Are elementary symmetric polynomials evaluated in powers of double cosines of rational multiples of $pi$...
$begingroup$
For $n,k,mge 1$ integer, define
$$S(n,m,k)=sum_{Asubset{1,ldots,n},\#A=k}left(prod_{ain A}(-1)^a2^mcos^mleft(frac{api}nright)right).$$
In other words, $S(n,m,k)$ is the $k$'th elementary symmetric polynomial evaluated in
$$-2^mcos^mleft(frac{pi}nright),2^mcos^mleft(frac{2pi}nright),ldots,(-1)^n2^mcos^mleft(frac{ncdot npi}nright).$$
It seems like all $S(n,m,k)$ are integer. Can anyone prove or refute this?
If they are all integer, I'd like to find some 'combinatorical' formula for
$S(n,m,k)$, containing only polynomials, powers of integers and factorials.
The closest I've gotten is the following identity:
$$2+prod_{j=1}^{2n}left(X-2cosleft(frac{pi k}nright)right)=2sum_{j=0}^nX^{2n-2j}(-1)^jfrac{n(2n-j-1)!}{(2n-2j)!j!}$$
combinatorics trigonometry polynomials symmetric-polynomials
asked Dec 21 '18 at 9:26
MastremMastrem
3,79211230
$endgroup$
add a comment |
$begingroup$
For $n,k,mge 1$ integer, define
$$S(n,m,k)=sum_{Asubset{1,ldots,n},\#A=k}left(prod_{ain A}(-1)^a2^mcos^mleft(frac{api}nright)right).$$
In other words, $S(n,m,k)$ is the $k$'th elementary symmetric polynomial evaluated in
$$-2^mcos^mleft(frac{pi}nright),2^mcos^mleft(frac{2pi}nright),ldots,(-1)^n2^mcos^mleft(frac{ncdot npi}nright).$$
It seems like all $S(n,m,k)$ are integer. Can anyone prove or refute this?
If they are all integer, I'd like to find some 'combinatorical' formula for
$S(n,m,k)$, containing only polynomials, powers of integers and factorials.
The closest I've gotten is the following identity:
$$2+prod_{j=1}^{2n}left(X-2cosleft(frac{pi k}nright)right)=2sum_{j=0}^nX^{2n-2j}(-1)^jfrac{n(2n-j-1)!}{(2n-2j)!j!}$$
combinatorics trigonometry polynomials symmetric-polynomials
asked Dec 21 '18 at 9:26
MastremMastrem
3,79211230
$endgroup$
1
$begingroup$
$2cos(q)$ is an algebraic integer for any rational $q$, hence so is the value of the symmetric polynomial. It is also rational, because the Galois group of whatever extension contains all the cosines permutes them. Hence this is an integer.
$endgroup$
– Wojowu
Dec 21 '18 at 9:54
$begingroup$
... meant $2cos(qpi)$.
$endgroup$
– metamorphy
Dec 21 '18 at 10:14
add a comment |
$begingroup$
For $n,k,mge 1$ integer, define
$$S(n,m,k)=sum_{Asubset{1,ldots,n},\#A=k}left(prod_{ain A}(-1)^a2^mcos^mleft(frac{api}nright)right).$$
In other words, $S(n,m,k)$ is the $k$'th elementary symmetric polynomial evaluated in
$$-2^mcos^mleft(frac{pi}nright),2^mcos^mleft(frac{2pi}nright),ldots,(-1)^n2^mcos^mleft(frac{ncdot npi}nright).$$
It seems like all $S(n,m,k)$ are integer. Can anyone prove or refute this?
If they are all integer, I'd like to find some 'combinatorical' formula for
$S(n,m,k)$, containing only polynomials, powers of integers and factorials.
