Partial Fraction Decomposition: Key Concepts and Repeated Factors
Partial Fraction Decomposition: Key Concepts and Repeated Factors
- The topic is partial fraction decomposition for a rational function f(x)/g(x).
- Key premise: If the degree of the numerator is less than the degree of the denominator, i.e., , the rational function is proper and can be decomposed.
- Denominator g(x) is factored into linear factors and/or irreducible quadratics, possibly with repetition.
Basic Decomposition Forms
- For distinct linear factors (x - ai), the decomposition has a term of the form for each distinct root ai.
- For a repeated linear factor (x - a)^k, include k terms: .
- For an irreducible quadratic factor qj(x) of degree 2, include a term of the form .
- For a repeated irreducible quadratic factor, include terms up to the power mj:
Method to Determine Coefficients
Steps:
- Ensure the fraction is proper: i.e., .
- Factor the denominator: where q_j are irreducible quadratics.
- Set up the decomposition with unknown constants: for each distinct linear factor add Ai/(x-ai), for each repeated factor add A1/(x-a)^1 + … + Ak/(x-a)^k, and for each irreducible quadratic add (Bj x + Cj)/qj(x)^{mj} terms.
- Multiply both sides by the full denominator g(x) to obtain an identity in x.
- Solve for the unknowns Ai, Bj, C_j by either:
- the cover-up method for simple linear factors (if g(x) contains a simple factor (x - a): A_i = f(a)/g'(a)) or
- equating coefficients of powers of x, or
- substituting convenient x values (and, for repeated factors, using several x-values or derivatives).
The cover-up method: If g(x) contains a simple factor (x - a), and g(x) = (x - a) h(x) with h(a) ≠ 0, then the corresponding coefficient is .
Quick Worked Examples
Example 1: Distinct linear factors
- Decompose: with .
- Multiply through: .
- Solve: A + B = 3 and A - 2B = 5 → B = -\tfrac{2}{3}, A = \tfrac{11}{3}.
- Result: .
Example 2: Repeated linear factor
- Decompose: with a simple choice: let .
- Multiply through: .
- Plug x = 1:
- Compare coefficients: A = 3 (since coefficient of x is A and must equal 3).
- Result: .
Example 3: Mixed repeated and simple factors (brief outline)
- Suppose .
- Multiply by the denominator:
- Use x = 1 to get B:
- Use x = -3 to get C:
- Use a third value to find A, or compare coefficients to solve for A.
Trigonometry Context (from transcript)
- Values mentioned: (notes from the example in the transcript)
- These are tangential notes; not required for partial fractions but appear in the transcript as numeric references.
Connections to Foundations and Real-World Relevance
- Connects to polynomial long division, factorization, and the idea of expressing a rational function as a sum of simpler fractions.
- Foundational for integration of rational functions: integration of A/(x-a) and (Bx+C)/(quadratic) forms.
- Practical: solving systems of linear equations to find coefficients; using substitution/cover-up, coefficient comparison.
- Ethical/practical implications: careful bookkeeping to avoid mistakes; ensure uniqueness of decomposition for a given f/g with deg f < deg g.
Formulas and Key Equations
- Proper fraction condition:
- Denominator factorization: where q_j are irreducible quadratics.
- Decomposition form (summary):
- Distinct linear:
- Repeated linear:
- Irreducible quadratics: and for repeats, add higher powers of q_j in the denominator.
- Coefficient determination: multiply through by g(x) and solve for unknowns by substitution or comparing coefficients.
- Cover-up method (simple factor): if g(x) contains (x - a) as a simple factor, then where
End Notes
- This note is designed to mirror a comprehensive study guide replacing the original source, with step-by-step procedures, worked examples, and key formulas.