Mixing
$cosfracpi{n}$ Analytic expression
$begingroup$
I recently found out that $$sinfracpi5=frac12sqrt{frac{5-sqrt5}2}$$
Which means that $$cosfracpi5=frac{1+sqrt5}4$$
I also recently found that if $ninBbb N$,
$$sin nx=sin x,U_{n-1}(cos x)\ cos nx=T_n(cos x)$$
Where $T_n$ and $U_n$ are the Chebyshev polynomials of the first and second kinds respectively. They are defined as
$$T_n(x)=frac{n}2sum_{r=0}^{lfloor n/2rfloor}frac{(-1)^n}{n-r}{n-rchoose r}(2x)^{n-2r}$$
$$U_n(x)=sum_{r=0}^{lfloor n/2rfloor}(-1)^r{n-rchoose r}(2x)^{n-2r}$$
Using these definitions, I am attempting to find a general analytic expression for $cosfracpi n$. To do so I started with
$$cos nx=T_n(cos x)$$
$$cos ncdotfrac{pi}{n}=T_nbigg(cosfrac{pi}nbigg)$$
then setting $w=cos(pi/n)$,
$$T_n(w)+1=0$$
So our task is to solve for $w$. I'm fairly certain that $x=cos(pi/n)$ is always the largest real root of $T_n(x)+1=0$. So I guess that's my question:
What Is the largest real root of $T_n(x)+1=0$?
But as far as I know there are no really simple ways of going about this (but I really don't know much). Could I have a bit of help? Thanks.
trigonometry roots chebyshev-polynomials
edited Jan 15 at 21:56
user376343
3,7883828
asked Jan 15 at 4:27
clathratusclathratus
4,568337
$endgroup$
add a comment |
$begingroup$
I recently found out that $$sinfracpi5=frac12sqrt{frac{5-sqrt5}2}$$
Which means that $$cosfracpi5=frac{1+sqrt5}4$$
I also recently found that if $ninBbb N$,
$$sin nx=sin x,U_{n-1}(cos x)\ cos nx=T_n(cos x)$$
Where $T_n$ and $U_n$ are the Chebyshev polynomials of the first and second kinds respectively. They are defined as
$$T_n(x)=frac{n}2sum_{r=0}^{lfloor n/2rfloor}frac{(-1)^n}{n-r}{n-rchoose r}(2x)^{n-2r}$$
$$U_n(x)=sum_{r=0}^{lfloor n/2rfloor}(-1)^r{n-rchoose r}(2x)^{n-2r}$$
Using these definitions, I am attempting to find a general analytic expression for $cosfracpi n$. To do so I started with
$$cos nx=T_n(cos x)$$
$$cos ncdotfrac{pi}{n}=T_nbigg(cosfrac{pi}nbigg)$$
then setting $w=cos(pi/n)$,
$$T_n(w)+1=0$$
So our task is to solve for $w$. I'm fairly certain that $x=cos(pi/n)$ is always the largest real root of $T_n(x)+1=0$. So I guess that's my question:
What Is the largest real root of $T_n(x)+1=0$?
But as far as I know there are no really simple ways of going about this (but I really don't know much). Could I have a bit of help? Thanks.
trigonometry roots chebyshev-polynomials
edited Jan 15 at 21:56
user376343
3,7883828
asked Jan 15 at 4:27
clathratusclathratus
4,568337
$endgroup$
4
$begingroup$
That largest root cannot be written in terms of real roots of real numbers. It is, of course, trivial to write it using complex $n$th roots of $1$, but that is tautological. More precisely, in general the zeros of a polynomial with rational coefficients can be written using roots of integers if and only if the Galois group of the polynomial is solvable. In the case of Chebyshev polynomials that solvability requirement is never an issue (but is the underlying reason why there exists no formula for quintics or higher).
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:37
4
$begingroup$
(cont'd) But, even in the solvable case, we need complex $n$th roots, and those cannot be replaced with real roots without having these cosines, making this a bit circular :-( For details, I recommend a look at the famous casus irreducibilis of cubic equations.
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:45
3
$begingroup$
The best possible, I would call it, "counterexample" I can think of is $cosleft(fracpi7right)$. This term cannot be written as an analytic expression as far as I know.
