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| {{more footnotes|date=September 2012}}
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| In [[algebra]], the '''partial fraction decomposition''' or '''partial fraction expansion''' of a [[rational fraction]] (that is a [[fraction (mathematics)|fraction]] such that the numerator and the denominator are both [[polynomial]]s) is the operation that consists in expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.
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| The importance of the partial fraction decomposition lies in the fact that it provides an [[algorithm]] for computing the [[antiderivative]] of a [[rational function]].
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| In symbols, one can use ''partial fraction expansion'' to change a rational fraction in the form
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| : <math> \frac{f(x)}{g(x)} </math>
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| where ''ƒ'' and ''g'' are polynomials, into an expression of the form
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| : <math> \sum_j \frac{f_j(x)}{g_j(x)} </math>
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| where ''g''<sub>''j''</sub> (''x'') are polynomials that are factors of ''g''(''x''), and are in general of lower degree.
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| Thus, the partial fraction decomposition may be seen as the inverse procedure of the more elementary operation of addition of [[rational fraction]]s, which produces a single rational fraction with a numerator and denominator usually of high degree.
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| The ''full'' decomposition pushes the reduction as far as it will go: in other words, the factorization of ''g'' is used as much as possible. Thus, the outcome of a full partial fraction expansion expresses that fraction as a sum of fractions, where:
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| * the [[denominator]] of each term is a [[Exponentiation|power]] of an [[irreducible polynomial|irreducible]] (not factorable) [[polynomial]] and
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| * the [[numerator]] is a polynomial of smaller degree than that irreducible polynomial. To decrease the degree of the numerator directly, the [[Euclidean division of polynomials|Euclidean division]] can be used, but in fact if ''ƒ'' already has lower degree than ''g'' this isn't helpful.
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| == Basic principles ==
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| The basic principles involved are quite simple; it is the algorithmic aspects that require attention in particular cases. On the other hand, the existence of a decomposition of a certain kind is an assumption in practical cases, and the principles should explain which assumptions are justified.
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| Assume a rational function ''R''(''x'') = ''ƒ''(''x'')/''g''(''x'') in one [[Indeterminate (variable)|indeterminate]] ''x'' has a [[denominator]] that factors as
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| :<math> g(x) = P(x) \cdot Q(x) \, </math>
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| over a [[field (mathematics)|field]] ''K'' (we can take this to be [[real number]]s, or [[complex number]]s). If ''P'' and ''Q'' have no common factor, then ''R'' may be written as
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| :<math> \frac{A}{P} + \frac{B}{Q}</math>
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| for some polynomials ''A''(''x'') and ''B''(''x'') over ''K''. The ''existence'' of such a decomposition is a consequence of the fact that the [[polynomial ring]] over ''K'' is a [[principal ideal domain]], so that
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| :<math>CP + DQ = 1 \, </math>
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| for some polynomials ''C''(''x'') and ''D''(''x'') (see [[Bézout's identity]]).
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| Using this idea inductively we can write ''R''(''x'') as a sum with denominators powers of [[irreducible polynomial]]s. To take this further, if required, write:
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| :<math>\frac {G(x)}{F(x)^n}</math>
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| as a sum with denominators powers of ''F'' and [[numerator]]s of degree less than ''F'', plus a possible extra polynomial. This can be done by the [[Euclidean algorithm]], polynomial case. The result is the following [[theorem]]:
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| {{quotation|1=Let ''ƒ'' and ''g'' be nonzero polynomials over a field ''K''. Write ''g'' as a product of powers of distinct irreducible polynomials :
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| : <math>g=\prod_{i=1}^k p_i^{n_i}.</math>
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| There are (unique) polynomials ''b'' and ''a''<sub>''ij''</sub> with deg ''a''<sub>''ij''</sub> < deg ''p''<sub>''i''</sub> such that
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| : <math>\frac{f}{g}=b+\sum_{i=1}^k\sum_{j=1}^{n_i}\frac{a_{ij}}{p_i^j}.</math>
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| If deg ''ƒ'' < deg ''g'', then ''b'' = 0.}}
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| Therefore, when the field ''K'' is the complex numbers, we can assume that each ''p''<sub>''i''</sub> has degree 1 (by the [[fundamental theorem of algebra]]) the numerators will be constant. When ''K'' is the real numbers, some of the ''p''<sub>''i''</sub> might be quadratic, so in the partial fraction decomposition a quotient of a linear polynomial by a power of a quadratic might occur.
