A rational fraction is one which is a ratio of two polynomials. That is, a function of the form
$f(x) = \frac{P_m(x)}{Q_n(x)}$
where $P_m(x) = a_mx^m + a_{m-1}x^{m - 1} + \ldots + a_2x^2 + a_1x + a_0$ is a polynomial of degree $m$, and $Q_n(x) = b_nx^n + b_{n-1} x^{n-1} + \ldots + b_2x^2 + b_1 x + b_0$ is a polynomial of degree $n$. Here, $a_m, a_{m-1}, \ldots, a_2, a_1, b_m, b_{m-1}, \ldots, b_2, b_1, b_0$ are constants, and $a_m \not=0, b_n \not=0$.
Depending upon the degrees of the numerator and the denominator, we can categorize these rational fractions into two types :
The fraction $f(x)$ is said to be an improper fraction, if $m < n$. For example,
$\frac{3x^3 + 4x^2 - 11}{x^2 + 3x -2}$
The fraction $f(x)$ is said to be a proper fraction, if $m < n$. For example,
$\frac{3x +4}{x^2 - 7x + 1}$
If $f(x) = \frac{P_m(x)}{Q_n(x)}$ is an improper fraction (i.e., $m\gen), then it can be expressed as a sum of a polynomial and a proper fraction. That is,
$\frac{P_m(x)}{Q_n(x)} = A_{m-n} (x) + \frac{R}{Q_n(x)}$
where, $A_{m-n}(x)$ is an polynomial of degree $m -n$, and $R$ is another polynomial of degree less than $n$. For illustration, consider the following improper fraction:
$\frac{2x^3 + 4x^2 -11}{x^2 + 3x + 2} $ $= (2x -2) +$ $\frac{2x - 7}{x^2 + 3x + 2}$
Note that we can perform addition and subtraction on a group of rational fractions to get a rational fraction whose denominator is the lowest common denominator of the given fractions. But the converse process of separating a fraction into a group of simple, or partial fractions is often required. It is natural to inquire about how far the decomposition of a given rational fraction into partial fractions.
Since we can always, by ordinary division, represent an improper fraction as the sum of an integral function and a proper fraction, we need to consider the resolution of a proper fraction into partial fractions.
Remark:
If the denominator of a proper fraction can be factorized into two integral factors, which are algebraically prime to each other, then we can always decompose the fraction into sum of two proper fractions.
Three cases along with the method of solving a rational fraction into a sum of partial fractions is given hereunder:
Partial Fraction Rules 1:
Case 1:If
$f(x) = \frac{A_m(x)}{B_n(x)} = \frac{A_m(x)}{b (x - \alpha_1)(x - \alpha_2) \ldots \ldots (x - \alpha_n)}$
is a proper rational fraction (i.e., m
$f(x) = \frac{A_1}{x- \alpha_1} + \frac{A_2}{x - \alpha_2} + \ldots \ldots + \frac{A_n}{x - \alpha_n}$
where $A_1, A_2, \ldots \ldots, A_n$ are constants.
Illustrative Example 1: Resolving $\frac{8x-4}{3x^2 - 2x -1}$ into a sum of partial fractions.
Solution
We know that
$3x^2 - 2x -1 = 3x^2 - 3x + x -1$
$=(3x^2 - 3x) + (x -1)$
$= 3x(x -1) + (x -1)$
$= (3x + 1)(x -1)$
Thus,
$\frac{8x -4}{3x^2 - 2x -1} = \frac{8x -4}{(3x+1)(x - 1)}$
Let us assume that
$\frac{8x -4}{(3x+1)(x-1)} = \frac{A}{3x+1} + \frac{B}{x-1}$
$\Longrightarrow \quad \frac{8x -4}{(3x+1)(x-1)} = \frac{A(x-1) + B(3x + 1)}{(3x+1)(x -1)}$
$\Longrightarrow \quad \frac{8x -4}{(3x+1)(x - 1)} = \frac{(A + 3B)x + (B - A)}{(3x + 1)(x -1)}$
Since the denominators are equal, we can equate the numerators. Thus, we have
$8x - 4 = (A + 3B)x + (B - A)$
I am planning to write more post on List of Fibonacci Numbers and What is a Fraction. Keep checking my blog.
Comparing x coefficients and constants on both sides, we have
$A + 3B = 8$ ---------------(1)
$A - B = 4$ ----------------(2)
Solving (1) and (2), we have
$A = 5$ and $B = 1$.
