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math104-s21:s:vpak

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vpak

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Summary of Material

1. Numbers, Sets, and Sequences

Rational Zeros Theorem. For polynomials of the form cnxn + … + c0 = 0 , where each coefficient is an integer, then the only rational solutions have the form $\frac{c}{d}$ where c divides cn and d divides c0; rational root r must divide c0.

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 (sn) is a function mapping from $\N$ to $\R$. It converges to s if $\forall$ $\epsilon$ $>$ 0 there exists N such that N $>$ n $\implies$ |(sn)-s| $<$ $\epsilon$
In other words, $\lim$(sn) $=$ s

Important limit theorems include:
$\lim$(sn)(tn) $=$ ($\lim$(sn))($\lim$(tn))
$\lim$(sn)+(tn) $=$ ($\lim$(sn)) + ($\lim$(tn))
$\lim$($\frac{1}{n^p}$) $=$ 0 for p > 0
$\lim$ n(1/n) $=$ 1

A subsequence (sn(k)) of (sn) 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 (sn) and let S be the set of subsequential limits of (sn). Define:
$\lim$ $\sup$ (sn) $=$ $\lim\limits_{N \to \infin}$ $\sup${(sn): n > N} $=$ $\sup$ S
$\lim$ $\inf$ (sn) $=$ $\lim\limits_{N \to \infin}$ $\inf${(sn): n > N} $=$ $\inf$ S

$\lim$ $\inf$ |sn+1| / |sn| $\leq$ $\lim$ $\inf$ |sn|^(1/n) $\leq$ $\lim$ $\sup$ |sn|^(1/n) $\leq$ $\lim$ $\sup$ |sn+1| / |sn|

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.

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 $\R$n 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 $\Sigma$ sn $=$ M for some M $\in$ $\reals$.
A series $\Sigma$ sn satisfies is Cauchy if for any $\epsilon$ $>$ 0, there exists N such that n $\geq$ m $>$ N $\implies$ |$\displaystyle\sum_{i=m}^n$ si| $<$ $\epsilon$
If $\Sigma$ sn converges, then $\lim$ sn $=$ 0. (Note this is a one-way statement)

Comparison Test
If $\Sigma$ an converges and |bn| $\leq$ an $\forall$n, then $\Sigma$ bn converges.
If $\Sigma$ an $=$ $\infin$ and |bn| $\geq$ an $\forall$n, then $\Sigma$ bn $=$ $\infin$.

Ratio Test $\Sigma$ an of nonzero terms
(i) converges if $\lim$ $\sup$ |an+1 / an| $<$ 1
(ii) diverges if $\lim$ $\inf$ |an+1 / an| $>$ 1
(iii) else inconclusive test

Root Test Let $\alpha$ $=$ $\lim$ $\sup$ |an|(1/n). Then $\Sigma$ an
(i) converges if $\alpha$ $<$ 1
(ii) diverges if $\alpha$ $>$ 1
(iii) else inconclusive test

Alternating Series Theorem If an is monotonically decreasing and $\lim$ an $=$ 0, then $\Sigma$ (-1)n an 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 (sn) in domain(f) that converge to x, $\lim$ f(sn) $=$ f(x)
(3) Let $\bold{f}$ be mapping between metric spaces X $\to$ Y. $\bold{f}$ is continuous if the preimage $\bold{f}$-1 of any open set in Y is open in X. Similarly if the preimage $\bold{f}$-1 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 sn is a Cauchy sequence in S, then $\bold{f}$(sn) 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 $\bold{f}$n converges to f pointwise if for each x in the domain, $\lim\limits_{n \to \infin}$ $\bold{f}$n(x) $=$ f(x).

A function $\bold{f}$n converges to f uniformly if $\lim$ $\sup${|$\bold{f}$n - f|} $=$ 0.
In other words, for any fixed $\epsilon$ $>$ 0 there exists N such that n $>$ N $\implies$ |$\bold{f}$n(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 $\bold{f}$ $\ =$ $\textstyle\sum_{i=1}^\infin$ $\bold{f}$n. If $\exists$Mn $>$ 0 such that $\sup$ |$\bold{f}$n| $\leq$ Mn and $\textstyle\sum_{i=1}^\infin$ Mn $<$ $\infin$, then $\textstyle\sum_{i=1}^\infin$ $\bold{f}$n converges uniformly.

Let K be a compact set, and the following:

  • {$\bold{f}$n} is sequence of monotonically decreasing, continuous functions
  • {$\bold{f}$n} $\to$ $\bold{f}$ pointwise

Then $\bold{f}$n $\to$ $\bold{f}$ 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 $-$ $\alpha$)k
Then there exists a point x between $\alpha$ and $\beta$ such that
$\bold{f(\beta)}$ $=$ P($\beta$) $+$ $\frac{\bold{f^n(x)}}{n!}$ ($\beta$ $-$ $\alpha$)n
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$=$x0$\leq$x1$\leq$…xn$=$b
Let $\alpha$ be a weight function that is monotonically increasing. Define
U(P, f, $\alpha$) $=$ $\displaystyle\sum_{i=0}^{n}$ Mi $\Delta{\alpha}$i
L(P, f, $\alpha$) $=$ $\displaystyle\sum_{i=0}^{n}$ mi $\Delta{\alpha}$i
where Mi is the $\sup$ and mi 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

math104-s21/s/vpak.txt · Last modified: 2022/01/11 10:57 by pzhou