$endgroup$
– mrtaurho
Jan 15 at 5:41
$begingroup$
What about $cosleft(frac{pi}{17}right)$ ?
$endgroup$
– Aleksas Domarkas
Jan 15 at 9:24
1
$begingroup$
@AleksasDomarkas One of Gauss's most famous results was showing that the regular 17-gon is constructible, or in other words $cos(2pi/17)$ (and therefore $cos(pi/17)$ as well) can be written using iterated square roots. In modern terms, the reason for this is that the Galois group of $mathbb{Q}(e^{2pi i/17})$ over $mathbb{Q}$ has order 16, so you can write a chain of degree 2 (normal) subgroups.
$endgroup$
– Daniel Schepler
Jan 15 at 22:17
add a comment |
$begingroup$
I recently found out that $$sinfracpi5=frac12sqrt{frac{5-sqrt5}2}$$
Which means that $$cosfracpi5=frac{1+sqrt5}4$$
I also recently found that if $ninBbb N$,
$$sin nx=sin x,U_{n-1}(cos x)\ cos nx=T_n(cos x)$$
Where $T_n$ and $U_n$ are the Chebyshev polynomials of the first and second kinds respectively. They are defined as
$$T_n(x)=frac{n}2sum_{r=0}^{lfloor n/2rfloor}frac{(-1)^n}{n-r}{n-rchoose r}(2x)^{n-2r}$$
$$U_n(x)=sum_{r=0}^{lfloor n/2rfloor}(-1)^r{n-rchoose r}(2x)^{n-2r}$$
Using these definitions, I am attempting to find a general analytic expression for $cosfracpi n$. To do so I started with
$$cos nx=T_n(cos x)$$
$$cos ncdotfrac{pi}{n}=T_nbigg(cosfrac{pi}nbigg)$$
then setting $w=cos(pi/n)$,
$$T_n(w)+1=0$$
So our task is to solve for $w$. I'm fairly certain that $x=cos(pi/n)$ is always the largest real root of $T_n(x)+1=0$. So I guess that's my question:
What Is the largest real root of $T_n(x)+1=0$?
But as far as I know there are no really simple ways of going about this (but I really don't know much). Could I have a bit of help? Thanks.
trigonometry roots chebyshev-polynomials
edited Jan 15 at 21:56
user376343
3,7883828
asked Jan 15 at 4:27
clathratusclathratus
4,568337
$endgroup$
I recently found out that $$sinfracpi5=frac12sqrt{frac{5-sqrt5}2}$$
Which means that $$cosfracpi5=frac{1+sqrt5}4$$
I also recently found that if $ninBbb N$,
$$sin nx=sin x,U_{n-1}(cos x)\ cos nx=T_n(cos x)$$
Where $T_n$ and $U_n$ are the Chebyshev polynomials of the first and second kinds respectively. They are defined as
$$T_n(x)=frac{n}2sum_{r=0}^{lfloor n/2rfloor}frac{(-1)^n}{n-r}{n-rchoose r}(2x)^{n-2r}$$
$$U_n(x)=sum_{r=0}^{lfloor n/2rfloor}(-1)^r{n-rchoose r}(2x)^{n-2r}$$
Using these definitions, I am attempting to find a general analytic expression for $cosfracpi n$. To do so I started with
$$cos nx=T_n(cos x)$$
$$cos ncdotfrac{pi}{n}=T_nbigg(cosfrac{pi}nbigg)$$
then setting $w=cos(pi/n)$,
$$T_n(w)+1=0$$
So our task is to solve for $w$. I'm fairly certain that $x=cos(pi/n)$ is always the largest real root of $T_n(x)+1=0$. So I guess that's my question:
What Is the largest real root of $T_n(x)+1=0$?
