Expressing $zeta^k+zeta^{-k}$ as a polynomial in $zeta+zeta^{-1}$.
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Let $zeta$ be an $n$-th root of unity and let $chi:=zeta+zeta^{-1}$. Then $zeta^k+zeta^{-k}=P_k(chi)$ where $P_kinBbb{Z}[X]$ is a polynomial not depending on $n$. For example we have
begin{eqnarray*}
zeta^2+zeta^{-2}&=&chi^2-2,
qquad&text{ so }&qquad
P_2&=&X^2-2,\
zeta^3+zeta^{-3}&=&chi^3-3chi,
qquad&text{ so }&qquad
P_3&=&X^3-3X,\
zeta^4+zeta^{-4}&=&chi^4-4chi^2+2,
qquad&text{ so }&qquad
P_4&=&X^4-4X^2+2,\
zeta^5+zeta^{-5}&=&chi^5-5chi^3+5chi,
qquad&text{ so }&qquad
P_5&=&X^5-5X^3+5,\
zeta^6+zeta^{-6}&=&chi^6-6chi^4+9chi^2+18,
qquad&text{ so }&qquad
P_6&=&X^6-6X^4+9X^2+18.
end{eqnarray*}
It isn't hard to see that $P_{ab}=P_acirc P_b=P_bcirc P_a$ for all positive integers $a$ and $b$, and that we have a recurrence relation
$$P_a=X^a-sum_{i=1}binom{a}{i}P_{a-2i},$$
where we take the convention that $P_k=0$ for all $k<0$, and $P_0=1$. My question is:
Is there a simple explicit expression for $P_k$?
algebraic-number-theory roots-of-unity cyclotomic-fields
asked Nov 15 at 16:38
Servaes
20.6k33789
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0
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Let $zeta$ be an $n$-th root of unity and let $chi:=zeta+zeta^{-1}$. Then $zeta^k+zeta^{-k}=P_k(chi)$ where $P_kinBbb{Z}[X]$ is a polynomial not depending on $n$. For example we have
begin{eqnarray*}
zeta^2+zeta^{-2}&=&chi^2-2,
qquad&text{ so }&qquad
P_2&=&X^2-2,\
zeta^3+zeta^{-3}&=&chi^3-3chi,
qquad&text{ so }&qquad
P_3&=&X^3-3X,\
zeta^4+zeta^{-4}&=&chi^4-4chi^2+2,
qquad&text{ so }&qquad
P_4&=&X^4-4X^2+2,\
zeta^5+zeta^{-5}&=&chi^5-5chi^3+5chi,
qquad&text{ so }&qquad
P_5&=&X^5-5X^3+5,\
zeta^6+zeta^{-6}&=&chi^6-6chi^4+9chi^2+18,
qquad&text{ so }&qquad
P_6&=&X^6-6X^4+9X^2+18.
end{eqnarray*}
It isn't hard to see that $P_{ab}=P_acirc P_b=P_bcirc P_a$ for all positive integers $a$ and $b$, and that we have a recurrence relation
$$P_a=X^a-sum_{i=1}binom{a}{i}P_{a-2i},$$
where we take the convention that $P_k=0$ for all $k<0$, and $P_0=1$. My question is:
Is there a simple explicit expression for $P_k$?
algebraic-number-theory roots-of-unity cyclotomic-fields
asked Nov 15 at 16:38
Servaes
20.6k33789
2
This may help you: en.wikipedia.org/wiki/Chebyshev_polynomials
– Seewoo Lee
Nov 15 at 16:40
@SeewooLee Can you turn that comment into an answer?
– Pedro Tamaroff♦
Nov 15 at 16:43
@SeewooLee Thank you for the link. So if I understand correctly we have $P_k(X)=2T_k(frac{X}{2})$, where $T_k$ is the $k$-th Chebyshev polynomial of the first kind?
– Servaes
Nov 15 at 16:46
@SeewooLee Ok all is clear, thanks again. If you care to write up an answer, I'll gladly accept.
