Differences

This shows you the differences between two versions of the page.

--- math104-s21:s:ryanpurpura [2021/05/12 22:26]
135.180.146.149
+++ math104-s21:s:ryanpurpura [2026/02/21 14:41] (current)
@@ Line 657: / Line 657: @@
 show there exists a sequence ${x_n}$ which converges to $s$.
-Consider expanding $\epsilon$ bounds. By definition, $(s - \epsilon, s)$
+Consider expanding $\epsilon$ bounds. By definition, $[s - \epsilon, s]$
 must contain a point in $S$, otherwise $s - \epsilon$ is a better upper bound
 than the supremum! Thus we can make a sequence of points by starting
 with an epsilon bound of say, $1$ and sampling a point within it,
 and then shrinking the epsilon bound to $1/2$, $1/4$, $1/8$, etc.
+: Show that $f(x) = 0 $ if $x$ is rational and $1$ if $x$ is irrational
+has no limit anywhere.
+A: Use the fact that $\mathbb{Q}$ is dense on $\mathbb{R}$.
+Given an $\epsilon < 1/2$, no matter what $\delta$ you pick you're always going to get both
+rational and irrational numbers within that epsilon bound, which means the
+function will take on both $1$ and $0$ within that bound, which
+exceeds the $\epsilon$ bound.
+: What exactly does it mean for a function to be convex?
+A: A function is convex if $\forall x,y$, $0 \le \alpha \le 1$ we have
+$$f(\alpha x + (1-\alpha) y) \le \alpha f(x) + (1-\alpha) f(y)$$.
+. 5.2: Show that the value of $\delta$ in the definition of continuity is not unique.
+A: Given some satisfactory $\delta > 0$, we can just choose $\delta / 2$.
+The set of numbers in the input space implied by $\delta / 2$ is a subset
+of those from the old $\delta$, so clearly all of the points must satisfy
+$|f(x) - f(c)| < \epsilon$ required for continuity.
+: Why is the wrong statement as presented in lecture for the Fundamental Theorem of Calculus Flawed?
+A: The reason is that the derivative of a function is not necessarily Riemann integral.
+: What function is Stieltjes integrable but not Riemann integrable?
+A: Imagine a piecewise function that is the
+rational indicator function for $0\le x \le 1$ and 0 elsewhere.
+This is obviously not Riemann integrable but we can assign $\alpha$ to be
+constant from $0$ to $1$ (i.e. assigning no weight to that part) to make it
+Stieltjes integrable.
+https://math.stackexchange.com/questions/385785/function-that-is-riemann-stieltjes-integrable-but-not-riemann-integrable
+: Why do continuous functions on a compact metrix space $X$ achieve their
+$\sup$ and $\inf$?
+A: $f(X)$ is compact, which implies that it contains its $\sup$ and $\inf$.
+. What's a counterexample to the converse of the Intermediate Value Theorem?
+A: Imagine piecewise function
+$f(x) = x$ for $0 \le x \le 5$ and $f(x) = x - 5$ for $5 \le x \le 10$.
+: Sanity check: why is $f(x) = x^2$ continuous at $x = 3$?
+A: $\lim_{x\to 3} x^2 = 9 = f(3)$ (Definition 3).
+Proving with Definition 1 is annoying but you can see the proof in the problem book.
+: Prove the corellary to the 2nd definition of continuity: $f: X \to Y$ is continuous
+iff $f^{-1}(C)$ is closed in $X$ for every closed set $C$ in $Y$.
+A: A set is closed iff its complement is open.
+We can then use the fact that $f^{-1}(E^c} = [f^-1(E)]^c$ for every $E \subset Y$.
+(Basically, imagine the 2nd definition but you just took the complement of everything).
+: What's a function that is not uniformly continuous but is continuous?
+A: A simple example is $f(x) = x^2$ which is clearly continuous but
+finding a single value for $\delta$ for a given $\epsilon$ is impossible.
+: If we restrict $x^2$ to the domain $[0,1]$, why does it become uniformly continuous?
+A: This is because we restricted the domain to a compact set!
+: Give an example where $E'$ is a subset of $E$.
+A: Consider the classic set $\{0\} \union \{1, 1/2, 1/3, 1/4, \dots}$.
+$E' = \{0\}$.
+: Give an example where it's a superset of $E$.
+A: Consider $E = (0, 1)$ then $E' = [0, 1]$.
+: Give an example where it's neither.
+A: Consider $E = \{1, 1/2, 1/3, 1/4, \dots \}$.
+$E' = \{0\}$. But this is neither subset nor superset of $E$.
+: Why is continuity defined at a point and uniform continuity defined on an interval?
+A: Continuity allows you to pick a different $\delta$ for each point,
+allowing you to have continuity at a single point. Uniform continuity
+needs to have a interval share the same value of $\delta$.
+: What is a smooth function, and give an example one.
+A: A smooth function is infinitely differentiable.
+All polynomials are smooth.
+: Prove differentiability implies continuity.
+A: Differentiability
+implies $\lim_{t \to x} \frac{f(t) - f(x)}{t-x} = L$.
+Now multiply both sides by $\lim_{t \to x} (t -x)$, taking advantage of limit rules.
+so we get $\lim_{t\to x} \frac{f(t) -f(x)}{t-x}(t-x) = f'(x) \cdot 0 = 0$
+which means $\lim_{t\to x} f(t) = f(x)$.
+: Use L'Hospital's Rule to evaluate $\lim_{x\to \infty} \frac{x}{e^x}$.
+A: We clearly see that the limits of the top and bottom go to infinity,
+so we can use L'Hospital's Rule.
+Taking derivative of top and bottom, we get $1/e^x$ which goes to $0$.
+: 5.1 Prove that if $c$ is an isolated point in $D$, then $f$ is automatically continuous at $c$.
+A: Briefly, no matter what $\epsilon$ is chosen, we can just choose a small
+enough neighborhood such $c$ is within the neighborhood.
+Then all $x$ (i.e., just c) in the neighborhood have $|f(x)-f(c)| = 0$.

Lecture Notes

User Tools

Site Tools

Differences

Page Tools