ENGR 2422 Engineering Mathematics 2
Two More Examples of
Partial Fractions

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1. Express   F (s)   in partial fractions:

                  2
               6 s  + 2 s - 38
       F(s) =  ---------------
               (s-1)(s+1)(s+2)
   

   ---

          2
       6 s  + 2 s - 38      a     b     c
       ---------------  =  --- + --- + ---
       (s-1)(s+1)(s+2)     s-1   s+1   s+2
   
   

   All three denominators are linear factors that are not repeated.
   The cover-up rule may be used in all three cases.

                 2
             6(1)  + 2(1) - 38     -30
       a  =  -----------------  =  ---  =  -5
              *** (1+1)(1+2)         6
   
                  2
             6(-1)  + 2(-1) - 38     -34
       b  =  -------------------  =  ---  =  +17
             (-1-1) *** (-1+2)        -2
   
                  2
             6(-2)  + 2(-2) - 38     -18
       c  =  -------------------  =  ---  =  -6
             (-2-1)(-2+1) ***          3
   

   Therefore

   
               2
            6 s  + 2 s - 38     -5     17    6
   F(s)  =  ---------------  =  --- + --- - ---
            (s-1)(s+1)(s+2)     s-1   s+1   s+2
   

   If one does not wish to employ the cover-up rule, then

       2
    6 s  + 2 s - 38  =  a(s+1)(s+2) + b(s-1)(s+2) + c(s-1)(s+1)
   

   Setting   s = 1   yields   6 + 2 - 38 = 6a + 0 + 0
   Setting   s = -1   yields   6 - 2 - 38 = 0 - 2b + 0
   Setting   s = -2   yields   24 - 4 - 38 = 0 + 0 + 3c
   The same values for   a, b and c   then follow.

2. Express   F (s)   in partial fractions:

                  3      2
               5 s  + 7 s  - 3 s + 1
       F(s) =  ---------------------
                        2   2
                   (s+1)  (s +1)
   

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   All lower positive integer powers of a repeated factor must appear.
   In this case, there must be separate terms for (s+1) and (s+1)².
   Where the [non-repeated] denominator of a partial fraction is an n^th order polynomial, the corresponding numerator must be an (n-1)^th order polynomial.
   The simple cover-up rule cannot be used in either case, (although a modified version can be used to find b).

              3      2
           5 s  + 7 s  - 3 s + 1       a      b      cs + d
           ---------------------   =  --- + ------ + ------
                    2   2                        2     2
               (s+1)  (s +1)          s+1   (s+1)     s +1
   
           3      2                       2         2                 2
   ==>  5 s  + 7 s  - 3 s + 1  =  a(s+1)(s +1) + b(s +1) + (cs+d)(s+1)
   

   Setting   s = -1   yields   -5 + 7 + 3 + 1 = 0 + 2b + 0     Þ     b = 3
   Setting   s = 0   (or, equivalently, matching coefficients of s⁰) yields
   0 + 0 + 0 + 1 = a + 3 + d     Þ     d = -a - 2
   At this point we have several choices:
   Match two of the remaining three coefficients, or
   Substitute two more values of s, or
   Some combination of the above two methods.
   In this example I shall choose to match coefficients of s³ and s².
   s³:     5 = a + 0 + c     Þ     c = 5 - a
   s²:     7 = a + 3 + d + 2c = a + 3 - a - 2 + 10 - 2a     Þ     a = 2
   Þ     c = 5 - 2 = 3
   and     d = -2 - 2 = -4
   Therefore

   
              3      2
           5 s  + 7 s  - 3 s + 1       2      3      3s - 4
           ---------------------   =  --- + ------ + ------
                    2   2                        2     2
               (s+1)  (s +1)          s+1   (s+1)     s +1
   

Why does the simple cover-up rule work (for non-repeated linear factors)?
Consider the following more general case:

   f(s)         a      b(s)
----------  =  ---  +  ----
(s-k) g(s)     s-k     g(s)

where a, k are constants, g(s) is an (n)^th order polynomial in s that that does not have (s-k) as a factor, (so that g(k) is not zero), f(s) is a polynomial in s of order at most (n) and b(s) is a polynomial in s of order (n-1) containing n coefficients to be determined.

==>  f(s)  =  a.g(s)  +  (s-k).b(s)    for all s

Let   s = k  then

==>  f(k)  =  a.g(k)  +  (k-k).b(k) 

           f(k)
==>  a  =  ----
           g(k)

which is just the original fraction evaluated at s = k with the zero factor (s-k) covered up.