121616949-math.276 - ≤ 2 so we can use the series to...

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262 Chapter 10 Sequences and Series for polynomials, the obvious question is whether the same is true for infinite series. The answer is yes: THEOREM 10.46 Suppose the power series f ( x ) = X n =0 a n ( x - a ) n has radius of convergence R . Then f ( x ) = X n =0 na n ( x - a ) n - 1 , Z f ( x ) dx = C + X n =0 a n n + 1 ( x - a ) n +1 , and these two series have radius of convergence R as well. EXAMPLE 10.47 Starting with the geometric series: 1 1 - x = X n =0 x n Z 1 1 - x dx = - ln | 1 - x | = X n =0 1 n + 1 x n +1 ln | 1 - x | = X n =0 - 1 n + 1 x n +1 when | x | < 1. The series does not converge when x = 1 but does converge when x = - 1 or 1 - x = 2. The interval of convergence is [ - 1 , 1), or 0
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Unformatted text preview: ≤ 2, so we can use the series to represent ln( x ) when 0 < x ≤ 2. For example ln(3 / 2) = ln(1- -1 / 2) = ∞ X n =0 (-1) n 1 n + 1 1 2 n +1 and so ln(3 / 2) ≈ 1 2-1 8 + 1 24-1 64 + 1 160-1 384 + 1 896 = 909 2240 ≈ . 406 . Because this is an alternating series with decreasing terms, we know that the true value is between 909 / 2240 and 909 / 2240-1 / 2048 = 29053 / 71680 ≈ . 4053, so correct to two decimal places the value is 0 . 41....
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