Section 1.7 Linear Inequalities and Absolute Value Inequalities.
MTH4100 Calculus Iklages/teaching/mth4100/2008/week1web.… · Rules for inequalities some useful...
Transcript of MTH4100 Calculus Iklages/teaching/mth4100/2008/week1web.… · Rules for inequalities some useful...
MTH4100 Calculus I
Week 1 (mainly Thomas’ Calculus Section 1.1)
Rainer Klages
School of Mathematical SciencesQueen Mary, University of London
Autumn 2008
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Lecture 1
What is Calculus?
“advanced algebra and geometry”:setting up Mathematics as a formal language
fundamental: real numbers
study of functions of real variablesone real variable (Calculus I)many variables (Calculus II)
geometric view: graph of a functioncontinuity propertiesslope ↔ derivativearea ↔ integral
many techniques, based on algebraic manipulations
many applications in all branches of modern society
(next level of mathematical abstraction is called analysis)
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Lecture 1
Real numbers and the real line
think of the real numbers, e.g., as all decimals
examples: −34 = −0.7500 . . . ; 1
3 = 0.333 . . . ;√
2 = 1.4142 . . .
The real numbers R can be represented as points on the real line:
-3 -2 -1 0 1 2 3 4-3/4 1/3 π2
three fundamental properties of real numbers
algebraic: formalisation of rules of calculation (addition, subtraction,multiplication, division)example: 2(3 + 5) = 2 · 3 + 2 · 5 = 6 + 10 = 16
order: inequalities (geometric picture: see the real line!)example: − 3
4 <13 ⇒ − 1
3 <34
completeness: there are “no gaps” on the real line
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Lecture 1
Algebraic properties
1. algebraic properties of the reals for addition (a, b, c ∈ R):
(A1) a + (b + c) = (a + b) + c
(A2) a + b = b + a
(A3) there is a 0 such that a + 0 = a
(A4) there is an x such that a + x = 0
associativity
commutativity
identity
inverse
why these rules? They define an algebraic structure (commutative group).define analogous algebraic properties for multiplication:
(M1) a(bc) = (ab)c
(M2) ab = ba
(M3) there is a 1 such that a 1 = a
(M4) there is an x such that a x = 1 (for a 6= 0)
connect multiplication with addition: (D) a(b + c) = ab + ac distributivity(These 9 rules define an algebraic structure called a field.)
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Lecture 1
Order: the real number line
2. Order properties of the reals:
(O1) for any a, b ∈ R, a ≤ b or b ≤ a totality of ordering I
(O2) if a ≤ b and b ≤ a then a = b totality of ordering II
(O3) if a ≤ b and b ≤ c then a ≤ c transitivity
(O4) if a ≤ b then a + c ≤ b + c order under addition
(O5) if a ≤ b and 0 ≤ c then a c ≤ b c order under multiplication
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Lecture 1
Rules for inequalities
some useful rules for calculations with inequalities (see exercises!):
these rules can all be “proven” by using (O1) to (O5): 1. to 3. followstraightforwardly, 4. to 6. require more work
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Lecture 1
Subsets of the real numbers R
3. Completeness property can be understood by the followingconstruction of the real numbers: (! using set notation !)
Start with “counting numbers” 1, 2, 3, . . .
N = {1, 2, 3, 4, . . .} natural numbers
→ can we solve a + x = b for x?
Z = {. . . ,−2,−1, 0, 1, 2, . . .} integers
→ can we solve ax = b for x?
Q = {pq|p, q ∈ Z, q 6= 0} rational numbers
→ can we solve x2 = 2 for x?
R real numbersexample: positive solution to the equation x2 = 2 is
√2
This is an irrational number whose decimal representation is noteventually repeating:
√2 = 1.414 . . . Another example is
π = 3.141 . . .⇒ N ⊂ Z ⊂ Q ⊂ R
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Lecture 1
Q has “holes”
In fact, one has to ”prove” this:
Theorem
x2 = 2 has no solution x ∈ Q
The real numbers R are complete in the sense that they correspond to allpoints on the real line, i.e., there are no “holes” or “gaps”, whereas therationals have ”holes” (namely the irrationals)
(see your textbook Appendix 4 for details; ”proof”of completeness of R
covered in MTH5104 Convergence and Continuity, 2nd year ”analysis”module)
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Lecture 2
Revision of Lecture 1
Properties of real numbers R:
algebraic: rules of calculation
order: inequalities
completeness: “no gaps”
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Lecture 2
√2 is irrational
Theorem
x2 = 2 has no solution x ∈ Q
Proof.
Assume there is an x ∈ Q with x2 = 2. This must be of the formx = p
q, p, q ∈ Z, q 6= 0, and we can assume that p and q have no
common factors (otherwise cancel them).
x2 = 2 then implies that (pq)2 = 2, or p2 = 2q2 , so p2 is even.
However, p2 even implies that p is even (...requires proof...).
Write p = 2p1, so that p2 = (2p1)2, or 4p2
1 = 2q2, or 2p21 = q2 .
This implies that q2 is even, so q is even as well.We have now shown that both p and q must be even, so they share a
common factor 2.This is a contradiction! Therefore the assumption must be wrong.
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Lecture 2
Definitions, Theorems, Proofs, . . .
You have just seen a Theorem with Proof.
University mathematics is built upon
basic properties (Definitions, Axioms)
statements deduced from these(Lemma, Proposition, Theorem, Corollary, . . .)in form of proofs!
example: The technique in the previous proof is called Proof byContradiction.
