Xirius-PARTIALFRACTIONS20267-MTH101105.pdf
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This document, titled "PARTIAL FRACTIONS" for the course MTH101/105, provides a comprehensive guide to the decomposition of rational algebraic fractions into simpler fractions. It serves as a fundamental resource for students to understand and apply the techniques of partial fraction decomposition, which are crucial for various mathematical operations, particularly in integral calculus.
The document begins by defining proper and improper fractions, establishing the prerequisite for decomposition (the fraction must be proper, or converted to one). It then systematically categorizes and explains four main cases of partial fraction decomposition based on the nature of the factors in the denominator: distinct linear factors, repeated linear factors, irreducible quadratic factors, and repeated irreducible quadratic factors. For each case, it outlines the general form of the decomposition and demonstrates the method for finding the unknown constants through detailed worked examples. The document also addresses the handling of improper fractions by first performing polynomial long division.
The overall aim of this material is to equip students with the necessary tools to break down complex rational expressions into a sum of simpler fractions, making them more manageable for further mathematical analysis, such as integration. It emphasizes a step-by-step approach, ensuring clarity and practical application of the concepts.
MAIN TOPICS AND CONCEPTS
The document introduces the concept of partial fractions as a method to decompose a complex rational algebraic fraction into a sum of simpler fractions. This technique is particularly useful in calculus for integrating rational functions.
* Proper Fraction: A rational fraction $\frac{P(x)}{Q(x)}$ where the degree of the numerator $P(x)$ is strictly less than the degree of the denominator $Q(x)$.
* Example: $\frac{x+1}{x^2-1}$
* Improper Fraction: A rational fraction $\frac{P(x)}{Q(x)}$ where the degree of the numerator $P(x)$ is greater than or equal to the degree of the denominator $Q(x)$.
* Example: $\frac{x^3+x^2+1}{x^2-1}$
* Prerequisite: Partial fraction decomposition can only be directly applied to proper fractions. If the given fraction is improper, polynomial long division must be performed first to express it as a sum of a polynomial and a proper fraction.
Case 1: Denominator with Distinct Linear FactorsThis case deals with rational fractions where the denominator can be factored into unique linear expressions.
* Key Points:
* Each distinct linear factor $(ax+b)$ in the denominator corresponds to a partial fraction of the form $\frac{A}{ax+b}$, where $A$ is a constant to be determined.
* The number of partial fractions will be equal to the number of distinct linear factors.
* General Form:
For a proper fraction $\frac{P(x)}{(a_1x+b_1)(a_2x+b_2)...(a_nx+b_n)}$, the decomposition is:
$$ \frac{P(x)}{(a_1x+b_1)(a_2x+b_2)...(a_nx+b_n)} = \frac{A_1}{a_1x+b_1} + \frac{A_2}{a_2x+b_2} + ... + \frac{A_n}{a_nx+b_n} $$
* Method for finding constants:
1. Multiply both sides of the equation by the common denominator to clear the denominators.
2. Substitute the roots of each linear factor (i.e., $x = -b_1/a_1$, $x = -b_2/a_2$, etc.) into the resulting equation to directly solve for the constants $A_1, A_2, ..., A_n$.
* Example: Decompose $\frac{x+1}{(x-1)(x+2)}$
1. Set up the decomposition: $\frac{x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$
2. Clear denominators: $x+1 = A(x+2) + B(x-1)$
3. Substitute $x=1$: $1+1 = A(1+2) + B(1-1) \Rightarrow 2 = 3A \Rightarrow A = \frac{2}{3}$
4. Substitute $x=-2$: $-2+1 = A(-2+2) + B(-2-1) \Rightarrow -1 = -3B \Rightarrow B = \frac{1}{3}$
5. Result: $\frac{x+1}{(x-1)(x+2)} = \frac{2/3}{x-1} + \frac{1/3}{x+2}$
Case 2: Denominator with Repeated Linear FactorsThis case applies when the denominator contains one or more linear factors raised to a power greater than one.
* Key Points:
* For each repeated linear factor $(ax+b)^n$, there will be $n$ partial fractions associated with it: $\frac{A_1}{ax+b}, \frac{A_2}{(ax+b)^2}, ..., \frac{A_n}{(ax+b)^n}$.