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| In the preceding theorem, one may replace "distinct irreducible polynomials" by "[[pairwise coprime]] polynomials that are coprime with their derivative". For example, the ''p''<sub>''i''</sub> may be the factors of the [[square-free factorization]] of ''g''. When ''K'' is the field of the rational numbers, as it is typically the case in [[computer algebra]], this allows to replace factorization by [[polynomial greatest common divisor|greatest common divisor]] to compute the partial fraction decomposition.
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| ==Application to symbolic integration==
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| For the purpose of [[symbolic integration]], the preceding result may be refined into
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| {{quotation|1=Let ''ƒ'' and ''g'' be nonzero polynomials over a field ''K''. Write ''g'' as a product of powers of pairwise coprime polynomials which have no multiple root in an algebraically closed field:
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| : <math>g=\prod_{i=1}^k p_i^{n_i}.</math>
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| There are (unique) polynomials ''b'' and ''c''<sub>''ij''</sub> with deg ''c''<sub>''ij''</sub> < deg ''p''<sub>''i''</sub> such that
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| : <math>\frac{f}{g}=b+\sum_{i=1}^k\sum_{j=2}^{n_i}\left(\frac{c_{ij}}{p_i^{j-1}}\right)' +
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| \sum_{i=1}^k \frac{c_{i1}}{p_i}.</math>
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| where <math> X'</math> denotes the derivative of <math>X.</math>}}
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| This reduces the computation of the [[antiderivative]] of a rational function to the integration of the last sum, with is called the ''logarithmic part'', because its antiderivative is a linear combination of logarithms. In fact, we have
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| :<math>\frac{c_{i1}}{p_i}=\sum_{\alpha_j:p_i(\alpha_j)=0}\frac{c_{i1}(\alpha_j)}{p'_i(\alpha_j)}\frac{1}{x-\alpha_j}.</math>
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| There are various methods to compute above decomposition. The one that is the simplest to describe is probably the so-called [[Charles Hermite|Hermite]]'s method. As the degree of ''c''<sub>''ij''</sub> is bounded by the degree of ''p''<sub>''i''</sub>, and
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| the degree of ''b'' is the difference of the degrees of ''f'' and ''g'' (if this difference is non negative; otherwise, ''b''=0), one may write these unknowns polynomials as polynomials with unknown coefficients. Reducing the two members of above formula to the same denominator and writing that the coefficients of each power of ''x'' are the same in the two numerators, one gets a [[system of linear equations]] which can be solved to obtain the desired values for the unknowns coefficients.
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| == Procedure ==
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| Given two polynomials <math>P(x)</math> and <math>Q(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n)</math>, where the ''α''<sub>''i''</sub> are distinct constants and deg ''P'' < ''n'', partial fractions are generally obtained by supposing that
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| : <math>\frac{P(x)}{Q(x)} = \frac{c_1}{x-\alpha_1} + \frac{c_2}{x-\alpha_2} + \cdots + \frac{c_n}{x-\alpha_n}</math> | |
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| and solving for the ''c''<sub>''i''</sub> constants, by substitution, by [[equating the coefficients]] of terms involving the powers of ''x'', or otherwise. (This is a variant of the [[method of undetermined coefficients]].)
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| A more direct computation, which is strongly related with [[Lagrange interpolation]] consists in writing
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| : <math>\frac{P(x)}{Q(x)} = \sum_{i=1}^n \frac{P(\alpha_i)}{Q'(\alpha_i)}\frac{1}{(x-\alpha_i)} </math>
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| where <math>Q'</math> is the derivative of the polynomial <math>Q</math>.
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| This approach does not account for several other cases, but can be modified accordingly:
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| * If deg ''P'' <math> \ge </math> deg ''Q'', then it is necessary to perform the [[Polynomial#Divisibility|Euclidean division]] of ''P'' by ''Q'', using [[polynomial long division]], giving ''P''(''x'') = ''E''(''x'') ''Q''(''x'') + ''R''(''x'') with deg ''R''</sub> < ''n''. Dividing by ''Q''(''x'') this gives
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| :: <math>\frac{P(x)}{Q(x)} = E(x) + \frac{R(x)}{Q(x)},</math>
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| :and then seek partial fractions for the remainder fraction (which by definition satisfies deg ''R''</sub> < deg ''Q'').
