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 ====== Ryan Purpura's Final Notes =======
-Questions can be found at the very bottom.
-===== Number systems =====
-a====== Ryan Purpura's Final Notes =======
 Questions can be found at the very bottom.
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 : Give an example where it's neither.
-A: Consider $E = \{1, 1/2, 1/3, 1/4, \dots}$.
+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 $\delta$.
-$\mathbb{N}$ is the set of natural numbers $\{1, 2, 3, \dots\}$.
+: What is a smooth function, and give an example one.
-Key properties of $\mathbb{N}$ are that $1 \in \mathbb{N}$ and
-$n \in \mathbb{N} \implies n + 1 \in \mathbb{N}$.
-This natually leads to the idea of mathematical induction,
-which allows us to prove statements for all $\mathbb{N}$.
-**Mathematical induction** works as follows: given a proposition $P_n$,
+A: A smooth function is infinitely differentiable.
-prove for the base case $P_1$ (or some other starting place)
+All polynomials are smooth.
-and then show that $P_k \implies P_{k+1}$. This shows that
-$P_n$ is true for all $\mathbb{N}$ (or subset if
-your base case is $n = 4$ for instance).
-Integers $\mathbb{Z}$ extend $\mathbb{N}$ to include $0$ and negative numbers.
+: Prove differentiability implies continuity.
-Rationals $\mathbb{Q}$ are ratios of integers.
-Property: If $r = \frac{c}{d} \in \mathbb{Q}$ ($c$ and $d$ are coprime) and $r$ satisfies $\sum_{i=0}^{n} c_i x^i$
+A: Differentiability
-with $c_i \in \mathbb{Z}$, $c_n \ne 0$, $c_0 \ne 0$, then $d$ divides $c_n$ and $c$ divides $c_0$.
+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 $\lim_{t\to x} f(t) = f(x)$.
-A natural corollary to this can be found by taking the contrapositive, stating that
+: Use L'Hospital's Rule to evaluate $\lim_{x\to \infty} \frac{x}{e^x}$.
-if $d$ does not divide $c_n$ or $c$ does not divide $c_0$, then $\frac{c}{d}$
-cannot satisfy the equation. This allows us to enumerate all possible
-rational solutions to equations of that form.
-Example: prove $\sqrt{2} \notin \mathbb{Q}$.
+A: We clearly see that the limits of the top and bottom go to infinity,
-$\sqrt{2}$ solves $x^2 - 2 = 0$. By the rational root theorem, the only possible
+so we can use L'Hospital's Rule.
-rational roots are $\pm \{1, 2\}$. Plugging these in, we can clearly see that
+Taking derivative of top and bottom, we get $1/e^x$ which goes to $0$.
-none of them solve the equation. Thus the equation has no rational solution,
-and so $\sqrt{2}$ cannot be rational.
-===== Sets and sequences =====
+: 5.1 Prove that if $c$ is an isolated point in $D$, then $f$ is automatically continuous at $c$.
-**Maximum** of a set of $S \in \mathbb{R}$ is $\alpha \in S$ such that $\alpha \ge \beta$ $\forall \beta \in S$.
+A: Briefly, no matter what $\epsilon$ is chosen, we can just choose a small
-**Minimum** is defined similarly.
+enough neighborhood such $c$ is within the neighborhood.
-Maxima and minima need not exist. However, they must exist for finite, nonempty
+Then all $x$ (i.e., just c) in the neighborhood have $|f(x)-f(c)| = 0$.
-subsets of $\mathbb{R}$.
-Upper and lower bounds are defined similarly, but now $\alpha$ need not exist in $S$.
-These also need not exist, e.g. $S = \mathbb{R}$ has no upper/lower bounds.
-We define the least upper bound of S to be the $\sup{S}$ and the greatest lower bound to be
- $\inf{S}$.
-Once again, these need not exist. However, we assume the Completeness Axiom to state that
-if $S$ is bounded from above, $\sup{S}$ exists, and likewise for $\inf{S}$.
-This allows us to prove the **Archimedian Property**, which states $a, b > 0 \implies \exists n \in N \text{ s.t. } na > b$.
-A quick proof sketch: assume for sake of contradiction that $a,b > 0$ but $\forall n \in \mathbb{N}$, $na \le b$.
