This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MAT 330: Fourier Series and Partial Differential Equations Professor Sergiu Klainerman Problem Set 1 Rik Sengupta rsengupt @ princeton.edu February 6, 2010 1. Stein and Shakarchi, p. 24, problem 4 For z C , we define the complex exponential by e z = X n =0 z n n ! . (a) Prove that the above definition makes sense, by showing that the series converges for every complex number z . Moreover, show that the convergence is uniform on every bounded subset of C . Solution. First we will show that the definition makes sense. We will use the ratio test to show that, once we fix z C , the function e z as defined in the problem is welldefined, i.e. the series is convergent. Recall that the ratio test states that the series X n =0 a n with nonzero terms a j C converges and diverges according as the quantity L = lim n a n +1 a n is less than or greater than 1. For L = 1, the ratio test is inconclusive. So here, first of all, it is easy to see that e is simply 1, from the power series definition, and therefore the case z = 0 is welldefined. If z 6 = 0, then the premises of the ratio test are satisfied, i.e. for all terms are nonzero complex numbers. Then, lim n a n +1 a n = lim n z n +1 z n n ! ( n + 1)! = lim n z n + 1 = lim n  z  n + 1 = 0 < 1 . 1 So, the series converges uniformly for fixed z C , and therefore is welldefined as a function. So, the definition makes sense. We will show uniform convergence on the bounded sets of C by means of the Weierstrass Mtest. Recall that the Weierstrass Mtest states that, given a sequence of real or complex valued functions { f n } on a set A , with positive constants M n R satisfying  f n ( x )  M n for all x A , for every n 1, the series X n =1 f n ( x ) converges uniformly on A if the series X n =1 M n ( x ) converges. So now, notice that any bounded subset of C by definition is a subset of some neighbor hood N r of radius r of the origin in the complex plane. We are looking for an upper bound M n on each term of the series, with M n independent of the position in the subset considered, i.e. a term M n for each term satisfying z n n ! M n , z N r . But such terms are easy to find, because we have z n n ! =  z  n n ! r n n ! , so we can simply take M n = r n n ! . So, from the Weierstrass Mtest, it is now enough to check that M n is convergent. Again, this is trivial, and this can be done most easily with the ratio test once again. It is easy to see that lim n M n +1 M n = lim n r n +1 r n n ! ( n + 1)! = lim n r n + 1 = 0 < 1 . So, the series M n is convergent, and therefore, by the Weierstrass Mtest, the original series e z = X n =0 z n n !...
View Full
Document
 Spring '10
 SERGIUKLAINERMANN
 Differential Equations, Equations, Partial Differential Equations, Fourier Series

Click to edit the document details