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 →∞ ² ² ² ² ² x m +1 5( m +1)+1 ³ x m 5 m +1 ² ² ² ² ² = lim m →∞ ´ 5 m +1 5 m +6 µ | x | = | x | is less than 1. That is, | x | < 1. Thus, the interval of convergence is | x | < 1o r ( - 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 →∞ ² ² ² ² ² ( m +1) m +1 x m +1 ( m + 1)! ³ m m x m m ! ² ² ² ² ² = lim m →∞ ( m +1) m ( m +1) ( m + 1)! · m ! m m x m ·| x | = | x | lim m →∞ ´ m +1 m µ m = e | x | . We have used the limit lim m →∞ ´ m +1 m µ m = lim m →∞ ´ 1+ 1 m µ 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
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Unformatted text preview: /e . One way to show lim m m + 1 m m = e is to show that the natural logarithm of the limit is 1: ln m + 1 m m = m ln m + 1 m = 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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