-Bounded above, so use completeness axiom to show that $\sup_n{na}$ exists but is not an upper bound.
-A sequence is a function $\mathbb{N} \to \mathbb{R}$, typically denoted as $a_n$.
-We define the **limit of a sequence** to be the number $\alpha \in \mathbb{R}$ if
-$\forall \epsilon > 0$, $\exists N > 0$ such that $\forall n > N$, we have
-$|a_n - \alpha| < \epsilon$.
-An important theorem is **that all convergent sequences are bounded** (but not all bounded
-sequences are convergent!!). Proof sketch: fix an $\epsilon$, which
-gives us an $N$ for which $n > N \implies a_n$ within $\epsilon$ of the limit.
-Then the function is either bounded by the $\epsilon$ bound or by
-the maximum of $|a_n|$ for $n < N$, which is a finite set.
-Limits of addition, multiplication, and division are equal to limits of
-addition, multiplication, and division of limits (assuming the individual limits exist
-and for division, taking care not to let the denominator equal 0).
-Note that $a_n \ne 0$ does not imply $\lim a_n \ne 0$.
-Examples: $a_n = 1/n$, which $\ne 0 \forall n$ but $lim a_n = 0$.
-We define $\lim a_n = +\infty$ if $\forall M > 0, \exists N > 0$ such that $a_n > M$ $\forall n > N$.
-For example, $a_n = n + \sqrt{n} \sin n$ goes to $+\infty$ because $n + \sin n \ge n - \sqrt{n}$.
-Informally, because $n$ grows faster than $\sqrt(n)$, $n - \sqrt{n}$ goes to infinity.
-**Monotone sequences**: if a sequence is increasing (resp. decreasing) and bounded, then it is convergent.
-Proof sketch: consider the set $S$ of all $a_n$. By completeness theorem,
-$\sup$ (resp. $\inf$) exists, so there is a sequence element greater than $\sup(S) - \epsilon$.
-Monotonicity implies that all elements after that satisfy the $\epsilon$ bound.
-===== Lim sup, Lim inf, Cauchy Sequences =====
-Given a sequence $a_n$, define $S_N = \sup {a_n | n \ge N}$.
-Notably, $N < M \implies S_N \ge S_M$.
-So decreasing and has a (possibly infinite) limit.
-We define $\lim \sup a_n = \lim_{N\to \infty} S_N = \lim_{N\to\infty}\left( \sup_{n>N} a_n \right)$
-Example: $a_n = 3 + (-1)^n = \{2, 4, 2, 4, \dots\}$.
-$\lim\sup{a_n} = 4$, $\lim\inf{a_n} = 2$.
-Some properties of these: $$\lim\inf(a_n) \le \lim\sup(a_n)$$, since
-nonstrict inequalities are preserved under limits.
-$$\lim\sup s_{n_k} \le \lim\sup s_n$$
-Proof sketch: use the fact that subsets have smaller (or equal)
-$\sup$ compared to the superset.
-**Cauchy sequence**: $a_n$ is a Cauchy sequence if $\forall \epsilon > 0$, $\exists N > 0$
-such that $\forall n, m > N$, we have $|a_n - a_m| < \epsilon$.
-**Cauchy sequence $\iff a_n$ converges. **
-Proof sketch for reverse direction: consider N for bound $\epsilon / 2$. Then use
-triangular inequality.
-Proof sketch for forward direction: first show (a_n) converges iff $\lim\sup(a_n) = \lim\inf a_n$.
-Then show that $\lim\sup a_n = \lim\sup a_n$ exist and are equal, taking advantage of the
-fact that Cauchy implies bounded.
-**Recursive sequences**: if $S_n = f(S_{n-1})$, then if the limit exists and equals $\alpha$, then
-$\alpha = f(\alpha)$.
-Example: If $S_n = \frac{s_{n-1} + 5}{2 S_{n-1}}$ with $S_1 =5$,
-$\lim_{n\to\infty} a_n = \alpha$ \implies  $\frac{\alpha^2 +5}{2\alpha} = \alpha$
-which means $\alpha = \pm \sqrt{5}$. Be careful though, because the
-initial condition matters. Given out initial condition, we can bound
-$S_n$ between $\sqrt{5}$ and $5$ inclusive for all relevant $n$ using induction,
-which implies $\sqrt{5}$ is the correct answer.