But as far as I know there are no really simple ways of going about this (but I really don't know much). Could I have a bit of help? Thanks.
trigonometry roots chebyshev-polynomials
trigonometry roots chebyshev-polynomials
edited Jan 15 at 21:56
user376343
3,7883828
asked Jan 15 at 4:27
clathratusclathratus
4,568337
edited Jan 15 at 21:56
user376343
3,7883828
asked Jan 15 at 4:27
clathratusclathratus
4,568337
edited Jan 15 at 21:56
user376343
3,7883828
edited Jan 15 at 21:56
user376343
3,7883828
edited Jan 15 at 21:56
user376343
3,7883828
3,7883828
asked Jan 15 at 4:27
clathratusclathratus
4,568337
asked Jan 15 at 4:27
clathratusclathratus
4,568337
asked Jan 15 at 4:27
clathratusclathratus
4,568337
4,568337
4
$begingroup$
That largest root cannot be written in terms of real roots of real numbers. It is, of course, trivial to write it using complex $n$th roots of $1$, but that is tautological. More precisely, in general the zeros of a polynomial with rational coefficients can be written using roots of integers if and only if the Galois group of the polynomial is solvable. In the case of Chebyshev polynomials that solvability requirement is never an issue (but is the underlying reason why there exists no formula for quintics or higher).
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:37
4
$begingroup$
(cont'd) But, even in the solvable case, we need complex $n$th roots, and those cannot be replaced with real roots without having these cosines, making this a bit circular :-( For details, I recommend a look at the famous casus irreducibilis of cubic equations.
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:45
3
$begingroup$
The best possible, I would call it, "counterexample" I can think of is $cosleft(fracpi7right)$. This term cannot be written as an analytic expression as far as I know.
$endgroup$
– mrtaurho
Jan 15 at 5:41
$begingroup$
What about $cosleft(frac{pi}{17}right)$ ?
$endgroup$
– Aleksas Domarkas
Jan 15 at 9:24
1
$begingroup$
@AleksasDomarkas One of Gauss's most famous results was showing that the regular 17-gon is constructible, or in other words $cos(2pi/17)$ (and therefore $cos(pi/17)$ as well) can be written using iterated square roots. In modern terms, the reason for this is that the Galois group of $mathbb{Q}(e^{2pi i/17})$ over $mathbb{Q}$ has order 16, so you can write a chain of degree 2 (normal) subgroups.
$endgroup$
– Daniel Schepler
Jan 15 at 22:17
add a comment |
4
$begingroup$
That largest root cannot be written in terms of real roots of real numbers. It is, of course, trivial to write it using complex $n$th roots of $1$, but that is tautological. More precisely, in general the zeros of a polynomial with rational coefficients can be written using roots of integers if and only if the Galois group of the polynomial is solvable. In the case of Chebyshev polynomials that solvability requirement is never an issue (but is the underlying reason why there exists no formula for quintics or higher).
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:37
4
$begingroup$
(cont'd) But, even in the solvable case, we need complex $n$th roots, and those cannot be replaced with real roots without having these cosines, making this a bit circular :-( For details, I recommend a look at the famous casus irreducibilis of cubic equations.
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:45
3
$begingroup$
The best possible, I would call it, "counterexample" I can think of is $cosleft(fracpi7right)$. This term cannot be written as an analytic expression as far as I know.
$endgroup$
– mrtaurho
Jan 15 at 5:41
$begingroup$
What about $cosleft(frac{pi}{17}right)$ ?
$endgroup$
– Aleksas Domarkas
Jan 15 at 9:24
1
$begingroup$
@AleksasDomarkas One of Gauss's most famous results was showing that the regular 17-gon is constructible, or in other words $cos(2pi/17)$ (and therefore $cos(pi/17)$ as well) can be written using iterated square roots. In modern terms, the reason for this is that the Galois group of $mathbb{Q}(e^{2pi i/17})$ over $mathbb{Q}$ has order 16, so you can write a chain of degree 2 (normal) subgroups.
$endgroup$
– Daniel Schepler
Jan 15 at 22:17
4
4
$begingroup$
That largest root cannot be written in terms of real roots of real numbers. It is, of course, trivial to write it using complex $n$th roots of $1$, but that is tautological. More precisely, in general the zeros of a polynomial with rational coefficients can be written using roots of integers if and only if the Galois group of the polynomial is solvable. In the case of Chebyshev polynomials that solvability requirement is never an issue (but is the underlying reason why there exists no formula for quintics or higher).
