An unreliable decimal approximation using N(prec=)

[John Spitzer]

When I run the following code on sagecell.sagemath.org, there seem to be problems in evaluating an expression I have generated in the course of my investigations:

for k in range(1,101):
    print k,N(1/6583885550*(1212730912710731739903840*e^2 - 1209871310179356975377887)*e^(2/3) - 1/6583885550*(186835325866398077159953*e^3 - 3081962205442654687288080*e)*e^(1/3) - 291669964562807277931896/658388555*e^2,prec=k)
print "Done"

The result I get is:

1 -2.3e15
2 1.1e15
3 -5.6e14
4 0.00
5 0.00
6 7.0e13
7 0.00
8 1.8e13
9 0.00
10 4.4e12
11 0.000
12 0.000
13 -5.50e11
14 2.75e11
15 0.0000
16 -6.872e10
17 0.0000
18 0.00000
19 0.00000
20 -4.2950e9
21 -2.14748e9
22 0.000000
23 5.36871e8
24 0.000000
25 0.0000000
26 0.0000000
27 0.0000000
28 1.6777216e7
29 -8.3886080e6
30 4.1943040e6
31 0.000000000
32 0.000000000
33 -524288.000
34 262144.000
35 0.0000000000
36 0.0000000000
37 32768.00000
38 16384.000000
39 0.00000000000
40 0.00000000000
41 0.000000000000
42 0.000000000000
43 0.000000000000
44 0.000000000000
45 0.0000000000000
46 64.00000000000
47 0.0000000000000
48 0.00000000000000
49 -8.0000000000000
50 4.0000000000000
51 2.00000000000000
52 0.000000000000000
53 0.000000000000000
54 0.500000000000000
55 0.2500000000000000
56 0.3750000000000000
57 0.3125000000000000
58 0.32812500000000000
59 0.32812500000000000
60 0.32031250000000000
61 0.318359375000000000
62 0.321289062500000000
63 0.319824218750000000
64 0.319091796875000000
65 0.3190917968750000000
66 0.3191528320312500000
67 0.3192138671875000000
68 0.31919860839843750000
69 0.31920623779296875000
70 0.31921005249023437500
71 0.319202423095703125000
72 0.319206237792968750000
73 0.319205284118652343750
74 0.319206237792968750000
75 0.3192059993743896484375
76 0.3192060589790344238281
77 0.3192059695720672607422
78 0.31920593976974487304688
79 0.31920596957206726074219
80 0.31920596957206726074219
81 0.319205965846776962280273
82 0.319205966778099536895752
83 0.319205967243760824203491
84 0.319205966545268893241882
85 0.3192059665452688932418823
86 0.3192059665452688932418823
87 0.3192059665161650627851486
88 0.31920596650161314755678177
89 0.31920596650888910517096519
90 0.31920596650525112636387348
91 0.319205966505251126363873482
92 0.319205966506160621065646410
93 0.319205966507070115767419338
94 0.319205966506615368416532874
95 0.3192059665063879947410896420
96 0.3192059665065016815788112581
97 0.3192059665065016815788112581
98 0.31920596650650168157881125808
99 0.31920596650649457615145365708
100 0.31920596650650168157881125808
Done

Although we get a good result eventually one does wonder if this issue happens elsewhere.

Have I stumbled upon an issue here?

Cheers,

John

[Vincent Delecroix]

This is not an issue. The numerical approximation function N simply
evaluates the expression at a given precision. There is no guarantee
on the overall result as cancellation may occur. If you want accurate
results, use interval or ball arithmetic. If you note your symbolic
expression "expr" you can use

for prec in [30,40,50,60,70,80,90]:
print(RealBallField(prec)(expr))

and get

[+/- 6.80e+7]
[+/- 6.64e+4]
[+/- 66.4]
[0.3 +/- 0.0578]
[0.3192 +/- 6.33e-5]
[0.3192060 +/- 8.68e-8]
[0.3192059665 +/- 4.33e-11]

which is correct at each step.

Vincent