1873_solutions

1873_solutions

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Unformatted text preview: for which the following series converge and when they converge absolutely: a. b. c. sin nx cos nx n sin nx cos nx 1 sin 2nx This series is just and so it follows from this earlier example that n n 2 converges for every number x and converges conditionally unless x is an integer multiple of  . 2 1 n cos nx n The series we are considering here can be expressed as cos n   x n and from the case considered above we see that the series converges if and only if   x is not an integer multiple of 2. In other words, the series converges if and only if x is not an odd multiple of . cos 2 nx n Hint: Use the identity cos 2 nx  1  1 cos 2nx 2 2 and show that the series is divergent for every number x. We see from the identity cos 2nx 1  2 cos 2 nx n n n that, unless x is an integer multiple of , the convergence of 1 guarantee that the series n cos 2nx and the divergence of n cos 2 nx n is divergent. On the other hand, if x is an integer multiple of  then the given series reduces to 1 which diverges. Therefore the series n cos 2 nx n diverges for every number x. d. |cos nx | n 343 Since |cos nx | cos 2 nx n n for all n and x we deduce from Part c that the series |cos nx | n diverges for every number x. 0 e. cos 3 nx n Hint: Use the identity cos 3 nx  3 cos nx  1 cos 3nx. 4 4 If x is an integer multiple of 2 then the series cos 3 nx n is 1 n which diverges. If 3x is an integer multiple of 2 but x is not then, since cos nx n is convergent and cos 3nx n is divergent, the series cos 3 nx  3 cos nx  1 cos 3nx n n 4 4n is divergent. If 3x is not an integer multiple of 2 then nor is x and, since both cos 3nx are convergent, so is n cos 3 nx . n f. cos nx and n cos 4 nx n Given any number x positive integer n we have cos 4 nx  3  1 cos 2nx  1 cos 4nx . n n n 8n 2 8 In the event that x is an integer multiple of /2 then the series cos 4 nx n is either 1 1 or n 2n and is therefore divergent. In the event that x is not an integer multiple of /2, both of the series cos 2nx cos 4nx and n n are convergent and, since 3 8n is divergent, the series cos 4 nx n is divergent. We conclude that the series 344 cos 4 nx n diverges for every number x. 6. With an eye on the preceding exercise give an example of a convergent series a 3 is divergent. n The series cos 2n 3 3n a n such that the series is convergent but the series cos 3 2n 3 3 n is divergent. 7. Find the values of x and  for which the binomial series  xn n is convergent.  n If  is a nonnegative integer then, since   0 whenever n   the series n n x converges for every number x. From now on we assume that  fails to be a nonnegative integer. Since  x n 1 n 1 lim  |x | nÝ |  xn | n we know that the series  xn n is divergent whenever |x |  1 and is absolutely convergent when |x |  1. Now we need to consider the cases x  1. Now we saw earlier that the series  n is absolutely convergent when   0. We also saw that the nth term of this series fails to approach 0 as n Ý when  1. Finally we s...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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