-**Subsequences**: If $s_n$ is a subsequence, let $n_k$ be a strictly increasing
-sequence in $\mathbb{N}$. Then we can define a new sequence
-$t_k = \left(S_{n_k}\right)_k$, called a subsequence of $s_n$.
-An important property of subsequences is that even if a sequence does not converge to a value,
-a subsequence of that sequence may. Specifically,
-let $(s_n)$ be any sequence. $(s_n)$ has a subsequence converging to $t$ iff $\forall \epsilon > 0$,
-the set $A_{\epsilon} = \{ n \in \mathbb{N} | |s_n-t| < \epsilon\}$ is infinite.
-In other words, there are infinite elements within the epsilon bound of $t$ in $s_n$.
-Forward direction proof sketch: just use the definition of convergence.
-Reverse direction proof sketch: Consider taking $\epsilon$
-smaller and smaller and taking an element from each $\epsilon$ shrinkage
-to create the convergent subsequence (make sure that each element
-has a greater index than the one before it).
-**Every sequence has a monotone subsequence.** Proof sketch:
-Define a dominant term to be $s_n$ such that $\forall m > n$, $s_n > s_m$.
-Either there are infinitely many dominant terms (then dominant terms form a
-monotone subsequence), or there are finitely many. In the second case,
-then after the last dominant term, each element has some element greater than
-it with a higher index. Use this to form a monotone subsequence.
-This theorem allows us to prove that every bounded sequence has a convergent
-subsequence, because a monotone subsequence must exist and it will be bounded,
-implying that it is convergent.
-Given a sequence (s_n), we say $t \in \mathbb{R}$ is a **subsequence limit**
-if there exists a subsequence that converges to $t$. For example,
-$\lim\inf s_n$ and $\lim\sup s_n$ are subsequence limits. Proof
-sketch: Use definition of $\lim\sup$ and $\lim\inf$, specifically
-that it is a limit of $\sup$ values of the tail. Use the limit to
-produce an $\epsilon$ bound on the $\sup$ of tails for large enough $n$.
-Then show that this implies infinite elements of the original sequence
-in that  $\epsilon$ bound.
-Containing a single subsequence limit is necessary and sufficient for the existence
-of a convergent sequence.
-Closed subsets: $S \subset \mathbb{R}$ is closed if for all convergent sequences
-in $S$, the limit also belongs to $S$.
-Example: $(0, 1)$ is not closed, because the sequence $a_n = 1/n$ converges
-to $0$, which is not in the set.
-An extremely important result is that for positive sequences,
-$$\lim\inf \frac{s_{n+1}}{s_n} \le \lim\inf \left(s_n\right)^{1/n} \le \lim\sup \left(s_n\right)^{1/n}
-\le \lim\sup \frac{s_{n+1}}{s_n}.$$
-Proof sketch for last inequality (first is similar):
-Use fact that if $S_{n+1}/S_n \le L + \epsilon$, then $S_{n+k}/S{n} \le (L + \epsilon)^k$.
-For $n > N$, we have $S_n^{1/n} \le \left( S_N (1+\epsilon)^{n-N} \right)^{1/n}$.
-Take $\lim\sup$ of both sides, and the right side can be massaged to show $\lim\sup(S_n)^{1/n} \le L + \epsilon$.
-===== Metric Spaces =====
-A **metric space** is a set $S$ and a function $d: S \cross S \to \mathbb{R}^{0+}$.
-The function $d$ must satisfy nonnegativity, $d(x,y) = 0 \iff x = y$, commutativity,
-and the triangular inequality.
-We can generalize sequences to use $S$ instead of $R$. Specifically, if $s_n$ is a sequence in $S$,
-then we define $s_n$ to be **Cauchy** if $\forall \epsilon > 0, \exists N > 0$ s.t. $\forall n,m > N$, $d(s_n, s_m) \le \epsilon$. Also, we say $s_n$ converges to $s \in S$ if $\forall \epsilon > 0$, $\exists N > 0$ s.t. $\forall n > N$,
-$d(s_n, s) < \epsilon$. Just like the real number case, these two notions are equivalent.
