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--- math104-s21:s:vpak [2021/05/09 18:59]
68.186.63.173 [Summary of Material]
+++ math104-s21:s:vpak [2026/02/21 14:41] (current)
@@ Line 12: / Line 12: @@
 The maximum of a set S is the largest element in the set.\\ The minimum is the smallest element in the set.\\  The $\inf$ of S is the greatest lower bound.\\  The $\sup$ of S is the smallest upper bound.\\ S is bounded if $\forall$s $\in$ S, s$\leq$M for some M $\in$ $\reals$\\ **Completeness Axiom.** If S is a nonempty bounded set in $\reals$, then $\inf$ S and $\sup$ S exist.\\ **Archimedean Property.** If a, b $\gt$ 0, then $\exists$n such that na $\gt$ b.
-A //sequence// (s<sub>n</sub>) is a function mapping from $\N$ to $\R$. It converges to s if $\forall$ $\epsilon$ > 0 there exists N such that N > n $\implies$ |(s<sub>n</sub>)-s| < $\epsilon$\\ In other words,
+A //sequence// (s<sub>n</sub>) is a function mapping from $\N$ to $\R$. It converges to s if $\forall$ $\epsilon$ $>$ 0 there exists N such that N $>$ n $\implies$ |(s<sub>n</sub>)-s| $<$ $\epsilon$\\ In other words,
-$\lim$(s<sub>n</sub>) = s
+$\lim$(s<sub>n</sub>) $=$ s
-Important limit theorems include:\\ $\lim$(s<sub>n</sub>)(t<sub>n</sub>) = ($\lim$(s<sub>n</sub>))($\lim$(t<sub>n</sub>))\\ $\lim$(s<sub>n</sub>)+(t<sub>n</sub>) = ($\lim$(s<sub>n</sub>)) + ($\lim$(t<sub>n</sub>))\\ $\lim$($\frac{1}{n^p}$) = 0 for p > 0\\ $\lim$ n<sup>(1/n)</sup> = 1
+Important limit theorems include:\\ $\lim$(s<sub>n</sub>)(t<sub>n</sub>) $=$ ($\lim$(s<sub>n</sub>))($\lim$(t<sub>n</sub>))\\ $\lim$(s<sub>n</sub>)+(t<sub>n</sub>) $=$ ($\lim$(s<sub>n</sub>)) + ($\lim$(t<sub>n</sub>))\\ $\lim$($\frac{1}{n^p}$) $=$ 0 for p > 0\\ $\lim$ n<sup>(1/n)</sup> $=$ 1
 A //subsequence// (s<sub>n(k)</sub>) of (s<sub>n</sub>) is a sequence that is a subset of the elements in the original sequence with relative order preserved.\\ **Bolzano-Weierstrass Theorem.** Every bounded sequence has a convergent subsequence, having some //subsequential limit//.
-Given any (s<sub>n</sub>) and let S be the set of subsequential limits of (s<sub>n</sub>). Define:\\ $\lim$ $\sup$ (s<sub>n</sub>) = $\lim\limits_{N \to \infin}$ $\sup${(s<sub>n</sub>): n > N} = $\sup$ S\\
+Given any (s<sub>n</sub>) and let S be the set of subsequential limits of (s<sub>n</sub>). Define:\\ $\lim$ $\sup$ (s<sub>n</sub>) $=$ $\lim\limits_{N \to \infin}$ $\sup${(s<sub>n</sub>): n > N} $=$ $\sup$ S\\
-$\lim$ $\inf$ (s<sub>n</sub>) = $\lim\limits_{N \to \infin}$ $\inf${(s<sub>n</sub>): n > N} = $\inf$ S
+$\lim$ $\inf$ (s<sub>n</sub>) $=$ $\lim\limits_{N \to \infin}$ $\inf${(s<sub>n</sub>): n > N} $=$ $\inf$ S
 $\lim$ $\inf$ |s<sub>n+1</sub>| / |s<sub>n</sub>| $\leq$ $\lim$ $\inf$ |s<sub>n</sub>|^(1/n) $\leq$ $\lim$ $\sup$ |s<sub>n</sub>|^(1/n) $\leq$ $\lim$ $\sup$ |s<sub>n+1</sub>| / |s<sub>n</sub>|
@@ Line 26: / Line 26: @@
 **2. Topology**
-**Metric Space:** A set S with a //metric//, distance function d. For any x,y,z $\in$ S \\ (1) d(x,y) > 0 if x $\not =$ y, d(x,x) = 0 \\ (2) d(x,y) = d(y,x)\\ (3) d(x,z) $\leq$ d(x,y) + d(y,z) \\ **Important* **A metric is only valid if it outputs a real number for any inputs, ie. d(x,y) = $\infin$ is not valid.
