Discrete Mathematics with Graph Theory (3rd Edition) 153

Discrete Mathematics with Graph Theory (3rd Edition) 153 -...

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Section 5.4 U . . If' -3 + 13x A B d . smg partIa ractIons we set (1 _ 2x)(1 _ 3x) = 1 _ 2x + 1 _ 3x an ,equatIng numerators, A(1 - 3x) + B(1 - 2x) (A + B) + (-3A - 2B)x -3 + 13x -3 + 13x. Hence, A + B = -3, -3A - 2B = 13 and A = -7, B = 4. Therefore, -7 4 1-2x + 1- 3x f(x) 151 (-7)(1 + 2x + 4x 2 + ... + 2 n x n + ... ) + 4(1 + 3x + 9x 2 + ... + 3 n xn + ... ) -3 - 2x + 8x 2 + ... + (-7· 2 n + 4· 3 n )xn + ... and, hence, an = -7(2n) + 4(3 n ) is the solution, continning the result of Problem 20. 6. f(x) ao + alX + a2x2 + + anx n + + 4an_lX n + + 4an_2 Xn + 4xf(x) = 4aox + 4alX2 + 4x 2 f(x) = 4aox2 + Therefore, f(x) - 4xf(x) + 4x 2 f(x) = ao + (al - 4ao)x + (a2 - 4al + 4ao)x2 + ... + (an - 4an -l + 4an_2)X n + .... Since ao = 1, al - 4ao = 4 - 4(1) = 0 and an - 4an -l + 4an-2 = 0 for n > 2, we get f(x)(1 - 4x + 4x 2 ) = 1. Therefore, f(x) 1 = 1 + 2(2x) + 3(2x)2 + ... + (n + 1)(2x)n + ... (1 - 2x)2 1 + 4x + 12x2 + ... + (n + 1)2nxn
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Unformatted text preview: + ... ; hence, an = (n + 1)2n is the solution. Our solution is exactly the same as the sequence in Problem 21 and so our answer here verities the result obtained earlier. 7. [BB] f(x) = xf(x) = ao + alX + a2x2 + + anx n + aox + alx 2 + ... + an_lX n + Subtracting gives f(x) -xf(x) (1- x)f(x) ao + (al -ao)x + ... + (an -an_l)X n + ... 5 + 2x + 4x 2 + ... + 2 n x n + . . . (1 + 2x + 4x 2 + ... + 2 n x n + ... ) + 4 since ao = 5 and an -an-l = 2n. Thus, f(x) = _1_(1 + 2x +4x 2 + ... + 2 n x n + ... ) +4(_1_) I-x I-x = [1 + (2 + l)x + (4 + 2 + l)x 2 + ... + (2n + 2 n-1 + 2 n-2 + ... + 2 + l)xn + ... J + 4(1 + x + x 2 + ... + xn + ... ) = (1 + 3x + 7x 2 + ... + (2 n + 1 -l)xn + ... ) + (4 + 4x + 4x 2 + ... + 4xn + ... ) Note that 1 + 2 + 22 + ... + 2 n = 2n2+~11 = 2 n + 1 -1 as the sum of a geometric sequence. Thus, an = 2 n + 1 -1 +4 = 2 n + 1 +3....
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