n}(x)=P_{m-1}(x)$$
answered Nov 27 at 6:29
Claude Leibovici
117k1156131
The polynomial $P_{m-1}$ can be identified using the identity (see <a href="functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/…) $$, _2F_{1}(a,b;c;z) = (1-z)^{c-a-b} , _2F_{1}(c-a,c-b;c;z)$$ as $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
– sirron
Nov 29 at 15:51
add a comment |
up vote
1
down vote
If you want the generic function$$f_{m,n}(x)=sum_{k=1}^n {n choose k} {m+k-1 choose m} x^k$$ $$f_{m,n}(x)=n, x ,, _2F_1(m+1,1-n;2;-x)$$ where appears the Gaussian or ordinary hypergeometric function (see here).
A few expressions
$$left(
begin{array}{cc}
m & f_{m,n}(x) \
0 & (x+1)^n-1 \
1 & n x (x+1)^{n-1} \
2 & frac{1}{2} n x (x+1)^{n-2} ((n+1) x+2) \
3 & frac{1}{6} n x (x+1)^{n-3} left((n^2 +3 n +2) x^2+6 (n+1) x+6right)
end{array}
right)$$ which are just polynomials of degree $n$.
If you define
$$g_{m,n}(x)=frac{m!} {n ,x, (x+1)^{n-m}}, f_{m,n}(x)=P_{m-1}(x)$$
answered Nov 27 at 6:29
Claude Leibovici
117k1156131
The polynomial $P_{m-1}$ can be identified using the identity (see <a href="functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/…) $$, _2F_{1}(a,b;c;z) = (1-z)^{c-a-b} , _2F_{1}(c-a,c-b;c;z)$$ as $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
– sirron
Nov 29 at 15:51
add a comment |
up vote
1
down vote
up vote
1
down vote
If you want the generic function$$f_{m,n}(x)=sum_{k=1}^n {n choose k} {m+k-1 choose m} x^k$$ $$f_{m,n}(x)=n, x ,, _2F_1(m+1,1-n;2;-x)$$ where appears the Gaussian or ordinary hypergeometric function (see here).
A few expressions
$$left(
begin{array}{cc}
m & f_{m,n}(x) \
0 & (x+1)^n-1 \
1 & n x (x+1)^{n-1} \
2 & frac{1}{2} n x (x+1)^{n-2} ((n+1) x+2) \
3 & frac{1}{6} n x (x+1)^{n-3} left((n^2 +3 n +2) x^2+6 (n+1) x+6right)
end{array}
right)$$ which are just polynomials of degree $n$.
If you define
$$g_{m,n}(x)=frac{m!} {n ,x, (x+1)^{n-m}}, f_{m,n}(x)=P_{m-1}(x)$$
answered Nov 27 at 6:29
Claude Leibovici
117k1156131
If you want the generic function$$f_{m,n}(x)=sum_{k=1}^n {n choose k} {m+k-1 choose m} x^k$$ $$f_{m,n}(x)=n, x ,, _2F_1(m+1,1-n;2;-x)$$ where appears the Gaussian or ordinary hypergeometric function (see here).
A few expressions
$$left(
begin{array}{cc}
m & f_{m,n}(x) \
0 & (x+1)^n-1 \
1 & n x (x+1)^{n-1} \
2 & frac{1}{2} n x (x+1)^{n-2} ((n+1) x+2) \
3 & frac{1}{6} n x (x+1)^{n-3} left((n^2 +3 n +2) x^2+6 (n+1) x+6right)
end{array}
right)$$ which are just polynomials of degree $n$.
If you define
$$g_{m,n}(x)=frac{m!} {n ,x, (x+1)^{n-m}}, f_{m,n}(x)=P_{m-1}(x)$$
answered Nov 27 at 6:29
Claude Leibovici
117k1156131
edited Nov 27 at 9:21
answered Nov 27 at 6:29
Claude Leibovici
117k1156131
answered Nov 27 at 6:29
Claude Leibovici
117k1156131
answered Nov 27 at 6:29
Claude Leibovici
117k1156131
117k1156131
The polynomial $P_{m-1}$ can be identified using the identity (see <a href="functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/…) $$, _2F_{1}(a,b;c;z) = (1-z)^{c-a-b} , _2F_{1}(c-a,c-b;c;z)$$ as $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
– sirron
Nov 29 at 15:51
add a comment |
The polynomial $P_{m-1}$ can be identified using the identity (see <a href="functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/…) $$, _2F_{1}(a,b;c;z) = (1-z)^{c-a-b} , _2F_{1}(c-a,c-b;c;z)$$ as $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
– sirron
Nov 29 at 15:51
The polynomial $P_{m-1}$ can be identified using the identity (see <a href="functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/…) $$, _2F_{1}(a,b;c;z) = (1-z)^{c-a-b} , _2F_{1}(c-a,c-b;c;z)$$ as $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
– sirron
Nov 29 at 15:51
The polynomial $P_{m-1}$ can be identified using the identity (see <a href="functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/…) $$, _2F_{1}(a,b;c;z) = (1-z)^{c-a-b} , _2F_{1}(c-a,c-b;c;z)$$ as $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
– sirron
Nov 29 at 15:51
add a comment |
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It would be interesting (and possibly helpful) to know what motivates your Question. As you define it, each fixed positive integer $m$ will give rise to a family of polynomials with degrees $n=1,2,3,ldots$. Checking a few examples might help by ruling out some of the well-known polynomial families.
– hardmath
Nov 26 at 21:45
2
The expression is the result of an integral based on a conformal transformation, $x$ is a parameter in the mapping, $m$ and $n$ are Fourier coefficients in the original and mapped domains.
– sirron
Nov 26 at 21:59