Polynomial defined by products of binomial coefficients












2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – hardmath
    Nov 26 '18 at 21:45






  • 2




    $begingroup$
    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.
    $endgroup$
    – sirron
    Nov 26 '18 at 21:59
















2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – hardmath
    Nov 26 '18 at 21:45






  • 2




    $begingroup$
    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.
    $endgroup$
    – sirron
    Nov 26 '18 at 21:59














2












2








2


0



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 '18 at 21:35









sirronsirron

111




111












  • $begingroup$
    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.
    $endgroup$
    – hardmath
    Nov 26 '18 at 21:45






  • 2




    $begingroup$
    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.
    $endgroup$
    – sirron
    Nov 26 '18 at 21:59


















  • $begingroup$
    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.
    $endgroup$
    – hardmath
    Nov 26 '18 at 21:45






  • 2




    $begingroup$
    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.
    $endgroup$
    – sirron
    Nov 26 '18 at 21:59
















$begingroup$
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.
$endgroup$
– hardmath
Nov 26 '18 at 21:45




$begingroup$
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.
$endgroup$
– hardmath
Nov 26 '18 at 21:45




2




2




$begingroup$
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.
$endgroup$
– sirron
Nov 26 '18 at 21:59




$begingroup$
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.
$endgroup$
– sirron
Nov 26 '18 at 21:59










1 Answer
1






active

oldest

votes


















2












$begingroup$

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
$$g_{m,n}(x)=frac{m!} {n ,x, (x+1)^{n-m}}, f_{m,n}(x)=P_{m-1}(x)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
    $endgroup$
    – sirron
    Nov 29 '18 at 15:51











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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
$$g_{m,n}(x)=frac{m!} {n ,x, (x+1)^{n-m}}, f_{m,n}(x)=P_{m-1}(x)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
    $endgroup$
    – sirron
    Nov 29 '18 at 15:51
















2












$begingroup$

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
$$g_{m,n}(x)=frac{m!} {n ,x, (x+1)^{n-m}}, f_{m,n}(x)=P_{m-1}(x)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
    $endgroup$
    – sirron
    Nov 29 '18 at 15:51














2












2








2





$begingroup$

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
$$g_{m,n}(x)=frac{m!} {n ,x, (x+1)^{n-m}}, f_{m,n}(x)=P_{m-1}(x)$$






share|cite|improve this answer











$endgroup$



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
$$g_{m,n}(x)=frac{m!} {n ,x, (x+1)^{n-m}}, f_{m,n}(x)=P_{m-1}(x)$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 '18 at 9:21

























answered Nov 27 '18 at 6:29









Claude LeiboviciClaude Leibovici

120k1157132




120k1157132












  • $begingroup$
    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 $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
    $endgroup$
    – sirron
    Nov 29 '18 at 15:51


















  • $begingroup$
    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 $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
    $endgroup$
    – sirron
    Nov 29 '18 at 15:51
















$begingroup$
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 $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
$endgroup$
– sirron
Nov 29 '18 at 15:51




$begingroup$
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 $$P_{m-1}(x) = m! , _2F_{1}(1-m,n+1; 2; -x)$$
$endgroup$
– sirron
Nov 29 '18 at 15:51


















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