Page 39 - Applied Numerical Methods Using MATLAB

28 MATLAB USAGE AND COMPUTATIONAL ERRORS
that case, it is not the computer, but yourself as the user or the programmer, who
is to blame for the wrong result. In this context, we should always be careful not
to let the computer produce a farfetched output. In this section we will see how
the computer represents and stores the numbers. Then we think about the cause
and the propagation effect of computational error in order not to be deceived by
unintentional mistakes of the computer and, it is hoped, to be able to take some
measures against them.

1.2.1 IEEE 64-bit Floating-Point Number Representation
MATLAB uses the IEEE 64-bit floating-point number system to represent all
numbers. It has a word structure consisting of the sign bit, the exponent field,
and the mantissa field as follows:

63 62 52 51 0
S Exponent Mantissa

Each of these fields expresses S, E,and M of a number f in the way described
below.

ž Sign bit

0 for positive numbers
S = b 63 =
1 for negative numbers
ž Exponent field (b 62 b 61 b 60 ··· b 52 ): adopting the excess 1023 code
11
E = Exp − 1023 ={0, 1,..., 2 − 1 = 2047}− 1023
={−1023, −1022,..., +1023, +1024}
 −1022
 −1023 + 1 for |f | < 2 (Exp = 00000000000)
= −1022 ∼+1023 for 2 −1022 ≤|f | < 2 1024 (normalized ranges)
+1024 for ±∞

ž Mantissa field (b 51 b 50 ... b 1 b 0 ):
In the un-normalized range where the numbers are so small that they can be
represented only with the value of hidden bit 0, the number represented by the
mantissa is
M = 0.b 51 b 50 ··· b 1 b 0 = [b 51 b 50 ·· · b 1 b 0 ] × 2 −52 (1.2.1)

You might think that the value of the hidden bit is added to the exponent, instead
of to the mantissa.
In the normalized range, the number represented by the mantissa together with
the value of hidden bit b h = 1is
M = 1.b 51 b 50 ·· · b 1 b 0 = 1 + [b 51 b 50 ··· b 1 b 0 ] × 2 −52
= 1 + b 51 × 2 −1 + b 50 × 2 −2 +· · · + b 1 × 2 −51 + b 0 × 2 −52