c = continued_fraction(pi, bits=33); c
[3, 7, 15, 1, 292, 2] [3, 7, 15, 1, 292, 2] |
c.convergents()
[3, 22/7, 333/106, 355/113, 103993/33102, 208341/66317] [3, 22/7, 333/106, 355/113, 103993/33102, 208341/66317] |
[c.pn(n) for n in range(len(c))]
[3, 22, 333, 355, 103993, 208341] [3, 22, 333, 355, 103993, 208341] |
[c.qn(n) for n in range(len(c))]
[1, 7, 106, 113, 33102, 66317] [1, 7, 106, 113, 33102, 66317] |
for n in range(-1, len(c)):
print c.pn(n)*c.qn(n-1) - c.qn(n)*c.pn(n-1),
1 -1 1 -1 1 -1 1 1 -1 1 -1 1 -1 1 |
for n in range(len(c)):
print c.pn(n)*c.qn(n-2) - c.qn(n)*c.pn(n-2),
3 -7 15 -1 292 -2 3 -7 15 -1 292 -2 |
c = continued_fraction(golden_ratio,bits=15)
v = [(i, c.pn(i)/c.qn(i)) for i in range(len(c))]
eps=0.5
G = plot(golden_ratio,0,len(v))+line(v,color='grey') + points(v,pointsize=50)
show(G,ymin=golden_ratio-eps, ymax=golden_ratio+eps)
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x = 92902834/292034827; x
92902834/292034827 92902834/292034827 |
continued_fraction(x)
[0, 3, 6, 1, 33, 1, 14, 1, 4, 1, 1, 2, 1, 1, 6, 4, 1, 4, 2] [0, 3, 6, 1, 33, 1, 14, 1, 4, 1, 1, 2, 1, 1, 6, 4, 1, 4, 2] |
We do repeated long division using the Euclidean algorithm and output the partial quotients. As we proved, they are the convergents in the above continued fraction.
(a, b) = (x.numerator(), x.denominator())
while True:
q, r = a.quo_rem(b)
print q,
a, b = b, r
if r == 0: break
0 3 6 1 33 1 14 1 4 1 1 2 1 1 6 4 1 4 2 0 3 6 1 33 1 14 1 4 1 1 2 1 1 6 4 1 4 2 |
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