Thus,
$\frac{8x -4}{3x^2 - 2x -1} = \frac{5}{3x+1} + \frac{1}{x -1}$.
Partial Fraction Rules 2:
Case 2:
When some of the zeros are repeating, we can write
$f(x) = \frac{A_m(x)}{b (x - \alpha_1)^{k_1} (x - \alpha_2)^{k_2} \ldots \ldots (x - \alpha_r)^{k_r}}$
where $r \leq n$; and $k_1, k_2, \ldots \ldots, k_r$ are positive integers such that $k_1 + k_2 + \ldots + k_r = n$ and $\alpha_1, \alpha_2, \ldots \ldots, \alpha_r$ are distinct zeros of $B_n(x)$. In this case, the decomposition takes the following form:
$f(x) = \frac{A_1}{x - \alpha_1} + \frac{A_2}{(x-\alpha_1)^2} + \ldots + \frac{A_{k_1}}{(x-\alpha_1)^{k_1}} + \frac{A_{k_1 +1}}{x - \alpha_2} + \ldots + \frac{A_{n - k_r +1}}{x - \alpha_r}+ \ldots + \frac{A_n}{(x - \alpha_r)^{k_r}}$
Illustrative Example 2: Resolving $\frac{3x^2 + x -2}{(x-2)^2 (1-2x)}$ into a sum of partial fractions.
Solution
Assume that
$\frac{3x^2 + x -2}{(x-2)^2 (1-2x)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{1-2x}$
$\Longrightarrow \quad \frac{3x^2 + x -2}{(x-2)^2 (1-2x)} =\frac{A(x-2)(1-2x) + B(1-2x) + C(x-2)^2}{(x-2)^2 (1-2x)}$
Since the denominators are equal, we can equate the numerators. Thus, we have
$3x^2 + x -2 = A (x-2x^2 -2 + 4x)+ B(1-2x) + C(x^2 - 4x + 4)$
$\Longrightarrow \quad 3x^2 + x -2 = (-2A + C) x^2 +(5A - 2B -4C) x -(2A - B-4C)$
Comparing on both sides, we have
$-2A + C = 3 -------------------- (1)$
$5A - 2B - 4C = 1 --------------(2)$
$2A -B-4C = 2 ------------------(3)$
From (1), (2) and (3), we get
$ A = - \frac{5}{3}, B = -4, and C = -\frac{1}{3}$
Thus, we have
$\frac{3x^3 + x -2}{(x-2)^2 (1 - 2x)} = - \frac{5}{3(x-2)} -\frac{4}{(x-2)^2} - \frac{1}{3(1-2x)}$.
Partial Fraction Rules 3:
When some of the roots of the denominator $B_n(x)$ of the proper rational fraction $f(x) = \frac{P_m(x)}{Q_n(x)}$ are complex, we can no longer factor $B_n(x)$ in the form of $b(x-\alpha_1)^{k_1} (x - \alpha_2)^{k_2} \ldots (x - \alpha_r)^{k_r}$ with real factors. However, if the coefficients are real numbers, then the complex roots always come in conjuage pairs of the form $\alpha \pm \beta i$ and that, corresponding to each such a pair of complex conjugate zeros, there is a quadratic factor of $B_n(x)$.
Illustrative Example 3: Resolving $\frac{2x+1}{(x -1)(x^2+1)}$ into a sum of partial fractions.
Assume that
$\frac{2x+1}{(x -1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$
$\Longrightarrow \quad \frac{2x+1}{(x -1)(x^2+1)} =\frac{A(x^2 + 1) + (Bx + C)(x-1)}{(x -1)(x^2+1)}$
Since the denominators equal, we can equate the numerators. Thus, we have
$2x+1= A(x^2 + 1) + (Bx + C)(x-1)$
$\Longrightarrow \quad 2x + 1 = (A+B)x^2 + (C- B)x + (A -C)$
Comparing on both sides, we have
$A + B = 0 -------------- (1)$
$C - B = 2 --------------- (2)$
$A - C = 1 --------------- (3)$
From (1), (2), (3), we have
$A = \frac{3}{2}, B = - \frac{3}{2}$, and $C = \frac{1}{2}$.
Thus, we have
$\frac{2x+1}{(x -1)(x^2+1)} = \frac{3}{2(x-1)} - \frac{3x-1}{2(x^2+1)}$
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