1873_solutions

Test that 1 n a n is convergent the interval of

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: e that x and y are real numbers that are not both zero and that   arccos x x2  y2 . We see that sin   1 cos 2   1 cos 2 arccos x2 x  y2 2  We define r  x x2  y2 1 x 2  y 2 and we observe that 380  y x2  y2 . x  r cos  and y  r sin . In the event that the latter equation says that y  r sin  we define    and if the equation y  r sin  is false (in which case y  r sin ) we define     . In either event we have the two equations x  r cos  y  r sin  Finally we observe that since 0   we must have 0  2. However, if    then the equation y  r sin  is true and     . Thus the case   2 cannot occur and we conclude that 0   2. Exercises on Analytic Functions Prove that the function arctan is analytic on R. Suppose that c is any number. Using the fact that the rational function whose value at every number x is 1 2 is analytic, choose   0 and a sequence a n such that 1 x 1  1  x2 whenever |x Ý an x Ý arctan c n0 for |x n n0 c |  . If we now define f x  arctan x c an x n1 c n 1 c |  1 then we see at once that f is the constant 0 and so Ý arctan x  arctan c  n0 whenever |x an x n1 c n 1 c |  . 1. Given that f x  3x for every number x, prove that f is analytic on R 0 but is not analytic on R. Suppose that c 0. The equation f x  cx c 1/3  c 1/3 1  x c c 1/3 Ý  c 1/3 n0 1 cn n x c n holds whenever |x c |  |c |. Therefore f is analytic on the set R 0 . Since f is not differentiable at 0, the function f can’t be analytic on any open interval that contains 0. 2. Prove that the function tan is analytic in some neighborhood of the number 0. The function tan, being the quotient of the two analytic functions sin and cos, is analytic on the  interval , . 22 3. Given that fx  exp 0 1 x2 if x 0 , if x  0 prove that f is analytic on R 0 but is not analytic on R. In spite of the fact that this function f has derivatives of all orders at every number, we saw in an 381 earlier example that f can’t be expressed as the sum of it’s Taylor series center 0 in any neighborhood of 0. On the other hand, the composition theorem for analytic functions guarantees that f is analytic on the set R 0. 4. Prove that a rational function is analytic on any open set in which its denominator does not vanish. This fact follows at once from the fact that the quotient of two analytic functions is analytic as long as the denominator is not zero. 5. Given that 1x fx  1/x if x 0 if x  0 e prove that f is analytic on the interval 1, 1 . Use Scientific Notebook to work out the first few terms of some of the series expansions of this function. To do so, point at the expression 1  x 1/x , open the Maple menu and click on the Power Series option. We begin by observing that Ý 1 n n 1 x n1 log 1  x  n0 whenever |x |  1. We now define Ý 1n n x n1 gx  n0 and we observe that, since g 0  1 and gx  for every x 1, 1 log 1  x x 0 we have f x  exp g x whenever |x |  1. Since g is analytic on the interval funct...
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