Functions, Lecture Notes - Mathematics, Study notes of Calculus

Download Functions, Lecture Notes - Mathematics and more Study notes Calculus in PDF only on Docsity! Functions Adrian Down 16779577 July 8, 2005 1 Review So far, we have looked at R and the properties thereof. Tools: supremum, infimum, sequences (series), lim inf, lim sup Now we begin analysis: the study of functions and their properties 2 Functions Def: a function with domain S ⊆ R is a choice of number f(x) ∈ R corre- sponding to any x ∈ S. Call S the domain of f , and denote S = dom(f). Examples: 1) f(x) = 1 x , dom(f) = {x ∈ R|x 6= 0} 2) f(x) = √ 4− x2, dom(f) = {x ∈ R| − 2 ≤ x ≤ 2} Def: Given a function f , dom(f), and some fixed point x0 ∈ dom(f), we say that f is continuous at x0 if, given any sequence (xn) with lim(xn) = x0, lim f(xn) = f(x0). If f is continuous at x, ∀x ∈ dom(f), then f is a continuous function. Idea: f is continuous if it has no jumps ↔ if x is close to y, then f(x) should be close to f(y). Have to define what it means to be “close,” using sequences. Thm: f is continuous at x0 ↔ f has the property that ∀ > 0, ∃δ > 0 such that |x− x0| < δ → |f(x)− f(x0)| < (∗). Pf: Assume (∗) is true. Want to show that f is continuous. Let (xn) be a sequence such that lim xn = x0. Show that lim f(xn) = f(x0). Let  > 0. By assumption, ∃δ > 0 such that |x − x0| < δ → |f(x) − f(x0)| < . Since 1 xn converges to x0, and δ > 0 → ∃N such that n > N → |xn − x0| < δ. so for n > N , |f(xn)− f(x0)| < . Therefor by definition, lim f(xn) = f(x0). Now assume f is continuous and use proof by contradiction. Assume that (∗) is not true. Means that ∃ > 0 such that ∀δ > 0, “|x − x0| < δ → |f(x) − f(x0)| < ” is false → ∃ > 0 such that ∀δ > 0,∃x ∈ dom(f) such that |x− x0| < δ but |f(x)− f(x0)| ≥ . Let δ = 1 n , get xn ∈ dom(f) such that |xn−x0| < 1n but |f(x)−f(x0)| ≥ . Repeat with all n ∈ N , get a sequence (xn) such that lim xn = x0 but (f(xn)) can not converge to f(x) (because |f(xn) − f(x0)| ≥ ). Therefore f is not continuous at x0, in contradiction to the assumption. 3 Examples Why we need both definitions 1) f(x) = k,∀x ∈ R. Then f is continuous. Pf: let x0 ∈ R, for lim xn = x0, because → lim k = k, lim f(xn) = f(x0). Also, if  > 0, |f(x) − f(x0)| = |k − k| = 0 < ,∀x ∈ R. Let δ be anything greater than 0. In this case, both definitions are easy. 2)f(x) = 2x2 + 1. If lim xn = x0, the using the limit theorems from the beginning of the class, lim(2x2 + 1) = 2x20 + 1 → lim f(xn) = f(x0), so f is continuous. Using the , δ definition. Let  > 0. Want δ > 0 such that |x − x0| < δ → |f(xn) − f(x0)| <  ↔ |2x2 + 1 − (2x20 + 1)| <  ↔ 2 · |x2 − x20| <  ↔ 2|x − x0| · |x + x0| < . Make the first term small and keep a bound on the second. Note: |x− x0| < 1 → |x| < |x0|+ 1. So |x + x0| ≤ |x|+ |x0| < 2|x0|+ 1. Therefor let δ = min{1,  2(2|x0|+1)}. Then |x − x0| ≤  2(2|x0|+1) → |f(x) − f(x0)| ≤ 2 · |x− x0| · |x + x0| ≤ 2 · 2(2|x0|+1) · (2|x0|+ 1) = . 3) f(x) = x2 sin 1 x for x 6= 0 and 0 for x = 0. Then f is continuous at 0. Pf: Let  > 0. We want δ such that |x| < δ → |f(x)| < . |f(x)| = |x2 sin 1 x | = |x2| · | sin 1 x | (1) | sin a| ≤ 1 → |f(x)| ≤ |x2| (2) (3) Let δ = √ . If |x| < δ, then |x| < √  → |x|2 < , so |f(x)| < . 2