The closest I've gotten is the following identity:
$$2+prod_{j=1}^{2n}left(X-2cosleft(frac{pi k}nright)right)=2sum_{j=0}^nX^{2n-2j}(-1)^jfrac{n(2n-j-1)!}{(2n-2j)!j!}$$
combinatorics trigonometry polynomials symmetric-polynomials
asked Dec 21 '18 at 9:26
MastremMastrem
3,79211230
$endgroup$
For $n,k,mge 1$ integer, define
$$S(n,m,k)=sum_{Asubset{1,ldots,n},\#A=k}left(prod_{ain A}(-1)^a2^mcos^mleft(frac{api}nright)right).$$
In other words, $S(n,m,k)$ is the $k$'th elementary symmetric polynomial evaluated in
$$-2^mcos^mleft(frac{pi}nright),2^mcos^mleft(frac{2pi}nright),ldots,(-1)^n2^mcos^mleft(frac{ncdot npi}nright).$$
It seems like all $S(n,m,k)$ are integer. Can anyone prove or refute this?
If they are all integer, I'd like to find some 'combinatorical' formula for
$S(n,m,k)$, containing only polynomials, powers of integers and factorials.
The closest I've gotten is the following identity:
$$2+prod_{j=1}^{2n}left(X-2cosleft(frac{pi k}nright)right)=2sum_{j=0}^nX^{2n-2j}(-1)^jfrac{n(2n-j-1)!}{(2n-2j)!j!}$$
combinatorics trigonometry polynomials symmetric-polynomials
combinatorics trigonometry polynomials symmetric-polynomials
asked Dec 21 '18 at 9:26
MastremMastrem
3,79211230
asked Dec 21 '18 at 9:26
MastremMastrem
3,79211230
edited Dec 21 '18 at 18:39
Mastrem
asked Dec 21 '18 at 9:26
MastremMastrem
3,79211230
asked Dec 21 '18 at 9:26
MastremMastrem
3,79211230
asked Dec 21 '18 at 9:26
MastremMastrem
3,79211230
3,79211230
1
$begingroup$
$2cos(q)$ is an algebraic integer for any rational $q$, hence so is the value of the symmetric polynomial. It is also rational, because the Galois group of whatever extension contains all the cosines permutes them. Hence this is an integer.
$endgroup$
– Wojowu
Dec 21 '18 at 9:54
$begingroup$
... meant $2cos(qpi)$.
$endgroup$
– metamorphy
Dec 21 '18 at 10:14
add a comment |
1
$begingroup$
$2cos(q)$ is an algebraic integer for any rational $q$, hence so is the value of the symmetric polynomial. It is also rational, because the Galois group of whatever extension contains all the cosines permutes them. Hence this is an integer.
$endgroup$
– Wojowu
Dec 21 '18 at 9:54
$begingroup$
... meant $2cos(qpi)$.
$endgroup$
– metamorphy
Dec 21 '18 at 10:14
1
1
$begingroup$
$2cos(q)$ is an algebraic integer for any rational $q$, hence so is the value of the symmetric polynomial. It is also rational, because the Galois group of whatever extension contains all the cosines permutes them. Hence this is an integer.
$endgroup$
– Wojowu
Dec 21 '18 at 9:54
$begingroup$
$2cos(q)$ is an algebraic integer for any rational $q$, hence so is the value of the symmetric polynomial. It is also rational, because the Galois group of whatever extension contains all the cosines permutes them. Hence this is an integer.
$endgroup$
– Wojowu
Dec 21 '18 at 9:54
$begingroup$
... meant $2cos(qpi)$.
$endgroup$
– metamorphy
Dec 21 '18 at 10:14
$begingroup$
... meant $2cos(qpi)$.
$endgroup$
– metamorphy
Dec 21 '18 at 10:14
add a comment |
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$begingroup$
$2cos(q)$ is an algebraic integer for any rational $q$, hence so is the value of the symmetric polynomial. It is also rational, because the Galois group of whatever extension contains all the cosines permutes them. Hence this is an integer.
$endgroup$
– Wojowu
Dec 21 '18 at 9:54
$begingroup$
... meant $2cos(qpi)$.
$endgroup$
– metamorphy
Dec 21 '18 at 10:14