4.2 KiB
Math 4111 Final Review
Weierstrass M-test
Let \sum_{n=1}^{\infty} f_n(x) be a series of functions.
The weierstrass M-test goes as follows:
\exists M_n \geq 0such that\forall x\in E, |f_n(x)| \leq M_n.\sum M_nconverges.
Then \sum_{n=1}^{\infty} f_n(x) converges uniformly.
Example:
Ver.0
\forall x\in [-1,1),
\sum_{n=1}^{\infty} \frac{x^n}{n}
converges. (point-wise convergence on [-1,1))
\forall x\in [-1,1),
\left| \frac{x^n}{n} \right| \leq \frac{1}{n}
Since \sum_{n=1}^{\infty} \frac{1}{n} diverges, we don't know if the series converges uniformly or not using the weierstrass M-test.
Ver.1
However, if we consider the series on [-1,1],
\sum_{n=1}^{\infty} \frac{x^n}{n^2}
converges uniformly. Let M_n = \frac{1}{n^2}. This satisfies the weierstrass M-test. And this series converges uniformly on [-1,1].
Ver.2
\forall x\in [-\frac{1}{2},\frac{1}{2}],
\sum_{n=1}^{\infty} \frac{x^n}{n}
converges uniformly. Since \left| \frac{x^n}{n} \right|=\frac{|x|^n}{n}\leq \frac{(1/2)^n}{n}\leq \frac{1}{2^n}=M_n, by geometric series test, \sum_{n=1}^{\infty} M_n converges.
M-test still not applicable here.
\sum_{n=1}^{\infty} \frac{x^n}{n}
converges uniformly on [-\frac{1}{2},\frac{1}{2}].
Comparison test:
For a series
\sum_{n=1}^{\infty} a_n, if
\exists M_nsuch that|a_n|\leq M_n\sum_{n=1}^{\infty} M_nconvergesThen
\sum_{n=1}^{\infty} a_nconverges.
Proving continuity of a function
If f:E\to Y is continuous at p\in E, then for any \epsilon>0, there exists \delta>0 such that for any x\in E, if |x-p|<\delta, then |f(x)-f(p)|<\epsilon.
Example:
Let f(x)=2x+1. For p=1, prove that f is continuous at p.
Let \epsilon>0 be given. Let \delta=\frac{\epsilon}{2}. Then for any x\in \mathbb{R}, if |x-1|<\delta, then
|f(x)-f(1)|=|2x+1-3|=|2x-2|=2|x-1|<2\delta=\epsilon.
Therefore, f is continuous at p=1.
You can also use smaller \delta and we don't need to find the "optimal" \delta.
Play of open covers
Example of non compact set:
\mathbb{Q} is not compact, we can construct an open cover G_n=(-\infty,\sqrt{2})\cup (\sqrt{2}+\frac{1}{n},\infty).
Every unbounded set is not compact, we construct an open cover G_n=(-n,n).
Every k-cell is compact.
Every finite set is compact.
Let p\in A and A is compact. Then A\backslash \{p\} is not compact, we can construct an open cover G_n=(\inf(A)-1,p)\cup (p+\frac{1}{n},\sup(A)+1).
If K is closed in X and X is compact, then K is compact.
Proof:
Let \{G_\alpha\}_{\alpha\in A} be an open cover of K.
Ais open inX, if and only ifX\backslash Ais closed inX.
Since X\backslash K is opened in X, \{G_\alpha\}_{\alpha\in A}\cup \{X\backslash K\} is an open cover of X.
Since X is compact, there exists a finite subcover \{G_{\alpha_1},\cdots,G_{\alpha_n},X\backslash K\} of X.
Since X\backslash K is not in the subcover, \{G_{\alpha_1},\cdots,G_{\alpha_n}\} is a finite subcover of K.
Therefore, K is compact.
Cauchy criterion
In sequences
Def: A sequence \{a_n\} is Cauchy if for any \epsilon>0, there exists N such that for any m,n\geq N, |a_m-a_n|<\epsilon.
Theorem: In \mathbb{R}, every sequence is Cauchy if and only if it is convergent.
In series
Let s_n=\sum_{k=1}^{n} a_k.
Def: A series \sum_{n=1}^{\infty} a_n converges if the sequence of partial sums \{s_n\} converges.
\forall \epsilon>0, there exists N such that for any m>n\geq N,
|s_m-s_n|=\left|\sum_{k=n+1}^{m} a_k\right|<\epsilon.
Comparison test
If |a_n|\leq b_n and \sum_{n=1}^{\infty} b_n converges, then \sum_{n=1}^{\infty} a_n converges.
Proof:
Since \sum_{n=1}^{\infty} b_n converges, \forall \epsilon>0, there exists N such that for any m>n\geq N,
\left|\sum_{k=n+1}^{m} b_k\right|<\epsilon.
By triangle inequality,
\left|\sum_{k=n}^{m}a_k\right|\leq \sum_{k=n+1}^{m} |a_k|\leq \sum_{k=n+1}^{m} b_k<\epsilon.
Therefore, \forall \epsilon>0, there exists N such that for any m>n\geq N,
|s_m-s_n|=\left|\sum_{k=n+1}^{m} a_k\right|<\epsilon.
Therefore, \{s_n\} is Cauchy, and \sum_{n=1}^{\infty} a_n converges.