-We call a metric space $(S, d)$ complete if every Cauchy sequence has a limit in $S$.
-For example, $\mathbb{Q}$ is not complete, but $\mathbb{R}$ is.
-The **Bolzano-Weierstrass Theorem** states that every bounded sequence in $\mathbb{R}$
-has a convergent subsequence.
-A topology on a set $S$ is a collection of open subsets such that
-$S, \varnothing$ are open; (possibly infinite) union of open subsets are open, and
-finite intersection of open sets are open. We can define a topology on a metric
-space by defining the ball $B_r(p) = \{ x \in S | d(p,x) < r \}$ to be open.
-We define $E \subset S$ to be **closed** iff $E^c$ is open, that is,
-$\forall x \notin E$, $\exists \delta > 0$ such that $B_\delta(x) \cap E = \varnothing$.
-We define the **closure** of a set $E \in S$ to be $$\bar{E} = \cap \{ F | F \subset S \text{ closed set, } F \subset E \}$.
-We similarly define the **interior** to be the union of all open subsets of $E$.
-The boundary of $E$ is the set difference bewteen the closure and the interior.
-We define the **set of limit points** $E'$ to be all such points $p$ where
-$\forall \epsilon > 0, \exists q \in E, q \ne p$, such that $d(p, q) < \epsilon$.
-Very importantly, $E \cup E' = \bar{E}$. (It's also possible
-to *define* $\bar E$ this way as Rudin does)
-An important property relating closedness and sequences is that
-$$E \subset S \text{ is closed } \iff  \forall \text{ convergent sequence } x_n in S,
-x = \lim x_n$$. The proof for both directions utilize proof by contradiction.
-===== Compactness =====
-We define an **open cover** of $E \subset S$ to be a collection of open sets $\{G_d\}_{\alpha\in A}$
-such that $E \subset \bigcup _\alpha G_\alpha$. From there, we can define compactness.
-A set $K \subset S$ is **compact** if for any open cover of $k$, there exists a finite subcover.
-For example, if $K$ is finite, we can produce a finite subcover by, for each $x \in K$, picking
-a $G_x$ such that $x \in G_x$. Then the subcover is finite.
-The **Heine-Borel Theorem** states that a subset of $\mathbb{R^n}$ is compact iff it is closed and bounded.
-Lemma: a closed subset in a compact subset is compact.
-**Sequential compactness** is an alternate definition of compactness, which states that
-$X$ is sequentially compact if every sequence of points in $X$ has a convergent
-subsequence converging to a point in $X$.
-===== Series =====
-Series is an infinite sum of a sequence: $\sum _{n=1}^\infty a_n$.
-We define convergence as the convergence of partial sums $S_n = \sum_{j=1}^n a_j$.
-We say that a series satisfies the **Cauchy Condition** if $\forall \epsilon > 0,$
-$\exists N > 0$ such that $\forall n,m > N$,
-$\left| \sum_{j=n}^m a_n < \epsilon \right|$.
-A Cauchy series is equivalent to the sequence of partial sums being Cauchy
-which is equivalent to the convergence of both.
-If the series converges, then $\lim a_n = 0$. Note that the opposite is
-not necessarily true.
-To determine whether series converge, we can use comparison tests.
-**Comparison test**: $\sum_n a_n \text{ converges } \wedge |b_n| < a_n  \implies b_n \text{converges} $
-Proof sketch: Show that $\sum_n b_n$ is Cauchy. Related, we say that $\sum_n b_n$
-**converges absolutely** if $\sum_n |b_n|$ converges.
-A classic example of a series that does not converge absolutely is the
-alternating harmonic series.
-For the following tests, we recall that
-$$\lim\inf \frac{s_{n+1}}{s_n} \le \lim\inf \left(s_n\right)^{1/n} \le \lim\sup \left(s_n\right)^{1/n}
-\le \lim\sup \frac{s_{n+1}}{s_n}.$$
-** Root Test **
-Let $\alpha = \lim\sup |a_n|^{1/n}$.
-. If $\alpha > 1$, then series diverges
-. If $\alpha < 1$, then series converges absolutely.
-. If $\alpha = 1$, then series could converge or diverge.
