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Ah Mate

In: Computers and Technology

Submitted By lifin
Words 2948
Pages 12
Analysis II Solutions: June 2010
April 2013
Question 1. a) (i) (ii) The sequence an converges to a if for any
1 1

> 0 there exists an N such that |xn − a| <

for n ≥ N

(3n + 22n ) n = (3n + 4n ) n → max {3, 4} = 4 by a standard result.
4n2 2n +2n3 4n 8n4 2n +n3 4n

=
1

4 ( n )( 1 )n +2 2 8n( 1 )n +1 2 1 4

→ 2 by the product, sum and quotient rule.
1 1 4 1 1 1 1 1 1

5 5 (iii) (5n3 +6n4 ) n = 6 n n n ( 6n +1) n and limn→∞ 6 n n n = limn→∞ 6 n (n n ×n n ×n n ×n n ) = 1 and limn→∞ ( 6n +1) n = 1 1 4 1 7 and hence limn→∞ 5 n n n ( 5n + 1) n = 1 by the product rule.

b)

A sequence an eventually has some property if ∃N such that aN +n has this property ∀n > 0. b Assume bn → b and choose = 2 then ∃N such that for n > N , |bn − b| < b 2

then

3b b < bn < 2 2
2 b2

b so eventually bn > 2 . If b > 0 then eventually

>

1 bbn

so 2 |bn − b| 1 1 |bn − b| > = | − | > 0. b2 |bbn | bn b

Hence by the sandwich theorem an a bn → b . c)

1 bn



1 b

and then if we let an → a then by the product rule an ×

1 bn

→a×

1 b

and hence

n+1 Ratio Lemma: If | aan | ≤ l where 0 ≤ l < 1 then an → 0. n+1 Want to prove that if | aan | ≤ l where 0 ≤ l < 1 then 0 < an+1 < ln a1 We will use proof by induction. First for n = 1, |a2 | ≤ l|a1 | which is true by the assumption on the sequence. Now assume that 0 ≤ |an+1 | ≤ ln |a1 |. Then for the induction step since |an+2 | < l|an+1 | then |an+2 | ≤ ln+1 |an+1 |. So since 0 ≤ l < 1, ln → 0 and then by the sandwich rule an → 0

d)

By definition lim sup an = lim sup am . n→∞ n→∞ m≥n

Fix

> 0 then as limn→∞ bn = b > 0 then ∃N s.t. for n ≥ N |bn − b| < .

So for n ≥ N sup am bm ≥ sup am (b − ) = (b − ) sup am . m≥n m≥n m≥n

Taking the limit as n → ∞ lim sup an bn ≥ (b − ) lim sup an . n→∞ n→∞

Since this is true ∀ > 0 we have that lim sup an bn ≥ b lim sup an . n→∞ n→∞

Likewise we can show that lim supn→∞ an bn ≤ b lim supn→∞ an . Hence lim sup an bn = b lim sup an . n→∞ n→∞

Question 2. 1

a)

A function f : E → R is continuous at c if for any > 0 there exists a δ > 0 such that if |x − c| < δ and x ∈ E then |f (x) − f (c)| < .

f is sequentially continuous at c if for every sequence xn ∈ E if xn → c then f (xn ) → f (c). The two are equivalent. b) By the triangle inequality, |f (x)g(x) − f (c)g(c)| = |f (x)g(x) − f (c)g(x) + f (c)g(x) − f (c)g(c)| ≤ |f (x)g(x) − f (c)g(x)| + |f (c)g(x) − f (c)g(c)| ≤ |g(x)||f (x) − f (c)| + |f (c)||g(x) − g(c)|. Let > 0 choose δ1 > 0 s.t. |g(x) − g(c)| < 2f (c) for |x − c| < δ1 . This δ1 exists since g is continuous. Now we choose δ2 > 0 such that if |x − c| < δ2 then |g(x) − g(c)| < 1 ⇒ |g(x)| < |g(c)| + 1. Since 2 (|g(c)| + 1) > 0 we can choose δ3 > 0 such that if |x − c| < δ3 then |f (x) − f (c)| < 2 (|g(c)| + 1). Finally choose δ = min{δ1 , δ2 , δ3 }, then if |x − c| < δ then |f (x)g(x) − f (c)g(c)| < and so f g is continuous. f (x) is continuous only at 0, π, 2π. Let c = 0, π or 2π and fix > 0. Since sin x is cts ∃δ > 0 s.t. for |x − c| < δ | sin x − sin c| = | sin x| < . So since |f (x) − f (c)| ≤ | sin(x)| < f is cts at c. Now let c ∈ Q \ {π, 2π} and let xn → c with xn ∈ Q. / Then f (xn ) = 0 = f (c) so f is discontinuous at c ∈ Q. / If c ∈ Q \ {0}, then let xn → c with xn ∈ Q so / f (xn ) = sin xn → sin c = 0 = f (c) so f is not cts at c. d) Let h(x) = sin x 0 for x = 0 for x = 0. c)