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| * If ''Q''(''x'') contains factors which are irreducible over the given field, then the numerator ''N''(''x'') of each partial fraction with such a factor ''F''(''x'') in the denominator must be sought as a polynomial with deg ''N'' < deg ''F'', rather than as a constant. For example, take the following decomposition over '''R''':
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| :: <math>\frac{x^2 + 1}{(x+2)(x-1)\color{Blue}(x^2+x+1)} = \frac{a}{x+2} + \frac{b}{x-1} + \frac{\color{OliveGreen}cx + d}{\color{Blue}x^2 + x + 1}.</math>
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| * Suppose ''Q''(''x'') = (''x'' − ''α'')<sup>''r''</sup>''S''(''x'') and ''S''(''α'') ≠ 0. Then ''Q''(''x'') has a zero ''α'' of [[Multiplicity (mathematics)#Multiplicity of a root of a polynomial|multiplicity]] ''r'', and in the partial fraction decomposition, ''r'' of the partial fractions will involve the powers of (''x'' − ''α''). For illustration, take ''S''(''x'') = 1 to get the following decomposition:
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| :: <math>\frac{P(x)}{Q(x)} = \frac{P(x)}{(x-\alpha)^r} = \frac{c_1}{x-\alpha} + \frac{c_2}{(x-\alpha)^2} + \cdots + \frac{c_r}{(x-\alpha)^r}.</math>
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| === Illustration ===
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| In an example application of this procedure, {{nowrap|(3''x'' + 5)/(1 − 2''x'')<sup>2</sup>}} can be decomposed in the form
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| : <math>\frac{3x + 5}{(1-2x)^2} = \frac{A}{(1-2x)^2} + \frac{B}{(1-2x)}.</math>
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| Clearing denominators shows that {{nowrap|1=3''x'' + 5 = ''A'' + ''B''(1 − 2''x'')}}. Expanding and equating the coefficients of powers of ''x'' gives
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| : 5 = ''A'' + ''B'' and 3''x'' = −2''Bx''
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| Solving for ''A'' and ''B'' yields ''A'' = 13/2 and ''B'' = −3/2. Hence,
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| : <math>\frac{3x + 5}{(1-2x)^2} = \frac{13/2}{(1-2x)^2} + \frac{-3/2}{(1-2x)}.</math>
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| === Residue method ===
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| {{see also|Heaviside cover-up method}}
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| Over the complex numbers, suppose ''ƒ''(''x'') is a rational proper fraction, and can be decomposed into
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| : <math>f(x) = \sum_i \left( \frac{a_{i1}}{x - x_i} + \frac{a_{i2}}{( x - x_i)^2} + \cdots + \frac{a_{i k_i}}{(x - x_i)^{k_i}} \right). </math>
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| Let
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| : <math> g_{ij}(x)=(x-x_{i})^{j-1}f(x),</math>
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| then according to the [[Laurent series#Uniqueness|uniqueness of Laurent series]], ''a''<sub>''ij''</sub> is the coefficient of the term (''x'' − ''x''<sub>''i''</sub>)<sup>−1</sup> in the Laurent expansion of ''g''<sub>''ij''</sub>(''x'') about the point ''x''<sub>''i''</sub>, i.e., its [[residue (complex analysis)|residue]]
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| : <math>a_{ij} = \operatorname{Res}(g_{ij},x_i).</math>
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| This is given directly by the formula
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| : <math>a_{ij}=\frac{1}{(k_{i}-j)!}\lim_{x\to x_i}\frac{d^{k_{i}-j}}{dx^{k_{i}-j}}\left((x-x_{i})^{k_{i}}f(x)\right),</math>
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| or in the special case when ''x''<sub>''i''</sub> is a simple root,
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| : <math>a_{i1}=\frac{P(x_{i})}{Q'(x_{i})},</math>
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| when
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| : <math>f(x)=\frac{P(x)}{Q(x)}.</math>
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| Note that ''P''(''x'') and ''Q''(''x'') may or may not be polynomials.
