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math104-s21:s:ryanpurpura [2021/05/12 13:43]
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math104-s21:s:ryanpurpura [2026/02/21 14:41] (current)
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 ====== Ryan Purpura's Final Notes ======= ====== Ryan Purpura's Final Notes =======
 +
 +Questions can be found at the very bottom.
  
 ===== Number systems ===== ===== Number systems =====
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 and $\alpha$ is monotonically increasing on $[a,b]$ and $f\in \mathcal R(\alpha)$ and $\alpha$ is monotonically increasing on $[a,b]$ and $f\in \mathcal R(\alpha)$
 on $[a,b]$. Define $\beta(y) = \alpha(\varphi(y))$ and $g(y) = f(\phi(y))$. on $[a,b]$. Define $\beta(y) = \alpha(\varphi(y))$ and $g(y) = f(\phi(y))$.
-Then $\int _A^g d\beta = \int_a^b f d\alpha$.+Then $\int _A^g d\beta = \int_a^b f d\alpha$. 
 + 
 +**Integration and Differentiation**. 
 + 
 +If $F(x) = \int_a^x f(t)dt$, then $F$ continuous on $[a,b]$, and  
 +if $f$ is continuous at $x_0$, $F$ is differentiable at $x_0$, 
 +and $F'(x_0) = f(x_0)$. 
 + 
 +Proof sketch: Suppose $|f(t)| \le M$. For $x < y$: 
 +$|F(y) - F(x)| = \left| \int_x^y f(t) dt \right| \le M(y-x)$. 
 +So we can bound $|F(y) - F(x)|$ by epsilon since we can make $|y-x|$ 
 +as small as we want, therefore uniformly continuous. 
 + 
 +If $f$ is continuous at $x_0$, then given $\epsilon$ choose $\delta$ such that 
 +$|f(t) - f(x_0)| < \epsilon|$ if $|t-x_0| < \delta$ and $a \le t \le b$. 
 +Hence $\left| \frac{F(t) -F(s)}{t-s} - f(x_0) \right| = \left| \frac{1}{t-s} \int_s^t [f(u) - f(x_0)] du\right| < \epsilon$. Therefore, $F'(x_0) = f(x_0)$. 
 + 
 +**Fundamental Theorem of Calculus**: 
 +If $f$ integrable and $\exists F$ such that $F' = f$ then 
 +$\int_a^b f dx = F(b) - F(a)$. 
 + 
 +====== Final Questions ====== 
 + 
 +1: Prove there is no rational number whose square is 12 (Rubin 1.2). 
 +     
 +A: Such a number would satify $x^2 - 12 = 0$. 
 +B the rational roots theorem, we can enumerate possible 
 +rational solutions $x = \pm \{1, 2, 3, 4, 6, 12 \}$. It can be verified 
 +that none of these satisfy the equation, so the equation has no rational 
 +solution, so no rational number has the square of 12. 
 + 
 + 
 +2. Why is $S = (0, \sqrt 2]$ open in $\mathbb Q$? 
 + 
 +A: For all points in S we can construct a ball such that the 
 +ball is entirely contained within the set S. This is because $\sqrt 2 \notin \mathbb Q$. 
 + 
 + 
 +3. Construct a bounded set of real numbers with exactly 3 limit points. 
 + 
 +A: ${1, 1/2, 1/3, 1/4, \dots} \cup ${2, 1+ 1/2, 1 + 1/3, 1+1/4, \dots}  \cup ${3, 2 + 1/2, 2+ 1/3, 2 + 1/4, \dots} $ 
 +with limit points 0, 1, 2 
 + 
 + 
 +4. Why is the interior open? 
 + 
 +A: It is defined to the the union  of all open subsets in $\E$, and union of open subsets are open. 
 + 
 +5. Does convergence of $|s_n|$ imply that $|s_n|$ converges? 
