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late again

June 27 2011 at 10:15 AM
David  (no login)


Response to Pi as a fraction

 

This uses the method of "continued fractions". I suspect qbguy's program dose as well, but his is more elegant. Change DP to get more (or less) decimal places.

DEFDBL A-B, X
DIM N(25) AS INTEGER
DP = 10
X = 4# * ATN(1#)
N(0) = INT(X)
A = X - N(0)
X2 = A
X9 = 5# / (10 ^ DP)
X8 = INT(X * 10 ^ (DP - 1) + .5) / (10 ^ (DP - 1))
WHILE ABS(X1 - X2) > X9
J = J + 1
b = 1 / A
N(J) = INT(b)
A = (b - N(J))
L = 0: K = 0
L = N(J) * N(J - 1) + 1
K = N(J)
FOR m = J - 2 TO 1 STEP -1
kk = L
L = L * N(m) + K
K = kk
NEXT m
X1 = CDBL(K) / CDBL(L)
WEND
PRINT X8; " = "; L * N(0) + K; "/"; L

 
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