MINIMAL POLYNOMIALS FOR TANGENTS
whei...@corunduminium.com
integers for sin(X) and cos(X) can be computed by
elementary methods. I am asking if anyone knows what
has been done with minimal polynomials for tan(X).
Using quotients (sines/cosines), one can construct
high degree polynomials which have these as roots, but
we want the irreducible factors of these.
In addition, I am interested in knowing what the
entire root sets of these irreducible factors would be -
are the results as nice as for the cosines (all values
corresponding to the same n denominator where k/n is
reduced to lowest terms).
Robert Israel
The degrees of the minimal polynomials for cot(pi/n), n >= 2
(and thus for tan(pi/n) if n >= 3) form sequence A089929 in the OEIS
<http://www.research.att.com/~njas/sequences/A089929>
Hmmm. It looks to me like (at least for 3 <= n <= 50) the degree
is phi(n)/2 if n is divisible by 4, and phi(n) otherwise.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Gerry Myerson
<25803884.1226629043...@nitrogen.mathforum.org>,
"whei...@corunduminium.com" <whei...@corunduminium.com> wrote:
Here are some papers that might have some references.
S Beslin, V de Angelis, Math Mag 77 (2004) 146-149
D H Lehmer, Amer Math Monthly 40 (1933) 165-166
D Surowski, P McCombs, Missouri J Math Sci 15 (2003) 4-14
W Watkins, J Zeitlin, Amer Math Monthly 100 (1993) 471-474
K W Wegner, Amer Math Monthly 66 (1959) 52-53.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
David Bernier
>> If X = 2k(pi)/n, the minimal polynomials over the
>> integers for sin(X) and cos(X) can be computed by
>> elementary methods. I am asking if anyone knows what
>> has been done with minimal polynomials for tan(X).
>> Using quotients (sines/cosines), one can construct
>> high degree polynomials which have these as roots, but
>> we want the irreducible factors of these.
>> In addition, I am interested in knowing what the
>> entire root sets of these irreducible factors would be -
>> are the results as nice as for the cosines (all values
>> corresponding to the same n denominator where k/n is
>> reduced to lowest terms).
>
> The degrees of the minimal polynomials for cot(pi/n), n >= 2
> (and thus for tan(pi/n) if n >= 3) form sequence A089929 in the OEIS
> <http://www.research.att.com/~njas/sequences/A089929>
>
> Hmmm. It looks to me like (at least for 3 <= n <= 50) the degree
> is phi(n)/2 if n is divisible by 4, and phi(n) otherwise.
There are the two identities:
tan^2(x) + 1 = sec^2(x) [ = 1/(cos^2(x)) ] and
cot^2(x) + 1 = csc^2(x) [ = 1/(sin^2(x)) ].
The two right-hand sides are rational functions in one of sin(x)
or cos(x) .
I don't know if knowing the algebraic degree or minimal polynomial
of sin(m*pi/n) with m, n being integers could be of use, along
with the 2nd identity above, in working on the minimal
polynomial of cot(m*pi/n). Maybe yes, maybe no...
David Bernier
Robert Israel
It puts bounds on...
If cos(2x) has degree n, then tan^2(x) = 2/(cos(2x)+1)-1 does too (except for
some trivial cases), so tan(x) has degree at most 2n.
Robert Israel
It looks to me like...
if n>=3 is odd, sin(pi/n) and tan(pi/n) have degree phi(n), cos(pi/n) has
degree phi(n)/2
if n is divisible by 4, sin(pi/n) and cos(pi/n) have degree phi(n), which
tan(pi/n) has degree phi(n)/2
if n == 2 mod 4 and n >= 6, cos(pi/n) and tan(pi/n) have degree phi(n),
sin(pi/n) has degree phi(n)/2
Robert Israel
See the thread "algebraic order of cyclotomic numbers" from December 2006
<http://groups.google.ca/group/sci.math/browse_thread/thread/f04a190dd7da1546>
As Keith Ramsay explained there, cos(2 pi/m) always has order phi(m)/2.
Note that phi(2n) = phi(n) if n is odd, 2 phi(n) if n is even. I suspect
a similar method can be used to get the result for sin(pi/n), and maybe
even for tan(pi/n).
Michael Press
Robert Israel <isr...@math.MyUniversitysInitials.ca> wrote:
sin(2.pi/n) = cos(pi/2 - 2.pi/n) = cos(2.pi(n - 4)/(4.n))
so the minimal polynomial of sin(2.pi/n) has degree
phi(4.n/gcd(n-4, 4.n))/2 = phi(2.n/gcd(n-4, 16))
(tan(x))^2 + 1 = 1/(cos(x))^2
(cos(2.pi/n))^2 = 1/2(1 + cos(4.pi/n))
Minimal polynomial of (tan(2.pi/n))^2 has degree phi(n/gcd(n,2))/2.