Then g(x) = f (x)h(x). 1 We claim that g is cts at c ∈ { kπ : k ∈ N} ∪ {0, π, 2π}. Pick c ∈ {0, π, 2π}, then note that g(c) = 0 and |g(x)| ≤ |f (x)| → 0 So f is cts at c. 1 Similarly pick c ∈ { kπ : k ∈ N} then g(c) = 0 and |g(x)| ≤ |h(x)| → 0 as x → c. as x → c.

So f is cts at c. Both proofs use the fact that f and h are bounded by 1. If c ∈ { kπ : k ∈ N} ∪ {0, π, 2π} then we can find two sequences s.t. xn ∈ Q with xn → c so / 1 g(xn ) = 0 → 0 and yn ∈ Q with yn → c with / g(yn ) = sin yn sin e) (i) False. f (x) = x + 2π x if x ≥ 0. x < 0. 1 1 → sin c sin = 0. yn c

f (x) is clearly discontinuous, however sin(x) = sin(x + 2π) so sin(f (x)) is continuous. (ii) False K(x) = 1, if x ∈ Q. −1, x ∈ Q. f (x) = 1 K is clearly bounded but Kf is not continuous. 2

(iii) True. Let h(x) = f (x) − g(x). Then h(x) = 0 for all irrational x and h is continuous. Let xn ∈ Q such that xn converges to c ∈ Q. Since h is continuous h(xn ) → h(c) and since h(xn ) = 0 for all n, h(c) = 0 and hence h(x) = 0 for all x. Hence f (x) = g(x) for all x. (iv) False 1 Let f (x) = x and xn = Question 3. a) Assume there exists a φ(x) such that f (x) = f (c) + φ(x)(x − c) then f (c) = lim f (x) − f (c) f (c) + φ(x)(x − c) − f (c) = lim = φ(x). x→c x−c x−c
1 n

→ 0. Then limn→∞ f (xn ) = ∞ however f (x) =

1 x

is not continuous at 0.

x→c

Hence f is differentiable at c and its derivative is equal to φ(c). Now assume that f is differentiable at a point c and let φ(x) = f (x)−f (c) , x−c

f (c),

if x = c. x = c.

Then f (x) = f (c) + φ(x)(x − c) and since f is differentiable at c, φ(x) is continuous at c.
1 b) Just need to find a φ(x) such that x2 sin( x ) = 0 + xφ(x) and that φ(x) is continuous at 0. By inspection φ(x) = 1 x sin( x ) which is defined as

φ(x) =

1 x sin( x ), if x = 0. 0, x = 0.

(1)

φ(x) is continuous at 0 hence f (x) is differentiable at 0. c) d) Let limx→∞ g(x) = l for l ∈ R if for all > 0 there exists an X such that for all x > X then |g(x) − l| < . limx→∞ g(x) = ∞ if for all M > 0 there exists an X such that for all x > X then g(x) > M . (i) True. > 0 ∃M such that for all x > M , | f (x) f (x) − 1| < ⇔ 1 − < < 1 + ⇔ x(1 − ) < f (x) < x(1 + ). x x

Then as x → ∞, f (x) → ∞ by the sandwich theorem. (ii) False Let f (x) = x + 1 then limx→∞ x+1 x

= 1 however limx→∞ (f (x) − x) = limx→∞ (x + 1 − x) = 1 = 0 > 0 there exists an M > 0 such that for all x > M , |f (x) − x| < ⇔ − < f (x) − x < .

