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# University of Illinois at Urbana Champaign Fall Description

Niveau: Supérieur, Doctorat, Bac+8
University of Illinois at Urbana-Champaign Fall 2006 Math 444 Group E13 Graded Homework X. Due Friday, November 17. 1.(a) Deﬁne a function f on R by setting f(x) = { f(x) = x sin( 1x ) if x 6= 0 0 else . Is this function continuous on R ? (you may use without proof the fact the the function x 7? sin(x) is continuous). (b) Let g : R? R be deﬁned by g(x) = { x if x ? Q 1? x else . At which points in R is g continuous ? Correction. (a) Since we are told that x 7? sin(x) is continuous on R, we see immediately that f is continuous on (?∞, 0)?(0,+∞) : indeed, on these two intervals f is a composition of continuous functions. Thus, our only problem is at x = 0 ; we need to see whether f has a limit at 0, and whether that limit is equal to f(0) = 0. As usual with sin, the important thing to remember is that | sin(x)| ≤ 1 for all x ? R ; thus, for all x 6= 0 we see that |f(x)| ≤ |x|.

• urbana-champaign fall

• must then

• squeeze theorem

• continuous

• functions

• there exists

Subjects

##### Existential quantification

Informations

(
1f(x) = xsin( ) x = 0
x1. f R f(x) =
0
R x7→ sin(x)(
x x∈Q
g:R→R g(x) = R g
1−x
x7→ sin(x) R f
(−∞,0)∪(0,+∞) f
x = 0 f 0 f(0) = 0
sin |sin(x)|≤ 1 x∈R x = 0
|f(x)|≤|x| lim f(x) = 0
x→0
lim f(x) = f(0) f 0 f
x→0
x∈R (q ) (α )n n
x = lim(q ) = lim(α ) g x g(x) = lim(f(q )) = lim(q ) = xn n n n
g(x) = lim(f(α )) = lim(1−α ) = 1−x g x x = 1−xn n
1
x =
2
1
g
2
ε,δ ε > 0 x
1 1
|x− | =|g(x)− | x
2 2
1 1 1 1
0 <|x− |≤ ε |g(x)− |≤ ε lim g(x) = = f( ) f
2 2 x→1/2 2 2
1
2
0 0 02. f: [a,b]→R a < b |f(x)−f(x )| <|x−x| x = x ∈ [a,b]
ε,δ f [a,b]
x ∈ [a,b] f(x) = x
f
[a,b] x∈ [a,b] ε > 0 y ∈ [a,b]
|x−y|≤ ε |f(x)−f(y)| <|x−y|≤|x−y| δ = ε
δ = ε f
f x x∈ I f I
g(x) = f(x)−x g(a) = f(a)−a≥ 0 a≤ g(x)≤ b x∈ [a,b]
g(b) = f(b)−b≤ 0 g(a) = 0 g(b) = 0 g a
b x∈ [a,b] g(x) = 0
f [a,b]
x ∈ [a,b] f(c) = x g(x) = 0 f(x) = x
0x = x ∈ [a,b] f(x) = x
0 0 0 0f(x ) = x f |f(x)−f(x )| < |x−x|
3. f: [0,1]→ [0,1] f(0) = f(1)
1 1
n∈N x∈ [0,1− ] f(x) = f(x+ )
n n
f((k +1)/n)−f(k/n) k = 0,...,n−1
n≥ 2 n = 1 f(0) = f(1)
1 1
f((k +1)/n) = f(k/n) k = 0,1,...,n−1 x∈ [0,1− ] f(x) = f(x+ )
n n
k/n f((k +1)/n) = f(k/n) k = 1,...,n−1
k∈ 0,...n−2 f((k +1))/n−f(k/n) f((k +2)/n)−f(k +1/n)
f(1/n) > f(0) f(2/n) > f(1/n)
f(3/n) > f(2/n) f(1) = f(n/n) > f((n− 1)/n) > ...f(0)
f(1) = f(0) f(1/n) < f(0)
k f((k +1))/n−f(k/n) f((k +2)/n)−f(k +1/n)
g(x) = f(x+1/n)−f(x) g [k/n,(k +1)/n]
k = 0,...,n− 2 g 0 0 g
x f(x) = f(x + 1/n) f
x∈ [0,1−1/n] f(x) = f(x+1/n)
4. f [a,b] f(x) > 0
x∈ [a,b] m > 0 f(x)≥ m x∈ [a,b]
f,g [0,1]→ [0,1] f(x) < g(x) x∈ [0,1]
m > 0 f(x)+m < g(x) x∈ [0,1]
M > 1 Mf(x) < g(x) x∈ [0,1]
f [a,b]
[a,b] c ∈ [a,b] f(c) ≤ f(x) x ∈ [a,b]
m = f(c) m > 0 f f(x)≥ m x∈ [a,b]
0g−f [0,1] m > 0
0m0g(x)−f(x)≥ m x∈ [0,1] m = m > 0 g(x)−f(x) > m
2
0x∈ [0,1] f(x)+m < g(x) x∈ [0,1] m
2
0 ≤ f(x) < g(x) x ∈ [0,1] g
f 0h = [0,1] h(x) < 1 x ∈ [0,1] m > 0
g
00≤ h(x)≤ 1−m x∈ [0,1] a x7→ 1−h(x) h
0m 0[0,1] m = h(x) < 1−m
2
100 0m 0 < m < x∈ [0,1] h 1−m > 0 g(x) > 0
2
1 1
f(x) < g(x) x∈ [0,1] M = M > 1
0 01−m 1−m
00 < m < 1 Mf(x) < g(x) x∈ [0,1]
x
f
5. f R R f(x+y) = f(x)+f(y) x,y∈R
f(q) = qf(1) q∈Q f(x) = xf(1)
x∈R
f(q) = qf(1) q x
(q ) lim(q ) = x fn n