University of Illinois at Urbana Champaign Fall

Description

University of Illinois at Urbana-Champaign Fall 2006 Math 444 Group E13 Graded Homework XI. Correction. 1. (a) Give an example of a function f : R? R which is not constant and satisfies f(x) = f(x2) for all x ? R. (b) Assume now that f is continuous at 0 and 1 and f(x) = f(x2) for all x ? R. Show that f must be constant. Hint : assume that |x| < 1 ; then what is the limit of the sequence (xn) defined by x1 = x, x2 = x2, . . . , xn+1 = x2n ? How about the sequence (f(xn)) ? Can you use a similar idea when |x| > 1 ? Correction. (a) There are many possible examples, one of them being the function f defined by f(0) = 0, and f(x) = 1 for all x ? R. (b) Following the indication, pick first x ? R such that |x| < 1. Then define a sequence (xn) by setting x1 = x, x2 = x2, . . . xn+1 = x2n, . . .. Then this sequences converges to 0, and one has f(xn+1) = f(x 2 n) = f(xn) for all n ? N ; thus an easy induction proof yields f(xn) = f(x) for all n

• between any


• real roots


• induction proof yields


• then


• bounded function


• di?erentiable both


• thus


• pick now

Sujets

• Bounded function

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 9
• Langue: English

Extrait

21. f:R→R f(x) =f(x ) x∈R
2f 0 1 f(x) =f(x ) x∈R f
2|x|< 1 (x ) x =x x =x ,...,x =n 1 2 n+1
2x (f(x )) |x|> 1nn
f f(0) = 0
f(x) = 1 x∈R
x∈R |x|< 1 (x ) x =xn 1
2 2 2x = x ,...x = x ,... 0 f(x ) = f(x ) = f(x )2 n+1 n+1 nn n
n∈N f(x ) = f(x) n∈N (x ) 0 fn n
0 limf(x ) =f(0) f(x) =f(0) x∈ (−1,1)n
f 1 f(1) = f(0) x∈R x > 1
n√ √ 1/2
(y ) y =x,y = x,...y = y ,... y =yn 1 2 n+1 n n 1
lim(y ) = 1 lim(f(y )) = f(1) = f(0) (y )n n n
2f(y ) =f(y ) =f(y ) n∈N f(y ) =f(y ) =f(x) n∈Nn n+1 n 1n+1
f(x) =f(0) x> 1 f f(x) =f(−x)
2 2f(x) =f(x ) =f((−x) ) f(x) =f(0) x<−1
f(x) =f(0) x∈R f R
2f(x) =f(x )
(x ) (y ) 0 1 f(x ) f(y )n n n n
2. f: [0,1]→ [0,1] f◦f =f (∗)
E ={x∈ [0,1]: f(x) =x} .f
Ef
E f([0,1])f
(∗)
E =f([0,1]) x∈E x =f(x) x∈f([0,1])f f
E ⊂ f([0,1]) x ∈ f([0,1] x = f(y) y ∈ [0,1]f
f(x) =f(f(y)) =f(y) =x (∗) f([0,1])⊂Ef
f([0,1]) = E f [0,1] Ef f
[a,b] 0≤a≤b≤ 1 f
• f [0,a] [a,b] f(a) =a
• f [b,1] [a,b] f(b) =b
• x∈ [a,b] f(x) =x
f a ≤ b ∈ [0,1] f (∗)
(
0 x = 0
3. f: R → R f(x) = f12x sin( )
x
0R f 0
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f (−∞,0) (0,+∞)
1 1 10 0 0f (x) =xsin( )−sin( ) f 0 f ( ) = 0
x x 2πn
1 1
f( ) = −1 −1 (x ) (y )n n
2πn+π/2 2πn+π/2
00 f(x ) f(y ) fn n
00 f 0
0f 0
f(x)−f(0) 10f (0) =xsin( )
x−0 x
00 x = 0 f (0) 0
(√
x 0≤x≤ 1
4. a,b∈R f: [0,+∞)→R f(x) =
2ax +bx+1
(0,+∞)
0 1√f [0,1) (1,+∞) f (x) = (0,1)
2 x
0f (x) = 2ax+b (0,1) x = 1 f
lim f(x) =f(1) a +b + 1 = 1 a +b = 0
x→1
f 1 a+b = 0 f
f(x)−f(1)
1 1 a+b = 0 x−1
f(x)−f(1) 10 √1 x< 1 =f (c) = c∈ (x,1)
x−1 2 c
f(x)−f(1) 1
x−1 2
f(x)−f(1) 11 2a+b 2a+b = f
x−1 2
1 1 11 a+b = 0 2a+b = a = b =−2 2 2
f x∈R
x
n5. f(x) =x +ax+b
a,b n
0 0f f f
0f f
0 0 n−1f f f (x) = nx + a
a0 n−1f (x) = 0⇔x =− f(x) = 0
n
a < a < a < a b ∈ (a ,a )1 2 3 4 1 1 2
0 0 0f (b ) = 0 b ∈ (a ,a ) f (b ) = 0 b ∈ (a ,a ) f (b ) = 0 b b b1 2 2 3 2 3 3 4 3 1 2 3
0f (x) = 0
+6. f: R = [0,+∞) → R l ∈ R f l +∞
+lim f(x) =l ε> 0 M ∈R x≥M ⇒|f(x)−l|≤ε
x→+∞
+f f:R →R
++∞ f R
+ +f:R →R f(0) = 1 lim f(x) = 0 f R
x→+∞
+R
0+ +f:R →R R lim f (x) =l l
x→+∞
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lim =l
x→+∞ x
f(x)−f(a) ε> 0 a> 0 x>a −l ≤ε