Partial Fraction Decomposition
Partial Fraction Decomposition
The goal of partial fraction composition is to be able to rewrite a rational expression as a sum of simpler fractions.
Conditions for application
Partial fraction decomposition works when we have a rational expression of the form:
\[f(x) = \frac{P(x)}{Q(x)}\]where both $P(x)$ and $Q(x)$ are polynomials and the degree of $P(x)$ is strictly less than the degree of $Q(x)$.
If the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$, perform long division first and work with the remainder term.
General Approach to Partial Fraction Decomposition
This approach always works as long as the conditions above are satisfied, but there are other ways; see below.
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First, factor the denominator as completely as possible.
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Handle linear factors of $Q(x)$
Let $x - r$ be a linear factor of $Q(x)$. Suppose that $(x - r)^m$ is the highest power of $x - r$ that divides $Q(x)$.
This factor will translate to the sum of $m$ partial fractions, as follows:
\[\frac{A_1}{x-r} + \frac{A_2}{(x-r)^2} + \cdots + \frac{A_m}{(x - r)^m}.\]Repeat this for each distinct linear factor of $Q(x)$.
- Handle irreducible quadratic factors of $Q(x)$
An irreducible quadtric factor is a quadratic expression that cannot be factored into linear factors with real coefficients.
Let $x^2 + px + q$ be an irreducible quadratic factor of $Q(x)$. Suppose $(x^2 +px + q)^n$ is the highest power of this factor thad divides $Q(x)$.
This factor will translate to the sum of $n$ partial fractions, as follows:
\[\frac{B_1 x + C_1}{x^2+px+q} + \frac{B_2 x + C_2}{(x^2+px+q)^2} + \cdots + \frac{B_n x + C_n}{(x^2+px+q)^n}\]Repeat this for each distinct irreducible quadratic factor of $Q(x)$.
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Set the original fraction $\frac{P(x)}{Q(x)}$ equal to the sum of all the partial fractions from steps 2 and 3.
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Clear the fractions by multiplying by $Q(x)$. Distribute. Arrange the terms in decreasing powers of $x$. Group coefficients of the same power of $x$ by factoring out that power of $x$.
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Equate the coefficients of corresponding powers of $x$ and solve the resulting equations for the undetermined coefficients.
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We can now write the equation in the form
note: why can we only deal with linear and irreducible quadratics? Can we always factor higher degree polynomials into terms that are either linear or quadratic?
Example
jmh: my material.
Express
\[\tag{a} \frac{3(1+x)}{1-x^3}\]as a sum of partial fractions.
Note that the degree of the numerator is 1 while the degree of the denominator is 3, so we satisfy the conditions for performing partial fraction decomposition and may proceed.
First, factor the denominator (recognizing that it’s a sum of cubes):
\[\tag{b} \frac{3(1+x)}{(1-x)(x^2+x+1)}\]The denominator is now composed of one linear factor in $x$: $1+x$, and one irreducible quadratic factor in $x$: $x^2+x+1$. Both are raised on the first power.
Write the partial fraction with an undetermined coefficient for the linear factor as:
\[\tag{c} \frac{A}{1-x}\]Write the partial fraction with undetermined coefficients for the irreducible quadratic factor as:
\[\tag{d} \frac{Bx + C}{x^2+x+1}\]Sum (c) and (d) and set it equal to (b):
\[\tag{e} \frac{3(1+x)}{(1-x)(x^2+x+1)} = \frac{A}{1-x} + \frac{Bx + C}{x^2+x+1}\]Clear the fractions by multiplying both sides by the denominator of the left hand side:
\[\tag{f} 3(1+x) = A(x^2+x+1) + (Bx+C)(1-x)\]Distribute
\[\tag{g} 3 + 3x = Ax^2 + Ax + A + Bx - Bx^2 +C - Cx\]Sort by descending powers of $x$:
\[\tag{h} 3x + 3 = Ax^2 - Bx^2 + Ax + Bx - Cx + A + C\]Group coefficients of like powers of $x$ by factoring out the common power of $x$:
\[\tag{i} 3x + 3 = x^2(A-B) + x(A+B-C) + (A+C)\]The equation will be an identity when the coefficients of respective powers of $x$ on both sides are equal. Set them up and solve for them.
\[\tag{j} \text{Coefficients of}~x^2: 0 = A - B\] \[\text{Coefficients of}~x^1: 3 = (A+B-C)\] \[\text{Coefficients of}~x^0: 3 = (A+C)\]Solving these simulataneously we get $A=2, B=2, C=1$
Substituting these values into (e) we get:
\[\tag{k} \frac{3(1+x)}{(1-x)(x^2+x+1)} = \frac{2}{1-x} + \frac{2x + 1}{x^2+x+1}\]which is the desired sum of partial fractions.
The Heaviside Cover-up Method
This method only applies when the factor of $Q(x)$ are:
- linear
- distinct
- raised to the first power
That is, they are of the form $(x-r_1),(x-r_2),\cdots,(x-r_n)$.
But when it does work, it’s a bit faster than the proper method above.
- First, factor the denominator as completely as possible. We now have a function in the form:
- For each of the $n$ factors, find the corresponding factor $A_n$ by setting $x$ to $r_n$ and covering the factor containing $r_n$. By covering, we mean remove it as it if weren’t there. For example, for $n = 1$:
- Repeat $n$ times to find the $n$ different factors, then you can write the function in decomposed format:
Differentiation
Like the Heaviside cover up method, this method only works when the roots are distinct. This method comes from Morris and Tenenbaum’s ODE book, lesson 26A.
A fraction of the form $\frac{1}{f(x)}$ will have a partial fraction expansion of the form:
\[\tag{a} \frac{1}{f(x)} = \frac{A_1}{x - r_1} + \frac{A_2}{x - r_2} + \cdots + \frac{A_k}{x - r_k} + \cdots + \frac{A_n}{x - r_n }\]If we multiply $(a)$ by $(x - r_k)$. Since $f(r_k) = 0$, there results:
\[\tag{b} \frac{x - r_k}{f(x) - f(r_k)} = \frac{A_1(x - r_k)}{x - r_1} + \cdots + A_k + \cdots + \frac{A_n(x - r_n)}{x - r_n}\]Let $x \longrightarrow r_k$. The left side of $(b)$ will approach $\frac{1}{f’(r_k)}$ and its right side will approach $A_k$. Hence,
\[\tag{c} A_k = \frac{1}{f'(r_k)}, \quad k = 1,2,\cdots,n.\]Substituting $(c)$ in $(a)$ we get:
\[\tag{d} \frac{1}{f(x)} = \frac{1}{f'(r_1)(x - r_1)} + \frac{1}{f'(r_2)(x - r_2)} + \cdots + \frac{1}{f'(r_n)(x - r_n)}.\]Other Methods
Thomas and Finney cover a couple of other methods of determining the coefficients:
- Differentiating
- Assigning small values to $x$ such as $x=0,\pm1,\pm2$ to get equations in $A$, $B$, and $C$.
References
Calculus by Thomas & Finney, 9th edition, pp 569-573