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| == Over the reals ==
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| Partial fractions are used in [[real number|real-variable]] [[integral calculus]] to find real-valued [[antiderivative]]s of [[rational function]]s. Partial fraction decomposition of real [[rational function]]s is also used to find their [[Inverse Laplace transform]]s. For applications of '''partial fraction decomposition over the reals''', see
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| * [[#Application to symbolic integration|Application to symbolic integration]], above
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| * [[Partial fractions in Laplace transforms]]
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| === General result ===
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| Let ''ƒ''(''x'') be any rational function over the [[real number]]s. In other words, suppose there exist real polynomials functions ''p''(''x'') and ''q''(''x'')≠ 0, such that
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| : <math>f(x) = \frac{p(x)}{q(x)}</math>
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| By dividing both the numerator and the denominator by the leading coefficient of ''q''(''x''), we may assume [[without loss of generality]] that ''q''(''x'') is [[monic polynomial|monic]]. By the [[fundamental theorem of algebra]], we can write
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| : <math>q(x) = (x-a_1)^{j_1}\cdots(x-a_m)^{j_m}(x^2+b_1x+c_1)^{k_1}\cdots(x^2+b_nx+c_n)^{k_n}</math>
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| where ''a''<sub>1</sub>,..., ''a''<sub>''m''</sub>, ''b''<sub>1</sub>,..., ''b''<sub>''n''</sub>, ''c''<sub>1</sub>,..., ''c''<sub>''n''</sub> are real numbers with ''b''<sub>''i''</sub><sup>2</sup> − 4''c''<sub>''i''</sub> < 0, and ''j''<sub>1</sub>,..., ''j''<sub>''m''</sub>, ''k''<sub>1</sub>,..., ''k''<sub>''n''</sub> are positive integers. The terms (''x'' − ''a''<sub>''i''</sub>) are the ''linear factors'' of ''q''(''x'') which correspond to real roots of ''q''(''x''), and the terms (''x''<sub>''i''</sub><sup>2</sup> + ''b''<sub>''i''</sub>''x'' + ''c''<sub>''i''</sub>) are the ''irreducible quadratic factors'' of ''q''(''x'') which correspond to pairs of [[complex number|complex]] conjugate roots of ''q''(''x'').
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| Then the partial fraction decomposition of ''ƒ''(''x'') is the following:
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| : <math>f(x) = \frac{p(x)}{q(x)} = P(x) + \sum_{i=1}^m\sum_{r=1}^{j_i} \frac{A_{ir}}{(x-a_i)^r} + \sum_{i=1}^n\sum_{r=1}^{k_i} \frac{B_{ir}x+C_{ir}}{(x^2+b_ix+c_i)^r}</math>
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| Here, ''P''(''x'') is a (possibly zero) polynomial, and the ''A''<sub>''ir''</sub>, ''B''<sub>''ir''</sub>, and ''C''<sub>''ir''</sub> are real constants. There are a number of ways the constants can be found.
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| The most straightforward method is to multiply through by the common denominator ''q''(''x''). We then obtain an equation of polynomials whose left-hand side is simply ''p''(''x'') and whose right-hand side has coefficients which are linear expressions of the constants ''A''<sub>''ir''</sub>, ''B''<sub>''ir''</sub>, and ''C''<sub>''ir''</sub>. Since two polynomials are equal if and only if their corresponding coefficients are equal, we can equate the coefficients of like terms. In this way, a system of linear equations is obtained which ''always'' has a unique solution. This solution can be found using any of the standard methods of [[linear algebra]]. It can also be found with [[limit (mathematics)|limits]] (see [[#Example 5 (limit method)|Example 5]]).