--
Michael Press
whei...@corunduminium.com
with this problem and have found some interesting
iterative schemes to generate polynomials whose root are
real, distinct, and equal to the tangents of the angles
of the form k(pi)/m, lying between -(pi)/2 and (pi)/2,
and being in lowest terms with m denominator. It
appears they have degree (phi)(m) (Euler totient
function). However, they may not always be irreducible
over Z. I am also checking the algebra before making any unequivocal assertions.
Explicit construction of the polynomials is revealing
some interesting patterns which may be important for the
analysis - like they (which contain only even powers)
are similar to "every other term" in the binomial expansion of (x+1)^n.
I am doing all this by hand, because I do not know
enough about computer programs that compute or factor
polynomials. Can anybody point me to a site which might
be used to compute them for larger values?
When this is in a little better shape, I will scan and
post thumbnailed images of the manuscript on my website
and post a notice with URL for anyone to go and see the
symbolically complex treatments in my best "Equation
3.0" rendition.
Thanks also for the historical references, all of which
I hope to find and read as time allows.
whei...@corunduminium.com
quartic polynomial is not irreducble. It appears
tan(+/-pi/8) and tan(+/-3pi/8)
are roots of
x^4 - 6x^2 + 1 = (x^2 + 2x - 1)(x^2 -2x -1).
This means that a suggestion the degree is [phi(m)]/2
when m is a multiple of 4 is not violated.
Happy Holidays everyone!
whei...@corunduminium.com
I may have it. It apparently depends upon whether the
following conjecture is true.
The polynomial
{(1-iu)^{phi(n)}*{psi sub n ([1+iu]/[1-iu]}
has real integer coefficients and is irreducible over
Z. Here, i is the imaginary unit, phi is the Euler
totient and psi sub n is the nth cyclotomic polynomial.
It may not be true when n is divisible by 4, because two
factors over Z may exist, each of degree half of phi
(n). I just found this, and have some work to do.
If so, I think we have all the answers - routinely
computable minimal polynomials (the ones above) for the
tangents at issue. If not, then it would be interesting
to see what modifications are necessary.
whei...@corunduminium.com
The bottom line for tangents (and therefore cotangents) seems to be that the degree of tan[k*pi/n] (reduced form) is phi(n) if n is not a multiple of 4 and [phi(n)]/2 if n is a multiple of 4. We have polynomials of these degrees for which these are the root sets; but as mentioned their sincerely believed irreducibility has not been established.
If anyone has ideas about completing the proof or can tell me where I can find a computer algorithm to generate more examples (high degree polynomial work), I would be grateful. Also, if any substantive errors are discovered, please let me know at williamh@wcjccedu (where the last "c" should be replaced by the inescapapble "dot")!
So, I offer this "Christmas present of questionable value" - Happy Holidays, everyone.
Will Heierman
The assertion was that the algebraic degrees of these values are phi(n) unless n is a multiple of 4 (in which case it is [phi(n)]/2).
Polynomials T(sub n) of degree phi(n) have been constructed which have these tangents as their root sets. By restricting the angles to the interval (0, pi), we see they are uniquely represented, and clearly there are phi(n) distinct such values.
When n is a multiple of 4, the tangents can be partitioned to form two sets (according to whether k is congruent to 1 or to 3 (mod 4)), each containing half of them, and these have been shown to be the root sets of two irreducible polynomials over the integers whose product is T(sub n). We also have an inductive way to construct these factors; meaning we actually have access to the minimal polynomials in this case.
The only remaining point to be rigorously established (yes, I am convinced it's true) is that T(sub n) is irreducible if n is not a multiple of 4. It will suffice to show T(sub n) is irreducible if n is odd, because in this case irreducibility of
T(sub 2n) is an elementary consequence.
I do have some nice closed forms and recursive relationships for these T(sub n) (and the polynomials themselves for n<128), but so far none of these have produced inspirations that lead to irreducibility. I am of the opinion that it is a peculiar property of the tangent function, rather than some Galois-theoretic property of polynomials in general, that will be the key.
I am also convinced that this nagging detail may be the reason the problem may in fact be unsolved (meaning literature searches have not revealed any published solution). If you have any ideas, please let me know. You may post here or send message to my college email which is williamh(at)wcjc(dot)edu. Assuming we get it all done, I will publish the full paper on my website (with URL given in an announcement on this string).
Oh, yeah:
T(sub n)(x) = [1+x^2]^{[phi(n)]/2}* C(sub n)([1-x^2]/[1+x^2]),
where C(sub n) is the minimal polynomial for cos(2k*pi/n).
Will Heierman
A complete solution to the problem posted here can be found in a lovely article by J. S. Calcut, University of Texas at Austin. It is at URL: http://www.ma.utexas.edu/users/jack/ .
Proposition 2 on page 5 states the consolidated result, and the proof given contains the point I was missing (it's steeped in Galois theory, so brace yourself if you're not familiar with that). I hope to rephrase it in more mundane terms, and publish the final result as stated before.
Thank you, everyone, for your valuable assistance ... gratification achieved!!