-** Ratio Test **
-Let $\alpha = \lim\sup \left| \frac{a_{n+1}}{a_n} \right|$.
-. If $\alpha \le 1$, the series converges absolutely.
-. If $\alpha > 1$, the series diverges.
-** Alernating Series Test **
-For "nonincreasing" (i.e. the absolute values of terms are nonincreasing)
-alternating series where the terms tend to 0, the series converges.
-Example: $\sum (-1)^n \frac{1}{\sqrt{n}}$ converges since terms tend to 0
-and $\frac{1}{\sqrt n}$ is nonincreasing.
-** Integral Test **
-If $f$ is continuous, positive, and decreasing such that $f(n) = a_n$,
-then convergence of the integral is same as convergence of series.
-===== Continuity and Uniform Convergence =====
-I already LaTaX'd notes on continuity which can be found here:
-https://rpurp.com/2021-03-02.pdf
-Now let's examine the link between compactness and continuity. Specifically,
-if $f$ is a continuous map from $X$ to $Y$, then $f(E) \subset Y$ is compact
-if $E$ is compact. This can be proven by sequential compactness: a quick
-proof sketch is that for a sequence in $f(E)$, we can reverse the map to find
-a sequence in $E$. Since $E$ is compact, there's a subsequence that converges
-to a value in $E$. Since continuity preserves limits, this is a
-subsequence limit for the sequence in $Y$ as well.
-A corollary to this is that if $f: x\to \mathbb{R}$ is continuous
-and $X \subset E$ is compact, then $\existsp, q \in E$ such that
-$f(p) = \sup f(E)$ and $f(q) = \inf f(E)$.
-However, a preimage of a compact set may not be compact.
-**Uniform Continuity**
-$f: X \to Y$ is **uniformly continuous** if $\forall \epsilon > 0$, $\exists \delta > 0$
-such that $\forall p,q \in X, d_X(p, q) < \delta \implies d_Y(f(p), f(q)) < \epsilon$.
-Compared with the normal definition of continuity, we notice that we must
-select a $\delta$ that works with *all* points. For regular continuity, we can
-choose a \delta per-point$.
-Theorem: For $f: X \to Y$, if $f$ is continuous and $X$ is compact, then $f$ is uniformly
-continuous.
-Example: $f(x) = 1/x$. Normally continuous except at $x= 0$. But if we restrict the
-domain to a compact interval $[1, 2]$, $f$ becomes uniformly continuous.
-Interesting theorem: Given $f: X \to Y$ continuous with $X$ compact, and $f$
-a bijection, then $f^{-1}: Y \to X$ is continuous.
-Proof sketch: Show the preimage of an open set in $f^{-1}$ is open.
-** Connectedness **
-We define connectedness as follows: $X$ is connected iff the only subset of $X$
-that is both open and closed are $X$ and $\varnothing$.
-An alternate definition is that $X$ is not connected iff
-$\forall U, V \subset X \text{ nonempty, open }$, $\U \cap V = \varnothing$ and $X = U \coprod V$.
-Or $\exists S \subset X, \varnothing \ne S \ne X$ such that $S$ is both open and closed.
-Then $X = S \coprod S^c$.
-Example: $X = [0,1] \union [2, 3]$.
-Then let $S = [0,1]$ and $S^c = [2,3]$.
-Theorem: Continuous functions preserve connectedness.
-Theorem: $E \subset \mathbb{R} \text{ connected } \iff \forall x,y \in E, x < y, [x,y] \subset E$.
-Rudin definition for connectedness:
-$S$ cannot be written as $A \cup B$, where $\bar A \cap B = A \cap \bar B = \varnothing$.
-**Intermediate value theorem**: This falls almost directly out of
-the fact that continuous functions preserve connectedness.
-If  $f: [a,b] \to \mathbb{R}$ is continuous, and $f(a) < f(b)$, then
-$\forall y \in (f(a), f(b)), \exists x \in (a,b)$ s.t. $f(x) = y$.
-**Discontinuities**:
-Given the left and right-hand limits of $f$ at $x_0$,
-if $f(x_0) = f(x_0^+) = f(x_0^-)$, we say the function is continuous at $x_0$.