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| == Examples ==
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| === Example 1 ===
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| : <math>f(x)=\frac{1}{x^2+2x-3}</math>
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| Here, the denominator splits into two distinct linear factors:
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| : <math>q(x)=x^2+2x-3=(x+3)(x-1)</math>
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| so we have the partial fraction decomposition
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| : <math>f(x)=\frac{1}{x^2+2x-3} =\frac{A}{x+3}+\frac{B}{x-1}</math>
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| Multiplying through by ''x''<sup>2</sup> + 2''x'' − 3, we have the polynomial identity
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| : <math>1=A(x-1)+B(x+3)</math>
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| Substituting ''x'' = −3 into this equation gives ''A'' = −1/4, and substituting ''x'' = 1 gives ''B'' = 1/4, so that
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| : <math>f(x) =\frac{1}{x^2+2x-3} =\frac{1}{4}\left(\frac{-1}{x+3}+\frac{1}{x-1}\right)</math>
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| '''
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| === Example 2 ===
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| : <math>f(x)=\frac{x^3+16}{x^3-4x^2+8x}</math>
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| After [[Polynomial long division|long-division]], we have
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| : <math>f(x)=1+\frac{4x^2-8x+16}{x^3-4x^2+8x}=1+\frac{4x^2-8x+16}{x(x^2-4x+8)}</math>
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| Since (−4)<sup>2</sup> − 4(8) = −16 < 0, ''x''<sup>2</sup> − 4''x'' + 8 is irreducible, and so
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| : <math>\frac{4x^2-8x+16}{x(x^2-4x+8)}=\frac{A}{x}+\frac{Bx+C}{x^2-4x+8}</math>
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| Multiplying through by ''x''<sup>3</sup> − 4''x''<sup>2</sup> + 8''x'', we have the polynomial identity
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| : <math>4x^2-8x+16 = A(x^2-4x+8)+(Bx+C)x</math>
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| Taking ''x'' = 0, we see that 16 = 8''A'', so ''A'' = 2. Comparing the ''x''<sup>2</sup> coefficients, we see that 4 = ''A'' + ''B'' = 2 + ''B'', so ''B'' = 2. Comparing linear coefficients, we see that −8 = −4''A'' + ''C'' = −8 + ''C'', so ''C'' = 0. Altogether,
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| : <math>f(x)=1+2\left(\frac{1}{x}+\frac{x}{x^2-4x+8}\right)</math>
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| The following example illustrates almost all the "tricks" one would need to use short of consulting a [[computer algebra system]]'''.
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| === Example 3 ===
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| : <math>f(x)=\frac{x^9-2x^6+2x^5-7x^4+13x^3-11x^2+12x-4}{x^7-3x^6+5x^5-7x^4+7x^3-5x^2+3x-1}</math>
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| After [[Polynomial long division|long-division]] and [[polynomial factorization|factoring]] the denominator, we have
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| : <math>f(x)=x^2+3x+4+\frac{2x^6-4x^5+5x^4-3x^3+x^2+3x}{(x-1)^3(x^2+1)^2}</math>
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| The partial fraction decomposition takes the form
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| : <math>\frac{2x^6-4x^5+5x^4-3x^3+x^2+3x}{(x-1)^3(x^2+1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}+\frac{Dx+E}{x^2+1}+\frac{Fx+G}{(x^2+1)^2}</math>
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| Multiplying through by (''x'' − 1)<sup>3</sup>(''x''<sup>2</sup> + 1)<sup>2</sup> we have the polynomial identity
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| : <math>
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| \begin{align}
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| & {} \quad 2x^6-4x^5+5x^4-3x^3+x^2+3x \\
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| & =A(x-1)^2(x^2+1)^2+B(x-1)(x^2+1)^2+C(x^2+1)^2+(Dx+E)(x-1)^3(x^2+1)+(Fx+G)(x-1)^3
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| \end{align}
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| </math>
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| Taking ''x'' = 1 gives 4 = 4''C'', so ''C'' = 1. Similarly, taking ''x'' = [[complex number|''i'']] gives 2 + 2''i'' = (''Fi'' + ''G'')(2 + 2''i''), so ''Fi'' + ''G'' = 1, so ''F'' = 0 and ''G'' = 1 by equating real and [[complex number|imaginary]] parts. With ''C'' = ''G'' = 1 and ''F'' = 0, taking ''x'' = 0 we get ''A'' − ''B'' + 1 − ''E'' − 1 = 0, thus ''E'' = ''A'' − ''B''.
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| We now have the identity
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| : <math>
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| \begin{align}
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| & {} 2x^6-4x^5+5x^4-3x^3+x^2+3x \\
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| & = A(x-1)^2(x^2+1)^2+B(x-1)(x^2+1)^2+(x^2+1)^2+(Dx+(A-B))(x-1)^3(x^2+1)+(x-1)^3 \\
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| & = A((x-1)^2(x^2+1)^2 + (x-1)^3(x^2+1)) + B((x-1)(x^2+1) - (x-1)^3(x^2+1)) + (x^2+1)^2 + Dx(x-1)^3(x^2+1)+(x-1)^3
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| \end{align}
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| </math>
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| Expanding and sorting by exponents of x we get
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| : <math>
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| \begin{align}
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| & {} 2 x^6 -4 x^5 +5 x^4 -3 x^3 + x^2 +3 x \\
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| & = (A + D) x^6 + (-A - 3D) x^5 + (2B + 4D + 1) x^4 + (-2B - 4D + 1) x^3 + (-A + 2B + 3D - 1) x^2 + (A - 2B - D + 3) x
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| \end{align}
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| </math>
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| We can now compare the coefficients and see that
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| : <math>
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| \begin{align}
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| A + D &=& 2 \\
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| -A - 3D &=& -4 \\
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| 2B + 4D + 1 &=& 5 \\
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| -2B - 4D + 1 &=& -3 \\
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| -A + 2B + 3D - 1 &=& 1 \\
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| A - 2B - D + 3 &=& 3 ,
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| \end{align}
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| </math>
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| with ''A'' = 2 − ''D'' and −''A'' −3 ''D'' =−4 we get ''A'' = ''D'' = 1 and so ''B'' = 0, furthermore is ''C'' = 1, ''E'' = ''A'' − ''B'' = 1, ''F'' = 0 and ''G'' = 1.
