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Bounded Functions

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Kernel: SageMath (stable)

Algebra Revisited 2018

Modern Algebra: A Logical Approach, Book Two,

circa 1966 [Allen, Pearson]

Chapter 6: Functions and Other Relations

Bounded Functions

p.325 (exercises 3, 5, 6) p.326 (exercise 8 proof)

(3)

M = {(x,y)y=x    x[0,)(x, y)\mid y = \sqrt{x}\; \land\; x\in[0, \infty)}

%display typeset x = var('x') f(x) = sqrt(x) domain = (x, 0, 1000) p = plot(f(x), domain) p.show(xmin=-10, xmax=50, ymin=-2, ymax=10)
Image in a Jupyter notebook

Since x0,  f(x)0x \ge 0,\; f(x) \ge 0 Thus M is bounded below. However, M is not bounded above.

limit(f(x), x=0, dir='+')
0\renewcommand{\Bold}[1]{\mathbf{#1}}0
limit(f(x), x=oo)
+\renewcommand{\Bold}[1]{\mathbf{#1}}+\infty

limx0+x=0\lim\limits_{x \to 0^+} \sqrt{x} = 0 limxx=\lim\limits_{x \to \infty} \sqrt{x} = \infty

(5)

S = {(x,y)y=1x4    x>4(x, y)\mid y = \frac{1}{x-4}\; \land\; x\gt 4}

%display typeset x = var('x') f(x) = 1/(x - 4) domain = (x, 4, 1000) p = plot(f(x), domain) p.show(xmin=0, xmax=10, ymin=-1, ymax=10)
Image in a Jupyter notebook

Since x>4,  f(x)>0x \gt 4,\; f(x) \gt 0 Thus S is bounded below. However, S is not bounded above.

limit(f(x), x=4, dir='+')
+\renewcommand{\Bold}[1]{\mathbf{#1}}+\infty
limit(f(x), x=oo)
0\renewcommand{\Bold}[1]{\mathbf{#1}}0

limx4+1x4=+\lim\limits_{x \to 4^+} \frac{1}{x-4} = +\infty limx1x4=0\lim\limits_{x \to \infty} \frac{1}{x-4} = 0

(6)

T = {(x,y)y=1x2+2(x, y)\mid y = \frac{1}{x^2+2}}

x = var('x') f(x) = 1/(x^2+2) p = plot(f(x), (x, -10, 10)) p.show(xmin=-10, xmax=10, ymin=0, ymax=1)
Image in a Jupyter notebook

Since xRx20x \in \mathbb{R} \land x^2 \ge 0, the Absolute Maximum is f(0)=12f(0) = \frac{1}{2} limx1x2+2=0\lim\limits_{x \to \infty} \frac{1}{x^2+2} = 0 limx1x2+2=0\lim\limits_{x \to -\infty} \frac{1}{x^2+2} = 0

(8)

Given that F = {(x,y)y=2xx2+1(x,y) \mid y = \frac{2x}{x^2+1}}, show that 1 is the absolute maximum of F.

We will need to use Theorem 8-3 and it's corollaries.

If a,bRa,b \in \mathbb{R}, then (ab>0ab \gt 0 \longleftrightarrow a and b have like signs) and (ab<0ab \lt 0 \longleftrightarrow a and b have opposite signs).

Corollary 1: (1) If aRa \in \mathbb{R}, then a20a^2 \ge 0 (2) If aRa0a \in \mathbb{R} \land a \ne 0, then a2>0a^2 \gt 0

Proof: If a = 0, then 00 = 0 If a 0\ne 0, then the two factors of aa are the same and have the same sign. It follows from Theorem 8-3 that a2>0a^2 \gt 0.

Corollary 2: 1 > 0 and -1 < 0 Proof that 1 > 0 follows from the fact that 1 = 1*1. Then the proof that -1 < 0 follows from the Corollary to Theorem 5-3. Theorem 5-3: If a,bRa, b \in \mathbb{R}, then a>bab<0a \gt b \longleftrightarrow a - b \lt 0

Corollary to Theorem 5-3: A real number is negative if and only if its opposite is positive, and a real number is negative if and only if its opposite is positive. $$

Corollary 3: If aRa \in \mathbb{R} and a0a \ne 0, then a and 1a\frac{1}{a} have like signs. $$

To prove that 1 is the absolute maximum of F = {(x,y)y=2xx2+1(x,y) \mid y = \frac{2x}{x^2 + 1}}, we need to prove (i) 2xx2+1=1\frac{2x}{x^2 + 1} = 1 is true when x=1x = 1, and (ii) 2xx2+11\frac{2x}{x^2 + 1} \le 1 is true.

We know that 2xx2+11\frac{2x}{x^2 + 1} \le 1 if 2xx2+12x \le x^2 + 1, and that 2xx2+12x \le x^2 + 1 is true if 0x22x+1=(x1)20 \le x^2 - 2x + 1 = (x-1)^2. This suggests a proof of the statement, "If xRx \in \mathbb{R}, then 2xx2+11\frac{2x}{x^2+1} \le 1"

(1)   \; xR(2)  (x1)R(3)  (x1)20(4)  x22x+10(5)  x2+12x(6)  12xx2+1x \in \mathbb{R} \longrightarrow (2)\; (x-1) \in \mathbb{R} \longrightarrow (3)\; (x-1)^2 \ge 0 \longrightarrow (4)\; x^2 - 2x + 1 \ge 0 \longrightarrow (5)\; x^2 + 1 \ge 2x \longrightarrow (6)\; 1 \ge \frac{2x}{x^2+1}. ParseError: KaTeX parse error: Undefined control sequence: \$ at position 1: \̲$̲ REASONS: (…;Given,(2) Given, (2);LaTeX\LaTeX\mathbb{R}isclosedunderaddition,(3) is closed under addition, (3);Theorem83Corollary1,(4) Theorem 8-3 Corollary 1, (4);Distributiveproperty,(5) Distributive property, (5);Additionpropertyofinequality,(6) Addition property of inequality, (6);$ Multiplication property of inequality.

NOTE: In step 6 we assumed that 1x2+1\frac{1}{x^2+1} was positive. This can be proved as follows: ParseError: KaTeX parse error: Undefined control sequence: \$ at position 1: \̲$̲ (1);LaTeX\LaTeXx \in \mathbb{R} \longrightarrow (2); x^2 \ge 0$, (3)  \; 1>0(4)  x2+1>0(5)  1x2+1>01 \gt 0 \longrightarrow (4)\; x^2 + 1 \gt 0 \longrightarrow (5)\; \frac{1}{x^2+1} \gt 0.

REASONS: (1)  \; Given, (2)  \; Theorem 8-3 Corollary 1, (3)  \; Theorem 8-3 Corollary 2, (4)  \; Addition propert of inequality, (5)  \; Theorem 8-3 Corollary 3.

x = var('x') f(x) = 2*x/(x^2+1) p = plot(f(x), (x, -10, 10)) p.show(xmin=-10, xmax=10, ymin=-2, ymax=2)
Image in a Jupyter notebook