University of Illinois at Urbana Champaign Fall

Description

University of Illinois at Urbana-Champaign Fall 2006 Math 444 Group E13 Training exercises : Correction. 5.2.7 : We more or less saw that example in class : define a function f : [0, 1] ? [0, 1] by setting, for all x ? [0, 1], f(x) = { ?1 if x ? Q 1 else . Pick a number x ? [0, 1], and recall that there exist a sequence (qn) of rational numbers in [0, 1 and a sequence (?n) of irrational numbers in [0, 1] such that lim(qn) = lim(?n) = x. One also has f(qn) = 0 for all n ? N, and f(?n) = 1 for all n inN ; this proves that f doesn't have a limit at x, so it cannot be continuous at that point. Thus f is discontinuous at every point of [0, 1], yet |f | is constant (equal to 1), so it is a continuous function. 5.2.8 : Recall again that any real number is the limit of a sequence of rational numbers ; pick x ? R, and a sequence of rational numbers (qn) that converges to x. Then one has lim f(qn) = f(x) since f is continuous at x, and for the same reason lim g(qn) = g(x).

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Sujets

• Existential quantification

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 10
• Langue: English

Extrait

f: [0,1] → [0,1](
−1 x∈Q
x ∈ [0,1] f(x) = x ∈ [0,1] (q )n
1 .
[0,1 (α ) [0,1] lim(q ) = lim(α ) = xn n n
f(q ) = 0 n∈N f(α ) = 1 n inN fn n
x f [0,1] |f|
1
x∈R
(q ) x limf(q ) =f(x) fn n
x limg(q ) = g(x) f(q ) = g(q ) f(x) = g(x)n n n
f,g
m
x∈R ε> 0 m∈Z n∈N x−ε≤ ≤x
n2
1 m 0n∈N ≤ε m∈Z <x A ms
n n2 2
m
m 1
m m∈ A m+16∈ A + ≥x
n n2 2
m 1
≥x−
n n2 2
m
x−ε≤ ≤x .
n2
m
S ={ : n∈N, m∈Z} R f
n2
f(s) = 0 s∈ S f
f(x) = 0 x∈R
g(x) = 0 x∈R g(y) = 0 y ∈R
g(y) =g(y−x+x) =g(y−x)g(x) = 0 .
g(x) = 0 x∈R g(0)g(x) = g(x) x inR
g(0) = 1
g 0 x∈R (x ) xn
limg(x ) = g(x) g 0n
0
(y ) = (x −x)n n
g(x ) =g(x −x+x) =g(x −x)g(x) =g(y )g(x).n n n n
limg(y ) = g(0) = 1 g 0 limg(x ) = g(x)n n
(x ) g x x ∈ R gn
R
g g(0) = 1 g
h(x) = ln(g(x)) h(x +y) = h(x) +h(y) x ∈ R
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√ √ √ √
f(1) = 2ln(1)+ 1−2 =−1 f(2) = 2ln(2)+ 2−2 = ln(4)+ 2−2≥ 1+ 2−2> 0
c ∈ [1,2] f(c) = 0
f(3/2) 0
[1,3/2] 5/4
f(5/4)≤ 0 [5/4,3/2]
11/8 f(11/8)≤ 0 [11/8,3/2]
−210
[189/128,190/128]
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√
K x ≤ Kx x ∈ [0,1]
K ≥ 1
1 1 K
x = 1 x = ≤
4 2 4K K K
K ≤ 1 K = 1r √
1 2 1
= ≤ K
2 2 2
g [0,1]
K x
x x
f R f(x+p) =f(x) p> 0 x∈R
f(x) = f(x +Kp) x∈R K ∈Z x∈R
K ∈Z x+Kp∈ [0,p]
x
E( ) f [0,p] m,M ∈R
p
m≤ f(y)≤ M y∈ [0,p] K m≤ f(x+Kp)≤ M
m≤f(x)≤M f R
f f
p
[0,2p] [0,p]
ε> 0 δ x,y∈ [0,2p] |x−y|≤δ⇒|f(x)−f(y)|≤
ε x,y ∈ R δ ≤ p
x,y∈R |x−y|≤δ K∈Z x+Kp y+Kp [0,2p]
0 0δ = min(δ,p) x,y∈R |x−y|≤δ K∈Z
0x+Kp y +Kp [0,2p] |(x+Kp)−(y +Kp)| =|x−y|≤ δ ≤ δ
|f(x+Kp)−f(y+Kp)≤ε| f p K∈Z f(x+Kp) =f(x) f(y+Kp) =f(y)
0x,y∈R |x−y|≤δ ⇒|f(x)−f(y)|≤ε f
R
[0,2p] [0,p]
f(x)−f(0) −2/3 −2/3=x x 0
x−0
f(x)−f(0)
x = 0 f 0
x−0
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−1 f(x) = f(−x)
0 0 0f (x) = (−1)f (−x) =−f (−x)
f(x) =−f(−x) f
0sin (x) = cos(x) x ∈ R x < y ∈ R
f(x)−f(y) = cos(c)(x−y) c∈ (x,y) |cos(c)|≤ 1 x∈R |f(x)−f(y)|≤|x−y|
x,y x < y x = y x,y
x>y |sin(x)−sin(y)|≤|x−y| x,y∈R
f(x)−f(a)0f (a) f
x−a
f(x)−f(a) 0 0=f (c) c∈ (a,x) lim f (x) =A
x→ax−a
0ε> 0 δ a<x≤a+δ⇒|f (x)−A|≤ε
f(x)−f(a)
x≤a+δ c≤a+δ a<x≤a+δ⇒ −A ≤ε
x−a
f(x)−f(a) 0lim =A f (a) A
x−a
√
f(x) = x f [0,1]
10(0,1) f (x) = √ (0,1)
2 x
0|f | (a,b) f
[a,b] f
[a,b] [a,b]
(a,b)
0x < y ∈ I f(y)−f(x) = f (c)(x−y) c ∈ (x,y)
f [x,y] f
0[x,y] [x,y] (x,y) f
f(y)−f(x)> 0 y >x f
0 0 0 0x,x ∈I f (x)≤ 0 f (x)≥ 0 f
0 0x x f 0 I
0 0f (x) > 0 x∈ I f (x) < 0
x∈I
f
f(x)> 0 f(x)> 0
0 0f (x) = 0 f
3f(x) =x
0 0g (x)−f (x)≥ 0 x
g−f g−f (g−f)(0) = g(0)−f(0) = 0
(g−f)(x) =g(x)−f(x)≥ 0 x≥ 0 (g−f)(x) =g(x)−f(x)≤ 0 x≤ 0
g(x)≥f(x) x≥ 0 g(x)≤f(x) x≤ 0
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