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| The partial fraction decomposition of ''ƒ''(''x'') is thus
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| : <math>f(x)=x^2+3x+4+\frac{1}{(x-1)} + \frac{1}{(x - 1)^3} + \frac{x + 1}{x^2+1}+\frac{1}{(x^2+1)^2}.</math>
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| Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at ''x=1'' and at ''x=i'' in the above polynomial identity. (To this end, recall that the derivative at ''x=a'' of ''(x−a)<sup>m</sup>p(x)'' vanishes if ''m > 1'' and it is just ''p(a)'' if ''m=1''.)
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| Thus, for instance the first derivative at ''x=1'' gives
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| : <math> 2\cdot6-4\cdot5+5\cdot4-3\cdot3+2+3 = A\cdot(0+0) + B\cdot( 2+ 0) + 8 + D\cdot0 </math> | |
| that is ''8 = 2B + 8'' so ''B=0''.
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| ===Example 4 (residue method)===
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| :<math> f(z)=\frac{z^{2}-5}{(z^2-1)(z^2+1)}=\frac{z^{2}-5}{(z+1)(z-1)(z+i)(z-i)}</math> | |
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| Thus, ''f''(''z'') can be decomposed into rational functions whose denominators are ''z''+1, ''z''−1, ''z''+i, ''z''−i. Since each term is of power one, −1, 1, −''i'' and ''i'' are simple poles.
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| Hence, the residues associated with each pole, given by
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| :<math>\frac{P(z_i)}{Q'(z_i)} = \frac{z_i^2 - 5}{4z_i^3}</math>,
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| are
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| :<math> 1, -1, \tfrac{3i}{2}, -\tfrac{3i}{2}</math>,
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| respectively, and
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| :<math> f(z)=\frac{1}{z+1}-\frac{1}{z-1}+\frac{3i}{2}\frac{1}{z+i}-\frac{3i}{2}\frac{1}{z-i}</math>.
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| ===Example 5 (limit method)===
| |
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| [[Limit (mathematics)|Limits]] can be used to find a partial fraction decomposition.<ref>{{cite book|last=Bluman|first=George W.|title=Problem Book for First Year Calculus|year=1984|publisher=Springer-Verlag|location=New York|pages=250–251}}</ref>
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| :<math>f(x) = \frac{1}{x^3 - 1}</math>
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| First, factor the denominator:
| |
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| :<math>f(x) = \frac{1}{(x - 1)(x^2 + x + 1)}</math>
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| The decomposition takes the form of
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| :<math>\frac{1}{(x-1)(x^2+x+1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}</math>
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| As <math>x \to 1</math>, the ''A'' term dominates, so the right-hand side approaches <math>\frac{A}{x - 1}</math>. Thus, we have
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| :<math>\frac{1}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1}</math>
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| :<math>A = \lim_{x \to 1}{\frac{1}{x^2 + x + 1}} = \frac{1}{3}</math>
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| As <math>x \to \infty</math>, the right-hand side is
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| :<math>\lim_{x \to \infty}{\frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}} = \frac{A}{x} + \frac{Bx}{x^2} = \frac{A + B}{x}.</math>
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| :<math>\frac{A + B}{x} = \lim_{x \to \infty}{\frac{1}{x^3 - 1}} = 0</math>
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| Thus, <math>B = -\frac{1}{3}</math>.
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| At <math>x=0</math>, <math>-1 = -A + C</math>. Therefore, <math>C = -\frac{2}{3}</math>.
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| The decomposition is thus <math>\frac{\frac{1}{3}}{x - 1} + \frac{-\frac{1}{3}x - \frac{2}{3}}{x^2 + x + 1}</math>.
