Am 15.04.16 um 02:36 schrieb Dennis Lee Bieber:
I should also have said that the square root of integer squares with
between 15 and 30 decimal digits will only be correct if the square
numbers themselves are exactly representable in 53 bits. So we can
expect failures for squares with 16 or more digits.
However, if a number with 31 or less digits is known to be the square of
an integer, the IEEE754 sqrt function will (I believe) give the correct
result.
How could it EXCEPT by having ~15 significant digits and an exponent --
since that is all the data that is provided by a double precision floating
point. That is, for example,
1000000000000000.0 * 1000000000000000.0
1e+30
import math
math.sqrt(1e30)
1000000000000000.0
only has ONE significant digit -- even though it has thirty 0s before the
decimal point.
No, you need to count the significant digits in binary. 1e30 is not an
even number in binary floating point.
Apfelkiste:AsynCA chris$ bc
bc 1.06
Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'.
obase=2
10^30
11001001111100101100100111001101000001000110011101001110110111101010\
01000000000000000000000000000000
sqrt(10^30)
11100011010111111010100100110001101000000000000000
Still, the floating point arithmetic should round to the correct answer
if both are converted to decimal.
Christian
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