– Servaes
Nov 15 at 17:09
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $zeta$ be an $n$-th root of unity and let $chi:=zeta+zeta^{-1}$. Then $zeta^k+zeta^{-k}=P_k(chi)$ where $P_kinBbb{Z}[X]$ is a polynomial not depending on $n$. For example we have
begin{eqnarray*}
zeta^2+zeta^{-2}&=&chi^2-2,
qquad&text{ so }&qquad
P_2&=&X^2-2,\
zeta^3+zeta^{-3}&=&chi^3-3chi,
qquad&text{ so }&qquad
P_3&=&X^3-3X,\
zeta^4+zeta^{-4}&=&chi^4-4chi^2+2,
qquad&text{ so }&qquad
P_4&=&X^4-4X^2+2,\
zeta^5+zeta^{-5}&=&chi^5-5chi^3+5chi,
qquad&text{ so }&qquad
P_5&=&X^5-5X^3+5,\
zeta^6+zeta^{-6}&=&chi^6-6chi^4+9chi^2+18,
qquad&text{ so }&qquad
P_6&=&X^6-6X^4+9X^2+18.
end{eqnarray*}
It isn't hard to see that $P_{ab}=P_acirc P_b=P_bcirc P_a$ for all positive integers $a$ and $b$, and that we have a recurrence relation
$$P_a=X^a-sum_{i=1}binom{a}{i}P_{a-2i},$$
where we take the convention that $P_k=0$ for all $k<0$, and $P_0=1$. My question is:
Is there a simple explicit expression for $P_k$?
algebraic-number-theory roots-of-unity cyclotomic-fields
asked Nov 15 at 16:38
Servaes
20.6k33789
Let $zeta$ be an $n$-th root of unity and let $chi:=zeta+zeta^{-1}$. Then $zeta^k+zeta^{-k}=P_k(chi)$ where $P_kinBbb{Z}[X]$ is a polynomial not depending on $n$. For example we have
begin{eqnarray*}
zeta^2+zeta^{-2}&=&chi^2-2,
qquad&text{ so }&qquad
P_2&=&X^2-2,\
zeta^3+zeta^{-3}&=&chi^3-3chi,
qquad&text{ so }&qquad
P_3&=&X^3-3X,\
zeta^4+zeta^{-4}&=&chi^4-4chi^2+2,
qquad&text{ so }&qquad
P_4&=&X^4-4X^2+2,\
zeta^5+zeta^{-5}&=&chi^5-5chi^3+5chi,
qquad&text{ so }&qquad
P_5&=&X^5-5X^3+5,\
zeta^6+zeta^{-6}&=&chi^6-6chi^4+9chi^2+18,
qquad&text{ so }&qquad
P_6&=&X^6-6X^4+9X^2+18.
end{eqnarray*}
It isn't hard to see that $P_{ab}=P_acirc P_b=P_bcirc P_a$ for all positive integers $a$ and $b$, and that we have a recurrence relation
$$P_a=X^a-sum_{i=1}binom{a}{i}P_{a-2i},$$
where we take the convention that $P_k=0$ for all $k<0$, and $P_0=1$. My question is:
Is there a simple explicit expression for $P_k$?
algebraic-number-theory roots-of-unity cyclotomic-fields
algebraic-number-theory roots-of-unity cyclotomic-fields
asked Nov 15 at 16:38
Servaes
20.6k33789
asked Nov 15 at 16:38
Servaes
20.6k33789
asked Nov 15 at 16:38
Servaes
20.6k33789
asked Nov 15 at 16:38
Servaes
20.6k33789
asked Nov 15 at 16:38
Servaes
20.6k33789
20.6k33789
2
This may help you: en.wikipedia.org/wiki/Chebyshev_polynomials
– Seewoo Lee
Nov 15 at 16:40
@SeewooLee Can you turn that comment into an answer?