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| == The role of the Taylor polynomial ==
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| The partial fraction decomposition of a rational function can be related to [[Taylor's theorem]] as follows. Let
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| :<math>P(x), Q(x), A_1(x),\dots, A_r(x)</math>
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| be real or complex polynomials; assume that
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| :<math>\textstyle Q=\prod_{j=1}^{r}(x-\lambda_j)^{\nu_j},</math>
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| that
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| :<math>\textstyle\deg(P)<\deg(Q)=\sum_{j=1}^{r}\nu_j,</math>
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| and that
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| :<math>\textstyle\deg A_j<\nu_j\text{ for }j=1,\dots,r.</math>
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| | |
| Define also
| |
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| :<math>\textstyle Q_i=\prod_{j\neq i}(x-\lambda_j)^{\nu_j}=\frac{Q}{(x-\lambda_i)^{\nu_i}} \text{ for }i=1,\dots,r.</math>
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| Then we have
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| :<math>\frac{P}{Q}=\sum_{j=1}^{r}\frac{A_j}{(x-\lambda_j)^{\nu_j}}</math>
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| if, and only if, for each <math>\textstyle i</math> the polynomial <math>\textstyle A_i(x)</math> is the Taylor polynomial of <math>\textstyle\frac{P}{Q_i}</math> of order <math>\textstyle\nu_i-1</math> at the point <math>\textstyle\lambda_i</math>:
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| :<math>A_i(x):=\sum_{k=0}^{\nu_i-1} \frac{1}{k!}\left(\frac{P}{Q_i}\right)^{(k)}(\lambda_i)\ (x-\lambda_i)^k.
| |
| </math>
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| Taylor's theorem (in the real or complex case) then provides a proof of the existence and uniqueness of the partial fraction decomposition, and a characterization of the coefficients.
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| '''Sketch of the proof:''' The above partial fraction decomposition implies, for each 1 ≤ ''i'' ≤ ''r'', a polynomial expansion
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| :<math>\frac{P}{Q_i}=A_i + O((x-\lambda_i)^{\nu_i})\qquad </math>, as <math>x\to\lambda_i;</math>
| |
| | |
| so <math>\textstyle A_i</math> is the Taylor polynomial of <math>\textstyle\frac{P}{Q_i}</math>, because of the unicity of the polynomial expansion of order <math>\textstyle\nu_i-1</math>, and by assumption <math>\textstyle\deg A_i<\nu_i</math>.
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| Conversely, if the <math>\textstyle A_i</math> are the Taylor polynomials, the above expansions at each <math>\textstyle\lambda_i</math> hold, therefore we also have
| |
| | |
| :<math>P-Q_i A_i = O((x-\lambda_i)^{\nu_i})\qquad </math>, as <math>x\to\lambda_i,</math>
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| | |
| which implies that the polynomial <math>\textstyle P-Q_iA_i</math> is divisible by <math>\textstyle (x-\lambda_i)^{\nu_i}.</math>
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| | |
| For <math>\textstyle j\neq i</math> also <math>\textstyle Q_jA_j</math> is divisible by <math>\textstyle (x-\lambda_i)^{\nu_i}</math>, so we have in turn that <math>\textstyle P- \sum_{j=1}^{r}Q_jA_j</math> is divisible by <math>\textstyle Q</math>. Since <math>\textstyle\deg\left( P- \sum_{j=1}^{r}Q_jA_j \right) < \deg(Q)</math> we then have
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| <math>\textstyle P- \sum_{j=1}^{r}Q_jA_j=0</math>, and we find the partial fraction decomposition dividing by <math>\textstyle Q</math>.
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| | |
| == Fractions of integers ==
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| The idea of partial fractions can be generalized to other [[integral domain]]s,
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| say the ring of [[integer]]s where [[prime numbers]] take the role of irreducible denominators.