 + 
 +A: No. Consider $-1, 1, -1, 1, \dots$. $|s_n|$ converges to $1$ but $s_n$ does not converge. 
 + 
 +6. Rudin 4.1: Suppose $f$ is a real function which satisfies  
 +$$\lim_{h\to 0} [f(x+h)-f(x-h)] = 0$$ for all $x \in \mathbb{R}$. Is $f$ necessarily continuous? 
 + 
 +A: No: a function with a simple discontinuity still passes the test. In fact, 
 +since limits imply approaching from both sides, $+h$ and $-h$ when approaching 
 +zero are the same thing, anyway. 
 + 
 +7. What's an example of a continuous function with a discontinuous derivative? 
 + 
 +A: Consider $f(x) = |x|$. The corner at $x= 0$ has different left and right-hand 
 +derivatives of $-1$ and $1$, respectively. This implies the derivative 
 +does not exist at $x=0$, and a type-1 discontinuity exists there. 
 + 
 +8. What's an example of a derivative with a type-2 discontinuity? 
 + 
 +A: An example would be $f(x) = x^2 \sin(1/x)$ with $f(0) := 0$ 
 +The derivative not zero is $f'(x) = 2x\sin(1/x) - \cos(1/x)$ 
 +which has a type-2 discontinuity at $x=0$. 
 +(Source: https://math.stackexchange.com/questions/292275/discontinuous-derivative) 
 + 
 +9: 3.3: Let $C$ be s.t. $|C| < 1$. Show $C^n \to 0$ as $n \to \infty$. 
 + 
 +A: Assume WOLOG $C$ is positive (this extends naturally 
 +to the negative case with a bit of finagling) the limit exists since the sequence is decreasing and bounded below. 
 +Using recursive sequence $C^{n+1} = C C^n$ we get $L = CL$ which implies $L = 0$. 
 + 
 + 
 +10: Can differentiable functions converge uniformly to a non-differentiable function? 
 + 
 +A: Yes! Consider $f_n(x) = \sqrt { x^2 + \frac{1}{n} }.$ 
 +It is clearly differentiable, and converges to $|x|$ uniformly. It develops 
 +a "kink" that makes it non-differentiable. 
 + 
 +11: 3.5: Let $S$ be a nonempty subset of $\mathbb{R}$ which is bounded above. If $s = \sup S$, 
 +show there exists a sequence ${x_n}$ which converges to $s$. 
 + 
 +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. 
 + 
 +12: 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. 
 + 
 +13: 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 
 +$$f(\alpha x + (1-\alpha) y) \le \alpha f(x) + (1-\alpha) f(y)$$. 
 + 
 +14. 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. 
 + 
 +15: 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. 
 + 
 +16: 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 
 + 
 +17: 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 $\inf$. 
 + 
 +18. 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 $5 \le x \le 10$. 
 + 
 +19: 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. 
 + 
 +20: 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). 
 + 
 +21: 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. 
 + 
 +22: 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! 
 + 
 +23: Give an example where $E'$ is a subset of $E$. 
 + 
 +A: Consider the classic set $\{0\} \union \{1, 1/2, 1/3, 1/4, \dots}$. 
 +$E' = \{0\}$. 
 + 
 +24: Give an example where it's a superset of $E$. 
 + 
 +A: Consider $E = (0, 1)$ then $E' = [0, 1]$. 
 + 
 +25: 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 $E$. 
 + 
 +26: 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 $\delta$. 
 + 
 +27: What is a smooth function, and give an example one. 
 + 
 +A: A smooth function is infinitely differentiable.  
 +All polynomials are smooth. 
 + 
 +28: 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 $\lim_{t\to x} f(t) = f(x)$. 
 + 
 +29: 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 $0$. 
 + 
 +30: 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 $|f(x)-f(c)| = 0$. 
 + 
  
  
math104-s21/s/ryanpurpura.1620826985.txt.gz · Last modified: 2026/02/21 14:44 (external edit)