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--- math104-s22:notes:lecture_4 [2022/01/27 09:12]
pzhou
+++ math104-s22:notes:lecture_4 [2022/01/27 10:46] (current)
pzhou [Lecture 4]
@@ Line 6: / Line 6: @@
   * Cauchy sequence.
-Discussion time: Ex 10.1, 10.6 in Ross
+Discussion time: Ex 10.1, 10.7, 10.8 in Ross
 ==== limit goes to $+\infty$? ====
@@ Line 15: / Line 15: @@
 **Theorem**: bounded monotone sequence is convergent.
-How do we construct such a limit? The limit is obviously the 'largest' number, but there is no largest number in the sequence. We should use 'sup' to get the largest number. But sup needs to applied to a set, not a sequence, so we need to first convert a sequence to its 'value set', $S = \{a_n \mid n \in \N\}$, and define $a = \sup S$. How do we show that $a_n \to a$? We need to show that for any $\epsilon > 0$, we have $|a_n - a| < \epsilon$ for all large enough $n$. Let's first show that, there exists **one** $n_0$, such that $|a_{n_0} - a| < \epsilon$, this holds by definition of the $\sup S$, namely, there is one element in $S$ that satisfies this condition, it must be $a_{n_0}$ for some $n_0$. Then, we can use monotone condition to say, for all $n > n_0$, we have $a_n \geq a_{n_0}$. Furthermore, since $a_n \leq a$, we know $a_n \in [a_{n_0} , a) \In (a-\epsilon, a)$ for all $n > n_0$, then we are done.
+Just consider the increasing case, the decreasing case is similar. How do we construct such a limit? The limit is obviously the 'largest' number, but there is no largest number in the sequence. We should use 'sup' to get the largest number. But sup needs to applied to a set, not a sequence, so we need to first convert a sequence to its 'value set', $S = \{a_n \mid n \in \N\}$, and define $a = \sup S$. How do we show that $a_n \to a$? We need to show that for any $\epsilon > 0$, we have $|a_n - a| < \epsilon$ for all large enough $n$. Let's first show that, there exists **one** $n_0$, such that $|a_{n_0} - a| < \epsilon$, this holds by definition of the $\sup S$, namely, there is one element in $S$ that satisfies this condition, it must be $a_{n_0}$ for some $n_0$. Then, we can use monotone condition to say, for all $n > n_0$, we have $a_n \geq a_{n_0}$. Furthermore, since $a_n \leq a$, we know $a_n \in [a_{n_0} , a) \In (a-\epsilon, a)$ for all $n > n_0$, then we are done.
+Let's look at the example 2 in Ross. Which is about recursive sequence
+$$ s_n = \frac{s_{n-1}^2 + 5}{2 s_{n-1}} $$
+(such is an example of deterministic 'evolution', how about a probabilist one? That would be a Markov Chain)
+Let's do the graphical trick. One draw the graph of $y = \frac{x^2+5}{2x}$ on the plane, and the diagonal $y = x$. Then, we can start from point $(x,y) = (s_1, s_1)$. Then, we
+  * go vertically to the graph, and we land on $(s_1, s_2)$ (since we go vertically, $x$ coordinate is fixed, and we move $y$ until we have $y = f(x)$)
+  * go horizontally to the diagonal, and we land on $(s_2, s_2)$
+  * go vertically to the graph, and we land on $(s_2, s_3)$
+  * ....
+This would show a zig-zag curve that goes to the intersection point of $y=x$ and $y=f(x)$, which is point $(s,s)$, where $s$ is the solution to equation $x=f(x)$.
+Now, to show that this converges, we apply the above theorem. Note, depending on initial condition, this will either be monotone increasing or decreasing.
+Unbounded monotone increasing (respectively, decreasing)  sequence converges to $+\infty$ (resp. $-\infty$)
+==== $\liminf$ and $\limsup$ ====
+Recall the definition of $\sup$. For $S$ a subset of $\R$, bounded above, we define $\sup(S)$ to be the real number $a$, such that $a$ is $\geq$ than any element in $S$, and for any $\epsilon>0$, there is some $s \in S$ such that $s > a-\epsilon$.
+Also, for a sequence $(a_n)_{n=m}^\infty$, we can define the 'value set' $\{a_n\}_{n=m}^\infty$, which is the 'foot print' of the 'journey'.
+Also, for a sequence $(a_n)_{n=1}^\infty$, we can define the tail $(a_n)_{n=N}^\infty$, and we only care about the tail of a sequence.
+We want to define a gadget, that captures the 'upper envelope' of a sequence, what does that mean? Let $(a_n)$ be a seq, we want define first an auxillary sequence
+$$ A_m = \sup_{n \geq m} a_n $$
+then we define
+$$ \limsup a_n = \lim A_m (= \inf A_m) $$
+Time for some examples, $a_n = (-1)^n (1/n)$.

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