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Step-by-Step Computation of a Partial Fraction Decomposition


All Maxima functions display a result, but not the computation steps that they perform to obtain the result.

The decomposition of a rational expression into partial fractions is performed with partfrac

partfrac ( 1/(x^2*(x^2 + 1)), x);

and returns this answer:

      1      1
      -- - ------
       2    2
      x    x  + 1

For a working mathematician this is sufficient, but it is not sufficient for a pupil that has to write down a detailed computation to satisfy a teacher. A user who needs a detailed computation has to perform the computation step by step. This is sometimes a bit tedious, but it is always a very good training.

To compute the partial fraction decomposition for the given rational expression manually, we have to know that the decomposition is a sum that may contain these three elementary fractions:

p1: a/x;
     a
     -
     x
p2: b/x^2;
    b
    --
     2
    x
p3: (c*x + d)/(x^2 + 1);
    c x + d
    -------
     2
    x  + 1

The partial fraction decomposition is the sum of these three fractions:

p1 + p2 + p3;
      c x + d   a   b
      ------- + - + --
       2        x    2
      x  + 1	    x

To compare the unknown numerator coefficients with the numerator of the given expression, we have to rewrite that sum on a common denominator:

ratsimp(%);
               3            2
      (c + a) x  + (d + b) x  + a x + b
      ---------------------------------
                    4    2
                   x  + x

Now, we access the numerator of this fraction:

n: num(%);
              3            2
     (c + a) x  + (d + b) x  + a x + b

Now we can set up and solve the equations. We equate coefficients of equal powers in x:

 solve ([coeff (n, x, 3) = 0,
         coeff (n, x, 2) = 0,
         coeff (n, x, 1) = 0,
         coeff (n, x, 0) = 1],
        [a, b, c, d]);

We obtain:

    [[a = 0, b = 1, C = 0, d = - 1]]

Now we substitute this result into the sum

p1 + p2 + p3:

 at(p1 + p2 + p3, first(%));

and obtain

   1      1
   -- - ------
    2    2
   x    x  + 1

What we have learned:



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