OFFSET
0,2
COMMENTS
This entry originated from a proposal by Ozgur Cem Birler for the integral of the fourth power of sin(x). Peter Bala suggested the use of triangles to cover all even and odd powers. See A254932.
The signed triangle T(n, k) appears in the formula
2^(2*n)*(2*n+1)!!*int(sin^{2*n+1}x dx) = Sum_{k=0..n} T(n, k)*cos((2*k+1)*x), n >= 0.
See the Gradstein - Ryshik reference, 2.513 2., p. 168. Here (2*n+1)!! = A001147(n+1), n >= 0.
The unsigned triangle T(n, k) appears in the formula
2^(2*n)* (2*n+1)!!*int(cos^{2*n+1}x dx) = Sum_{k=0..n} |T(n, k)|*sin((2*k+1)*x), n >= 0.
See the Gradstein & Ryshik reference, 2.513 4., p. 169.
REFERENCES
I. S. Gradstein and I. M. Ryshik, Tables of series, products, and integrals, Volume 1, Verlag Harri Deutsch, 1981.
LINKS
Wolfdieter Lang, The first 9 rows and more.
FORMULA
T(n, k) = (-1)^(k+1)*binomial(2*n+1,n-k)*(2*n+1)!!/(2*k+1), 0 <= k <= n, with the double factorials (2*n+1)!! = A001147(n+1).
EXAMPLE
The triangle T(n, k) begins:
n\k 0 1 2 4 5 6...
0: -1
1: -9 1
2: -150 25 -3
3: -3675 735 -147 15
4: -119070 26460 -6804 1215 -105
5: -4802490 1143450 -343035 81675 -12705 945
...
For more rows see the link.
n=2: 2^4*(1*3*5)*int((sin x)^5 dx) = -150*cos(x) + 25*cos(3*x) - 3*cos(5*x),
that is: int((sin x)^5 dx) = -(5/8)*cos(x) + (5/48)*cos(3*x) - (1/80)*cos(5*x).
2^4*(1*3*5)*int((cos x)^5 dx) = +150*sin(x) + 25*sin(3*x) + 3*sin(5*x),
that is: int((cos x)^5 dx) = (5/8)*sin(x) + (5/48)*sin(3*x) + (1/80)*sin(5*x).
MATHEMATICA
T[n_, k_] := (-1)^(k+1)*Binomial[2*n+1, n-k]*(2*n+1)!!/(2*k+1); Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 18 2015 *)
CROSSREFS
KEYWORD
AUTHOR
Wolfdieter Lang, Feb 15 2015
STATUS
approved