I said you could expand (integral of) sqrt(1-x^2) anywhere with a 4x4 matrix product. As CNWH sez, there are three kinds of mathematicians ... / 2 [ 2 asin(x) x sqrt(1 - x ) I sqrt(1 - x ) dx = ------- + -------------- ] 2 2 / [ a (2 j - 1) (x - a) 2 ] [ ------------------- 1 sqrt(1 - a ) (x - a) ] inf [ 2 ] /===\ [ (1 - a ) (j + 2) ] | | [ ] = UR( | | [ 2 ]) | | [ (j - 2) j (x - a) ] j = 0 [ ------------------------ 0 0 ] [ 2 ] [ (1 - a ) (j + 1) (j + 2) ] [ ] [ 0 0 1 ] where UR means upper right. This *does* induce a four term recurrence, 2 2 2 F := ((j - 4) (j - 2) F y - F y ((j - 6 j + 8) y - 2 a j + 7 a j - 5 a) j j - 3 j - 2 2 2 - (j - 1) F ((2 a j - 5 a) y + (1 - a ) j))/((a - 1) (j - 1) j), j - 1 2 2 2 a y F = 0, F = sqrt(1 - a ) y, F = sqrt(1 - a ) y - --------------, 0 1 2 2 2 sqrt(1 - a ) where y:=x-a and F_j is the expansion to j terms. Now we can choose endpoints x, say, 0 and sin(pi/6) with a the midpoint (1/4) to kill the even powers, getting 2 %pi 1 ---------- + ------- = 3 sqrt(15) sqrt(5) [ 2 ] oo [ 2 (19 j - 4) 4 j - 1 ] /===\ [ --------------------- ---------- 1 ] | | [ 225 (j + 1) (2 j + 3) 30 (j + 1) ] UR( | | [ ]) | | [ 2 (j - 1) j (4 j + 1) 2 (j - 1) j ] j = 0 [ ------------------------------- -------------------- 0 ] [ 225 (j + 1) (2 j + 1) (2 j + 3) 15 (j + 1) (2 j + 1) ] [ ] [ 0 0 1 ] with term ratio approaching 1/25. Again, use a four-term recurrence. Derivation: write [A[j+1],B[j+1],C[J+1]] = [A[j],B[j],C[J]].M[j] for three consecutive j, then eliminate from the nine equations the eight As and Bs. The term ratio for x_0 = sin pi/12, a = x_0/2 would be much smaller, but the matrix would be full of sqrt(2 and 3). --rwg Port au Prince porcupine rat