Page 40 - Applied Numerical Methods Using MATLAB

COMPUTER ERRORS VERSUS HUMAN MISTAKES 29
52
={1, 1 + 2 −52 , 1 + 2 × 2 −52 ,..., 1 + (2 − 1) × 2 −52 }
={1, 1 + 2 −52 , 1 + 2 × 2 −52 ,...,(2 − 2 −52 )}
52
={1, 1 + , 1 + 2 , ... , 1 + (2 − 1) = 2 − } ( = 2 −52 ) (1.2.2)
The set of numbers S, E,and M, each represented by the sign bit S, the
exponent field Exp and the mantissa field M, represents a number as a whole

f =±M · 2 E (1.2.3)
We classify the range of numbers depending on the value (E) of the exponent
and denote it as
E
R E = [2 , 2 E+1 ) with − 1022 ≤ E ≤+1023 (1.2.4)

In each range, the least unit—that is, the value of LSB (least significant bit) or
the difference between two consecutive numbers represented by the mantissa of
52 bits—is
E
E
E = × 2 = 2 −52 × 2 = 2 E−52 (1.2.5)
Let us take a closer look at the bitwise representation of numbers belonging
to each range.

0. 0(zero)
63 62 52 51 0
S 000 . . 0000 0000 0000 . . . 0000 0000
1. Un-normalized Range (with the value of hidden bit b h = 0)

R −1023 = [2 −1074 , 2 −1022 ) with Exp = 0,E = Exp − 1023 + 1 =−1022
( ) ( )

( )
( ) ( )
Value of LSB: −1023 = −1022 = 2 −1022−52 = 2 −1074
2. The Smallest Normalized Range (with the value of hidden bit b h = 1)
R −1022 = [2 −1022 , 2 −1021 ) with Exp = 1,E = Exp − 1023 =−1022
E
S 000 . . . 0001 0000 0000 . . . . 0000 0000 (1 + 0) × 2 = (1 + 0) × 2 −1022
S 000 . . . 0001 0000 0000 . . . . 0000 0001 (1 + 2 −52 ) × 2 −1022
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
S 000 . . . 0001 1111 1111 . . . . 1111 1111 {(1 + (2 52 − 1) 2 −52 ) = (2 − 2 −52 )} × 2 −1022

Value of LSB: −1022 = 2 −1022−52 = 2 −1074
3. Basic Normalized Range (with the value of hidden bit b h = 1)