UTK - Math 171 - Alexiades
** Floating Point Numbers - Roundoff **

** Machine numbers **

There are **finitely many** floating point numbers on any computer.
They are NOT evenly spaced (same # between consecutive powers of 2)
so spread out further and further apart.
Any given real number x must be approximated by a nearest
machine number, fl(x), resulting in ** roundoff error**.

** *** Important consequencies:*

1. There is an inherent error in representing real numbers,
called ** round-off**.

2. Machine arithmetic is not necessarily commutative,
i.e. **the order of operations may affect the result!**

3. ** Loss of significant digits in arithmetic** may occur:

• * unbalanced addition:*
when adding a very small to a very large real number

• * subtractive cancellation:*
when subtracting nearly equal reals

• * unbalanced multiplication:*
when nultiplying by a very small number
(or dividing by a very large number).

4. Number of decimal digits that can be trusted:
up to 7 in single precision, up to 15 in double precision.

5. Should NEVER ask for equality of computed real numbers,
use * if( abs(a − b) ≤ TOL )* , for some TOLerance

( with TOL no smaller than 1.e-15, for double precision )

** Machine epsilon ε**
**ε**_{mach} is the smallest positive number such that ** 1+ε**_{mach} > 1
ε_{mach} = 2^{−p} , p=precision = # of bits in mantissa.
In double precision, p=53, so ε_{mach} = 2^{−53} ≈ 10^{−15}
For any real x, ** fl(x) = x(1 + ε)**,
so ** ε = (fl(x)−x)/x**
= *relative error in representing x by fl(x)*,
with **ε ≤ ε**_{mach}
Thus, the * absolute error * in representing x by fl(x)
is ** roundoff = fl(x)−x = εx ** ,
so it is proportional to x .

So, the larger the number the larger the roundoff error!
In Matlab: **eps(x)** gives the machine number nearest to x
for that computer.

** eps(1) ** or just ** eps** gives ε_{mach} itself.
For example: eps(1) = 2.2204e-16 , eps(100) = 1.4211e-14 ,
eps(1000) = 1.1369e-13 , eps(1.e6) = 1.1642e-10 , ...

losing more and more accuracy...
For example: 1+eps == 1 gives false, 1+eps/2 == 1 gives true

So, any (positive) number less than ε is effectively zero
in computation!

** Some suggestions **
Guard against loss of digits,
(subtraction of nearly equal numbers,
multiplying by very small number, etc)

Try to re-write the expression somehow...
The quadratic formula is subject to loss of significant digits
when 0 < 4ac << b^{2} (much less than)

A safe way to find roots of a quadratic:
q = −( b + sign(b)√(b^{2}−4ac) ) / 2 ,
then x_{1} = q/a , x_{2} = c / q
Always evaluate polynomials in nested form:
saves operations and may reduce roundoff.

e.g. ax^{2} + bx + c = c + x*(b+a*x) ,
e.g. ax^{3} + bx^{2} + cx + d = d + x*(c+x*(b+a*x)) , etc
Enter very big and very small numbers in scientific notation:
e.g. 3.14e6 , 3.14e-6
Print very big and very small numbers with '%e' formating.