By Constantin Caratheodory

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**Extra resources for Theory of Functions of a Complex Variable. **

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0 2. We’ve shown geometrically that b xm dx = 0 bm+1 , m+1 b ≥ 0. Extend this result geometrically (not quoting a calculus result) to show that b am+1 bm+1 − , a ≤ b. xm dx = m+1 m+1 a Start with 0 ≤ a ≤ b and then consider the case a ≤ 0 ≤ b. 3. Denote by Ln and Rn respectively the left and right endpoint Riemann sums for the integral b f (x) dx. a Assume that the interval [a, b] is divided into n subintervals of equal length. If the function f (x) is decreasing, then b Rn ≤ f (x) dx ≤ Ln . a (a) Determine which rectangular areas appear in both left and right endpoint sums, and use this observation to show that Ln − Rn = b−a [f (a) − f (b)].

Now bring back the ejected horse, toss out another one, repeat the argument, and all K + 1 horses are white. Since there is a white horse somewhere in the world, all horses are white!! 6. Show that for any positive integer n the number n2 is the sum of the ﬁrst n odd numbers, n2 = n (2k − 1). k=1 Discrete Calculus 21 7. Suppose that for nonnegative integers m the function T satisﬁes the recurrence formula T (2m ) ≤ aT (2m−1 ) + b2m , m ≥ 1, T (1) ≤ b. Here a and b are nonnegative numbers. Use induction to show that for every positive integer m, m T (2m ) ≤ b2m (a/2)k = b2m k=0 1 − (a/2)m+1 .

That simpliﬁcation is our next order of business. 46 A Concrete Introduction to Real Analysis Notice that for x ≥ 0 log(1 + x) ≤ x. This follows from log(1) = 0 and 1 d log(1 + x) = dx 1+x so that d d log(1 + x) ≤ 1 = x, dx dx x ≥ 0. A simple logarithmic calculation now gives (n + 1/2) log(1 + 1 1 1 ) ≤ (n + 1/2) = 1 + , n n 2n or (1 + 1 n+1/2 ) ≤ e1+1/(2n) . 5 dt 4 Selected Area Computations 47 Similarly, the calculation 1+x log(1 + x) = 1 x 1 dt ≥ , t 1+x x≥0 gives (n + 1/2) log(1 + 2n + 1 1 2n 1 )≥ = 1/2, 2n 2 2n 2n + 1 or 1 n+1/2 ) .