tg

tg
tgt

Monday, 4 August 2014

CHAPTER 10 - Expansion Using Partial Fractions

This section deals with the integration of general algebraic rational functions, of the form \dfrac{{f(x)}}{{g(x)}}, where f(x)\, and g(x) are both polynomials. We already have seen some examples of this form. For example, we know how to integrate functions of the form \dfrac{1}{{Q(x)}} or \dfrac{{L(x)}}{{Q(x)}} or \dfrac{{P(x)}}{{Q(x)}} where L(x) is a linear factor, Q(x) is a quadratic factor and P(x) is a polynomial of degree n \ge 2. We intend to generalise that previous discussion in this section.
We are assuming the scenario g(x) where (the denominator) is decomposible into linear or quadratic factors. These are the only cases relevant to us right now. Any linear or quadratic factor in g(x) might also occur repeatedly.
Thus, g(x) could be of the following general forms.
  • g(x) = {L_1}(x)\,{L_2}(x)\,\ldots\,{L_n}(x)(n linear factors)
  • g(x) = {L_1}(x)\ldots L_r^k(x).\ldots {L_n}(x)(n linear factors; the {r^{th}}factor is repeated k times)
  • g(x) = L_1^{{k_1}}(x)L_2^{{k_2}}(x)\ldots L_n^{{k_n}}(x)(n linear factors, the {i^{th}} factor is repated {k_i} times)
  • g(x) = {L_1}(x){L_2}(x)\ldots {L_n}(x)\,{Q_1}(x)\,{Q_2}(x)\ldots {Q_m}(x)(n linear factors and mquadratic factors)
  • g(x) = \ldots Q_r^k(x)\ldots(a particular quadratic factors repeat more than once)
  • A combination of any of the above.
Suppose that the degree of g(x)\, is n and that of f(x) is m. If m \ge n, we can always divide f(x) by g(x) to obtain a quotient q(x) and a remainder r(x)whose degree would be less than n.
\dfrac{{f(x)}}{{g(x)}} = q(x) + \dfrac{{r(x)}}{{g(x)}}\ldots(1)
If m < n,\,\,\dfrac{{f(x)}}{{g(x)}} is termed a proper rational function.
The partial dfraction expansion technique says that a proper rational function can be expressed as a sum of simpler rational functions each possessing one of the factors of g(x). The simpler rational functions are called partial dfractions.
From now one, we consider only proper rational functions. If \dfrac{{f(x)}}{{g(x)}} is not proper, we make it proper \left( {\dfrac{{r(x)}}{{g(x)}}} \right) by the procedure described in (1) above.
Let us consider a few examples.
Let g(x) be a product of non-repeated, linear factors:
g(x) = {L_1}(x){L_2}(x)\ldots {L_n}(x)
Then, we can expand \dfrac{{f(x)}}{{g(x)}} in terms of partial dfractions as
\dfrac{{f(x)}}{{g(x)}} = \dfrac{{{A_1}}}{{{L_1}(x)}} + \dfrac{{{A_2}}}{{{L_2}(x)}} + \ldots + \dfrac{{{A_n}}}{{{L_n}(x)}}
where the {A_i}^\prime s are all constants that need to be determined
Suppose f(x) = x + 1 and g(x) = (x - 1)(x - 2)(x - 3). Let us write down the partial dfraction expansion of \dfrac{{f(x)}}{{g(x)}}:
\dfrac{{f(x)}}{{g(x)}} = \dfrac{{x + 1}}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{A}{{x - 1}} + \dfrac{B}{{x - 2}} + \dfrac{C}{{x - 3}}
We need to determine AB  and C. Cross multiplying in the expression above, we obtain:
(x + 1) = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)
ABC can now be determined by comparing coefficients on both sides. More simply since this relation that we’ve obtained should held true for all x, we substitute those values of x that would straight way give us the required values of AB and C. These values are obviously the roots of g(x).
x = 1  \,\,\, \Rightarrow   2 = A( - 1)( - 2) + B(0) + C(0)
\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,A = 1
x = 2  \,\,\, \Rightarrow \,\,\,\,  3 = A(0) + B(1)( - 1) + C(0)
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow  \,\,\,\, B =  - 3
x = 3\,  \,\,\, \Rightarrow  \,\,\,\, 4 = A(0) + B(0) + C(2)(1)
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow  \,\,\,\, C = 2
Thus, A = 1,\,\,B =  - 3 and C = 2.
We can therefore write \dfrac{{f(x)}}{{g(x)}} as a sum of partial fractions.
\dfrac{{f(x)}}{{g(x)}} = \dfrac{1}{{x - 1}} - \dfrac{3}{{x - 2}} + \dfrac{2}{{x - 3}}
Integrating \dfrac{{f(x)}}{{g(x)}} is now a simple matter of integrating the partial dfractions. This was our sole motive in writing such an expansion, so that integration could be carried out easily. In the example above:
\int {\dfrac{{f(x)}}{{g(x)}}} \,\,dx = \ln (x - 1) - 3\ln (x - 2) + 2\ln (x - 3) + C
Now, suppose that g(x) contains all linear factors, but a particular factor, say {L_1}(x), is repeated k times.
Thus,
g(x) = L_1^k(x)\,{L_2}(x)\ldots {L_n}(x)
\dfrac{{f(x)}}{{g(x)}} can now be expanded into partial dfractions as follows:
\dfrac{{f(x)}}{{g(x)}} = \underbrace {\dfrac{{{A_1}}}{{{L_1}(x)}} + \dfrac{{{A_2}}}{{L_1^2(x)}} + \dfrac{{{A_3}}}{{L_1^3(x)}} +\ldots \dfrac{{{A_k}}}{{L_1^k(x)}}}_{Comment} + \dfrac{{{B_2}}}{{{L_2}(x)}} + \ldots + \dfrac{{{B_n}}}{{{L_n}(x)}}
Comment: k partial dfractions corresponding to {L_1}(x)
This means that we will have k  terms corresponding to {L_1}(x). The rest of the linear factors will have single corresponding terms in the expansion. Here are some examples.
{\dfrac{1}{{{{(x - 1)}^2}(x - 2)}}}can be expanded as{\dfrac{A}{{x - 1}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{x - 2}}}
{\dfrac{1}{{{{(x - 1)}^3}(x - 2)(x - 3)}}}can be expanded as{\dfrac{A}{{x - 1}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{{{(x - 1)}^3}}} + \dfrac{D}{{(x - 2)}} + \dfrac{C}{{(x - 3)}}}
{\dfrac{1}{{{{(x - 1)}^2}{{(x + 5)}^3}}}}can be expanded as{\dfrac{A}{{x - 1}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{(x + 5)}} + \dfrac{D}{{{{(x + 5)}^2}}} + \dfrac{E}{{{{(x + 5)}^3}}}}
Post a Comment