We present an efficient and elementary method to find the partial fraction decomposition of a rational function when the denominator is a product of two highly powered linear factors.
Partial fraction decomposition is a classic topic with applications in calculus, differential equations, control theory, and other fields of mathematics. Theoretically, it is well-known that every rational function has a unique partial fraction decomposition as it is an easy exercise in abstract algebra. However, actually decomposing a rational function into partial fractions is computationally intensive. From the aspect of computation, there has been recent developments in this topic for general rational functions [
The case when the denominator is a power of a single linear factor has been treated in [
Our method is built on top of their methods with the observation that when the denominator is of the form
We separate the case when
Case 1: The denominator is a product of
In this case, we find the constants backward from
Then we subtract the last term from Equation (1) to get
where
where fi's are successive quotients from synthetic division. In the end, a function of the form
coefficient of
Example 1. We demonstrate how to decompose the following function.
By the Heaviside cover-up method,
and subtract
The numerator is divisible by
Repeat the process to get
Then
The whole process can be done as shown in
Remark 1. The remainder theorem says that the evaluation of
Case 2: The denominator is a product of
In this case, we take two steps. The first step is to make a substitution
We substitute the linear factor with a higher degree because it would reduce the amount of work in the second step.
We can get the coefficients of
Note that
The algorithm for this case is presented below for implementation in a computer.
14 − (−1) = 15 | ||
---|---|---|
−7 − (−2) = −5 | ||
1 − 3 = −2 | ||
3 − 2 = 1 |
Algorithm
Input: numerator
Output: partial fraction constants
Procedure: Step 1. Substitution
for
for
end for
end for
Step 2. Partial Fraction Decomposition
for
for
end for
end for
Example 2. We show how the method works for the following function.
Let
Then apply the method in Case 1 to get the answer
The whole process is described in
Substitution | Partial Fraction Decomposition |
---|---|
Case 3. The denominator is a product of
The method presented above also works when one of the factors in the denominator is a power of an irreducible quadratic function even though the computation could be challenging when it is done by hand.
The first step is to make a substitution
Let us elaborate on how to find
We reduce the right hand side modulo
We reduce
the repeated squaring method. Then we multiply two linear forms and reduce it again to finally get
Example 3. We demonstrate how the method works for the following function.
Let
The inverse of
Then the constants of the partial fractions are obtained as in
We get the final answer when we replace
We count the number of operations required for the method described in this article as follows. The synthetic division requires n multiplications and n additions where n is the degree of the polynomial. In the substitution step, we perform
Reductions Modulo u2 + u + 1 | Partial Fraction Decomposition |
---|---|
multiplications and additions. In the second step of partial fraction decomposition, we use less number of synthetic divisions. For the evaluation of functions through synthetic division, the cost is the same. Therefore, the total computational cost is
This method is not the best algorithm in terms of asymptotic speed as the algorithm in [
We thank the Editor and the referee for their comments.
Youngsoo Kim,Byunghoon Lee, (2016) Partial Fraction Decomposition by Repeated Synthetic Division. American Journal of Computational Mathematics,06,153-158. doi: 10.4236/ajcm.2016.62016