M2: Analysis II Flashcards
Proposition
Limit Points via Sequences
p ∈ ℝ is a limit point of E ⊆ ℝ iff
∃ sequence (pₙ) pₙ ∈ E\{p}: pₙ → p
Proof
(Limit Points via Sequences)
p ∈ ℝ is a limit point of E ⊆ ℝ iff
∃ sequence (pₙ) pₙ ∈ E\{p}: pₙ → p
2 points
- (⇒) Sandwich with δ = 1/n
- (⇐) Use convergence definition
Proposition
Limits via Sequences
If f: E → ℝ, p is a limit point of E, L ∈ ℝ then f(x) → L as x → p iff
For every sequence (pₙ) pₙ ∈ E\{p} pₙ → p
The sequence f(pₙ) → L as n → ∞.
Proof
(Limits via Sequences)
If f: E → ℝ, p is a limit point of E, L ∈ ℝ then f(x) → L as x → p iff
For every sequence (pₙ) pₙ ∈ E{p} pₙ → p
The sequence f(pₙ) → L as n → ∞.
3 points
- (⇒) Use δ in limit definition as ε in convergence definition
- (⇐) Use contrapositive
- Let δ = 1/n to find sequence
Theorem
AOL for Functions
4 points
Let p ∈ ℝ be a limit point of E ⊆ ℝ
f: E → ℝ and g: E → ℝ are functions s.t. f(x) → a and g(x) → b as x → p
Then f(x) ± g(x) → a ± b
f(x)g(x) → ab
f(x)/g(x) → a/b if b ≠ 0
|f(x)| → |a|
as x → p
Proof
(AOL for Functions)
Let p ∈ ℝ be a limit point of E ⊆ ℝ
f: E → ℝ and g: E → ℝ are functions s.t. f(x) → a and g(x) → b as x → p
Then f(x) ± g(x) → a ± b
f(x)g(x) → ab
f(x)/g(x) → a/b if b ≠ 0
|f(x)| → |a|
as x → p
2 points
- Pick pₙ ∈ E\{p} pₙ → p
- Use Analysis I AOL
Theorem
Extended AOL
4 points
Let p ∈ ℝ be a limit point of E ⊆ ℝ
f: E → ℝ and g: E → ℝ are functions s.t. f(x) → a and g(x) → b as x → p
a and/or b = ±∞
Then f(x) ± g(x) → a ± b except for ∞ - ∞ or -∞ + ∞
f(x)g(x) → ab except for a = 0 or b = 0
f(x)/g(x) → a/b if b ≠ 0 except ∞/∞ or b = 0
|f(x)| → |a|
as x → p
Theorem
Limits Preserve Weak Inequalities
f(x) → a, g(x) → b as x → p
∀ x: f(x) ≤ g(x) ⇒ a ≤ b
Theorem
Sandwiching
f(x), g(x) → a as x → p
f(x) ≤ h(x) ≤ g(x) ⇒ h(x) → a as x → p
Theorem
Algebra of Continuous Functions
If f, g: E → ℝ, f, g continouous at p ∈ E
⇒ f(x) ± g(x), f(x)g(x), |f(x)| are continuous at p
f(x)/g(x) is continuous at x = p provided g(p) ≠ 0
Proof
(Algebra of Continuous Functions)
If f, g: E → ℝ, f, g continuous at p ∈ E
⇒ f(x) ± g(x), f(x)g(x), |f(x)| are continuous at p
f(x)/g(x) is continuous at x = p provided g(x) ≠ 0
2 points
- Immediate for p isolated
- Use AOL for p limit point
Theorem
Continuous Functions Commute with Limits
If f: E → ℝ, g: E’ → ℝ, f(E) ⊆ E’
p ∈ ℝ is a limit point of E (or ±∞ E unbounded)
f(x) → L ∈ E’ as x → p, g continuous at L
Then lim[x→p] g(f(x)) = g(lim[x → p] f(x)) = g(L)
Proof
(Continuous Functions Commute with Limits)
If f: E → ℝ, g: E’ → ℝ, f(E) ⊆ E’
