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Chem Differential Eq HW Solutions Fall 2011 187

Chem Differential Eq HW Solutions Fall 2011 187 - /e One...

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Section A.4 The Power Series Method, Part I A187 Solutions to Exercises A.4 1. Using the ratio test, we have that the series m =0 x m 5 m + 1 converges whenever the limit lim m →∞ x m +1 5( m + 1) + 1 x m 5 m + 1 = lim m →∞ 5 m + 1 5 m + 6 | x | = | x | is less than 1. That is, | x | < 1. Thus, the interval of convergence is | x | < 1 or ( - 1 , 1). Since the series is centered at 0, the radius of convergence is 1. 5. Using the ratio test, we have that the series m =1 m m x m m ! converges whenever the following limit is < 1: lim m →∞ ( m + 1) m +1 x m +1 ( m + 1)! m m x m m ! = lim m →∞ ( m + 1) m ( m + 1) ( m + 1)! · m ! m m x m · | x | = | x | lim m →∞ m + 1 m m = e | x | . We have used the limit lim m →∞ m + 1 m m = lim m →∞ 1 + 1 m m = e (see the remark at the end of the solution). From | x | e< 1 we get | x | < 1 /e . Hence the interval of convergence is ( - 1 /e, 1 /e ). It is centered at 0 and has radius 1
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Unformatted text preview: /e . One way to show lim m →∞ ´ m + 1 m µ m = e is to show that the natural logarithm of the limit is 1: ln ´ m + 1 m µ m = m ln ´ m + 1 m µ = m [ln( m + 1)-ln m ] . By the mean value theorem (applied to the function f ( x ) = ln x on the interval [ m, m + 1]), there is a real number c m in [ m, m + 1] such that ln( m + 1)-ln m = f ± ( c m ) = 1 c m Note that 1 m + 1 ≤ 1 c m ≤ 1 m . So m m + 1 ≤ m [ln( m + 1)-ln m ] ≤ 1 . As m → ∞ , m m +1 → 1, and so by the sandwich theorem, m [ln( m + 1)-ln m ] → 1 . Taking the exponential, we derive the desired limit....
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