Read e-book online A First Course in Analysis PDF

By Donald Yau

ISBN-10: 9814417858

ISBN-13: 9789814417853

This publication is an introductory textual content on genuine research for undergraduate scholars. The prerequisite for this publication is an exceptional history in freshman calculus in a single variable. The meant viewers of this e-book contains undergraduate arithmetic majors and scholars from different disciplines who use actual research. when you consider that this ebook is geared toward scholars who should not have a lot previous adventure with proofs, the velocity is slower in past chapters than in later chapters. There are 1000's of workouts, and tricks for a few of them are incorporated.

Readership: Undergraduates and graduate scholars in research.

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Example text

F (ki ) are elements in T . In particular, we have {f (k1 ), . . , f (ki )} = T. The function g∶ {1, . . , i} → T defined by g(j) = f (kj ) ∈ T is injective because f is injective. Moreover, g is surjective because every element in T is of the form f (kj ) for some j = 1, . . , i. Thus, g is a bijection, which means that T is a finite set. This is a contradiction, and hence S is an infinite set. 6. Let T be a subset of a finite set S. Then T is a finite set. Proof. 8. Indeed, if T is an infinite set, then S is an infinite set as well, contradicting the assumption.

3! n! Prove directly from the definition that {an } is a Cauchy sequence. (10) Consider the sequence with an = 1 + 1 1 1 + +⋯+ . 2 3 n Prove that {an } is not a Cauchy sequence, so {an } is divergent. (11) Give an example of a divergent sequence {an } such that lim(an+N − an ) = 0 for every positive integer N . (12) Let a1 = 1, a2 = 2, and an = 21 (an−1 + an−2 ) for n ≥ 3. an = 1 + n−1 (a) For n ≥ 2 prove that an+1 − an = (− 21 ) . (b) Prove that {an } is a Cauchy sequence, hence a convergent sequence.

4) that ∣ ∣an ∣ K∣b∣ 1 ∣b∣ M K∣b∣ an a − ∣< ⋅ + ⋅ < ⋅ + = . bn b ∣bn ∣∣b∣ 2M ∣b∣ 2 K∣b∣ 2M 2 This shows that lim abnn = ab . 5in 42 analysis-yau A First Course in Analysis The above theorem is very useful in computing limits, as illustrated in the following examples. 13. 2) and an = ( n1 )p . n (2) Consider the sequence with 1 n → 0 and 1 n2 converges to 0 because 2n2 + 7 . 3n2 − n − 1 an = Since 1 np → 0, we have an = 2 + n72 2 2+0 2n2 + 7 = = . → 1 1 2 3n − n − 1 3 − n − n2 3−0−0 3 (3) Similarly, the sequence with an = 1 − 4n + 3n2 n2 satisfies 1 − 4n + 3n2 1 4 = 2 − + 3 → 0 − 0 + 3 = 3.

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A First Course in Analysis by Donald Yau


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