p ∈ ℝ is a limit point of E (or ±∞ E unbounded)
f(x) → l ∈ E’ as x → p, g continuous at l
Then lim[x→p] g(f(x)) = g(lim[x → p] f(x)) = g(l)
2 points
- Use continuiuty and limit definitions
- Use δ in first definition as ε in second
Theorem
Composition of Continuous Functions
f: E → ℝ, g: E’ → ℝ, f(E) ⊆ E’
If f(x) is continuous at p and g(x) is continuous at f(p)
Then g(f(x)) is continuous at p
Proof
(Composition of Continuous Functions)
f: E → ℝ, g: E’ → ℝ, f(E) ⊆ E’
If f(x) is continuous at p and g(x) is continuous at f(p)
Then g(f(x)) is continuous at p
2 points
- Immediate for p isolated
- Use continuous functions commute with limits for p limit point
Theorem
Boundedness Theorem
Suppose f: [a, b] → ℝ is continuous (a < b)
Then f is bounded on [a, b] and:
∃ x₁ ∈ [a, b]: f(x₁) = sup f
∃ x₂ ∈ [a, b]: f(x₂) = inf f
Proof
(Boundedness Theorem)
Suppose f: [a, b] → ℝ is continuous (a < b)
Then f is bounded on [a, b] and:
∃ x₁ ∈ [a, b]: f(x₁) = sup f
∃ x₂ ∈ [a, b]: f(x₂) = inf f
6 points
- Suppose f has no upper bound
- Find a sequence (pₙ) s.t. f(pₙ) ≥ n
- Use Bolzano-Weierstrass and continuity to show f(pₛ₍ₙ₎) converges
- Contradiction as f(pₛ₍ₙ₎) ≥ sₙ
- 1/(sup f - f(x)) bounded on [a, b]
- Find contradiction of sup definition
Theorem
Intermediate Value Theorem IVT
Assume f: [a, b] → ℝ is continuous and f(a) ≤ c ≤ f(b) [or f(a) ≥ c ≥ f(b)]
Then ∃ ξ ∈ [a, b]: f(ξ) = c
Proof
(Intermediate Value Theorem IVT)
Assume f: [a, b] → ℝ is continuous and f(a) ≤ c ≤ f(b) [or f(a) ≥ c ≥ f(b)]
Then ∃ ξ ∈ [a, b]: f(ξ) = c
2 points
- Construct [aₙ, bₙ] s.t. f(aₙ) ≤ c ≤ f(bₙ) and aₙ increasing, bₙ decreasing, bₙ - aₙ → 0 by divide and conquer
- Use continuity to show f(L) = c where aₙ, bₙ → L
Theorem
Continuous Inverse Function Theorem CIFT
If I is an interval, f: I → ℝ strictly monotonic, continuous
Then f(I) is an interval and f⁻¹: f(I) → I is also strictly monotonic and continuous
Proof
(Continuous Inverse Function Theorem CIFT)
If I is an interval, f: I → ℝ strictly monotonic, continuous
Then f(I) is an interval and f⁻¹: f(I) → I is also strictly monotonic and continuous
5 points
- Pick x = f(a), z = f(b) where a, b ∈ I
- Use IVT on y where x ≤ y ≤ z
- Show f(x) < f(y) ⇒ f⁻¹(f(x)) < f⁻¹(f(y)) as f strictly increasing and by trichotomy
- Set δ = min(f(x) - f(x - ε), f(x + ε) - f(x)) > 0
- Show |f⁻¹(q) - f⁻¹(p)| < ε where q ∈ f(I) and p = f(x)
Proposition
Uniform Continuity via Sequences
f: E → ℝ is uniformly continuous
⇔ ∀ xₙ, yₙ ∈ E, |xₙ - yₙ| → 0: |f(xₙ) - f(yₙ)| → 0
Proof