* Other distinct linear or quadratic factors are handled as in their respective cases.
* General Form:
For a proper fraction $\frac{P(x)}{(a_1x+b_1)^n Q(x)}$, where $Q(x)$ represents other factors:
$$ \frac{P(x)}{(a_1x+b_1)^n Q(x)} = \frac{A_1}{a_1x+b_1} + \frac{A_2}{(a_1x+b_1)^2} + ... + \frac{A_n}{(a_1x+b_1)^n} + \text{terms for } Q(x) $$
* Method for finding constants:
1. Multiply both sides by the common denominator.
2. Substitute the root of the repeated linear factor (e.g., $x = -b_1/a_1$) to find one of the constants (usually the one corresponding to the highest power, $A_n$).
3. For the remaining constants, either:
* Substitute other convenient values of $x$ (e.g., $x=0, 1, -1$).
* Expand the equation and equate coefficients of like powers of $x$ on both sides.
* Example: Decompose $\frac{x^2+1}{(x-1)^3}$
1. Set up the decomposition: $\frac{x^2+1}{(x-1)^3} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}$
2. Clear denominators: $x^2+1 = A(x-1)^2 + B(x-1) + C$
3. Substitute $x=1$: $1^2+1 = A(0)^2 + B(0) + C \Rightarrow 2 = C$
4. Substitute $x=0$: $0^2+1 = A(-1)^2 + B(-1) + C \Rightarrow 1 = A - B + C$. Since $C=2$, $1 = A - B + 2 \Rightarrow A - B = -1$.
5. Substitute $x=2$: $2^2+1 = A(2-1)^2 + B(2-1) + C \Rightarrow 5 = A + B + C$. Since $C=2$, $5 = A + B + 2 \Rightarrow A + B = 3$.
6. Solve the system:
$A - B = -1$
$A + B = 3$
Adding the equations: $2A = 2 \Rightarrow A = 1$.
Substituting $A=1$ into $A+B=3$: $1+B=3 \Rightarrow B=2$.
7. Result: $\frac{x^2+1}{(x-1)^3} = \frac{1}{x-1} + \frac{2}{(x-1)^2} + \frac{2}{(x-1)^3}$
Case 3: Denominator with Irreducible Quadratic FactorsThis case involves quadratic factors in the denominator that cannot be factored into real linear factors (i.e., their discriminant $b^2-4ac < 0$).
* Key Points:
* For each distinct irreducible quadratic factor $(ax^2+bx+c)$, the corresponding partial fraction is of the form $\frac{Ax+B}{ax^2+bx+c}$, where $A$ and $B$ are constants.
* Since irreducible quadratic factors do not have real roots, the substitution method (used in Case 1) is generally not applicable for finding all constants.
* General Form:
For a proper fraction $\frac{P(x)}{(a_1x^2+b_1x+c_1)(a_2x+b_2)...}$:
$$ \frac{P(x)}{(a_1x^2+b_1x+c_1)(a_2x+b_2)...} = \frac{A_1x+B_1}{a_1x^2+b_1x+c_1} + \frac{A_2}{a_2x+b_2} + ... $$
* Method for finding constants:
1. Multiply both sides by the common denominator.
2. Expand the right-hand side and collect terms by powers of $x$.
3. Equate the coefficients of corresponding powers of $x$ on both sides of the equation to form a system of linear equations.
4. Solve the system of equations for the unknown constants.
5. (Optional) If there are linear factors present, their roots can be substituted to find some constants directly, simplifying the system of equations.
