Number of digits
Correct digits
An approximation of a real number can be written
as a power serie in the basis b (10 for decimal digits):
Digit j is correct if:
If the first digit is digit n (the coëfficiënt of the highest
power of b), let k be the index of the last correct digit. Thus c(n)....c(i)
are the correct digits of the approximation. Then we have:
The number of correct digits is: n-i+1
Examples:
-
x* = 1, x = .999999, |x - x*| = 0.000001... <=
0.000005 = 0.5*10^(-5) #correct digits is -1-(-5)+1 = 5.
-
x* = 2.718281824..., x = 2.718281, |x - x*|
= 0.000000824... <= 0.000005 = 0.5*10^(-5) the number of correct
digits is 0-(-5)+1 = 6.
-
x* = 2.718281824..., x = 2, |x - x*| =
0.718281824... <= 5 = 0.5*10^1 has 0-1+1 = NO correct digit.
-
x* = 2.718281824..., x = 3, |x - x*| =
0.281718176... <= 0.5*10^0 has 0-(-0)+1 = one correct digit.
-
x* = .03141..., x = .031,
|x - x*| = 0.00041... <= 0.0005 = 0.5*10^(-3) has -1-(-3)+1 =
3 correct digits.
Correct digits after decimal point
Let c(n)...c(0)...c(i) be the correct digits of the approximation. The
number of correct digits after dp is i, defined as in the previous section.
And the number of correct digits after the decimal
point is:
Examples:
-
x* = 2.718281824..., x = 2.71828, |x - x*|
= 0.000001824... <= 0.000005 = 0.5*10^(-5) the number of
correct digits after the decimal point is 5.Using the formula givin above,
we get: i = log(1/(2*0.000001824)) = 5.43..
Significant digits
Not all correct digits are significant. For example: .031
has 3 correct, but only 2 significant digits. All the digits, starting
from the first non-zero digit are significant digits.
If k is the first significant digit and i is the last correct digit,
then we have:
The number of correct significant digits is: q = k-i+1 and we have approximately:
And the number of correct significant digits is:
By Jeans (21/7/98).
Back home: HOME