Download PDF by Serge Lang: A first course in calculus

By Serge Lang

This 5th version of Lang's booklet covers the entire issues regularly taught within the first-year calculus series. Divided into 5 elements, each one part of a primary direction IN CALCULUS comprises examples and purposes in terms of the subject lined. moreover, the rear of the e-book comprises certain options to loads of the workouts, permitting them to be used as worked-out examples -- one of many major advancements over prior variations.

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Example text

We leave that proof to the reader. Our first use of the existence of suprema and infima will be to establish a useful substitute for Axiom 2. There are times when it’s convenient to define a real number as the one point common to all the intervals in some collection when the collection is not a nested sequence. The theorem below 14 CHAPTER I THE REAL NUMBER SYSTEM is a real help in such situations. In particular, we’ll use it in Chapter 5 to help develop the theory of integration. 2: Let I be a nonempty family of closed intervals such that every pair of intervals in I has a nonempty intersection.

We should note that the intermediate value theorem gives only a property that continuous functions have, not a definition of continuity. The standard example of a discontinuous function that satisfies the conclusion of the intermediate value theorem is f (x) = sin (1/x) , x > 0 0, x ≤ 0. In every interval (0, δ), f takes on every value in [−1, 1], and that makes it easy to see that intermediate values will always exist even though f is not continuous at 0. An even more striking example is given in the book by R.

Then each In is a closed interval containing the next interval in the sequence and the lengths satisfy bn − an → 0. Once again, there is exactly one real number common to all these intervals; let’s call it s. Now let’s see why s = sup E. For any given x > s, there is an interval In in the sequence with length less than x − s, so its right endpoint is to the left of x. That shows that x is greater than an upper bound for E, so x∈ / E when x > s. Hence every x ∈ E satisfies x ≤ s. On the other hand, given any y < s, there is an interval In with its left endpoint to the right of y.

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A first course in calculus by Serge Lang


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