(iii) True If limx→∞ (f (x) − x) = 0 then for all

Hence

x− x

<

f (x) x

<

x+ x

so by the sandwich theorem limx→∞ f (x) = 1. x then x − < f (x) < x + . So

(iv) True. Let > 0 then there exists an M > 0 such that for all x > M , |f (x) − x| < (x − )2 < f (x)2 < (x + )2 , =⇒ −2x + 2 < f (x)2 − x2 < 2 x + 2 . So by the sandwich theorem f (x) − x2 → 0. Question 4.

a) Rolle’s Theorem: If f : [a, b] → R is continuous and f (a) = f (b) then there exists a point c ∈ (a, b) such that f (c) = 0. b) Mean Value Theorem: If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b) then there exists a point c ∈ (a, b) such that f (c) = f (b)−f (a) b−a Proof. Let f : [a, b] → R be continuous. Then let g(x) = f (x) − f (c) = f (b)−f (a) b−a f (b)−f (a) (x b−a

− a) so g(a) = g(b) and hence we can apply f (b)−f (a) b−a

Rolle’s Theorem. So there exists a c ∈ (a, b) such that g (c) = 0 and therefore g (c) = f (c) − = 0. 3

= 0 and hence

c) If |f (x)| ≤ 1 for all x and f is differentiable for all f then for any region [a, b] ⊂ R there is a c ∈ [a, b] such that 2 1 f (c) = f (b)−f (a) by the mean value theorem. Since |f (x)| ≤ 2 for all x then | f (b)−f (a) | ≤ 1 so b−a b−a 2 |f (b) − f (a)| ≤ 1 |b − a|. 2

Since our choice of a and b were arbitrary this holds for any a and b. d) Intermediate Value theorem: If f : [a, b] → R is continuous and assume f (a) < f (b) then ∀v ∈ (f (a), f (b)) there exists a c ∈ (a, b) such that f (c) = v. e) Let a ∈ [−1, 1], choose c s.t. sin(c) = a. We claim that g is cts at c. Note that g(c) = a. Let > 0, the since sin(x) is cts ∃δ > 0 s.t. |x − c| < δ So if |x − c| < δ then |g(x) − g(c)| ≤ max {|a − a|, | sin(x) − sin(c)|} = | sin(x) − sin(c)| < . So g is cts at c. f) Let h(x) = f (x) − x then h(x) is continuous. h(a) = f (a) − a ≥ a − a = 0 and h(b) = f (b) − b ≤ b − b = 0. Then if h(a) = 0, a is our wanted point. If h(b) = 0, b is our wanted point. Otherwise we have h(a) > 0 and h(b) < 0 then by the intermediate value theorem there is a point c ∈ [a, b] such that h(c) = 0 which means f (c) = c. Question 5. a) Choose = f (c) > 0 then since f is continuous at c there exists a δ > 0 such that if |x − c| < δ then |f (x) − f (c)| < 2 (c) ⇔ 0 < f (c) < f (x) < 3f2 so if x ∈ (c − δ, c + δ) then f (x) > 0. 2 b) If f (c) > 0 then by part a) ∃ a δ > 0 s.t. f (x) > 0 ∀x ∈ (c − δ, c + δ). Now let x, y ∈ (c − δ, c + δ) with x < y then by the mean value theorem ∃c ∈ (x, y) ⊂ (c − δ, c + δ) s.t. f (y) − f (x) = f (c) > 0. y−x Which implies f (y) > f (x) so f is increasing in (c − δ, c + δ). c) Extreme Value Theorem: If f : [a, b] → is continuous then there exists m, M ∈ R such that m ≤ f (x) ≤ M for all x ∈ [a, b]. Furthermore there exists x, x ∈ [a, b] such that f (x) ≤ f (x) ≤ f (¯) for all x ∈ [a, b]. ¯ x d) f (x) = limx→0 f (x) = ∞ and limx→1 f (x) = −∞. e) f (x) = arctan(x) f is bounded by f) (i) −π, π 2 2 however never attains these values.
1 x, 1 x−1 ,

=⇒

| sin(x) − sin(c)| < .

if 0 < x ≤ 0.5. 0.5 < x < 1.

Since f (x) → 0 as x → ±∞ there exists X s.t. |f (x)| < f (0) 2 for |x| > X.

Now consider f on [−X, X]. f is bounded on [−X, X] by EVT so let |f (x)| ≤ Aon[−X, X] and note that A ≥ f (0). Since |f (x)| < f (0) ∀|x| > X then |f (x)| ≤ A ∀x ∈ R. 2 (ii) By above f achieves its maximum on [−X, X]. sup f (x) = x∈R sup x∈[−X,X] f (x) so by EVT ∃x0 ∈ [−X, X] s.t. f (x0 ) = A.

