# From Calculus to Analysis by Steen Pedersen

By Steen Pedersen

Workouts embedded within the textual content with recommendations on the finish of every section
Approaches the genuine numbers via limitless decimals
Classroom proven with either arithmetic and arithmetic schooling majors

This textbook gains functions together with an explanation of the basic Theorem of Algebra, area filling curves, and the idea of irrational numbers. as well as the traditional result of complicated calculus, the ebook comprises a number of fascinating functions of those results.

The textual content is meant to shape a bridge among calculus and research. it really is in response to the authors lecture notes used and revised approximately each year over the past decade. The e-book comprises quite a few illustrations and move references all through, in addition to routines with suggestions on the finish of every section

Keywords » Derivatives - Fourier sequence - quantity conception - genuine Variables - Set Theory

Related matters » Algebra - research

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Extra resources for From Calculus to Analysis

Example text

4 Pseudo-sine function Fig. 5 A part of the graph of σ (1/x) with some important values of x indicated Some Consequences of the Existence of a Limit The results in this subsection may appear simple. But they are very useful when we want to use the existence of some limit(s) to establish the existence of related limits. They also serve to give us some experience using the assumption f (x) → L as x → a in proofs. In the following D is some subset of C and a is an accumulation point of D. We say f : D → C has a property P near a, if there is a δ > 0, such that f (x) satisfies P for all x in D ∩ Bδ (a).

Proof. Suppose a is irrational. Divide the closed interval [0, 1] into a finite number of subintervals each with length < ε . If j = k are integers, then { ja} = {ka} , since a is irrational. Hence, the set {{ka} | k ∈ N} is infinite. Hence, one of the subintervals must contain an infinite number of the fractional parts {ka} , k ∈ N. In particular, there are two integers 0 < m < n such that |{na} − {ma}| < ε . Hence, |(na − na ) − (ma − ma )| < ε . Setting k = na − ma and dividing by n − m gives a− ε k .

12) can be written as {a} + {b} = {a + b} 1 + {a + b} if {a} + {b} < 1 . 13) Also, if {a} > {b}, then {a} − {b} = {a − b} 1 − {b − a} if a ≥ b . 11) of the fractional part. 3. Show {a + {na}} = {(n + 1) a} for all n ∈ N. 4. Show {m {na}} = {mna} for all m, n ∈ N. 3, due to Liouville. 5 (Baby Liouville). Let a be a real number. Then a is irrational iff for any ε > 0 there is an m ∈ Z and an n ∈ N such that a = mn and a − mn < εn . Proof. Suppose a is irrational. Divide the closed interval [0, 1] into a finite number of subintervals each with length < ε .