(Uniform Continuity via Sequences)
f: E → ℝ is uniformly continuous
⇔ ∀ xₙ, yₙ ∈ E, |xₙ - yₙ| → 0: |f(xₙ) - f(yₙ)| → 0
4 points
- (⇒) Use uniform continuity and convergence definitions
- (⇐) Use contradiction
- Let δ = 1/n
- Construct (xₙ), (yₙ) s.t. |xₙ - yₙ| < 1/n but |f(xₙ) - f(yₙ)| ≥ ε
Theorem
Continuity Implies Uniform Continuity on Closed Bounded Intervals
If f: [a, b] → ℝ is continuous then it is uniformly continuous
Proof
(Continuity Implies Uniform Continuity on Closed Bounded Intervals)
If f: [a, b] → ℝ is continuous then it is uniformly continuous
4 points
- Suppose f is not uniformly continuous and find such sequences
- Use Bolzano-Weierstrass
- Use continuity of f at p where xₙ₍ₖ₎ → p
- Use triangle inequality to show |f(xₙ₍ₖ₎) - f(yₙ₍ₖ₎)| → 0
Theorem
Uniform Limits Preserve Continuity
If fₙ, f: E → ℝ and fₙ is continuous for each n and fₙ → f uniformly on E
Then f is continuous
Proof
Uniform Limits Preserve Continuity
If fₙ, f: E → ℝ and fₙ is continuous for each n and fₙ → f uniformly on E
Then f is continuous
2 points
- |f(x) - f(p)| ≤ |f(x) - fₙ(x)| + |fₙ(x) - fₙ(p)| + |fₙ(p) - f(p)|
- Use uniform convergence and continuity to bound each term
Theorem
Cauchy Convergence Criterion for Uniform Convergence of Sequences
Let fₙ: E → ℝ be a sequence of fₙs
Then fₙ converges uniformly to some f
⇔ ∀ ε > 0: ∃ N: ∀ n, m ≥ N: ∀x: |fₙ(x) - fₘ(x)| < ε
Proof
(Cauchy Convergence Criterion for Uniformly Convergent Series)
Let fₙ: E → ℝ be a sequence of fₙs
Then fₙ converges uniformly to some f
⇔ ∀ ε > 0: ∃ N: ∀ n, m ≥ N: ∀x: |fₙ(x) - fₘ(x)| < ε
3 points
- (⇒) Use triangle inequality on |fₙ(x) - fₘ(x)|
- (⇐) Fix x to show fₙ(x) → f(x) for some x
- Let m → ∞ in Cauchy convergence definition
Theorem
Weierstrass M-Test
Suppose uₖ: E → ℝ are such that
∀ k: ∃ Mₖ: ∀ x: |uₖ(x)| ≤ Mₖ
and Σ Mₖ converges
Then Σ uₖ converges uniformly on E
Proof
(Weierstrass M-Test)
Suppose uₖ: E → ℝ are such that
∀ k: ∃ Mₖ: ∀ x: |uₖ(x)| ≤ Mₖ
and Σ Mₖ converges
Then Σ uₖ converges uniformly on E
1 point
- Use Cauchy convergence definition
Theorem
Uniform Convergence and Continuity of Power Series
Suppose Σ aₖxᵏ has ROC R > 0 and suppose 0 < ρ < R
Then Σ aₖxᵏ converges uniformly on {x: |x| ≤ ρ}
Proof
(Uniform Convergence and Continuity of Power Series)
Suppose Σ aₖxᵏ has ROC R > 0 and suppose 0 < ρ < R
Then Σ aₖxᵏ converges uniformly on {x: |x| ≤ ρ}
1 point
- Apply M-Test with Mₖ = |aₖ|ρᵏ
Theorem
Algebraic Properties of Differentiation
3 points
Suppose f, g: E → ℝ are differentiable at x₀
