427K.07.20 - Tr“. omefiw‘j Coq'wlc Law Cuilfi~3K...

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Unformatted text preview: Tr“. omefiw‘j Coq'wlc Law Cuilfi~3K Misfire 2*§ 3031:“ i985 % ) ’—\ as“ R swam W‘”? 246 Polynomial and Rational Functions 5 E C T l O N PARTIAL FRACTIONS (OPTIONAL) A. Introduction 3 _ X _3(X2—X+2)—X(X+l) X+l Xz—X+2_ (X+l)(X2—X+2) _2X2—4X+6 — X3+X+2. (24.1) In this section we consider how to reverse that process—how to begin with an expression like that on the right in (24.1) and end with an expression like that on the left. This reversed process is needed in calculus, for example. The method we use rests on the following fact, whose proof is omitted: With real coefficients, every polynomial can be written as a product of linear and irreducible quadratic factors. That fact, together with the division al- gorithm for polynomials, can be used to justify the following statement. Every rational function can be written uniquely as a sum of a polynomial (which may be identically zero) and fractions of the form A and BX + C (an + b)m (ax2 + bX + C)“ where ax2 + bX + c is irreducible. The resulting form is called the partial fraction decomposition of the original rational function. The linear (ax + b) and quadratic (ax2 + bX + (3) factors in the decomposition are the irreducible factors of the original denominator. The first step is always to factor the original denominator into powers of irreduci- ble factors. (Remember that a quadratic factor is irreducible iff its discriminant is negative.) B. Distinct Linear Factors We call an irreducible factor of the denominator distinct (as opposed to I9“ peated) if it occurs only with exponent one when we factor the denominator. Partial Fractions (Optional) 247 Each distinct linear factor ax + b contributes a term of the form A aX+b to the decomposition. X+12 Example 24.1 Decompose into partial fractions. Solution The factors X — 2 and X + 5 will contribute terms A and B X—2 x+5 respectively. If X+12 _ A B (X—2)(X+5)-X—2 X+5 (24.2) then on multiplying by (X — 2)(X + 5) to clear the denominators, we obtain X + 12 = A(X + 5) + B(X —- 2). (24.3) The last equation must be true for all values of X, so in particular it must be true for X = — 5 and X = 2, the zeros of the two linear factors. By substituting these in turn we can determine A and B. Use X = —5 in (24.3): —5 +12 = A(O) + B(—7) B = --1. Use X = 2 in (24.3): 2 + 12 = A(7) + B(O) A7: 2. With A = 2 and B : — 1, Equation (24.2) gives the answer: X+l2 2 l (x—2)(x+5)=x—2 x+5' To check the answer perform the subtraction indicated on the right and show that the result simplifies to the expression on the left. 248 Polynomial and Rational Functions C. Repeated Linear Factors Each repeated linear factor (ax + b)m contributes a sum of the form Al A2 A ax+b+(ax+b)2+'”+(ax+b)m to the decomposition. 2 __ _ Example 24.2 Decompose X 5X 2 ' Solution First, factor the denominator completely. X3'+2X2+X=X(X2+2X+1)=X(X+1)2 The distinct factor X contributes é X. The repeated linear factor (X + l)2 contributes ———-———— 1nto artial fraction . x3 + 2x2 + x p S B C X + 1 + (X + I)” If X2 — 5X -— 2 A B C — _ 4.4 X(X+1)2 X+X+l+(X+l)2 (2) then x2 — 5x — 2 = A(X + 1)2 + BX(X + 1) + CX. (24.5) With X = O in (24.5) we will get A = —2. With X = —1 in (245) we will get C = —4. Now we use any other value of X in (24.5), along with A = — 2 and C = —4, to get B. Withx = lwe will get B = 3. If we use A = - 2, B = 3, and C = — 4 in Equation (24.5), and revert to the original form of the denominator on the left, we get T‘artial Fractions (Optional) 249 fli-_2+ 3 ___4__ X3+2X2+X— X X+l (X+l)2' D. Distinct Quadratic Factors Each distinct quadratic‘ factor (1X2 + bX + c contributes a term of the form AX+B £1X2+bX+C to the decomposition. 3X2 + 2X E le 24.3 h ' ‘ ' . xamp Decompose (X + 1)(X2 + X + 1)mto partial fractions Solution The factor X2 + X + 1 is irreducible. The factors X + 1 and X2 + X + 1 contribute terms A BX + C X + 1 and m respectively. If 2 M then 3X2 + 2X = A(X2 + X + l) + (BX + C)(X + 1). (24.7) WithX = —1 in (24.7) we will getA = l. WithX = 0 and A = 1 in (24.7) we will get C = —l. WithX =1,A = l, and C = —1 in (24.7) we will getB = 2. Thus 3X2+2X l 2X-l (X+l)(XZ+X+l) X+1+X1+X+l' $0 Polynomial and Rational Functions E. Repeated Quadratic Factors Each repeated quadratic factor (ax2 + bx + c)“ contributes a sum of the form Alx + B1 AZX + 32 Anx + Bn + +~ ax2 + bx + c (ax2 + bx + c)2 . “Flax2 + bx + c)“ to the decomposition. In the previous examples the coefficients in the partial fraction decom- positions have been determined by substituting appropriate numbers for X. The next example uses a method based on the fact that a polynomial in x is identically zero (that is, zero for every value of x) iff each of its coefficients is zero. (This will be proved in Section 26.) l . Exam Ie 24.4 -—-—~—— ' . p Decompose X X2 + 2p into partial fractions. Solution The factors x and (x7- + 2)2 contribute éand BX+C+DX+E X X2 + 2 (X2 + 2)2 respectively. If 1 _ é BX + C Dx + E X(X2 + 2V — X + x2 + 2 + (x2 + 2)2 (24'8) then 1 = A(x2 + 2)2 + (Bx + C)x(x2 + 2) + (Dx + E)x 1 = A(x4 + 4x7- + 4) + B(x4 + 2x2) + C(x3 +2x) + Dx2 + Ex (A + B)x4 + Cx3 + (4A + 28 + D)x2 + (2C + E)x + 4A — 1 = 0. (24.9) Since Equation (24.9) is to be satisfied for every value of x, each of the coefficients must be zero. (See the remark preceding this example.) x4zA+B=O x32C=0 x2:4A+2B+D=0 X22C+E=0 X014A—l=0 Partial Fractions (Optional) 251 The equation for X0 yields A = i. The equation for X4 then yields B = — i The equation for X3 yields C = 0. The equation for X then yields E = 0. With A = i and B = —%, the equation for X2 yields D = — %. With these values for A, B, C, D, and E, Equation (24.8) becomes 1 1 X X x(x2 + 2)2 4x ‘ 4(x2 + 2) 208 + 2)2‘ F. Improper Fractions A rational function is called a proper fraction if the degree of the numerator is less than the degree of the denominator; otherwise it is called an improper fraction. By division, an improper fraction can be written as a sum of a polynomial and a proper fraction. The preceding methods can then be applied to decompose the proper fraction into partial fractions. Example 24.5 Decompose f(X)_X4+3X3—11X2—2X+22 — X2+3x—10 into partial fractions. Solution Division yields X+12 _ 2_ H foo—X 1+X2+3X—10‘ The proper fraction on the right is the same as the fraction in Example 24.1. Therefore, using the solution of that example, we have f(X)=X2-l+ X—Z x+5‘ ...
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This note was uploaded on 05/10/2008 for the course M 427K taught by Professor Fonken during the Spring '08 term at University of Texas at Austin.

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427K.07.20 - Tr“. omefiw‘j Coq'wlc Law Cuilfi~3K...

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