+**Metric Space:** A set S with a //metric//, distance function d. For any x,y,z $\in$ S \\ (1) d(x,y) $>$ 0 if x $\not =$ y, d(x,x) $=$ 0 \\ (2) d(x,y) $=$ d(y,x)\\ (3) d(x,z) $\leq$ d(x,y) + d(y,z) \\ **Important* **A metric is only valid if it outputs a real number for any inputs, ie. d(x,y) $=$ $\infin$ is not valid.
+A //limit point// p of a set S is such that for some $\epsilon$ radius ball around p, there exists an element q $\not =$ p such that q $\notin$ S. Note that limit points may or may not lie in the set. \\ A set S is //open// if for every point p in S is interior in S. Think open ball of $\epsilon$ radius in S centered at p. \\ A set S is //closed// if every limit point of S is a point in S. \\ A set S is //perfect// if it is closed and every interior point is a limit point. \\ A set S is //dense// in a metric space X if every point in X is either a limit point of S or in S itself. \\ The //closure// of a set S is the union of S and the set of its limit points. It can also be thought of as the intersection of all closed sets containing S.
+An //open cover// for S is a collection of open sets that covers S. \\
+A set S is //compact// if for all open covers {G} of S, there exists a finite subcover of {G} that covers S. \\ **Heine-Borel Theorem.** A subset E of $\R$<sup>n</sup> is compact //if and only if// it is closed and bounded.
+A set S is //connected// if it **cannot** be written as disjoint union of nonempty, open sets
+**3. Series**
+A series //converges// if its partial sum $\Sigma$ s<sub>n</sub> $=$ M for some M $\in$ $\reals$. \\
+A series $\Sigma$ s<sub>n</sub> satisfies is //Cauchy// if for any $\epsilon$ $>$ 0, there exists N such that n $\geq$ m $>$ N $\implies$ |$\displaystyle\sum_{i=m}^n$ s<sub>i</sub>| $<$ $\epsilon$  \\
+If $\Sigma$ s<sub>n</sub> converges, then $\lim$ s<sub>n</sub> $=$ 0. (Note this is a one-way statement)
+**Comparison Test** \\
+If $\Sigma$ a<sub>n</sub> converges and |b<sub>n</sub>| $\leq$ a<sub>n</sub> $\forall$n, then $\Sigma$ b<sub>n</sub> converges. \\ If $\Sigma$ a<sub>n</sub> $=$ $\infin$ and |b<sub>n</sub>| $\geq$ a<sub>n</sub> $\forall$n, then $\Sigma$ b<sub>n</sub> $=$ $\infin$.
+**Ratio Test**  $\Sigma$ a<sub>n</sub> of nonzero terms \\
+(i) converges if $\lim$ $\sup$ |a<sub>n+1</sub> / a<sub>n</sub>| $<$ 1 \\
+(ii) diverges if $\lim$ $\inf$ |a<sub>n+1</sub> / a<sub>n</sub>| $>$ 1 \\
+(iii) else inconclusive test
+**Root Test** Let $\alpha$ $=$ $\lim$ $\sup$ |a<sub>n</sub>|<sup>(1/n)</sup>. Then $\Sigma$ a<sub>n</sub> \\
+(i) converges if $\alpha$ $<$ 1 \\
+(ii) diverges if $\alpha$ $>$ 1 \\
+(iii) else inconclusive test
+**Alternating Series Theorem**  If a<sub>n</sub> is monotonically decreasing and $\lim$ a<sub>n</sub> $=$ 0, then $\Sigma$ (-1)<sup>n</sup> a<sub>n</sub> converges.
