By Neil A. Watson

ISBN-10: 9810215916

ISBN-13: 9789810215910

All however the most simple proofs are labored out intimately sooner than being provided officially during this booklet. therefore lots of the rules are expressed in other ways: the 1st encourages and develops the instinct and the second one supplies a sense for what constitutes an explanation. during this approach, instinct and rigour look as companions instead of rivals. The casual discussions, the examples and the routines may possibly think a few familiarity with calculus, however the definitions, theorems and formal proofs are provided within the right logical order and suppose no past wisdom of calculus. hence a few simple wisdom of calculus is mixed into the presentation instead of being thoroughly excluded.

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Also, if x > a then x ¢ S, so that f (x) > q. We can then use the continuity of f at or to deduce that f (Q) = q. Looking at points on both sides of a assumes, of course, that a Ela, b[. This can be verified: since f is continuous at a and f (a) < q, we will have f (x) < q for all x near a, so Boundedness 39 that sup S > a; and since f is continuous at b and f (b) > q, we will have f (x) > q for all x near b, so that sup S < b. Now we've got all the pieces, we'll check them rigorously and in the correct logical order.

Since it is valid when n = 1, it is true for all n E N, by induction. 6) that {stn } is unbounded, so that {s} is unbounded and hence sn -+ 00 (since {S} is increasing). Thus > is divergent. Now consider the case p < 1. We can deduce that E(l/kP) is divergent by comparing its partial sums with those of >(1/k). If p < 1, then k < k for all k E N, so that 1 1 kP k Hence kp- > k=1 E k=1 for all n. Since the sum on the right tends to infinity with n, so does the one on the left. Hence >(1/kP) is divergent if p < 1.

If f is continuous on [a, b], then f is bounded on [a, b]. For this result, it is helpful to use the intuitive idea (familiar from calculus) that the graph of a continuous function on [a, b] can be interpreted as the path of a particle in the plane from (a, f (a)) to (b, f (b)). We start at a. Since f (a) is the sole element of {f(x) : a < x < a}, that set is bounded. Also, since f is continuous at a, the values of f (x) for x near a will be near f (a), so that there will be Si > 0 such that {f(x) : a < x < a + 61 1 is bounded.

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