Partial Fractions - Cover Up Rule
In partial fraction decomposition, the cover-up rule is a technique to find the coefficients of linear terms in a partial fraction decomposition. It is a faster technique in finding constants in a partial fraction. We can only apply this rule when the denominator is a product of linear factors.
To clearly understand this wiki, you should already know some elementary methods of breaking a rational function into its appropriate partial fractions. The cover-up rule applies to computing coefficients of linear terms when the denominator in the partial fraction decomposition consists of linear factors only or a combination of linear and irreducible factors.
Contents
Introduction
To compute the coefficients using the cover-up method, first set up a partial fraction decomposition with one term for each of the factors in the denominator. For example, if the denominator has three distinct linear terms, we have the decomposition
\[ \frac{f(x)}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}.\]
Then by the cover-up method, \(A\) can be computed by covering up the term \((x-a)\) in the denominator of the left-hand side and substituting \(x=a\) in the remaining expression. This works because the computation is equivalent to multiplying the expression throughout by the term \((x-a)\) and then making the substitution \(x=a\). This gives
\[ A = \frac{f(a)}{(a-b)(a-c)}.\]
Similarly, by substituting \(x=b\) and \(x=c\), we can compute \(B\) and \(C\):
\[B = \frac{f(b)}{(b-a)(b-c)}, \quad C = \frac{f(c)}{(c-a)(c-b)}.\]
Note: Keep in mind that in order to apply partial fractions, the degree of the polynomial in the numerator must be strictly smaller than the degree of the polynomial in the denominator. If this is not the case, then it is necessary to first apply polynomial division to obtain a quotient polynomial and a remainder, for which the degree of the numerator is strictly smaller than that of the denominator. Partial fractions may then be applied to the remainder.
Here is a basic example on how to use the partial fraction rule for factorization.
Given the partial fraction
\[\frac{3x}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2} ,\]
what is the value of \(A+B?\)
Note that this partial fraction has two distinct linear factors. To obtain \(A\), cover up the factor \((x-1)\) on the left-hand side and substitute \(x=1\) into the remaining terms to obtain
\[A = \frac{3(1)}{1+2} = \frac{3}{3} = 1.\]
Similarly, to compute \(B\), substitute \(x = -2\) into \(\frac{3x}{x - 1}\) to get
\[B = \frac{3(-2)}{-2 - 1} = \frac{-6}{-3} = 2.\]
Hence, \(A + B = 1 + 2 = 3\). \(_\square\)
Try the following problem based on the understanding of applying partial fraction.
Given that
\[1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+ \cdots = \dfrac{\pi}{4}, \]
find the value of \(n\) that satisfy the equation below:
\[\dfrac{1}{1 \times 3}+\dfrac{1}{5 \times 7}+\dfrac{1}{9 \times 11}+ \cdots = \dfrac{\pi}{n}.\]
Applications
This section is associated to the scenarios where the cover-up rule can be applied. Here are a couple of examples followed by some problems based on the applications of the rule:
Find the partial factor decomposition of
\[\frac{x^2 + x - 1}{x(x^2-1)}.\]
Observe that the denominator is \(x(x^2-1) = x(x-1)(x+1)\), which is a product of distinct linear factors. Then the partial factor decomposition can be written as
\[ \frac{x^2 + x -1}{x(x+1)(x-1)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-1}. \]
To compute \(A\), cover up the term \(x\) in the denominator of the left hand side and substitute \(x=0\) in the remaining terms to obtain \(A= \frac{0^2+0-1}{(0+1)(0-1)} = \frac{-1}{-1} = 1\). Similarly, cover up \((x+1)\) and substitute \(x=-1\) to obtain \(B = \frac{-1}{2}\) and cover up \((x-1)\) and substitute \(x=1\) to obtain \(C=\frac{1}{2}\). Therefore, the partial fraction decomposition is
\[ \frac{x^2 + x - 1}{x(x^2-1)} = \frac{1}{x} - \frac{1}{2(x+1)} + \frac{1}{2(x-1)} . \ _\square \]
Given the partial fraction
\[\frac{6x}{(x^2 +2)(x - 1)} = \frac{Ax+B}{x^2 + 2} + \frac{C}{x - 1} ,\]
what is the value of \(A+B+C?\)
This partial fraction decomposition has one irreducible quadratic factor and one linear factor. We calculate the linear term coefficient \(C\) by covering up the term \(x-1\) in the left-hand side and substitute \(x=1\) into the remaining terms. This gives \(C = \frac{6}{1^2+2} =2.\) To calculate \(A\) and \(B\), we substitute this value of \(C\) into the equation to obtain
\[ \begin{align} \frac{6x}{(x^2 +2)(x - 1)} & = \frac{Ax+B}{x^2 + 2} + \frac{2}{x - 1} \\ \frac{6x}{(x^2 +2)(x - 1)} - \frac{2}{x - 1} &= \frac{Ax+B}{x^2 + 2} \\ \frac{6x - 2(x^2 + 2)}{(x^2+2)(x-1)} &= \frac{Ax+B}{x^2 + 2}\\ \frac{-2(x-1)(x-2)}{(x^2+2)(x-1)} &= \frac{Ax+B}{x^2 + 2}\\ \frac{-2(x-2)}{(x^2+2)} & = \frac{Ax+B}{x^2 + 2}. \end{align}\]
Therefore, \(A=-2, B=4\), which gives \(A + B +C = -2 + 4 +2 =4\). \(_\square\)
This problem checks the understanding of the usage of cover-up rule in factorial terms.