+**4. Continuity and Convergence**
+There are three main definitions for a continuous function $\bold{f}$: \\
+(1) $\bold{f}$ continuous at x if for each $\epsilon$ $>$ 0, there exists $\delta$ $>$ 0 such that |x-y| $<$ $\delta$ where y $\in$ domain(f) $\implies$ |f(x) - f(y)| $<$ $\epsilon$ \\
+(2) $\bold{f}$ continuous at x if for all sequences (s<sub>n</sub>) in domain(f) that converge to x, $\lim$ f(s<sub>n</sub>) $=$ f(x) \\
+(3) Let $\bold{f}$ be mapping between metric spaces X $\to$ Y. $\bold{f}$ is continuous if the preimage $\bold{f}$<sup>-1</sup> of any open set in Y is open in X. Similarly if the preimage $\bold{f}$<sup>-1</sup> of any closed set in Y is closed in X.
+A  function $\bold{f}$ is //uniformly continuous// if for **all x in domain(f)**, for each $\epsilon$ $>$ 0, there exists $\delta$ $>$ 0 such that |x-y| $<$ $\delta$ where y $\in$ domain(f) $\implies$ |f(x) - f(y)| $<$ $\epsilon$.
+Generally, a continuous function $\bold{f}$ sends a compact set to another compact set. In this case, $\bold{f}$ is bounded, and $\sup$ $\bold{f}$ and $\inf$ $\bold{f}$ exists. \\ A continuous function acting on a compact set is uniformly continuous on this interval. \\
+//Cauchy relation:// If $\bold{f}$ is uniformly continuous on a set S and s<sub>n</sub> is a Cauchy sequence in S, then $\bold{f}$(s<sub>n</sub>) is a Cauchy sequence.
+Generally, a continuous function $\bold{f}$ sends connected set to another connected set.
+**Intermediate Value Theorem.** Let $\bold{f}$ be a continuous real function on interval [a,b]. Then for all y between f(a) and f(b), there exists at least one x $\in$ (a,b) such that f(x) $=$ y.
+If a function $\bold{f}$ is //discontinuous// at x, and both $\bold{f}$(x+) and $\bold{f}$(x-) exist, then $\bold{f}$ is said to have discontinuity of the //first kind// at x, or //simple// discontinuity.
+A function $\bold{f}$<sub>n</sub> converges to f //pointwise// if for each x in the domain, $\lim\limits_{n \to \infin}$ $\bold{f}$<sub>n</sub>(x) $=$ f(x).
+A function $\bold{f}$<sub>n</sub> converges to f //uniformly// if $\lim$ $\sup${|$\bold{f}$<sub>n</sub> - f|} $=$ 0.\\  In other words, for any fixed $\epsilon$ $>$ 0 there exists N such that n $>$ N $\implies$ |$\bold{f}$<sub>n</sub>(x) - f(x)| $<$ $\epsilon$ for all x in the domain. This is a stronger restraint than pointwise convergence, where we only cared about each x individually. (Compare with regular vs uniform continuity) \\ Uniform Cauchy $\iff$ Uniform Convergence
+**Weierstrass M-Test.** Let $\bold{f}$ $\ =$ $\textstyle\sum_{i=1}^\infin$ $\bold{f}$<sub>n</sub>. If $\exists$M<sub>n</sub> $>$ 0 such that $\sup$ |$\bold{f}$<sub>n</sub>| $\leq$ M<sub>n</sub> and $\textstyle\sum_{i=1}^\infin$ M<sub>n</sub> $<$ $\infin$, then $\textstyle\sum_{i=1}^\infin$ $\bold{f}$<sub>n</sub> converges uniformly.
