By Daniel W. Stroock

ISBN-10: 3319244671

ISBN-13: 9783319244679

ISBN-10: 3319244698

ISBN-13: 9783319244693

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**Sample text**

Finally, show that arctan x = 1+x 2 for x ∈ R. 13 Show that lim x→0 sin x x 1 1−cos x 1 = e− 3 . 40 1 Analysis on the Real Line In doing this computation, you might begin by observing that it suffices to show that lim x→0 sin x 1 log 1 − cos x x 1 =− . 3 At this point one can apply L’Hôpital’s rule, although it is probably easier to use Taylor’s Theorem. 14 Let f : (a, b) −→ R be a twice differentiable function. If f ≡ f (2) is continuous at the point c ∈ (a, b), use Taylor’s theorem to show that f (c) ≡ ∂ 2 f (c) = lim h→∞ f (c + h) + f (c − h) − 2 f (c) .

2 Functions of a Complex Variable 49 2n mθm ≤ C2 m=n+1 2C 2 nθn+1 . 1−θ Since θ ∈ (0, 1), α ≡ − log θ > 0, and so nθn+1 ≤ n → ∞. 2 . By the same reasoning as we used to show that ∞ m=0 m! ∞ ∞ m m has an infinite radius of convergence, one sees that m=0 am z and m=0 bm z do also. (n − m)! n! n m=0 n m n−m (z 1 + z 2 )n z1 z2 , = m n! 2 with z = 1, ∞ ez1 ez2 = n=0 (z 1 + z 2 )n = e z 1 +z 2 . n! Our next goal is to understand what e z really is, and, since e z = e x ei y , the problem comes down to understanding e z for purely imaginary z.

Thus if {z n : n ≥ 1} satisfies Cauchy’s criterion, so do both {xn : n ≥ 1} and {yn : n ≥ 1}. Hence, there exist x, y ∈ R such that xn → x and yn → y. Now let > 0 be given and choose n so that |xn − x| ∨ |yn − y| < √ for n ≥ n . Then 2 2 2 |z n − z|2 < 2 + 2 = 2 and therefore |z n − z| < for n ≥ n . 3 to show that every bounded sequence {z n : n ≥ 1} in C has a convergent subsequence. 46 2 Elements of Complex Analysis All the results in Sect. 2 about series and Sect. 9 about products extend more or less immediately to the complex numbers.

### A Concise Introduction to Analysis by Daniel W. Stroock

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