Topic 7: Continuity

Question 1

What is the definition for continuity (2,1)

Answer 1

Function f(x) is continuous at x = a if:
-lim_x->a f(x) exists
-lim_x->a f(x) = f(a)

-If not, the function is discontinuous at x = a

Question 2

What is the definition for a function being continuous over an interval (2)

Answer 2

-Function f(x) is continuous over an open interval (a,b) if it is continuous at x = c for all c ∈ (a,b)
-Function f(x) is continuous over closed interval [a, b] if it is continuous over (a,b) and lim_{x -> a⁺} f(x) = f(a) and lim_{x -> b^-} f(x) = f(b)

Question 3

What is a proposition on the sum, difference, product and quotient of continuous functions (1,1)

Answer 3

-The sum, difference, product and quotient of continuous functions are continuous

-However, be careful with domains

Question 4

What is a proposition of the continuity of composite functions (2)

Answer 4

-Composites of continuous functions are continuous
-if f: ℝ -> ℝ and g: ℝ -> ℝ are continuous at x = a, gof: ℝ -> ℝ is also continuous at x = a

Question 5

What is the extreme value theorem (2,2)

Answer 5

-Let f: ℝ -> ℝ be continuous over some closed interval [a, b]
-Then F is bounded on [a, b] and attains its bounds

-That is, there exists M₊, M_- ∈ ℝ such that M_- ≤ f(x) ≤ M₊ for all x ∈ [a, b]
-There also exists c, d ∈ [a, b] such that f(x) = M_- and f(d) = M₊

Question 6

How do limits work for composite continuous functions (1)

Answer 6

Suppose f: ℝ -> ℝ is continuous and g: ℝ -> ℝ
-lim_x->a f(g(x)) = f(lim_x->ag(x)), providing both limits exist

Question 7

What is the simple and general intermediate value theorem (2,2)

Answer 7

Simple:
-Let f:ℝ -> ℝ be continuous over closed interval [a, b]
-Then if f(a) and f(b) have opposite signs there exists c ∈ (a, b) such that f(c) = 0

General:
-Suppose f: ℝ -> ℝ is continuous over closed interval [a, b], and f(a) = α, f(b) = β.
-Then for any γ ∈ (α, β), there then exists c ∈(a, b) such that f(c) = γ

Question 8

What is Brouwers fixed point theorem (2)

Answer 8

-Let f: ℝ -> ℝ be continuous over a closed interval [a, b] and suppose that a ≤ f(x) ≤ b for all a ≤ x ≤ b
-The equation f(x) = x has a real root in [a, b]

Question 9

How can we prove Brouwers fixed point theorem (3)

Answer 9

-Define g: ℝ -> ℝ by g(x) = f(x) - x
-g(a) = f(a) - a ≥ 0
-g(b) = f(b) - b ≤ 0

Case 1:
-if g(a) = 0, then f(a) = a = 0 => f(a) = a, and hence a is our fixed point
Case 2:
-If g(b) = 0, f(b) - b = 0 => f(b) = b, b is our fixed point
Case 3:
-g(a) > 0, g(b) < 0, by the IVT we have some c ∈ (a,b) such that g(c) = 0 => f(c) - c = 0, f(c) = c, and hence c is our fixed point

Question 10

What are some numerical methods of solving f(x) = 0 (3)

Answer 10

-Bisection method (direct application of the IVT)
-Newton-Raphson method
-Direct iteration

Question 11

How can we use the bisection method to solve p(x) = x⁵ - 5x + 2 for the root in (0,1) (1,4,1)

Answer 11

-We know p(0) = 2>0, and p(1) = -2 < 0, so the IVT tells us there exists c ∈ (0,1) such that f(c) = 0

-Try x₀ = 0.5, then p(0.5)<0, so by the IVT c ∈ (0, 0.5)
-Try x₁ = 0.25, p(0.25) > 0, c ∈ (0.25, 0.5)
-Try x₂ = 0.375, p(0.375) > 0, c ∈ (0.375, 0.5)
-Try x₃ = 0.4375, p(0.4375) < 0, c ∈ (0.375, 0.4375)

-This is gradually approaching x ≈ 0.4

Question 12

What is the formula for the Newton-Raphson method (1)

Answer 12

-x_n+1 = x_n - (f(x_n)/f’(x_n))

Question 13

How can we use the newton-raphson method to solve x⁵ - 5x + 2 to find the root in (0,1) (2,3)

Answer 13

-Remember the N-R formula as x_n+1 = x_n - (f(x_n)/f’(x_n))
-x_n+1 = x_n - (x⁵-5x+2)/f’(5x⁴-5))

-Start with x₀ = 0.5, x₁ = 0.4
-x₂ = 0.4021
-We can thus stop here with x ≈ 0.4

Question 14

How can we use direct iteration to solve x⁵ - 5x + 2 to find the root in (0,1) (4)

Answer 14

-Rewrite x⁵ - 5x + 2 = 0 -> x = (1/5)(x⁵ + 2)
-let x₀ = 0.5, x₁ = 0.871, x₂ = 1.186
-This method isn’t tending to the root in (0,1), but the root in (1,2)
-If we choose a different value for x₀ we might/should get the root in (0,1)