Let:

A₀ = 0
A₁ = √(1/2 + 1/2*A₀)
A₂ = √(1/2 + 1/2*A₁)
A₃ = √(1/2 + 1/2*A₂)
...
A_n = √(1/2 + 1/2*A_n-1)
...

Evaluate, analytically, the infinite product

P = A₁ * A₂ * A₃ * ...

 i     A(i)              partial P
 1 0.7071067811865476 0.7071067811865476
 2 0.9238795325112867 0.6532814824381883
 3 0.9807852804032304 0.6407288619353766
 4 0.9951847266721969 0.6376435773361455
 5 0.9987954562051724 0.6368755077217536
 6 0.9996988186962042 0.6366836927259824
 7 0.9999247018391445 0.6366357516148734
 8 0.9999811752826011 0.6366237671267633
 9 0.9999952938095762 0.6366207710540869
10 0.9999988234517019 0.6366200220390021
11 0.9999997058628822 0.6366198347854237
12 0.9999999264657179 0.6366197879720412
13 0.9999999816164293 0.6366197762686964
14 0.9999999954041073 0.6366197733428602
15 0.9999999988510269 0.6366197726114012
16 0.9999999997127567 0.6366197724285364
17 0.9999999999281892 0.6366197723828202
18 0.9999999999820472 0.6366197723713911
19 0.9999999999955118 0.6366197723685338
20 0.9999999999988780 0.6366197723678195
21 0.9999999999997196 0.6366197723676410
22 0.9999999999999298 0.6366197723675964
23 0.9999999999999825 0.6366197723675852
24 0.9999999999999956 0.6366197723675824
25 0.9999999999999989 0.6366197723675817
26 0.9999999999999998 0.6366197723675816
27 1.0000000000000000 0.6366197723675816
28 1.0000000000000000 0.6366197723675816

DEFDBL A-Z
p = 1
a = 0
FOR i = 1 TO 40
 a = SQR(.5 + .5 * a)
 p = p * a
 PRINT USING "## "; i;
 PRINT USING "#.################ "; a; p
NEXT

Posted by Charlie on 2008-06-05 12:27:29