I didn't update study notes for awhile because I realized something. I don't what what something is but I realized something. If could be one of the following
- I don't understand what the book is saying
- I don't understand what the book is saying
- I'm sick
- The problems are too hard
- I don't understand what the book is saying
- I want to do olympiad
- I don't want to do olympiad
- I want to play puzzles
- Analysis and Linear Algebra is too hard
- Analysis and Linear Algebra is very hard
Honestly I don't really get what analysis is doing. I hope someone can tell what analysis is all about. I want to read Complex Analysis but I realize my analysis skills is too weak so lets review some analysis.
I think i can solve the problems in the book but I just don't get the idea?
I like to read Linear Algebra because its quite practical. But I don't really again understand that Jordan thing and how thats useful. I really don't know how these things are useful. That chapter about the Jordan thing completely wiped me out so I will do some lighter stuff like analysis which doesn't seems light so maybe do number theory? Or should I just do my personal statement?
I think I prefer olympiad questions more. Even if it uses linear algebra group theory analysis or whatever.
I don't seem practical uses of each idea... maybe I read the wrong book?
I think the plan is to review forier transform and begin reading complex analysis. And in the mean time I'll take some rest because there is no more olympiad for me.
数学分析中的具体问题与方法 Chapter 1
Limits Definition
If \(\lim_{n\to \infin} x_n = A\), then by definition, \(\forall \epsilon > 0\), \(\exist \delta > 0\), such that when \(n>\delta\), \(|x_n-A|<\epsilon\).
Example
Show that
\[ \lim_{n \to \infin} (n+1)^{1/n}=1\]
By Definition, we need to find some finite \(n\) such that
\[ |(n+1)^{1/n}-1|<\epsilon\]
Which is hard \[ a = (n+1)^{1/n}-1\]
So the idea is to enlarge \(a\)
\[ (1+n) = (1+a)^n = \sum_{k}\binom{n}{k}a^{k} < \frac{n(n-1)}{2}a^2 \]
Hence \[ a < \frac{2}{\sqrt{n-1}}\]
So
\[|(n+1)^{1/n}-1| = |a| < |\frac{2}{\sqrt{n-1}}|<\epsilon \]
So when \(n\) is super large, namely \(n > \frac{4}{\epsilon^2}+1\), inequality holds. By definition, let \(\delta = \frac{4}{\epsilon^2}+1\) and inequality is proved.
Cauchy's Rule
Don't need to know what is the limit.
\(x_n\) converges iff \(\forall \epsilon > 0\), \(\exist \delta > 0\), \(n>N\),\(|x_{n+p}-x_n|<\epsilon\).
Example
\[x_n = \sum_{i=1}^n \frac{\sin i}{2^i}\]
\[|x_{n+p}-x_n|=\sum_{i=n+1}^{n+p} \frac{\sin i}{2^i} \leq \sum_{i=n+1}^{n+p} \frac{1}{2^i} \leq \frac{1}{2^{n}}<\frac{1}{n}<\epsilon\]
So let
\[\epsilon > \frac{1}{n}\]
\[N = \frac{1}{\epsilon}\]
Reversed Cauchy's Rule
\(x_n\) diverges iff \(\exist \epsilon > 0\), \(\forall \delta > 0\), \(n>N\),\(|x_{n+p}-x_n|<\epsilon\).
Simple Taylor Expansion
\[ x \sim \sin x \sim \tan x \sim \arcsin x \sim \arctan x\]
\[x \sim \ln(1+x) \sim e^x-1 \sim \frac{a^x-1}{\ln a} \sim \frac{(1+x)^a-1}{a}\]
\[1-\cos x \sim \frac{1}{2} x^2\]
Used only in multiplication and division
Important Limits
if the limit for both functions are suitable for each case. I didn't write the condition because i have a stomach ache and i think the condition is obvious
\[\lim_{x \to a} f(x)^{g(x)} = \lim_{x \to a} f(x)^{\lim_{x \to a}g(x)}\]
\[\lim_{x \to a} (1+f(x))^{g(x)} = \exp{(f(x)g(x))}\]
L'Hospital's Rule
same, condition is omitted
\[\lim_{x}\frac{f(x)}{g(x)}=\lim_{x}\frac{f'(x)}{g'(x)}\]
Using Integration
Omitted
Stolz Formula
inf/inf, hope you get what i mean D for difference
\[\lim_{n}\frac{Df_n}{Dg_n}=A \to \lim_{n}\frac{f_n}{g_n}=A\]