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| For example:
| |
| | |
| : <math>\frac{1}{18} = \frac{1}{2} - \frac{1}{3} - \frac{1}{3^2}. </math>
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| == Notes ==
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| {{Reflist}}
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| == References ==
| |
| *{{cite news
| |
| |first1=K. R. | last1=Rao
| |
| |first2=N. | last2=Ahmed
| |
| |title=Recursive techniques for obtaining the partial fraction expansion of a rational function
| |
| |year=1968
| |
| |volume=11 | number=2
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| |journal=IEEE Trans. Educ.
| |
| |pages=152–154
| |
| |doi=10.1109/TE.1968.4320370
| |
| }}
| |
| *{{cite news | first1=Peter | last1=Henrici
| |
| |title=An algorithm for the incomplete decomposition of a rational function into partial fractions
| |
| |journal=Z. f. Angew. Mathem. Physik
| |
| |year=1971
| |
| |volume=22 | number=4 | pages=751–755
| |
| |doi=10.1007/BF01587772 }}
| |
| *{{cite news | first1=Feng-Cheng | last1=Chang
| |
| |title=Recursive formulas for the partial fraction expansion of a rational function with multiple poles
| |
| |year=1973
| |
| |journal = Proc. IEEE
| |
| |volume=61 | number=8 |pages=1139–1140
| |
| |doi=10.1109/PROC.1973.9216 }}
| |
| *{{cite doi|10.1137/0206042|noedit}}
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| * {{cite news | first1=Dan |last1=Eustice
| |
| |first2=M. S.|last2=Klamkin
| |
| |title=On the coefficients of a partial fraction decomposition
| |
| |year=1979| jstor=2320421 |volume=86
| |
| |number=6 | pages=478–480
| |
| }}
| |
| *{{cite news
| |
| |first1=J. J. | last1=Mahoney
| |
| |first2=B. D. | last2=Sivazlian
| |
| |title=Partial fractions expansion: a review of computational methodology and efficiency
| |
| |journal=J. Comp. Appl. Math.
| |
| |year=1983
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| |doi=10.1016/0377-0427(83)90018-3
| |
| |volume=9
| |
| |pages=247–269
| |
| }}
| |
| *{{cite book |last1=Miller |first1=Charles D. |last2=Lial |first2=Margaret L. |last3=Schneider |first3=David I. |title=Fundamentals of College Algebra |edition=3rd ed. |year=1990|publisher=Addison-Wesley Educational Publishers, Inc. |isbn=0-673-38638-4|pages=364–370}}
| |
| *{{cite news
| |
| |first1=David | last1=Westreich
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| |title=partial fraction expansion without derivative evaluation
| |
| |year=1991
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| |journal=IEEE Trans. Circ. Syst.
| |
| |volume=38 | number=6
| |
| |pages=658–660
| |
| |doi=10.1109/31.81863
| |
| }}
| |
| *{{springer|id=u/u095160|title=Undetermined coefficients, method of|first=L. D.|last=Kudryavtsev}}
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| *{{cite news
| |
| |first1=Daniel J.
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| |last1=Velleman
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| |title=Partial fractions, binomial coefficients and the integral of an odd power of sec theta
| |
| |year=2002
| |
| |journal= Am. Math. Monthly
| |
| |volume=109
| |
| |number=8
| |
| |pages=746–749
| |
| |jstor=3072399 }}
| |
| *{{cite news | first1=Damian | last1=Slota | first2=Roman | last2=Witula
| |
| |year=2005 | series=Lect. Not. Computer Sci.
| |
| |title=Three brick method of the partial fraction decomposition of some type of rational expression
| |
| |pages=659–662 | volume=33516 |
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| doi=10.1007/11428862_89}}
| |
| *{{cite journal | first1=Sidney H. | last1=Kung
| |
| |journal=Coll. Math. J.
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| |title= Partial fraction decomposition by division
| |
| |year=2006 | volume=37 | number=2 | pages=132–134
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| |jstor=27646303
| |
| }}
| |
| *{{cite news | first2=Damian | last2=Slota | first1=Roman | last1=Witula
| |
| |year=2008 | journal=Appl. Math. Comput.
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| |title=Partial fractions decompositions of some rational functions
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| |pages=328–336 | volume=197 | doi=10.1016/j.amc.2007.07.048 | mr=2396331
| |
| }}
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| == External links ==
| |
| * {{MathWorld |urlname=PartialFractionDecomposition |title=Partial Fraction Decomposition}}
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| * {{cite web|first1= Sam | last1=Blake
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| |url=http://calc101.com/webMathematica/partial-fractions.jsp
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| |title=Step-by-Step Partial Fractions}}
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| * [http://cajael.com/eng/control/LaplaceT/LaplaceT-1_Example_2_6_OGATA_4editio.php] Make partial fraction decompositions with [[Scilab]].
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| [[Category:Algebra]]
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| [[Category:Elementary algebra]]
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| [[Category:Partial fractions]]
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