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A104203
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Expansion of the sine lemniscate function sl(x).
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8
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1, 0, 0, 0, -12, 0, 0, 0, 3024, 0, 0, 0, -4390848, 0, 0, 0, 21224560896, 0, 0, 0, -257991277243392, 0, 0, 0, 6628234834692624384, 0, 0, 0, -319729080846260095008768, 0, 0, 0, 26571747463798134334265819136, 0, 0, 0, -3564202847752289659513902717468672, 0, 0
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OFFSET
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1,5
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COMMENTS
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For the series expansion of the cosine lemniscate cl(x) see A159600. The lemniscatic functions sl(x) and cl(x) played a significant role in the development of mathematics in the 18th and 19th centuries. They were the first examples of elliptic functions. In algebraic number theory all abelian extensions of the Gaussian rationals Q(i) are contained in extensions of Q(i) generated by division values of the lemniscatic functions. - Peter Bala, Aug 25 2011
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LINKS
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FORMULA
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The function sl(x) satisfies the differential equation sl''(x) = -2*sl^3(x) with initial conditions sl(0) = 0, sl'(0) = 1.
Recurrence relation:
a(n+2) = -2*sum {i+j+k = n} n!/(i!*j!*k!)*a(i)*a(j)*a(k).
The inverse of the sine lemniscate function may be defined as the algebraic integral
sl^(-1)(x) := int {0..x} 1/sqrt(1-s^4)) = x + x^5/10 + x^9/24 + 5*x^13/208 + ....
Series reversion produces the expansion
sl(x) = x - 12*x^5/5! + 3024*x^9/9! - 4390848*x^13/13! + ....
The coefficients in this expansion can be calculated using nested derivatives as follows (see [Dominici, Theorem 4.1): Let f(x) = sqrt(1-x^4). Define the nested derivative D^n[f](x) by means of the recursion
D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0.
The coefficients in the expansion of D^n[f](x) in powers of f(x) are given in A145271. Then we have a(n) = D^(n-1)[f](0).
a(n) is divisible by 12^n and a(n)/12^n produces (a signed and aerated version of) A144853(n).
(End)
The function sl(x) satisfies the differential equation sl'(x)^2 + sl(x)^4 = 1 with initial conditions sl(0) = 0, sl'(0) = 1. - Michael Somos, Oct 12 2019
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EXAMPLE
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G.f. = x - 12*x^5 + 3024*x^9 - 4390848*x^13 + 21224560896*x^17 + ...
Example of the recurrence relation a(n+2) = -2*sum {i+j+k = n} n!/(i!*j!*k!)*a(i)*a(j)*a(k) for n = 13:
There are only 6 compositions of 13-2 = 11 that give a nonzero contribution to the sum, namely 11 = 9+1+1 = 1+9+1 = 1+1+9 and 11 = 5+5+1 = 5+1+5 = 1+5+5
and hence
a(13) = -2*(3*11!/(9!*1!*1*)*a(9)*a(1)*a(1)+3*11!/(5!*5!*1!)*a(5)*a(5)*a(1)) = -4390848.
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MATHEMATICA
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Drop[ Range[0, 37]! CoefficientList[ InverseSeries[ Series[ Integrate[1/(1 - x^4)^(1/2), x], {x, 0, 37}]], x], 1] (* Robert G. Wilson v, Mar 16 2005 *)
a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ JacobiSD[x, 1/2] 2^((n - 1)/2), {x, 0, n}]]; (* Michael Somos, Jan 17 2017 *)
a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ JacobiSN[x, -1], {x, 0, n}]]; (* Michael Somos, May 26 2021 *)
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PROG
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(PARI) x='x+O('x^66); Vec(serlaplace(serreverse( intformal(1/sqrt(1-x^4))))) \\ Joerg Arndt, Mar 24 2017
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CROSSREFS
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Taking every fourth term gives A283831.
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KEYWORD
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sign
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AUTHOR
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Troy Kessler (tkessler1977(AT)netzero.com), Mar 13 2005
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EXTENSIONS
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STATUS
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approved
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