-If both left and right-handed limits exist, but they disagree
-with each other or the value of the function, we call this a simple discontinuity.
-If one of the sides doesn't exist at all (classic example is $\sin(1/x)$ around
-$x=0$, then we call it a 2nd-type discontinuity.
-**Monotonicity and Continuous Functions**:
-If $f: (a,b) \to \mathbb{R}$ is monotone-increasing, then $f$ has countably
-many discontinuities.
-===== Convergence of Functions =====
-We say that a sequence of functions $f_n$ converges **pointwise** to $f$
-if $\forall x \in X$ we have
-$\lim f_n(x) =f(x)$.
-Example: $f_n(x) = 1 + \frac{1}{n} \cdot \sin x$.
-Then $f_n \to f$ pointwise where $f(x) = 1$.
-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.
-We can see here that pointwise convergence does not preserve integrals.
-Also, convergence of $f$ doesn't imply convergence of $f'$ (pointwise).
-Vectors converging pointwise is equivalent to them converging in a $d_2$ sense. (Actually,
-as long as it is a norm).
-We can also have uniform convergence of functions.
-We say that $f_n$ converges to $f$ **uniformly** if $\forall \epsilon > 0$, $\exists N > 0$
-such that $\forall n > N$, $\forall x \in X$, we have $|f_n(x)-f(x)| < \epsilon$.
-We can also express uniform convergence using the equivalent notion of **Uniformly Cauchy**:
-$\forall \epsilon > 0$ $\exists N > 0$ s.t. $\forall n,m > N$, $|f_n(x) -f_m(x) | < \epsilon$.o
-We also can talk about **series of functions** $\sum_{n=1}^\infty f_n$.
-We say that this converges uniformly to $f$ if the partial sums $F_n = \sum_{n=1}^N f_n$ converge
-uniformly to $f$.
-A test we can apply to see whether a series of functions converges uniformly is the
-**Weierstrass M-test**. If $M_n \in \mathbb{R}$ s.t. $M_n \ge \sup_{x\in X} |f_n(x)|$:
-if $\sum_{n=1}^\infty$, then $\sum_{n=1}^{\infty}$ converges uniformly.
-A proof sketch: Consider the absolute value of the series and use the
-triangular inequality.
-Unlike pointwise convergence, uniform convergence preserves continuity.
-Theorem: If $K$ is compact, and $f_n: K \to \mathbb{R}$ with the following conditions:
-. $f_n$ is continuous
-. $f_n\to f$ pointwise, $f$ continuous
-. $f_n(x) > f_{n+1}(x)$
-Then $f_n \to f$ uniformly.
-===== Derivatives =====
-We define the derivative of $f$ as $f'(x) = \lim_{t\to x} \frac{f(t) -f(x)}{t-x}$.
-Differentiability implies continuity. This can be verified by multiplying the
-above by $t-x$ and using limit rules.
-Chain rule: If $h(t) = g(f(t))$, then $h'(x) = g'(f(x))f'(x)$.
-We say f has a **local maximum** at $p \in X$ if $\exists \delta > 0$ such that
-$f(q) \le f(p)$ for all $q \in X$ with $d(p,q) < \delta$.
-If $f$ has a local maximum at $x \in (a,b)$, and if $f'(x)$ exists, then $f'(x) = 0$.
-This can be shown by bounding above and below by 0, proving it's 0.
-**Generalized MVT**:
-If $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ then $\exists x \in (a,b)$ such that
-$$[f(b) - f(a)]g'(x) = [g(b)-g(a)]f'(x)$$
-If we let $g(x) = x$, we get the regular MVT:
-$$f(b) - f(a) = (b-a) f'(x)$$
-**Intermediate Value Theorem for Derivatives**: If $f$ is differentiable on $[a,b]$ and
-$f'(a) < \lambda < f'(b)$ then $\exists x \in (a,b)$ such that $f'(x) = \lambda$.
-Proof sketch: Let $g(t) = f(t) - \lambda t$. $g'$ goes from negative to postive,
-so somewhere $g$ attains its minimum and $g'(x) = 0$.
-Correlary: If $f$ is differentiable on $[a,b]$, $f'$ cannot have simple discontinuities,
-only discontinuities of the second kind.