Question 6. 4

a) Taylor’s theorem: Let x > x0 . Suppose that f is n times continuously differentiable on [x0 , x] and n + 1 times differentiable on (x0 , x). Then n f (k) (x0 ) (x − x0 )k + Rn (x) f (x) = k! k=0 for some ξ ∈ (x0 , x). where Rn (x) = This also holds for x < x0 , if f is n times continuously differentiable on [x, x0 ] and n + 1 times differentiable on (x, x0 ). b) n (x−x0 )n+1 (n+1) (ξ) (n+1)! f

sin(1) = k=0 (−1)n + Rn (1) (2n + 1)! sin(ξ) (n+1)!

We want Rn (1) < 0.01 where Rn (1) = we choose n = 4.

sin(ξ) (n+1)!) .

We know sin(ξ) ≤ 1∀ξ so
4



1 (n+1)!

then we want

1 (n+1)!

<

1 100

so

sin(1) = k=0 (−1)n (2n + 1)!

to 0.01 precision. So sin(1) ≈

263 315

1 1 1 1 1 c) (i) f ( m = f ( n + n + · · · + n ) m times. This is equal to f ( n ) × · · · × f ( n ) m times by the property of the function n 1 m 1 1 1 which is equal to f ( n )m . Now f (1) = f ( n )n hence f ( n ) = f (1) n and so f ( m ) = f (1) n . n f (0 + y) = f (0)f (y) then f (y) = f (0)f (y) and hence f (0) = 1

(ii)

By completeness and continuity f (x) = (f (1))x ∀x ∈ R. So f (x) = (f (1))x log(f (1)).

As 1 = f (0) = log(f (1)) then f (x) = (f (1))x = f (x). By induction f (n) (x) = f (x) for all n. (iii) n f (x) = i=0 xn+1 f (0) i x + f (ξ) i! (n + 1)!

for ξ ∈ (0, x)

Now |Rn (x)| = | If we consider an =
|x|n x+1 , n! |f (1)|

xn+1 |x|n+1 (f (1))ξ | ≤ |f (1)|x+1 . (n + 1)! (n + 1)!

|

an+1 |x|n+1 n! 1 |x| |= |f (1)|x+1 n = →0 x+1 an (n + 1)! |x| |f (1)| n+1

then by the ratio test an → 0 and so |Rn (x) → 0 as n → ∞. d) Let f (x) = First assume that f (n) (0) = 0 ∀n then


e − x2 0

1

x=0 x = 0.

i=0

f (i) (0) i x = 0. i!

So f (x) = unless x = 0. Now we show that f (n) (0) = 0. e− x2 y lim = lim y2 = 0 + y→∞ e x x→o
1 where y = x . Also
1

f (i) (0) i i! x

x→0

lim−

e − x2 y = lim y2 = 0 y→∞ e x

1

hence f (0) exists and f (0) = 0. 1 P (x) Since f (n) (x) = Q(x) e− x2 for polynomials P, Q then we have that f (n) (0) = 0 ∀n. 5

Question 7. a) If the radius of convergence is R then a power series n=0 an xn converges for |x| < R. If R = 0 then the power series only converges for x = 0 and if R = ∞ then the power series converges for any x. b) Let lim sup |an | n = r. Then the radius of convergence is n→∞ r = ∞ then the radius of convergence is 0. c) R= d) (ii)

1



1 r.

If r = 0 then the radius of converges is ∞ and if

1 n+1 limn→∞ | aan |

(i)

By the ratio test R = ∞. 3n


x2n = an xn 2n+1 n−0 n=0 where a2n =
3n 2n+1

and a2n+1 = 0. So lim sup |an | n = lim | n→∞ n→∞ 2 3
1

3 (2n + 1)
1 n

= lim

n→∞

3 2(1 +

1 2n )

=

3 . 2

So by Hadamards test the radius of convergence is (iii) e)
1

lim sup |an | n = 4. So the radius of convergence is 1 . 4 n→∞ Consider lim | an+1 (n + 1)xn+1 (n + 1)an+1 xn+1 n + 1 an+1 an+1 an+1 xn+1 | = lim | | = lim | |. x| = lim | x| = lim | n→∞ n→∞ n→∞ n→∞ an nxn nan xn n an an an xn

n→∞

As lim f) and

n+1 = 1 then by the ratio test they have the same order of convergence. n→∞ n
∞ n=0

an xn converges so

∞ n=2

an xn converges so x−1
∞ ∞

∞ n=2

an xn−1 converges. Therefore as x → 0.

∞ n=2

an xn−1 converges

an xn−1 = x n=2 n=0

an+2 xn → 0

So

f (x) − a0 − a1 x →0 x lim

as x → 0.

Therefore x→0 f (x) − a0 = a1 . x

Since f (0) = a1 then f (0) = a1 .

6

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