(a) (Linearity) a, b constants a, b ∈ ℝ
af(x) + bg(x) is differentiable at x₀ with derivative af’(x₀) + bg’(x₀)
(b) (Product Rule) f(x)g(x) differentiable at x₀ with derivative f’(x₀)g(x₀) + f(x₀)g’(x₀)
(c) (Quotient Rule) If g(x₀) ≠ 0, f(x)/g(x) differentiable at x₀ with derivative (f’(x₀)g(x₀) - f(x₀)g’(x₀))/g(x₀)²
Proof
Algebraic Properties of Differentiation
Suppose f, g: E → ℝ are differentiable at x₀
(a) (Linearity) a, b constants a, b ∈ ℝ
af(x) + bg(x) is differentiable at x₀ with derivative af’(x₀) + bg’(x₀)
(b) (Product Rule) f(x)g(x) differentiable at x₀ with derivative f’(x₀)g(x₀) + f(x₀)g’(x₀)
(c) (Quotient Rule) If g(x₀) ≠ 0, f(x)/g(x) differentiable at x₀ with derivative (f’(x₀)g(x₀) - f(x₀)g’(x₀))/g(x₀)²
3 points
- (Linearity & Product Rule) Use f(x₀ + h) = f(x₀) + f’(x₀)h + ε₁(h)h, ε₁(h) → 0
- (Quotient Rule) Show d/dx 1/g(x) = -g’(x₀)/g(x₀)²
- Apply product rule
Theorem
Chain Rule
Assume f: E → ℝ, g: E’ → ℝ
f(E) ⊆ E’
f is differentiable at x₀ ∈ E, which is a limit point of E
g is differentiable at f(x₀) ∈ E’
Then g(f(x)) is differentiable at x₀ with derivative g’(f(x₀))f’(x₀)
Proof
(Chain Rule)
Assume f: E → ℝ, g: E’ → ℝ
f(E) ⊆ E’
f is differentiable at x₀ ∈ E, which is a limit point of E
g is differentiable at f(x₀) ∈ E’
Then g(f(x)) is differentiable at x₀ with derivative g’(f(x₀))f’(x₀)
4 point
- f(x₀ + h) = f(x₀) + f’(x₀)h + ε₁(h)
- g(f(x₀) + η) = g(f(x₀)) + g’(f(x₀))η + ε₂(η)η
- Define ε₂(0) = 0 so ε₂ continuous at 0
- Show g(f(x₀ + h)) = g(f(x₀)) + g’(f(x₀))f’(x₀)h + [ … ]h where [ … ] → 0 as h → 0
Theorem
Inverse Function Theorem IFT
Suppose I interval (non-trivial), f: I → ℝ strictly monotonic and continuous
Let g: f(I) → ℝ be the (continuous, strictly monotonic) inverse
Suppose x₀ ∈ I and f is differentiable at x₀ with f’(x₀) ≠ 0
Then g is differentiable at f(x₀)
g’(f(x₀)) = 1/f’(x₀)
Proof
(Inverse Function Theorem IFT)
Suppose I interval (non-trivial), f: I → ℝ strictly monotonic and continuous
Let g: f(I) → ℝ be the (continuous, strictly monotonic) inverse
Suppose x₀ ∈ I and f is differentiable at x₀ with f’(x₀) ≠ 0
Then g is differentiable at f(x₀)
g’(f(x₀)) = 1/f’(x₀)
1 point
- Use derivative definition
Theorem
Differentiation Theorem for Power Series
Suppose Σₙ₌₀ aₙxⁿ has ROC R > 0 (or R = ∞)
Define f(x) = Σₙ₌₀ aₙxⁿ for |x| < R
Then for |x| < R f is differentiable at x
f’(x) = Σₙ₌₁ naₙxⁿ⁻¹
Theorem
Fermat’s Theorem on Extrema
Let f: (a, b) → ℝ, x₀ ∈ (a, b) is extremum of f
Suppose f is differentiable at x₀
Then, f’(x₀) = 0
Proof
(Fermat’s Theorem on Extrema)