+Let K be a compact set, and the following: \\
+  * {$\bold{f}$<sub>n</sub>} is sequence of monotonically decreasing, continuous functions \\
+  * {$\bold{f}$<sub>n</sub>} $\to$ $\bold{f}$ pointwise \\
+Then $\bold{f}$<sub>n</sub> $\to$ $\bold{f}$ uniformly.
+**5. Differentiation and Integration**
+Let $\bold{f}$ be real-valued on [a,b]. Its //derivative// is defined as \\
+$\bold{f'(x)}$ $=$ $\lim\limits_{t \to x}$ $\frac{\bold{f(t)} - \bold{f(x)}}{\bold{t - x}}$
+If $\bold{f}$ is differentiable at x, then $\bold{f}$ is also continuous at x. \\
+If $\bold{f}$ is differentiable on interval $\bold{I}$, and $\bold{g}$ is differentiable on range($\bold{f}$),  then $\bold{h}$ $=$ $\bold{g(\bold{f})}$ is differentiable on $\bold{I}$
+A real function $\bold{f}$ has a //local maximum// at point p if there exists $\delta$ $>$ 0 such that $\bold{f(y)}$ $\leq$ $\bold{f(x)}$ for any y where d(x,y) $<$ $\delta$. \\
+If $\bold{f}$ has a local maximum at x, and if $\bold{f'(x)}$ exists, then $\bold{f'(x)}$ $=$ 0.
+**Mean Value Theorem.** If $\bold{f}$ is a real continuous function on [a,b], and is differentiable on (a,b), then there exists an x $\in$ (a,b) such that \\
+$\bold{f(b)}$ $-$ $\bold{f(a)}$ $=$ (b $-$ a) $\bold{f'(x)}$ \\ The generalized theorem for $\bold{f}$ and $\bold{g}$ continuous real functions on [a,b] is \\
+($\bold{f(b)}$ $-$ $\bold{f(a)}$) $\bold{g'(x)}$ $=$ ($\bold{g(b)}$ $-$ $\bold{g(a)}$) $\bold{f'(x)}$
+**Theorem 5.12.** Suppose $\bold{f}$ is real differentiable function on [a,b], and $\bold{f'(a)}$ $<$ $\lambda$ $<$ $\bold{f'(b)}$. Then there exists x $\in$ (a,b) such that $\bold{f'(x)}$ $=$ $\lambda$.
+A function $\bold{f}$ is said to be //smooth// on interval I if $\forall$ x $\in$ I, $\forall$ k $\in$ $\N$, $\bold{f^k}$ exists.
+**L'Hopital Rule.** $\lim\limits_{x \to a}$ $\frac{\bold{f(x)}}{\bold{g(x)}}$ $=$ $\lim\limits_{x \to a}$ $\frac{\bold{f'(x)}}{\bold{g'(x)}}$ if either
+  * $\bold{f(x)}$ $\to$ 0 and $\bold{g(x)}$ $\to$ 0 as x $\to$ a
+  * $\bold{g(x)}$ $\to$ $\infin$ as x $\to$ a
+**Taylor's Theorem.** Let $\bold{f}$ be a real function on [a,b], assume $\bold{f^{n-1}}$ is continuous and $\bold{f^n}$ exists, and for any distinct $\alpha$, $\beta$ $\in$ [a,b] define \\
+P(t) $=$ $\displaystyle\sum_{k=0}^{n-1}$ $\frac{\bold{f^k(\alpha)}}{k!}$ (t $-$ $\alpha$)<sup>k</sup> \\
+Then there exists a point x between $\alpha$ and $\beta$ such that \\
+$\bold{f(\beta)}$ $=$ P($\beta$) $+$ $\frac{\bold{f^n(x)}}{n!}$ ($\beta$ $-$ $\alpha$)<sup>n</sup> \\
+Note Taylor Series on smooth functions may not converge, and may not be equal to original function f(x).
