Encyclopedia > Loss of significance

  Article Content

Loss of significance

Loss of significance is an undesirable effect in calculations using floating-point arithmetic. It occurs when two nearly equal numbers are subtracted to produce a result much smaller than either of the original numbers. The effect is that the number of accurate (significant) digits in the result is reduced unacceptably. Ways to avoid this effect are studied in numerical analysis.

In floating-point arithmetic, only a small number of digits of the number are maintained; floating-point numbers can only approximate most real numbers.

Consider the real number

.1234567891234567890 .

A floating-point representation of this number on a machine that keeps 10 floating-point digits would be

.1234567891,

which seems pretty close--the difference is very small in comparison with either of the two numbers.

Now perform the calculation

.1234567891234567890 - .1234567890 .

The real answer, accurate to 10 digits, is

.0000000001234567890

But on the 10-digit floating-point machine, the calculation yields

.1234567891 - .1234567890 = .0000000001 .

Whereas the original numbers are accurate in all of the first (most significant) 10 digits, their floating-point difference is only accurate in its first digit. This amounts to loss of information.

It is possible to do all rational arithmetic keeping all significant digits, but to do so is often prohibitively slower than floating-point arithmetic. Furthermore, it usually only postpones the problem: What if the data is accurate to only 10 digits? The same effect will occur.

One of the most crucial parts of numerical analysis is to avoid or minimize loss of significance in calculations. If the underlying problem is well-posed, there should be a stable algorithm[?] for solving it. The art is in finding a stable algorithm.

Instability of the quadratic formula

For example, consider the venerable quadratic formula for solving a quadratic equation

<math>a x^2 + b x + c = 0</math> .

The quadratic formula gives the two solutions as

<math> x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} </math>

The case <math>a = 1</math>, <math>b = 200</math>, <math>c = -0.000015</math> will serve to illustrate the problem:

<math>x^2 + 200 x - 0.000015 = 0</math> .

We have

<math>\sqrt{b^2 - 4 a c} = \sqrt{200^2 + 4 \times 1 \times 0.000015} = 200.00000015...</math>

In real arithmetic, the roots are

<math>( -200 - 200.00000015 ) / 2 = -200.000000075</math> ,
<math>( -200 + 200.00000015 ) / 2 = .000000075</math> .

In 10-digit floating-point arithmetic,

<math>( -200 - 200.0000001 ) / 2 = -200.00000005</math> ,
<math>( -200 + 200.0000001 ) / 2 = .00000005</math> .

Notice that the bigger solution is accurate to ten digits, but the first nonzero digit of the smaller solution is wrong.

Because of the subtraction that occurs in the quadratic formula, it does not constitute a stable algorithm to calculate the two roots of a quadratic equation.

A better algorithm

A better algorithm for solving quadratic equations is based on the observations that one solution is always accurate when the other is not, and that given one solution of the quadratic, the other is easy to find.

If

<math> x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} </math>

and

<math> x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} </math>

then we have the identity

<math>x_1 x_2 = c / a</math> .

The algorithm is: use the quadratic formula to find the larger of the two solutions (the one that doesn't suffer from loss of precision). then use this identity to calculate the other root. Since no subtraction is involved, no loss of precision occurs.

Applying this algorithm to our problem, and using 10-digit floating-point arithmetic, the larger of the two solutions, as before, is <math>x_1 = -200.00000005</math>. The other solution is then

<math> x_2 = c / (-200.00000005) = 0.000000075</math> ,

which is accurate.



All Wikipedia text is available under the terms of the GNU Free Documentation License

 
  Search Encyclopedia

Search over one million articles, find something about almost anything!
 
 
  
  Featured Article
Photosynthesis

... others further oxidize it, producing sulfates. In general, photosynthesis requires a source of hydrogen with which to reduce carbon dioxide into carbohydrates. Van Neil's ...

 
 
 
This page was created in 25.1 ms