\[\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \frac{4}{5!} + \cdots = \ ?\]
The long equations in the following two problems can be easily reduced to a shorter form by applying the partial fraction rule.
Find the positive integer \(x\) such that
\[ \dfrac1{10x} =\dfrac1{x^2+x}+ \dfrac1{x^2+3x+2}+ \dfrac1{x^2+5x+6} .\]
\[ \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} + \frac{1}{(x-3)(x-4)} = \frac{1}{6} \]
What is the sum of all real values of \(x\) that satisfy the above equation?
Further problems also require wise usage of partial fraction rule to tackle them easily.
\[ \dfrac{1}{2^{2}-1}+\dfrac{1}{4^{2}-1}+\dfrac{1}{6^{2}-1}+\dfrac{1}{8^{2}-1}+\cdots+\dfrac{1}{1000^{2}-1}\ \]
The value of the sum above can be expressed as \(\frac{a}{b},\) where \(a\) and \(b\) are coprime positive integers. Find the value of \(a+b\).
Evaluate
\[\sum_{n=0}^{\infty}\frac{n}{n^4+n^2+1}.\]
Note: Refrain from using Wolfram Alpha to solve this problem.
Problem Solving
This section contains several problems trying to build problem-solving skills involving the usage of the partial fraction rule.
\[ x = 101! \times \left(\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} +\cdots+ \frac{99}{100!}\right) \]
Find \( 101! - x .\)
The sum
\[\displaystyle \frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\dots+\frac{1}{100\sqrt{99}+99\sqrt{100}}\]
can be expressed as \(\frac{a}{b},\) where \(a\) and \(b\) are coprime positive integers.
What is the value of \(a+b\)?
This problem is adapted from a past year KMC Contest.
\[ \sum_{n=2}^\infty \dfrac1{(n^2-1)n^2} = \dfrac{1}{3\cdot4}+\dfrac{1}{8\cdot 9}+\dfrac{1}{15\cdot16}+\dfrac{1}{24\cdot25}+\cdots \]
The series above is equal to \(\dfrac{A-B\pi^{C}}{D}\) where \(A,B,C\) and \(D\) are positive integers with \(A\) and \(B\) coprime.
Find the value of \(A+B+C+D\).
If \( A = \displaystyle \sum_{n=1}^\infty \frac {1}{n(n+1)(n+2)(n+3)(n+4)(n+5)} , \) what is the value of \(\frac {1}{A}?\)
\[ \sum_{ k=1 }^\infty{ \frac { 1 }{ { k }^{ 2 } } =\frac { { \pi }^{ 2 } }{ 6 } },\qquad { S }_{ i }=\sum_{ k=1 }^\infty\frac { i }{ { \big(36{ k }^{ 2 }-1\big) }^{ i } } \]
Given the above, \(S_{1}+S_{2}\) can be represented as \(\dfrac { { \pi }^{ 2 } }{ a } -\dfrac { b }{ c } \).
Find the value of \(a+b+c\) with \(c\) a prime number.