question on addition of numbers.
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Ashwini
el 15 de Ag. de 2016
Comentada: John D'Errico
el 15 de Ag. de 2016
I have 2 numbers. a=11.699124064044344*2^192 and b=-17.935606820123120*2^252. adding a and b in MATLAB gives me the c=-1.2980e+77 (c=a+b). c-b gives me zero instead of 'a' value.
How to get back 'a' from 'c'
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FirefoxMetzger
el 15 de Ag. de 2016
The simple answer is that a double is not precise enough to represent a + b .
A double is represented using IEEE® Standard 754 . Which (very roughly) stores variables as (sign) * 1.abcdef...*10^(XYZ) where all the numbers are binary.
Representing variable a as X * 10^252 means that X has a lot of leading zeros (0.000...00001169912...). At some point this takes more then the 52 bit to represent the number, so the standard tells us to round the 53nd bit, which is 0. All leading digits are also 0. Thus a = 0 * 10^252 (due to lack of precision). Logically a + b = b if we account for this lack of precision.
To still calculate correctly concider digits which are variable precision numbers. However be advised that they might be a lot slower when computing
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Azzi Abdelmalek
el 15 de Ag. de 2016
when I run
a=11.699124064044344*2^192
b=-17.935606820123120*2^254
c=a+b
isequal(c,b)
c and b are equal, that's why c-b is 0. To understand the reason, read this http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F
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John D'Errico
el 15 de Ag. de 2016
You CANNOT recover a from c, when using double precision. NEVER. A double lacks sufficient precision to do so.
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