* Example: Decompose $\frac{x^2+x+1}{(x^2+1)(x-1)}$
1. Set up the decomposition: $\frac{x^2+x+1}{(x^2+1)(x-1)} = \frac{Ax+B}{x^2+1} + \frac{C}{x-1}$
2. Clear denominators: $x^2+x+1 = (Ax+B)(x-1) + C(x^2+1)$
3. Substitute $x=1$ (root of $x-1$): $1^2+1+1 = (A(1)+B)(1-1) + C(1^2+1) \Rightarrow 3 = 2C \Rightarrow C = \frac{3}{2}$
4. Expand and equate coefficients (using $C=\frac{3}{2}$):
$x^2+x+1 = Ax^2 - Ax + Bx - B + Cx^2 + C$
$x^2+x+1 = (A+C)x^2 + (-A+B)x + (-B+C)$
Equating coefficients of $x^2$: $1 = A+C \Rightarrow 1 = A + \frac{3}{2} \Rightarrow A = -\frac{1}{2}$
Equating coefficients of $x$: $1 = -A+B \Rightarrow 1 = -(-\frac{1}{2})+B \Rightarrow 1 = \frac{1}{2}+B \Rightarrow B = \frac{1}{2}$
(Check constant term: $1 = -B+C \Rightarrow 1 = -\frac{1}{2}+\frac{3}{2} \Rightarrow 1 = 1$, which is consistent)
5. Result: $\frac{x^2+x+1}{(x^2+1)(x-1)} = \frac{-\frac{1}{2}x+\frac{1}{2}}{x^2+1} + \frac{\frac{3}{2}}{x-1}$
Case 4: Denominator with Repeated Irreducible Quadratic FactorsThis case combines the complexity of repeated factors with irreducible quadratic factors.
* Key Points:
* For each repeated irreducible quadratic factor $(ax^2+bx+c)^n$, there will be $n$ partial fractions associated with it: $\frac{A_1x+B_1}{ax^2+bx+c}, \frac{A_2x+B_2}{(ax^2+bx+c)^2}, ..., \frac{A_nx+B_n}{(ax^2+bx+c)^n}$.
* This case almost exclusively relies on equating coefficients.
* General Form:
For a proper fraction $\frac{P(x)}{(a_1x^2+b_1x+c_1)^n Q(x)}$:
$$ \frac{P(x)}{(a_1x^2+b_1x+c_1)^n Q(x)} = \frac{A_1x+B_1}{a_1x^2+b_1x+c_1} + \frac{A_2x+B_2}{(a_1x^2+b_1x+c_1)^2} + ... + \frac{A_nx+B_n}{(a_1x^2+b_1x+c_1)^n} + \text{terms for } Q(x) $$
* Method for finding constants:
1. Multiply both sides by the common denominator.
2. Expand the right-hand side and collect terms by powers of $x$.
3. Equate the coefficients of corresponding powers of $x$ on both sides to form a system of linear equations.
4. Solve the system of equations for the unknown constants.
5. (Optional) If there are linear factors, their roots can be substituted to find some constants, reducing the complexity of the system.
* Example: Decompose $\frac{x^3+x^2+1}{(x^2+1)^2(x-1)}$
1. Set up the decomposition: $\frac{x^3+x^2+1}{(x^2+1)^2(x-1)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2} + \frac{E}{x-1}$
2. Clear denominators: $x^3+x^2+1 = (Ax+B)(x^2+1)(x-1) + (Cx+D)(x-1) + E(x^2+1)^2$
3. Substitute $x=1$: $1^3+1^2+1 = (A(1)+B)(1^2+1)(1-1) + (C(1)+D)(1-1) + E(1^2+1)^2 \Rightarrow 3 = E(2)^2 \Rightarrow 3 = 4E \Rightarrow E = \frac{3}{4}$
4. Expand and equate coefficients (this will be a lengthy process, as shown in the document, involving solving a system of 5 equations for A, B, C, D, E). The document provides the final constants: $A=\frac{1}{4}, B=\frac{1}{4}, C=\frac{1}{2}, D=\frac{1}{2}, E=\frac{3}{4}$.
5. Result: $\frac{x^3+x^2+1}{(x^2+1)^2(x-1)} = \frac{\frac{1}{4}x+\frac{1}{4}}{x^2+1} + \frac{\frac{1}{2}x+\frac{1}{2}}{(x^2+1)^2} + \frac{\frac{3}{4}}{x-1}$
Improper FractionsWhen the degree of the numerator is greater than or equal to the degree of the denominator, the fraction is improper and requires an initial step before partial fraction decomposition.
* Key Points:
* Perform polynomial long division to express the improper fraction as a sum of a polynomial and a proper fraction.
* The proper fraction obtained from the remainder can then be decomposed using the appropriate partial fraction case.
* General Form:
For an improper fraction $\frac{P(x)}{Q(x)}$:
$$ \frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)} $$
where $S(x)$ is the quotient polynomial and $\frac{R(x)}{Q(x)}$ is a proper fraction.