+A //partition// P of [a,b] is the finite set of points where a$=$x<sub>0</sub>$\leq$x<sub>1</sub>$\leq$...x<sub>n</sub>$=$b \\
+Let $\alpha$ be a weight function that is monotonically increasing. Define \\
+U(P, f, $\alpha$) $=$ $\displaystyle\sum_{i=0}^{n}$ M<sub>i</sub> $\Delta{\alpha}$<sub>i</sub> \\
+L(P, f, $\alpha$) $=$ $\displaystyle\sum_{i=0}^{n}$ m<sub>i</sub> $\Delta{\alpha}$<sub>i</sub> \\
+where M<sub>i</sub> is the $\sup$ and m<sub>i</sub> is the $\inf$ over that subinterval. \\
+If $\inf$ U(P, f, $\alpha$) $=$ $\sup$ L(P, f, $\alpha$) over all partitions, then the //Riemann integral// of f with respect to $\alpha$ on [a,b] exists \\
+$\int_a^b f(x)d{\alpha}(x)$
+A //refinement// Q of P contains all the partition points in P, with additional points. \\
+If U(P, f, $\alpha$) $-$ L(P, f, $\alpha$) $<$ $\epsilon$, then U(Q, f, $\alpha$) $-$ L(Q, f, $\alpha$) $<$ $\epsilon$. In other words, refinements maintain the condition for integrability.
+**//Key Theorems: //** \\
+  * If f is continuous on [a,b], then f is integrable on [a,b].
+  * If f is monotonic on [a,b] and if $\alpha$ is continuous on [a,b], then f is integrable on [a,b].
+  * Suppose f is bounded and has finitely many discontinuities on [a,b]. If $\alpha$ is continuous at every point of discontinuity, then f is integrable.
+  * If f is integrable on [a,b] and g is continuous on the range of f, then h $=$ g(f) is integrable on [a,b].
+  * If a$<$s$<$b, f is bounded, f is continuous at s, and $\alpha$(x) $=$ I(x-s) where I is the //unit step function//, then $\int_a^b fd{\alpha}$ $=$ f(s)
+  * Suppose $\alpha$ increases monotonically, $\alpha$' is integrable on [a,b], and f is bounded real function on [a,b]. Then f is integrable with respect to $\alpha$ if and only if f$\alpha$' is integrable: \\ $\int_a^b fd{\alpha}$ $=$ $\int_a^b f(x){\alpha}'(x)d(x)$
+  * Let f be integrable on [a,b] and for a$\leq$x$\leq$b, let F(x) $=$ $\int_a^x f(t)dt$, then \\ (1) F(x) is continuous on [a,b] \\ (2) if f(x) is continuous at p $\in$ [a,b], then F(x) is differentiable at p, with F'(p) $=$ f(p)
+**Fundamental Theorem of Calculus.** Let $f$ be integrable on $[a,b]$ and $F$ be a differentiable function on [a,b] such that $F'(x)$ $=$ $f(x)$, then $\int_a^b f(x)dx$ $=$ $F(b)$ $-$ $F(a)$
+Let $\alpha$ be increasing weight function on $[a,b]$. Suppose $f_n$ is integrable, and $f_n$ $\to$ $f$ uniformly on $[a,b]$. Then $f$ is integrable, and \\
+$\int_a^b fd{\alpha}$ $=$ $\lim\limits_{n \to \infin}$ $\int_a^b f_{n}d{\alpha}$
+Suppose {$f_n$} is a sequence of differentiable functions on $[a,b]$ such that $f_n$ $\to$ $g$ uniformly and there exists p $\in$ $[a,b]$ where {$f_n(p)$} converges. Then $f_n$ converges to some $f$ uniformly, and \\
+$f'(x)$ $=$ $g(x)$ $=$ $\lim\limits_{n \to \infin}$ $f'_{n}(x)$ \\
+Note $f'_{n}(x)$ may not be continuous.
+==== Questions ====
+**1. What **

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