-**L'Hospital's Rule**:
-Suppose $f$ and $g$ are real and differentiable on $(a,b)$, and $g'(x) \ne 0$
-for all $x\in (a,b)$. Let $\lim_{x\to a} \frac{f'(x)}{g'(x)} = A$ if it exists.
-If $f(x) \to 0$ and $g(x) \to 0$ as $x \to a$, or if
-$g(x) \to \infty$ as $x \to a$, then
-$$\lim_{x\to a} \frac{f(x)}{g(x)} = A.$$
-The proof is quite involved and is included as a question later.
-**Taylor's Theorem**
-Let $$P(t) = \sum_{k=0}^{n-1} \frac{f^{(k)}(a)}{k!}(t-\alpha)^k$$.
-Then there exists a point $x$ between $\alpha$ and $\beta$ such that
-$$f(\beta) = P(\beta) + \frac{f^{(n)}(x)}{n!}(\beta - \alpha)$$.
-This is essentially a different way to generalize the mean value theorem.
-This proof will also be included as a question.
-===== Integrals =====
-Unless otherwise stated, assume $f$ is bounded and bounds of all integrals are from $a$ to $b$.
-We define a **partition** $P$ of $[a,b]$ as finite set of points $x_0, x_1, \dots, x_n$
-such that $a = x_0 \le x_1 \le \dots \le x_n = b$
-and define $\delta x_i = x_i - x_{i-1}$ for convenience.
-We further define $M_i = \sup f(x)$, $m_i = \inf f(x)$,
-$U(P,f) = \sum_{i=1}^n M_i \delta x_i$, and
-$L(P,f) = \sum_{i=1}^n m_i \delta x_i$.
-We define the upper and lower Riemann integrals of $f$ to be
-$ \overline { \int  } f dx = \inf U(P, f) $ and
-$ \underline {\int } f dx = \sup L(P, f)$.
-If the upper and lower integrals are equal, we say that $f$ is Riemann-integral
-on $[a,b]$ and write $f \in \mathcal{R}$.
-Since $\exists m, M$ s.t. $m \le f(x) \le M$, we have for all $P$,
-$$m(b-a) \le L(P,f) \le U(P,f) \le M(b-a).$$
-So $L(P,f)$ and $U(P,f)$ are bounded, so the upper and lower integrals are
-defined for every bounded function $f$.
-We now generalize a bit. Consider a partition $P$ and let
-$\alpha$ be a monotonically increasing function on $[a,b]$.
-We write $\delta \alpha_i = \alpha (x_i) - \alpha (x_{i-1})$.
-We define $U(P,f, \alpha) = \sum_{i=1}^n M_i \delta\alpha_i$
-and $L(P,f,\alpha) = \sum_{i=1}^n m_i \delta \alpha_i$.
-We define
-$ \overline { \int  } f d\alpha = \inf U(P, f, \alpha) $ and
-$ \underline {\int } f d\alpha = \sup L(P, f, \alpha)$.
-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 \subset P^*$.
-The **common refinement** of $P_1$ and $P_2$ is defined to be $P_1 \union P_2$.
-Theorem: $L(P, f, \alpha) \le L(P^*, f, \alpha)$
-and $U(P^*, f, \alpah) \le U(P,f, \alpha)$.
-Proof sketch: To start, assume $P^*$ contains one more point $x^*$ than $P$
-and consider the $\inf f(x)$ of the two "induced" intervals by $x^*$'s inclusion.
-Clearly both are larger than $m_i$ (corresponding to original interval containing $x^*$.
-Theorem:
-$ \overline { \int  } f d\alpha = \inf U(P, f, \alpha) \ge \underline {\int } f d\alpha = \sup L(P, f, \alpha)$.
-Proof sketch: $L(P_1, f, \alpha) \le U(P_2, f, \alpha)$.
-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$
-$\exists P$ such that $U(P,f,\alpha) - L(P,f,\alpha) < \epsilon$.
-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$.
-Only if direction: If $f \in \mathcal R$, $\exists P_1, P_2$ s.t.
-$U(P_2, f, \alpha) - \int f d\alpha < \epsilon/2$
-and $\int f d\alpha - L(P_1, f, \alpha) < \epsilon/2$.
-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!

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