* Method:
1. Divide $P(x)$ by $Q(x)$ using polynomial long division.
2. The result will be in the form $S(x) + \frac{R(x)}{Q(x)}$.
3. Apply partial fraction decomposition to the proper fraction $\frac{R(x)}{Q(x)}$.
* Example: Decompose $\frac{x^3+x^2+1}{x^2-1}$
1. Perform polynomial long division:
$(x^3+x^2+1) \div (x^2-1)$
$x(x^2-1) = x^3-x$
$(x^3+x^2+1) - (x^3-x) = x^2+x+1$
$1(x^2-1) = x^2-1$
$(x^2+x+1) - (x^2-1) = x+2$
So, $\frac{x^3+x^2+1}{x^2-1} = x+1 + \frac{x+2}{x^2-1}$
2. Decompose the proper fraction $\frac{x+2}{x^2-1}$. Note that $x^2-1 = (x-1)(x+1)$. This is Case 1.
$\frac{x+2}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$
$x+2 = A(x+1) + B(x-1)$
Substitute $x=1$: $1+2 = A(1+1) \Rightarrow 3 = 2A \Rightarrow A = \frac{3}{2}$
Substitute $x=-1$: $-1+2 = B(-1-1) \Rightarrow 1 = -2B \Rightarrow B = -\frac{1}{2}$
3. Result: $\frac{x^3+x^2+1}{x^2-1} = x+1 + \frac{3/2}{x-1} - \frac{1/2}{x+1}$
KEY DEFINITIONS AND TERMS
* Proper Fraction: A rational algebraic fraction $\frac{P(x)}{Q(x)}$ where the degree of the polynomial in the numerator, $P(x)$, is strictly less than the degree of the polynomial in the denominator, $Q(x)$. This is a prerequisite for direct partial fraction decomposition.
* Improper Fraction: A rational algebraic fraction $\frac{P(x)}{Q(x)}$ where the degree of the numerator $P(x)$ is greater than or equal to the degree of the denominator $Q(x)$. Such fractions must first be converted into a sum of a polynomial and a proper fraction using polynomial long division before partial fraction decomposition can be applied to the proper fractional part.
* Partial Fractions: Simpler rational algebraic fractions whose sum equals a more complex rational algebraic fraction. The process of breaking down a complex fraction into these simpler components is called partial fraction decomposition.
* Linear Factor: A polynomial factor of the form $(ax+b)$, where $a$ and $b$ are constants and $a \neq 0$.
* Repeated Linear Factor: A linear factor $(ax+b)$ that appears multiple times in the factorization of the denominator, typically expressed as $(ax+b)^n$ where $n > 1$.
* Irreducible Quadratic Factor: A quadratic polynomial factor of the form $(ax^2+bx+c)$ that cannot be factored into linear factors with real coefficients. This occurs when its discriminant, $b^2-4ac$, is negative.
* Repeated Irreducible Quadratic Factor: An irreducible quadratic factor $(ax^2+bx+c)$ that appears multiple times in the factorization of the denominator, typically expressed as $(ax^2+bx+c)^n$ where $n > 1$.
IMPORTANT EXAMPLES AND APPLICATIONS
The document is rich with examples illustrating each case of partial fraction decomposition. Here are some key examples that demonstrate the application of the concepts:
* Example for Distinct Linear Factors (Case 1):
Decompose $\frac{x+1}{(x-1)(x+2)}$.
This example clearly shows how to set up the decomposition as $\frac{A}{x-1} + \frac{B}{x+2}$ and then efficiently solve for $A$ and $B$ by substituting the roots of the linear factors ($x=1$ and $x=-2$) into the cleared equation $x+1 = A(x+2) + B(x-1)$. The solution yields $A=\frac{2}{3}$ and $B=\frac{1}{3}$. This method is the most straightforward for this case.
* Example for Repeated Linear Factors (Case 2):
Decompose $\frac{x^2+1}{(x-1)^3}$.
This example demonstrates the setup for repeated factors: $\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}$. It illustrates finding one constant ($C=2$) by substituting the root ($x=1$) and then using other arbitrary values ($x=0, x=2$) to form a system of equations to solve for the remaining constants ($A=1, B=2$). This highlights the need for a combination of substitution and equating coefficients or multiple substitutions.
