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A220673
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Coefficients of formal series in powers of (tan(x))^2 for tan(5*x)/tan(x).
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2
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5, 40, 376, 3560, 33720, 319400, 3025400, 28657000, 271443000, 2571145000, 24354235000, 230686625000, 2185095075000, 20697517625000, 196049700875000, 1857009420625000, 17589845701875000, 166613409915625000, 1578184870646875000
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OFFSET
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0,1
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COMMENTS
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Formally Sum_{n>=0} a(n)*(tan(x))^(2*n) = tan(5*x)/ tan(x).
Convergence holds for x from the two open intervals (1-sqrt(6/5),1-2/sqrt(5)) and (1+2/sqrt(5),1+sqrt(6/5), namely (-0.095445115, .1055728092) and (1.894427191, 2.095445115) (10 digits).
These intervals follow from the denominator of the o.g.f. G(5,x) = (5 - 10*x + x^2)/(1 - 10*x + 5*x^2).
If one replaces x by (tan(x))^2 in this o.g.f. one obtains the formula for tan(5*x)/tan(x) in terms of (tan(x))^2. This formula is the special (n=5) solution of a general recurrence derivable from the addition theorem for tan(n*x) = tan(x + (n-1)*x), namely, with Q(n,x) := tan(n,x)/tan(x), Q(n,x) = (1 + Q(n-1,x))/(1 - v*Q(n-1,x)), where v = v(x) =(tan(x))^2, and the input is Q(1,x) = 1. Read as function of v the solution for Q(5,x) is just G(5,v) with replaced v=v(x).
See the irregular triangles A034867 and A034839 whose row polynomials N(n,x) and D(n,x), respectively, give for n >= 1 the solution to the recurrences N(n,x) = D(n-1,x) + N(n-1,x), D(n,x) = D(n-1,x) + x*N(n-1,x), with inputs N(1,x) = 1 and D(1,x) = 1. The proof by the Pascal triangle A007318 recurrence is trivial. Therefore, Q(n,x) from the preceding comment is given by Q(n,x) = N(n,-v)/D(n,-v) with v=v(x) = (tan(x))^2.
One also has, with Chebyshev's S polynomials (see A049310) Q(n,x) = tan(n*x)/tan(x) = (S(n,y) + S(n-2,y))/(S(n,y) - S(n-2,y)) = y*S(n-1,y)/(S(n,y) - S(n-2,y)) = 1/(1 - (2/y)*S(n-2,y)/S(n-1,y)), where y = y(x) = 2/sqrt(1 + (tan(x))^2). This derives from sin(n*x)/cos(n*x) in terms of Chebyshev S polynomials with argument 2*cos(x) = y(x). Note that S(n-2,y)/S(n-1,y) has the continued fraction representation 1/(y-1/(y- ... -1/(y )..(n-1)brackets..), i.e. (n-1) y's.
These calculations have been motivated by e-mails from Thomas Olsen.
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LINKS
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FORMULA
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O.g.f.: G(5,x) = (5 - 10*x + x^2)/(1 - 10*x + 5*x^2).
a(n) = delta(n,0)/5 - 8*b(n) + 24*b(n+1)/5, n>=0,
with Kronecker's delta and b(n):= A190987(n).
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EXAMPLE
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Q(5,x=0.1) = tan(5.1)/tan(0.1) = 5.444802663 (Maple 10 digits);
G(5,tan(0.1)^2) = 5.444802664;
sum(a(n)*(tan(0.1))^(2*n),n=0..100)) = 5.444802664.
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MATHEMATICA
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CoefficientList[Series[(5-10*x+x^2)/(1-10*x+5*x^2), {x, 0, 50}], x] (* G. C. Greubel, Mar 06 2018 *)
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PROG
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(PARI) x='x+O('x^30); Vec((5-10*x+x^2)/(1-10*x+5*x^2)) \\ G. C. Greubel, Mar 06 2018
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((5-10*x+x^2)/(1-10*x+5*x^2))); // G. C. Greubel, Mar 06 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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