– Pedro Tamaroff♦
Nov 15 at 16:43
@SeewooLee Thank you for the link. So if I understand correctly we have $P_k(X)=2T_k(frac{X}{2})$, where $T_k$ is the $k$-th Chebyshev polynomial of the first kind?
– Servaes
Nov 15 at 16:46
@SeewooLee Ok all is clear, thanks again. If you care to write up an answer, I'll gladly accept.
– Servaes
Nov 15 at 17:09
add a comment |
2
This may help you: en.wikipedia.org/wiki/Chebyshev_polynomials
– Seewoo Lee
Nov 15 at 16:40
@SeewooLee Can you turn that comment into an answer?
– Pedro Tamaroff♦
Nov 15 at 16:43
@SeewooLee Thank you for the link. So if I understand correctly we have $P_k(X)=2T_k(frac{X}{2})$, where $T_k$ is the $k$-th Chebyshev polynomial of the first kind?
– Servaes
Nov 15 at 16:46
@SeewooLee Ok all is clear, thanks again. If you care to write up an answer, I'll gladly accept.
– Servaes
Nov 15 at 17:09
2
2
This may help you: en.wikipedia.org/wiki/Chebyshev_polynomials
– Seewoo Lee
Nov 15 at 16:40
This may help you: en.wikipedia.org/wiki/Chebyshev_polynomials
– Seewoo Lee
Nov 15 at 16:40
@SeewooLee Can you turn that comment into an answer?
– Pedro Tamaroff♦
Nov 15 at 16:43
@SeewooLee Can you turn that comment into an answer?
– Pedro Tamaroff♦
Nov 15 at 16:43
@SeewooLee Thank you for the link. So if I understand correctly we have $P_k(X)=2T_k(frac{X}{2})$, where $T_k$ is the $k$-th Chebyshev polynomial of the first kind?
– Servaes
Nov 15 at 16:46
@SeewooLee Thank you for the link. So if I understand correctly we have $P_k(X)=2T_k(frac{X}{2})$, where $T_k$ is the $k$-th Chebyshev polynomial of the first kind?
– Servaes
Nov 15 at 16:46
@SeewooLee Ok all is clear, thanks again. If you care to write up an answer, I'll gladly accept.
– Servaes
Nov 15 at 17:09
@SeewooLee Ok all is clear, thanks again. If you care to write up an answer, I'll gladly accept.
– Servaes
Nov 15 at 17:09
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
As I said, Chebyshev polynomials are exactly what you said.
answered Nov 15 at 17:55
Seewoo Lee
5,931826
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
accepted
As I said, Chebyshev polynomials are exactly what you said.
answered Nov 15 at 17:55
Seewoo Lee
5,931826
add a comment |
up vote
1
down vote
accepted
As I said, Chebyshev polynomials are exactly what you said.
answered Nov 15 at 17:55
Seewoo Lee
5,931826
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As I said, Chebyshev polynomials are exactly what you said.
answered Nov 15 at 17:55
Seewoo Lee
5,931826
As I said, Chebyshev polynomials are exactly what you said.
answered Nov 15 at 17:55
Seewoo Lee
5,931826
answered Nov 15 at 17:55
Seewoo Lee
5,931826
answered Nov 15 at 17:55
Seewoo Lee
5,931826
answered Nov 15 at 17:55
Seewoo Lee
5,931826
5,931826
add a comment |
add a comment |
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This may help you: en.wikipedia.org/wiki/Chebyshev_polynomials
– Seewoo Lee
Nov 15 at 16:40
@SeewooLee Can you turn that comment into an answer?
– Pedro Tamaroff♦
Nov 15 at 16:43
@SeewooLee Thank you for the link. So if I understand correctly we have $P_k(X)=2T_k(frac{X}{2})$, where $T_k$ is the $k$-th Chebyshev polynomial of the first kind?
– Servaes
Nov 15 at 16:46
@SeewooLee Ok all is clear, thanks again. If you care to write up an answer, I'll gladly accept.
– Servaes
Nov 15 at 17:09