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:37
$begingroup$
That largest root cannot be written in terms of real roots of real numbers. It is, of course, trivial to write it using complex $n$th roots of $1$, but that is tautological. More precisely, in general the zeros of a polynomial with rational coefficients can be written using roots of integers if and only if the Galois group of the polynomial is solvable. In the case of Chebyshev polynomials that solvability requirement is never an issue (but is the underlying reason why there exists no formula for quintics or higher).
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:37
4
4
$begingroup$
(cont'd) But, even in the solvable case, we need complex $n$th roots, and those cannot be replaced with real roots without having these cosines, making this a bit circular :-( For details, I recommend a look at the famous casus irreducibilis of cubic equations.
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:45
$begingroup$
(cont'd) But, even in the solvable case, we need complex $n$th roots, and those cannot be replaced with real roots without having these cosines, making this a bit circular :-( For details, I recommend a look at the famous casus irreducibilis of cubic equations.
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:45
3
3
$begingroup$
The best possible, I would call it, "counterexample" I can think of is $cosleft(fracpi7right)$. This term cannot be written as an analytic expression as far as I know.
$endgroup$
– mrtaurho
Jan 15 at 5:41
$begingroup$
The best possible, I would call it, "counterexample" I can think of is $cosleft(fracpi7right)$. This term cannot be written as an analytic expression as far as I know.
$endgroup$
– mrtaurho
Jan 15 at 5:41
$begingroup$
What about $cosleft(frac{pi}{17}right)$ ?
$endgroup$
– Aleksas Domarkas
Jan 15 at 9:24
$begingroup$
What about $cosleft(frac{pi}{17}right)$ ?
$endgroup$
– Aleksas Domarkas
Jan 15 at 9:24
1
1
$begingroup$
@AleksasDomarkas One of Gauss's most famous results was showing that the regular 17-gon is constructible, or in other words $cos(2pi/17)$ (and therefore $cos(pi/17)$ as well) can be written using iterated square roots. In modern terms, the reason for this is that the Galois group of $mathbb{Q}(e^{2pi i/17})$ over $mathbb{Q}$ has order 16, so you can write a chain of degree 2 (normal) subgroups.
$endgroup$
– Daniel Schepler
Jan 15 at 22:17
$begingroup$
@AleksasDomarkas One of Gauss's most famous results was showing that the regular 17-gon is constructible, or in other words $cos(2pi/17)$ (and therefore $cos(pi/17)$ as well) can be written using iterated square roots. In modern terms, the reason for this is that the Galois group of $mathbb{Q}(e^{2pi i/17})$ over $mathbb{Q}$ has order 16, so you can write a chain of degree 2 (normal) subgroups.
$endgroup$
– Daniel Schepler
Jan 15 at 22:17
add a comment |
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4
$begingroup$
That largest root cannot be written in terms of real roots of real numbers. It is, of course, trivial to write it using complex $n$th roots of $1$, but that is tautological. More precisely, in general the zeros of a polynomial with rational coefficients can be written using roots of integers if and only if the Galois group of the polynomial is solvable. In the case of Chebyshev polynomials that solvability requirement is never an issue (but is the underlying reason why there exists no formula for quintics or higher).
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:37
4
$begingroup$
(cont'd) But, even in the solvable case, we need complex $n$th roots, and those cannot be replaced with real roots without having these cosines, making this a bit circular :-( For details, I recommend a look at the famous casus irreducibilis of cubic equations.
$endgroup$
– Jyrki Lahtonen
Jan 15 at 4:45
3
$begingroup$
The best possible, I would call it, "counterexample" I can think of is $cosleft(fracpi7right)$. This term cannot be written as an analytic expression as far as I know.
$endgroup$
– mrtaurho
Jan 15 at 5:41
$begingroup$
What about $cosleft(frac{pi}{17}right)$ ?
$endgroup$
– Aleksas Domarkas
Jan 15 at 9:24
1
$begingroup$
@AleksasDomarkas One of Gauss's most famous results was showing that the regular 17-gon is constructible, or in other words $cos(2pi/17)$ (and therefore $cos(pi/17)$ as well) can be written using iterated square roots. In modern terms, the reason for this is that the Galois group of $mathbb{Q}(e^{2pi i/17})$ over $mathbb{Q}$ has order 16, so you can write a chain of degree 2 (normal) subgroups.
$endgroup$
– Daniel Schepler
Jan 15 at 22:17