Many different ones to come! Details about the logic behind proofs, e.g.,in MTH5117 Mathematical Writing.
This formal framework is illustrated in Calculus 1 by many examples,exercises, applications, . . .
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Lecture 2
Intervals
Definition
A subset of the real line is called an interval if it contains at least twonumbers and all the real numbers between any two of its elements.
examples:
x > −2 defines an infinite interval; geometrically, it corresponds to aray on the real line
3 ≤ x ≤ 6 defines a finite interval; geometrically, it corresponds to aline segment on the real line
So we can distinguish between two basic types of intervals – let’s furtherclassify:
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Lecture 2
Types of Intervals
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Lecture 2
Finding intervals of numbers
Solve inequalities to find intervals of x ∈ R:
(a) 2x − 1 < x + 3
2x < x + 4
x < 4
(b) − x
3< 2x + 1
− x < 6x + 3
− 3
7< x
(c) 6x−1 ≥ 5: must hold x > 1!
6 ≥ 5x − 511
5≥ x
solution sets on the real line:
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Lecture 2
Absolute Value
Definition
The absolute value (or modulus) of a real number x is
|x | =
{
x x ≥ 0−x x < 0 .
geometrically, |x | is the distance between x and 0
example:
|x − y | is the distance between x and y
example:
an alternative definition of |x | is
|x | =√
x2 ,
since taking the square root always gives a non-negative result!R. Klages (QMUL) MTH4100 Calculus 1 Week 1 15 / 24
Lecture 2
Inequalities with |x |
|x | in an inequality:
|x | < a ⇔ −a < x < a
distance from x to 0 is less than a > 0 ⇔ x must lie between a and −a
absolute value properties:
1 | − a| = |a|2 |ab| = |a| |b|3 | a
b| = |a|
|b| for b 6= 0
4 |a + b| ≤ |a| + |b|, the triangle inequality
prove these statements!
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Lecture 3
Revision of Lecture 2
Definitions, Theorems, Proofs:
theorem and proof - example: irrationality of√
2
definition: interval; and examples: (a, b), [a, b], [a,∞), etc.
another definition: absolute value |x | and some properties
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Lecture 3
Some simple proofs
key idea: use |x | =√
x2
1 Proof of | − a| = |a|:
| − a| =√
(−a)2 =√
a2 = |a|Note: we have used a direct proof: we started on the left hand side(LHS) of the equation and transformed it step by step until we havearrived at the right hand side (RHS)
2 Proof of |ab| = |a| |b|:
|ab| =√
(ab)2 =√
a2b2 =√
a2√
b2 = |a| |b|
3 Proof of | ab| = |a|
|b| for b 6= 0:
∣
∣
∣
a
b
∣
∣
∣=
√
(a
b
)2=
√
a2
b2=
√a2
√b2
=|a||b|
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Lecture 3
Proof of the triangle inequality
4 Proof of |a + b| ≤ |a| + |b|: use a little trick and prove instead:
|a + b|2 ≤ (|a| + |b|)2
|a + b|2 = (a + b)2 (|x | = ±x yields |x |2 = x2)
= a2 + 2ab + b2
≤ a2 + 2|a| |b| + b2 (because ab ≤ |ab| = |a||b|)= |a|2 + 2|a| |b| + |b|2 (see above)
= (|a| + |b|)2
now take the square root and observe that the arguments of both roots arepositive – we are done.
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Lecture 3
Further properties
note: “if and only if” is often abbreviated by the sign “⇔”examples
(a) |2x − 3| ≤ 1
(b) |2x − 3| ≥ 1
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Lecture 3
Three important inequalities
Triangle inequality
|a + b| ≤ |a| + |b|
arithmetic mean: 12 (a + b); geometric mean
√ab
Arithmetic-geometric mean inequality
√ab ≤ 1
2(a + b) for a, b ≥ 0
Cauchy-Schwarz inequality
(ac + bd)2 ≤ (a2 + b2)(c2 + d2)
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Lecture 3
Proof of the arithmetic-geometric mean inequality
multiply inequality by 2 and square:
√ab ≤ 1
2(a + b) ⇔ 4ab ≤ (a + b)2
use direct proof: start on RHS until the LHS is obtained
(a + b)2 = a2 + 2ab + b2
= a2 + 2ab+2ab − 2ab + b2
= 4ab + (a − b)2 , (a − b)2 ≥ 0 and therefore
≥ 4ab
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Lecture 3
Proof of the Cauchy-Schwarz inequality
Use direct proof: start on one side until the other side is obtained
(ac + bd)2 ≤ (a2 + b2)(c2 + d2)
decide which side:
LHS: (ac + bd)2 = a2c2 + 2abcd + b2d2
RHS: (a2 + b2)(c2 + d2) = a2c2 + b2c2 + a2d2 + b2d2
Start on RHS and work it out:
(a2 + b2)(c2 + d2) = a2c2+2abcd + b2d2
+b2c2−2abcd + a2d2
= (ac + bd)2 + (bc − ad)2
≥ (ac + bd)2
Q.E.D. (quod erat demonstrandum)R. Klages (QMUL) MTH4100 Calculus 1 Week 1 23 / 24
Lecture 3
Reading Assignment
Read
Thomas’ Calculus, Chapter 1.2:
Lines, Circles, and Parabolas
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