330 HW 1 - MAT 330 Fourier Series and Partial Differential...

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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 well-defined, 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 0 is simply 1, from the power series definition, and therefore the case z = 0 is well-defined. 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
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So, the series converges uniformly for fixed z C , and therefore is well-defined as a function. So, the definition makes sense. We will show uniform convergence on the bounded sets of C by means of the Weierstrass M-test. Recall that the Weierstrass M-test 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 M-test, 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 M-test, the original series e z = X n =0 z n n ! . converges uniformly on every bounded subset of C . 2
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(b) If z 1 , z 2 are two complex numbers, prove that e z 1 e z 2 = e z 1 + z 2 . Solution 1. We will multiply the two series e z 1 and e z 2 by means of the Cauchy product.
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