Let f: (a, b) → ℝ, x₀ ∈ (a, b) is extremum of f
Suppose f is differentiable at x₀
Then, f’(x₀) = 0
2 points
- Consider right and left derivatives
- Show one is ≤ 0 and the other is ≥ 0
Theorem
Rolle’s Theorem
Suppose f: [a, b] → ℝ, such that
(a) f is continuous on [a, b]
(b) f is differentiable on [a, b]
(c) f(a) = f(b)
Then ∃ ξ ∈ (a, b): f’(ξ) = 0
Proof
(Rolle’s Theorem)
Suppose f: [a, b] → ℝ, such that
(a) f is continuous on [a, b]
(b) f is differentiable on [a, b]
(c) f(a) = f(b)
Then ∃ ξ ∈ (a, b): f’(ξ) = 0
2 points
- Apply Boundedness Theorem
- Apply Fermat’s Theorem on Extrema
Theorem
Mean Value Theorem
Suppose f: [a, b] → ℝ such that
(a) f is countinuous on [a, b]
(b) f is differentiable on (a, b)
Then ∃ ξ ∈ (a, b): f’(ξ) = (f(b) - f(a))/(b - a)
Proof
(Mean Value Theorem)
Suppose f: [a, b] → ℝ such that
(a) f is countinuous on [a, b]
(b) f is differentiable on (a, b)
Then ∃ ξ ∈ (a, b): f’(ξ) = (f(b) - f(a))/(b - a)
2 points
- Define F(x) = f(x) - f(a) - ((f(b) - f(a))/(b - a))(x - a)
- Apply Rolle’s Theorem
Theorem
Cauchy’s MVT / Generalised MVT
2 points
Let f, g: [a, b] → ℝ such that
(a) f, g continuous on [a, b]
(b) f, g differentiable on (a, b)
Then ∃ ξ ∈ (a, b): f’(ξ)(g(b) - g(a)) = g’(ξ)(f(b) - f(a))
If g’ ≠ 0 for all x ∈ (a, b) then g(b) ≠ g(a) and f’(ξ)/g’(ξ) = (f(b) - f(a))/(g(b) - g(a))
Proof
(Cauchy’s MVT / Generalised MVT)
Let f, g: [a, b] → ℝ such that
(a) f, g continuous on [a, b]
(b) f, g differentiable on (a, b)
Then ∃ ξ ∈ (a, b): f’(ξ)(g(b) - g(a)) = g’(ξ)(f(b) - f(a))
If g’ ≠ 0 for all x ∈ (a, b) then g(b) ≠ g(a) and f’(ξ)/g’(ξ) = (f(b) - f(a))/(g(b) - g(a))
3 points
- For g(b) ≠ g(a) define F(x) = (f(x) - f(a)) - K(g(x) - g(a)) where K = (f(b) - f(a))/(g(b) - g(a))
- Apply Rolle’s Theorem
- For g(b) = g(a) apply Rolle’s Theorem to g
Theorem
Constancy Theorem
Suppose I is an interval and f: I → ℝ is such that
f’(x) = 0 for all x ∈ I
Then f is constant
Proof
(Constancy Theorem)
Suppose I is an interval and f: I → ℝ is such that
f’(x) = 0 for all x ∈ I
Then f is constant
1 point
- Apply Mean Value Theorem to two members of I
Theorem
Real Binomial Theorem
If -1 < x < 1, p ∈ ℝ
(1 + x)ᵖ = Σ (ᵖₖ)xᵏ
Proof
(Real Binomial Theorem)
If -1 < x < 1, p ∈ ℝ
(1 + x)ᵖ = Σ (ᵖₖ)xᵏ
4 points
- Let g(x) = eᵖˡᵒᵍ⁽¹ ⁺ ˣ⁾, h(x) = Σ (ᵖₖ)xᵏ
- g’(x) = (p/(1+x))g(x) and similarly for h
- Define F(x) = h(x)/g(x)
- Apply Constancy Theorem
Theorem
Taylor’s Theorem
Assume f: [a, b] → ℝ is such that f, f’, f’’, …, f⁽ⁿ⁾ exist and are continuous on [a, b] and f⁽ⁿ⁺¹⁾ exists on (a, b)
