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--- math104-s21:s:ryanpurpura [2021/05/12 12:40]
135.180.146.149
+++ math104-s21:s:ryanpurpura [2026/02/21 14:41] (current)
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 ====== Ryan Purpura's Final Notes =======
+Questions can be found at the very bottom.
 ===== Number systems =====
@@ Line 361: / Line 363: @@
 An extremely useful object is the bump function, which we typically denote as
 $\phi(x) = \begin{cases}0 & x<0 \\ 2x & x \in [0, 1/2] \\ 1 -2x & x \in [1/2, 1] \\ 0 & x > 1 \end{cases}$
 For example, $f_n(x) = \phi(x-n)$ converges pointwise to 0.
@@ Line 487: / Line 490: @@
 If they are equal, we have a Riemann-Stieltjes integral $\int f d\alpha$
-We define $P^*$ to be a **refinement** of $P$ if $P^* \superset P$.
+We define $P^*$ to be a **refinement** of $P$ if $P \subset P^*$.
 The **common refinement** of $P_1$ and $P_2$ is defined to be $P_1 \union P_2$.
@@ Line 502: / Line 505: @@
 Fix $P_2$ and take $\sup$ over $P_1$, then take $\inf$ over all $P_2$.
-**Theorem 6.6:** $f\in \mathcal{R}(\alpha) on $[a,b]$ iff $\forall \epsilon > 0$
+**Theorem 6.6:** $f\in \mathcal{R}(\alpha)$ on $[a,b]$ iff $\forall \epsilon > 0$
 $\exists P$ such that $U(P,f,\alpha) - L(P,f,\alpha) < \epsilon$.
-If direction: $L \le \underline \int \le \overline \int \le U$
+If direction: $L \le \underline{ \int } \le \overline \int \le U$
-so $0 \le \overline \int - \underline \int < \epsilon$, so they are equal and $f \in \mathcal r$.
+so $0 \le \overline { \int  }- \underline { \int } < \epsilon$, so they are equal and $f \in \mathcal r$.
 Only if direction: If $f \in \mathcal R$, $\exists P_1, P_2$ s.t.
@@ Line 513: / Line 516: @@
 Take common refinement $P$. Show $U(P, f, \alpha) \le L(P, f, \alpha) + \epsilon$.
+Fun facts about 6.6:
+. If holds for some $P$ and some $\epsilon$, then holds with same $\epsilon$ for refinement of $P$.
+. If $s_i, t_i$ \in $[x_{i-1},x_i]$, then $\sum_{i=1}^n |f(s_i) - f(t_i)| \delta \alpha_i < \epsilon$.
+. If f is integrable and hypotheses of $b$ hold, then
+$$\left| \sum_i f(t_i) \delta\alpha_i - \int f d\alpha \right| < \epsilon.$$
+Theorem: Continuitiy implies integrability. Proof: use fact uniformly continuous.
+Theorem: Monotonicity of $f$ and continuous $\alpha$ implies integrability.
+Theorem: If $\alpha$ is continuous where $f$ is discontinuous,
+and $f$ has only finite number of discontinuities, then f is integrable.
+Theorem: Suppose $f \in \mathcal {R}(\alpha)$ on $[a,b]$, $m \le f\le M$,
+$\phi$ continuous on $[m, M]$, and $h(x) = \phi(f(x))$ on $[a,b]$. Then
+$h \in \mathcal{R}{\alpha}$ on $[a,b]$.
+**Properties of the integral** (That you didn't learn in Calculus BC):
+$$\left| f d\alpha \right| \le M[\alpha(b) - \alpha(a)]$$
+$$\int f d(\alpha_1 + \alpha_2) = \int f d\alpha + \int f d\alpha_2$$.
+$f$,$g$ integrable implies $fg$ integrable.
+Proof sketch: let $\phi(t) = t^2$ and use identity $4fg = (f+g)^2 - (f-g)^2$.
+$$\left| \int f d\alpha \right| \le \int |f| d\alpha$$.
+Proof sketch: let $\phi(t) = |t|$.
+Given unit step function $I$, if $\alpha(x) = I(x-s)$ then
+$\int f d\alpha = f(s)$ (This is quite similar to using the Dirac delta function
+in signal processing).
+**Relating sums and integrals: **
+Given nonnegative $c_n$, $\sum_n c_n$ converges, and sequence $s_n$, and $f$ continuous,
+If $\alpha(x) = \sum_{n=1}^\infty c_n I(x-s_n)$
+then $\int f \d \alpha = \sum_{n=1}^\infty c_n f(s_n)$.
+**Relating Derivatives and Integrals**
+Assume $\alpha'$ is integrable. $f$ is integrable $\iff$ $f\alpha$ is integrable.
+In that case $\int f d\alpha = \int f(x) a'(x) dx$.
+Proof sketch: Use Theorem 6.6. Then use mean value theorem to
+obtain points $t_i$ such that $\delta \alpha_i = \alpha'(t_i) \delta x_i$.
+**Change of Variable Theorem**
+If $\varphi$ is strictly increasing continuous function that maps $[A,B]$ onto $[a,b]$,
+and $\alpha$ is monotonically increasing on $[a,b]$ and $f\in \mathcal R(\alpha)$
+on $[a,b]$. Define $\beta(y) = \alpha(\varphi(y))$ and $g(y) = f(\phi(y))$.
+Then $\int _A^B g d\beta = \int_a^b f d\alpha$.
+**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 ======
+: 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.
+. 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 $\sqrt 2 \notin \mathbb Q$.
+. 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
+. 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.
+. 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.
+. 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.
+. 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.
+. 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)
+: 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 $L = 0$.
+: 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.
+: 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.
+: 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 $|f(x)-f(c)| = 0$.

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