Practical numerical methods

Numerical experiments

by Nikolai Shokhirev

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Numerical experiments

All numerical experiments are performed on PC with Intel processor using single and double precision.

Test projects for Borland's C++Builder 6 and Delphi 7 are available for download.   

Is 10 times 1/10 iqual to 1?

The solution by summation is

  var i: integer;
      sum: double;
  
  sum = 0.0;
  for i := 1 to 100 do 
  begin
    sum := sum + 0.1;
    if sum = 1.0 then
      break;
  end;
  writeln(sum);

  double sum = 0.0;

  for(int i=1; i<=100; i++) 
  {
    sum += 0.1;
    if (sum == 1.0) 
    { 
      break; 
    } 
  }        
  cout << sum << endl;

The result is sum = 9.99999999999998. On the other hand, (10.0*(0.1) = 1.0 ) is true. However you should never rely on the equality of floating-point numbers.

Calculation of machine epsilon

The obvious solution is

  eps := 1.0;
  while (1.0 + eps) > 1.0 do 
  begin
    writeln(eps);
    eps := eps/2.0;
  end;  

  eps = 1.0;
  while ((1.0+eps)> 1.0) 
  {
    cout << eps << endl;
    eps /= 2.0; 
  } 

The last printed value must be the machine epsilon. However, for Intel processor regardless of the precision of eps it gives 1.08420217248550E-0019 ( see the console projects macheps). This is because this small piece of code was optimized and the internal processor precision wass used. The following code 

  eps := 1.0;
  repeat
    writeln(eps);
    eps := eps/2.0;
    sum := 1.0 + eps;
  until sum <= 1.0 ;

  eps = 1.0;
  do {
    cout << eps << endl;
    eps /= 2.0;
    sum = 1.0 + eps;
  } while (sum > 1.0);

gives ε ₀ = 1.19209289550781E-7 for single/float and ε ₀ = 2.22044604925031E-16 for double.

Round-off error example 1.

It is proven that the series

(3)

diverges (tends to infinity as ln n). Let us check this with a computer. The summation is terminates when the sum stops changing: 

  sum := 0.0;
  n   := 1.0;
  repeat
    sum1 := sum;
    sum := sum + 1.0/n; 
    n := n + 1.0;
  until sum1 = sum;

  sum = 1.0;
  n = 1.0;
  do {
    sum1 = sum;
    sum = sum + 1.0/n; 
    n += 1.0;
  } while (sum > sum1);

For the single/float numbers the summation stopped at n = 2097153 and S = 15.4036827087402. The result is much less than the infinity because 1/n < sum*ε ₀ . How lonf will it take to run with the double precision?

Round-off error example 2.

It is proven that the series

(4)

converges for any finite x. In this experiment the summation was terminated when a term became less than the threshold. 

We can see the single precision result for x = -9.5 has no correct digits. Switching to the double precision only moves the problem to larger values.  

For this problem there is much better cure. Let us use the identity

The results have all possible correct digits:

This trick is discussed in the section "Error propagation".   

To be continued ...

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