James Buddenhagen associated the polynomial:
x^3 - sqrt(7)*x^2 + x + 1/sqrt(7)
to the identity:
tan(3*pi/14) + cot(pi/7) - tan(pi/14) = sqrt(7).
Is there a polynomial that we can associate to the fact that
tan(3*pi/11) + 4*sin(2*pi/11) = sqrt(11)?
Warut
rwg> BCCee Dick Askey replied:
Here are two similar identities. tan(3pi/11) + 4sin(2pi/11) = 11^(1/2) sin(2pi/7) + sin(4pi/7) + sin(8pi/7) = (1/2)*7^(1/2). These are in Chapter 7 of Hobson's "A Treatise on Plane and Advanced Trigonometry". In case you do not know this book, it has a number of gems. Dover has it back in print. Dick
Mma 7.0 FullSimplify does these, too. Maybe it just guesses them numerically? No, it has a ferocious algebraic number simplifier:
In[39]:= ToRadicals[Tan[3*Pi/11] + 4*Sin[2*Pi/11]] Out[39]= 5/11 7/22 4/11 I (1 + (-1) ) -2 (-1) (-1 + (-1) ) + ---------------- 5/11 -1 + (-1)
Perhaps I should have moved the problem to Macsyma and interpolated some missing steps: (c1) (domain:complex, mma2mac( -2*(-1)^(7/22)*(-1+(-1)^(4/11))+(I*(1+(-1)^(5/11)))/(-1+(-1)^(5/11)))) 5/11 ((- 1) + 1) %i 7/22 4/11 (d1) ------------------ - 2 (- 1) ((- 1) - 1) 5/11 (- 1) - 1 Find polynomials for these two terms separately: (c2) multiply_conjugates(x-args(%)) 11 9 7 5 3 (d2) [x - 55 x + 330 x - 462 x + 165 x - 11 x, 22 20 18 16 14 12 x - 88 x + 3344 x - 71808 x + 957440 x - 8200192 x 10 8 6 4 + 45101056 x - 154632192 x + 309264384 x - 317194240 x 2 + 126877696 x ] The polynomial satisfied satisfied by the sums of pairs of roots of these two polynomials is (c3) factor(resultant(subst(y,x,%[1]),subst(x-y,x,%[2]),y)) 2 2 10 (d3) x (x - 11) 10 8 6 4 2 2 (x - 143 x + 2970 x - 15246 x + 29909 x - 20339) 10 8 6 4 2 2 (x - 143 x + 3674 x - 25102 x + 41877 x - 20339) 10 8 6 4 2 2 (x - 143 x + 4906 x - 60126 x + 224389 x - 49379) 10 8 6 4 2 2 (x - 143 x + 5962 x - 87934 x + 514789 x - 1036739) 10 8 6 4 2 2 (x - 143 x + 7194 x - 161678 x + 1594197 x - 5131379) 10 8 6 4 2 2 (x - 55 x + 330 x - 462 x + 165 x - 11) 10 8 6 4 2 4 (x - 55 x + 858 x - 5214 x + 11605 x - 5819) 10 8 6 4 2 4 (x - 55 x + 1034 x - 7502 x + 14949 x - 11) 10 8 6 4 2 2 (x - 44 x + 704 x - 4928 x + 14080 x - 11264)
In[40]:= RootReduce[%]
Out[40]= Sqrt[11]
I confess ignorance of how, other than numerically, to choose the correct factor. Perhaps that 10th power is trying to tell me something? --rwg
On Thu, Jul 30, 2009 at 11:30 AM, James Buddenhagen <jbuddenh@gmail.com> wrote:
tan(3pi/14), cot(pi/7), and -tan(pi/14) are the roots of
x^3 - sqrt(7) x^2+ x + 1/sqrt(7)
and so their sum is sqrt(7) and their product is -1/sqrt(7) and the sum of their products in pairs is 1. �I found the polynomial after the fact, so no proof here.
I have no idea how Mathematica simplifies the sum.
On Wed, Jul 29, 2009 at 10:33 PM, <rwg@sdf.lonestar.org> wrote:
Last night, Mathematica embarrassed Macsyma and me by simplifying tan(3*pi/14)+cot(pi/7)-tan(pi/14) to sqrt(7) in front of dozens of impressionable children. �Later I found a tricky proof: �The quotient of two specializations of the very handy formula
prod(a*sin(2*j*pi/n+g)+b*cos(2*j*pi/n+g),j,1,n) �= 2*(b^2+a^2)^(n/2)*(T[n](0)-T[n](-(a*sin(g)+b*cos(g))/sqrt(b^2+a^2)))/2^n,
(where T[n](x):=cos(n*acos(x)) = then nth Chebychev polynomial)
has the limit, with n=7,
(x^2-tan(pi/14)^2)*(x^2-cot(pi/7)^2)*(x^2-tan(3*pi/14)^2) �= x^6-5*x^4+3*x^2-1/7 ,
(giving that the product of these three tans is the negative reciprocal �of their sum).
Equating coefficients of x, adjoining tan(3*pi/14)+cot(pi/7)-tan(pi/14) = y, and eliminating the trigs gives
� � � �4*(y^2-7)*(7*y^6-91*y^4+245*y^2-169)
and the correct root can be selected numerically from the eight.
Can someone suggest a method more likely to be Mathematica's? --rwg
This is settled.