1873_solutions

# 1873_solutions

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online