# 0 1 n n which converges by leibnitz theorem for x 2

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=0 ( - 1) n n , which converges by Leibnitz’ theorem. For x = 2, we get n =0 1 n which diverges (see this either by the integral test or comparison with the harmonic series). Combining this together, we get that the interval of convergence is [0 , 2), and the radius of convergence is 1. Example 51. Find the interval of convergence and radius of convergence of n =1 n n (2 x + 5) n Solution. First, we have to express the given function as a power series: X n =1 n n (2 x + 5) n = X n =1 n n (2 n ) x + 5 2 n (in particular, note that a = - 5 2 , not - 5). Using then-th root test, we see (1) n and if x + 5 2 = - 1 2 , then x = - 3 and the corresponding series is n =1 n n ( - 1) n . Neither of these series has a n 0, so they both diverge. This means that the interval of convergence is ( - 3 , - 2). 7.2. Term-By-Term Calculus. If c n ( x - a ) n converges on some positive radius of convergence (i.e. it doesn’t diverge for every x 6 = a ), then f ( x ) = c n ( x - a ) n is a continuous function (not to difficult to show), and we can reduce taking derivatives and integrals of f to taking derivatives and integrals of the terms (more difficult to show): 7.2.1. Term-By-Term Differentiation. Theorem 7.3. If f ( x ) = n =0 c n ( x - a ) n converges on | x - a | < R , then: (1) f has derivatives of arbitrary order on ( a - R, a + R ) (i.e. f 0 ( x ) , f 00 ( x ) , f 000 ( x ) , . . . exist) (2) The derivatives are given by: f 0 ( x ) = X n =1 nc n ( x - a ) n - 1 , f 00 ( x ) = X n =2 ( n )( n - 1)( x - a ) n - 2 , . . . on ( a - R, a + R ) . We can use this to express some power series in other ways: Example 52. Recall that f ( x ) = n =0 x n converges for - 1 < x < 1 . The above theorem tells us that f 0 ( x ) = X n =1 nx n - 1 and f 00 ( x ) = X n =2 n ( n - 1) x n - 2 On the other hand, we know that f ( x ) = 1 1 - x , so f 0 ( x ) = 1 (1 - x ) 2 and f 00 ( x ) = 2 (1 - x ) 3 . This gives us formulae: X n =1 nx n - 1 = 1 (1 - x ) 2 , and X n =2 n ( n - 1) x n - 2 = 2 (1 - x ) 3 It is important to note here that c n don’t involve x in any way (otherwise the series would not be a power series). It is not true that if we take any series f n ( x ) of functions of x that we can differentiate term-by-term as above: Example 53. The series n =1 sin( n ! x ) n 2 converges absolutely for all x , but if we differentiate term- by-term we get n =1 n ! cos( n ! x ) n 2 which diverges for all x .
MATH 214: SEQUENCES AND SERIES 13 7.2.2. Term-By-Term Integration. Theorem 7.4. Given f ( x ) = n =0 c n ( x - a ) n that converges on | x - a | < R , the function F ( x ) = n =0 c n ( x - a ) n +1 n +1 converges on | x - a | < R and Z f ( x ) d x = F ( x ) + C We can use this to identify series with other functions as well: Example 54. Identify f ( x ) = n =0 - (1 - t ) n +1 n +1 on 0 < x < 2 . Solution. First, differentiate the given series term-by-term to get f 0 ( x ) = X n =0 (1 - t ) n = 1 1 - (1 - t ) = 1 t This means that f ( x ) = R f 0 ( x )d x + C = ln( x )+ C . Setting x = 1 we get C = 0, i.e. f ( x ) = ln( x ).