Application of Taylor theorem to polynomial and partial fraction

 

Taylor’s theorem

 

          Let aÎR, f(x) be a function that is continuous with its first (n + 1) derivatives on an interval containing a and x. Then

 

         

where the remainder:

         

A useful special case in which a = 0 is called the Maclaurin series:

         

where  x lies between 0 and x.

 

 

Partial Fraction

 

          Very often the Taylor’s theorem is used to expand infinite series such as sin x or ex. However, it can be applied to functions that can be expanded finitely. You may then find the theorem very useful in dealing with polynomials or partial fractions.

 

 

Example 1

 

  Express  x5 + 4x3 + 3x + 1 as a polynomial with variable (x – 1).

 

Solution

 

Let  f(x) = x5 + 4x3 + 3x + 1, then              f(1) = 9

          f’(x) = 5x4 + 12x2 + 3x,                        f’(1) = 20

          f(2)(x) = 20 x3 + 24x,                             f(2)(1) = 44

          f(3)(x) = 60x2 + 24                                 f(3)(1) = 84

          f(4)(x) = 120x                                         f(4)(1) = 120

          f(5)(x) = 120                                           f(5)(1) = 120

and   f(6)(x) = 0, and the series terminates.

 

By Taylor’s theorem in (1), we get:

 

\  x5 + 4x3 + 3x + 1 = 9 + 20(x - 1) + 22(x - 1)2 + 14(x - 1)3 + 5(x - 1)4 + (x - 1)5…… (3)

 

 

Example 2

 

Express in partial fraction:

                 

 

Solution

 

          Using the technique in the previous example for the numerator of the fraction, from (3) we get:

         

         

         

 

 

Example 3

 

Express in partial fraction:

                 

Solution

 

          This problem is more complicate since there is one more linear factor in the denominator. We can handle like this:

         

We can get :          x2 º f(x) (x – 2) + A (x –1)4        ……… (4)

Put x = 2 in (4),      we get  A = 4                              ……… (5)

Substitute (5) in (4) and change the subject to f(x), we have:

          

         

 

Now, since           f(x) = -4x3 + 8x2 – 7x + 2,          f(1) = 2

                                  f’(x) = -12x2 + 16x – 7                 f’(1) = -3

                                  f(2)(x) = -24x + 16                         f(2)(1) = -8

                                  f(3)(x) = -24                                    f(3)(1) = -24

By Taylor’s Theorem,

Finally we get: