PartialFractions

PartialFractions - PartialFractions.nb 1 Partial Fractions...

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Partial Fractions ü Copyright Brian G. Higgins (2003) Introduction The method of partial fractions allows one to simplify proper rational functions of the form P H s L ê Q H s L . Partial fraction decomposition of rational polynomials is very helpful for solving integrals and finding the inverse Laplace Transform of polynomial functions. The idea behind partial fraction decomposition is founded in the addition of fractions: 1 ÅÅÅÅ 2 + 1 ÅÅÅÅ 5 = 5 ÅÅÅÅÅÅÅ 10 + 2 ÅÅÅÅÅÅÅ 10 = 7 ÅÅÅÅÅÅÅ 10 The partial fraction decomposition of 7 ÅÅÅÅÅ 10 is 7 ÅÅÅÅÅÅÅ 10 = 1 ÅÅÅÅ 2 + 1 ÅÅÅÅ 5 The same concept carries over for algebraic fractions 2 ÅÅÅÅ s + s ÅÅÅÅÅÅÅÅÅÅÅÅÅ s 2 + 3 = 2 H s 2 + 3 L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ s H s + 3 L + s 2 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ s H s 2 + 3 L = 3 s 2 + 6 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ s H s 2 + 3 L Thus the partial fraction decomposition of H 3 s 2 + 6 L ê s H s 2 + 3 L is P H s L ÅÅÅÅÅÅÅÅÅÅÅÅÅ Q H s L = 3 s 2 + 6 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ s H s 2 + 3 L = 2 ÅÅÅÅ s + s ÅÅÅÅÅÅÅÅÅÅÅÅÅ s 2 + 3 In this notebook the theoretical basis for the method of partial fractions is reviewed using 4 key concepts from calculus/algebra. We also show how Mathematica 's Apart and Factor functions can be used to facilitate working with partial fractions. Mathematical Concepts Concept 1 The fundamental theorem of algebra(FTA) states that a n-th degree polynomial (1) Q H s L = a n s n + a n - 1 s n - 1 + a 1 s + a 0 has n roots, although some may be repeated. If s i is a root of Q(s), then H s - s i L is a factor of Q(s). We say s i is a root of multiplicity m i of Q(s) if m i is the largest positive integer such that H s - s i L m i is a factor of Q(s). These ideas can be expressed mathematically as (2) Q H s L = A i = 1 M H s - s i L m i where A is a constant. Eqn(2) Is a mathematical statement of the FTA. In general the roots s i are complex, and m i is the multiplicity of root s i such that (3) n = i = 1 M m i PartialFractions.nb 1
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m i = 1 for all i, and n = M so that (4) Q H s L = A i = 1 n H s - s i L Example 1: Roots of polynomials Show that the following polynomials can be expressed as (5) Q H s L = A i = 1 M H s - s i L m i (6) (i) s 3 - s , (ii) s 2 + s - 12 , (iii) 8 s 2 + 12 s 3 + 6 s 4 + s 5 (iv) 2 s 2 + 4 s + 8 , (v) 16 s + 8 s 3 + s 5 For each case determine A, n , and m i ü Solution Problem (i) The function for factoring a polynomial over the integers is Factor. We can use this function on the polynomial s 3 - s as follows: Factor @ s 3 - s D H - 1 + s L s H 1 + s L Thus A = 1 , n = 3, m i = 1 Alternatively, we can use Solve to find the factors of the polynomial equation: Solve @ s 3 - s ã 0 D 88 s Ø - 1 < , 8 s Ø 0 < , 8 s Ø 1 << However, using Solve does not determine the factor A! ü
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PartialFractions - PartialFractions.nb 1 Partial Fractions...

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