Then ∃ ξ ∈ (a, b) such that
f(b) = f(a) + f’(a)(b - a) + (f’‘(a)/2)(b - a)² + … + (f⁽ⁿ⁾(a)/n!)(b - a)ⁿ + (f⁽ⁿ⁺¹⁾(ξ)/(n + 1)!)(b - a)ⁿ⁺¹
Proof
(Taylor’s Theorem)
Assume f: [a, b] → ℝ is such that f, f’, f’’, …, f⁽ⁿ⁾ exist and are continuous on [a, b] and f⁽ⁿ⁺¹⁾ exists on (a, b)
Then ∃ ξ ∈ (a, b) such that
f(b) = f(a) + f’(a)(b - a) + (f’‘(a)/2)(b - a)² + … + (f⁽ⁿ⁾(a)/n!)(b - a)ⁿ + (f⁽ⁿ⁺¹⁾(ξ)/(n + 1)!)(b - a)ⁿ⁺¹
7 point
- Use induction on n
- n = 0 is MVT
- Define F(x) = f(x) - f(a) - f’(a)(x - a) - … - (f⁽ⁿ⁾(a)/n!)(x - a)ⁿ - (K/(n + 1)!)(x - a)ⁿ⁺¹ where K is s.t. F(b) = 0
- Apply Rolle’s Theorem to get c s.t. F’(c) = 0
- Apply n - 1 case to f’ on [a, c]
- F’(c) = 0 substituting in f’(c) ⇒ K = f⁽ⁿ⁺¹⁾(ξ)
- Subsitute x = b into F(x)
Theorem
Simple L’Hôpital
3 points
Suppose f, g: E → ℝ, p ∈ E limit point
(a) f(p) = g(p) = 0
(b) f’(p), g’(p) exist
(c) g’(p) ≠ 0
Then limₓ→ₚ f(x)/g(x) exists and = f’(p)/g’(p)
Proof
(Simple L’Hôpital)
Suppose f, g: E → ℝ, p ∈ E limit point
(a) f(p) = g(p) = 0
(b) f’(p), g’(p) exist
(c) g’(p) ≠ 0
Then limₓ→ₚ f(x)/g(x) exists and = f’(p)/g’(p)
2 points
- Rearrange f(x)/g(x)
- Use AOL
Theorem
L’Hôpital 0/0 Form
4 points
Suppose f, g real valued defined in some interval (p, p + δ) δ > 0
Also suppose:
(a) f, g differentiable on (p, p + δ)
(b) f(x), g(x) → 0 as x → p⁺
(c) g’(x) ≠ 0 on (p, p + δ)
(d) limₓ→ₚ₊ f’(x)/g’(x) exisits (possibly ±∞)
Then limₓ→ₚ₊ f(x)/g(x) = limₓ→ₚ₊ f’(x)/g’(x)
Also g(x) ≠ 0 on some (p, p + δ’) δ’ > 0
Proof
(L’Hôpital 0/0 Form)
Suppose f, g real valued defined in some interval (p, p + δ) δ > 0
Also suppose:
(a) f, g differentiable on (p, p + δ)
(b) f(x), g(x) → 0 as x → p⁺
(c) g’(x) ≠ 0 on (p, p + δ)
(d) limₓ→ₚ₊ f’(x)/g’(x) exisits (possibly ±∞)
Then limₓ→ₚ₊ f(x)/g(x) = limₓ→ₚ₊ f’(x)/g’(x)
Also g(x) ≠ 0 on some (p, p + δ’) δ’ > 0
4 points
- Define f(p) = g(p) = 0
- Fix x ∈ (p, p + δ)
- Use Cauchy MVT
- Let x → p⁺
Theorem
L’Hôpital ∞/∞ Form
4 points
If f, g: (a, a + δ) → ℝ
(a) f, g differentiable on (a, a + δ)
(b) |f|, |g| → ∞
(c) g’ ≠ 0 on (a, a + δ)
(d) f’(x)/g’(x) → L ∈ ℝ ∪ {±∞} as x → a⁺
Then f(x)/g(x) → L as x → a⁺
Also g ≠ 0 on (a, a + δ’) for some δ’ > 0
Proof
(L’Hôpital ∞/∞ Form)
If f, g: (a, a + δ) → ℝ
(a) f, g differentiable on (a, a + δ)
(b) |f|, |g| → ∞
(c) g’ ≠ 0 on (a, a + δ)
(d) f’(x)/g’(x) → L ∈ ℝ ∪ {±∞} as x → a⁺
Then f(x)/g(x) → L as x → a⁺
Also g ≠ 0 on (a, a + δ’) for some δ’ > 0
4 points
- Fix c ∈ (a, a + δ), x ∈ (a, c)
- Use Cauchy MVT on [x, c]
- |f’(ξ)/g’(ξ) - L| < ε for all ξ ∈ (a, c)
- Rearrange to |f(x)/g(x) - L|
