Lecture 29 Section 6.7 Thu, Mar 3, 2005

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Lecture 29 Section 6.7 Thu, Mar 3, 2005 The Binomial Theorem Lecture 29 Section 6.7 Thu, Mar 3, 2005

(a + b)n = k = 0..n C(n, k)an – kbk. The Binomial Theorem Theorem: Given any numbers a and b and any nonnegative integer n, (a + b)n = k = 0..n C(n, k)an – kbk. Proof: Use induction on n. Basic step: n = 0. (a + b)0 = 1. k=0..0 C(0, k)a0 – kbk = C(0, 0)a0b0 = 1. Therefore, the statement is true when n = 0.

Proof, continued Inductive step Suppose the statement is true for some n  0. Then

Proof, continued

Proof, continued Thus, the statement is true for all n  0. Therefore, the statement is true for n + 1. Thus, the statement is true for all n  0.

Example: Binomial Theorem Expand (a + b)8. C(8, 0) = C(8, 8) = 1. C(8, 1) = C(8, 7) = 8. C(8, 2) = C(8, 6) = 28. C(8, 3) = C(8, 5) = 56. C(8, 4) = 70.

Example: Binomial Theorem Therefore, (a + b)8 = a8 + 8a7b + 28a6b2 + 56a5b3 + 70a4b4 + 56a3b5 + 28a2b6 + 8ab7 + b8.

Example: Calculating 1.0016 Compute 1.0016. 1.0016 = (1 + 0.001)6 = 1 + 6(0.001) + 15(0.001)2 + 20(0.001)3 + 15(0.001)4 + 6(0.001)5 + (0.001)6 = 1 + .006 + .000015 + .000000020 + .000000000015 + .000000000000006 + .000000000000000001 = 1.006015020015006001.

Example: Approximating (1+x)n Theorem: For small values of x, (1 + x)n  1 + nx. (1 + x)n  1 + nx + C(n, 2)x2 = 1 + nx + n(n – 1)/2 x2. (1 + x)n  1 + nx + C(n, 2)x2 + C(n, 3)x3. = 1 + nx + n(n – 1)/2 x2 + n(n – 1)(n – 2)/6 x3. And so on.

Example For example, (1 + x)10  1 + 10x + 45x2 + 120x3 when x is small. Compute the value of (1 + x)10 and the approximation when x = .01.

Newton’s Generalization of the Binomial Theorem Theorem: Given any numbers a, b, and n, (a + b)n = k=0.. C(n, k)an – kbk. where C(n, k) = [n(n – 1)…(n – k + 1)]/k! Note that n need not be an integer nor positive.

Example Expand (a + b)-1 in a series, showing the first 5 terms. Compute the first 5 coefficients C(-1, 0) = 1. C(-1, 1) = -1. C(-1, 2) = (-1)(-2)/2! = 1. C(-1, 3) = (-1)(-2)(-3)/3! = -1. C(-1, 4) = (-1)(-2)(-3)(-4)/4! = 1.

Example, continued Therefore,

Example Expand (a + b)5/2 in a series, showing the first 5 terms. Compute the first 5 coefficients C(5/2, 0) = 1. C(5/2, 1) = 5/2. C(5/2, 2) = (5/2)(3/2)/2! = 15/8. C(5/2, 3) = (5/2)(3/2)(1/2)/3! = 5/16. C(5/2, 4) = (5/2)(3/2)(1/2)(-1/2)/4! = -5/128.

Example, continued Therefore,

Example Approximate (1.2)5/2. Let a = 1 and b = 0.2 = 1/5. Then b/a = 1/5.

The Multinomial Theorem Theorem: In the expansion of (a1 + … + ak)n, the coefficient of a1n1a2n2…aknk is n!/(n1!n2!…nk!) where n = n1 + n2 + … + nk.

Example: The Multinomial Theorem Expand (a + b + c + d)3. The terms are a3, b3, c3, d3, with coefficient 3!/3! = 1. a2b, a2c, a2d, ab2, b2c, b2d, ac2, bc2, c2d, ad2, bd2, cd2, with coefficient 3!/(1!2!) = 3. abc, abd, acd, bcd, with coefficient 3!/(1!1!1!) = 6.

Example: The Multinomial Theorem Therefore, (a + b + c + d)3 = a3 + b3 + c3 + d3 + 3a2b + 3a2c + 3a2d + 3ab2 + 3b2c + 3b2d + 3ac2 + 3bc2 + 3c2d + 3ad2 + 3bd2 + 3cd2 + 6abc + 6abd + 6acd + 6bcd. Find (a + b + c)4.