* Example for Irreducible Quadratic Factors (Case 3):
Decompose $\frac{x^2+x+1}{(x^2+1)(x-1)}$.
This example showcases the form $\frac{Ax+B}{x^2+1} + \frac{C}{x-1}$ for an irreducible quadratic factor. It effectively uses a combination of substituting the root of the linear factor ($x=1$ to find $C=\frac{3}{2}$) and then equating coefficients of powers of $x$ to find $A=-\frac{1}{2}$ and $B=\frac{1}{2}$. This mixed approach is often the most efficient for denominators with both linear and irreducible quadratic factors.
* Example for Improper Fractions:
Decompose $\frac{x^3+x^2+1}{x^2-1}$.
This example is crucial as it addresses the initial step for improper fractions. It demonstrates the polynomial long division process to transform the improper fraction into $x+1 + \frac{x+2}{x^2-1}$. Subsequently, it applies Case 1 decomposition to the proper fractional part $\frac{x+2}{x^2-1}$, yielding $\frac{3/2}{x-1} - \frac{1/2}{x+1}$. This emphasizes that partial fraction decomposition is only applied to the remainder term after division.
DETAILED SUMMARY
The "PARTIAL FRACTIONS" document for MTH101/105 provides a comprehensive and systematic guide to decomposing rational algebraic fractions into simpler forms. The core purpose of this technique is to simplify complex fractions, making them more amenable to further mathematical operations, most notably integration in calculus.
The document begins by establishing foundational definitions: a proper fraction has a numerator degree less than its denominator degree, while an improper fraction has a numerator degree greater than or equal to its denominator degree. A critical prerequisite is highlighted: partial fraction decomposition can only be directly applied to proper fractions. If an improper fraction is encountered, the first step is always to perform polynomial long division to express it as a sum of a polynomial and a proper fraction, with the decomposition then applied only to the proper fractional component.
The main body of the document is structured around four distinct cases of partial fraction decomposition, categorized by the nature of the factors in the denominator:
1. Case 1: Denominator with Distinct Linear Factors: When the denominator can be factored into unique linear terms like $(a_1x+b_1)(a_2x+b_2)...$, the fraction is decomposed into a sum of terms, each of the form $\frac{A}{ax+b}$. The constants ($A, B, ...$) are typically found by multiplying through by the common denominator and then substituting the roots of each linear factor into the resulting polynomial equation.
2. Case 2: Denominator with Repeated Linear Factors: If a linear factor $(ax+b)$ appears multiple times, e.g., $(ax+b)^n$, it contributes $n$ partial fractions: $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + ... + \frac{A_n}{(ax+b)^n}$. Finding the constants often involves a combination of substituting the root of the repeated factor (to find the constant associated with the highest power) and then substituting other convenient values for $x$ or equating coefficients of like powers of $x$ to solve for the remaining constants.
3. Case 3: Denominator with Irreducible Quadratic Factors: When the denominator contains a quadratic factor $(ax^2+bx+c)$ that cannot be factored into real linear terms (i.e., its discriminant $b^2-4ac < 0$), the corresponding partial fraction takes the form $\frac{Ax+B}{ax^2+bx+c}$. Since these factors have no real roots, constants are primarily determined by expanding the equation after clearing denominators and then equating coefficients of corresponding powers of $x$ on both sides, often combined with substituting roots of any accompanying linear factors.
4. Case 4: Denominator with Repeated Irreducible Quadratic Factors: This is the most complex case, involving an irreducible quadratic factor raised to a power, e.g., $(ax^2+bx+c)^n$. It contributes $n$ partial fractions: $\frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + ... + \frac{A_nx+B_n}{(ax^2+bx+c)^n}$. The determination of constants in this scenario almost exclusively relies on the method of equating coefficients, leading to a system of linear equations that must be solved.
Throughout the document, each case is thoroughly explained with its general form and detailed, step-by-step worked examples. These examples are crucial for understanding the practical application of the theoretical concepts and the various algebraic manipulations required to solve for the unknown constants. The systematic approach presented ensures that students can confidently tackle a wide range of rational algebraic fractions, preparing them for advanced topics in mathematics, particularly in integral calculus where partial fraction decomposition is an indispensable tool for integrating rational functions. The document effectively serves as a clear and comprehensive instructional resource for MTH101/105 students.