muzimuzhi/math-notes

Notes of self-taught undergraduate-level math

math-notes

Notes of self-taught undergraduate-level math

To render latex-markup eqations, install Chrome extension MathJax Plugin for Github.

• GitHub repo: https://github.com/orsharir/github-mathjax

math-notes's Issues

Physical examples in Apostol's Calculus

Volumn 1

• sec.

Exercise 1.2.11

For $x \in [a, b]$, $y \in [c, d]$, prove
$$
\lvert x-y \rvert = \frac12 [(b-a) + (d-c) + \lvert c-a \rvert + \lvert d-b \rvert].
$$

Link to free pdf of meijq's analysis (in Chinese): http://maths.nju.edu.cn/~meijq/

• analysis I, http://maths.nju.edu.cn/~meijq/calculus/math1.pdf
• analysis II, http://maths.nju.edu.cn/~meijq/calculus/math2.pdf

Examples 4.6

(a)

To simplify letter typing, denote $A$ as $\mathcal A$ and $\leq$ as $\subseteq$ in the original text, reminding $\mathcal A \subseteq \mathcal{P}(X)​$ is non-empty.

$\sup (A) = \bigcup A​$

• $A$ is bounded above, since $X \in \mathcal P(X)$ is an upper bound of $A$.
• $\bigcup A = \sup (A)​$, since for every upper bound $B​$ of $A​$, every element $a \in A​$, $a \leq B​$. Therefore $\bigcup A \leq B​$ is the least upper bound of $A​$.

$\inf (A) = \bigcap A​$

• $A$ is bounded below, since $\emptyset \in \mathcal P(X)$ is an lower bound of $A$.
• $\bigcap A = \inf(A)​$, since for every lower bound $B​$ of $A​$, every element $a \in A​$, $B \leq a​$. Therefore $B \leq \bigcap A​$, $\bigcap A​$ is the greatest lower bound of $A​$.

Exercise 8.2.6

Let $\sum_{n = 0}^{\infty}a_n$ be a series which is conditionally convergent, but not absolutely convergent. Show that there exists a bijection $f: \mathbf{N} \to \mathbf{N}$ such that $\sum_{n = 0}^{\infty}a_{f(m)}$ diverges to $+\infty$, or more precisely that

$$\displaystyle\liminf_{N \to \infty}\sum_{m = N}^{\infty}a_{f(m)} = \limsup_{N \to \infty}\sum_{m = N}^{\infty}a_{f(m)} = +\infty.$$

MathJax test

text $a + \sqrt b = \frac 12$ text

$$ \sum_a^b \int_a^b $$

text $a + \sqrt b = \frac 12$ text

$$
  \sum_a^b \int_a^b
$$

[test] chrome extension github-math

• https://github.com/orsharir/github-mathjax
• https://chrome.google.com/webstore/detail/mathjax-plugin-for-github/ioemnmodlmafdkllaclgeombjnmnbima

Exercises I.4-4

To simplify input, denote $\circ$ insdead of $\circledast$

$\circ$ is not commutative

• Prove by contradiction. Assume $\circ​$ is commutative.
• There exists $x \in X​$, $x \neq r​$.
• From (ii), $x \circ x = r$.
• By commutativity, $x \circ r = r \circ x = x$.
• Hence $r = \color{red}{x} \circ \color{blue}{x} = \color{red}{(x \circ r)} \circ \color{blue}{(r \circ x)}$.
• Using (ii), $x = r​$, which is a contradiction.
• Therefor $\circ$ is not commutative.

$\circ$ has no identity element

• Prove by contradiction. Assume $e$ is the identity element of $\circ$.
• From (ii), if $x \circ y = r$, then $x = y$.
• Since $e \circ r = r$, $e = r$.
• Hence $x \circ r = r \circ x = x$, which will lead to contradiction.

Exercise 4.4-12

Describe the fallacy in the following “proof” by induction:

Statement. Given any collection of $n$ blonde girls. If at least one of the girls has blue eyes, then all $n$ of them have blue eyes.

“Proof.” The statement is obviously true when $n = 1$. The step from $k$ to $k + 1$ can be illustrated by going from $n = 3$ to $n = 4$. Assume, therefore, that the statement is true when $n = 3$ and let $G_{1}$, $G_{2}$, $G_{3}$, $G_{4}$, be four blonde girls, at least one of which, say $G_{1}$ , has blue
eyes. Taking $G_{1}$, $G_{2}$, and $G_{3}$, together and using the fact that the statement is true when $n = 3$, we find that $G_{2}$ and $G_{3}$, also have blue eyes. Repeating the process with $G_{2}$, $G_{3}$ and $G_{4}$, we find that $G_{4}$, has blue eyes. Thus all four have blue eyes. A similar argument allows us to make the step from $k$ to $k + 1$ in general.

Corollary. All blonde girls have blue eyes.

Proof. Since there exists at least one blonde girl with blue eyes, we can apply the foregoing
result to the collection consisting of all blonde girls.

Note: This example is from G. Pólya, who suggests that the reader may want to test the
validity of the statement by experiment.

Exercise 4.4-11

Let n and d denote integers. We say that d is a divisor of n if n = cd for some integer c. An
integer n is called a prime if n > 1 and if the only positive divisors of n are 1 and n. Prove, by
induction, that every integer n > 1 is either a prime or a product of primes.

Can't understand how to proof Cantor's theorem

In P198, the proof of Cantor's theorem. Tao consider a set $A := {x \in X : x \notin f(x) }$ and then said "this set is well-defined since $f(x)$ is an element of $2^X$ and is hence a subset of $X$".
Why $f(x)$ is a subset of $X$?

Exercise 4.7-13

Prove that $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n - 1})$ if $n \geq 1$. Then use this to prove that
$$2\sqrt{m} < \sum_{n=1}^{m}\frac{1}{\sqrt{n}} < 2\sqrt{m} - 1$$
if $m \geq 2$. In particular, when $m = 10^{6}$, the sum lies between 1998 and 1999.

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I have proved out $2\sqrt{m} &lt; \sum_{n=1}^{m}\frac{1}{\sqrt{n}}$, but I can't prove $ \sum_{n=1}^{m}\frac{1}{\sqrt{n}} < 2\sqrt{m} - 1$, I can only prove $\sum_{n=1}^{m}\frac{1}{\sqrt{n}} &lt; 2\sqrt{m}$. Thank for answer.