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--- math104-s21:s:antonthan [2021/05/07 01:18]
45.48.153.5 [Problem 16]
+++ math104-s21:s:antonthan [2022/01/11 18:30] (current)
24.253.46.239 ↷ Links adapted because of a move operation
@@ Line 7: / Line 7: @@
 Problems from https://ywfan-math.github.io/104s21_final_practice.pdf.
+Solutions here vary in how detailed they are; they could be full solutions or only guiding steps.
 ==== Problem 1 ====
 === (a) ===
@@ Line 49: / Line 50: @@
 === (a) ===
-We show that Cauchy condition on  $B(X)$  implies that the sequence converges (uniformly) to a function  $f(x)$ , which can be found through pointwise convergence.  We can then use the uniform convergence to show that  $f(x)$  is bounded (and obviously, defined on every point in  $X$ ), so it is an element of  $B(X)$ .  So, every Cauchy sequence converges in  $B(X)$ , so the metric space is complete.
+We show that Cauchy condition on  $B(X)$  is the definition of being uniformly Cauchy.  Uniformly Cauchy implies uniform convergence, and our convergent function is bounded, so  $B(X)$  is complete.
 === (b) ===
@@ Line 56: / Line 57: @@
 Therefore, the set of continuous functions is closed.
+Easier way: consider that a set is closed iff it contains all its limit points, and consider a sequence of continuous functions that uniformly converges.
 === (c) ===
@@ Line 144: / Line 147: @@
 ==== Problem 14 ====
+We consider the function  $g(x) = f(x) - f(x + T/2)$ , which is continuous.  Show that  $g(0) = -g(T/2)$ , and then use IVT.
 ==== Problem 15 ====
+Following the hint,  $f'(0) = a_1 + 2a_2 + \ldots + na_n$ .  We then show that assuming $|f'(0)| > 1$ leads to a contradiction of the original inequality.  We can use the limit definition of the derivative for this.
 ==== Problem 16 ====
 False.
@@ Line 152: / Line 156: @@
 Let $A = \{1/\sqrt{p} \vert p \text{ is prime}\} $. Define$ f(x) = 1 $when$ x \in A $, and$ f(x)=0$ otherwise.
-Clearly, if we take $y_n = 1/\sqrt{p_n}$, where  $p_n$  is the  $n$ th prime,  $lim y_n = 0$ , yet  $lim f(y_n) = 1$ .
+Clearly, if we take $y_n = 1/\sqrt{p_n}$, where  $p_n$  is the  $n$ th prime,  $lim y_n = 0$ , yet $lim f(y_n) = 1 $, so$ \lim_{x \to 0} f(x) \neq 0$.
 We just have to prove that  $f$  satisfies the original function requirement.  We do this by showing the sequence  $x_n = r/n$  contains at most one element in  $A$  for all real  $r$ .  Proceed by contradiction, and say  $\{x_n\}$  contains  $y_a$  and  $y_b$  for distinct  $a, b$ .  Then for some  $m, n$  (note that  $r \neq 0$ ):
@@ Line 161: / Line 165: @@
 ==== Problem 17 ====
+Consider  $f(x)e^{g(x)}$ , which is continuous and differentiable on the same intervals.  We take its derivative and apply MVT, and we are done.
 ==== Problem 18 ====
 === (a) ===
+Let partition $P_n = \{0, 1/n, 2/n, \ldots, n/n\}$.  Note that  $mesh(P_n) = 1/n$  and  $L(f, P_n) \leq R_n \leq U(f, P_n)$ .
+The idea now is to use Ross 32.7.  Let  $\epsilon > 0$ , and since  $f$  is integrable, there exists a  $\delta > 0$  that satisfies  $mesh(P) < \delta \implies U(f, P) - L(f, P) < \epsilon$  for all partitions  $P$ .  We choose  $n$  large enough such that  $mesh(P_n) < \delta$ , so that we get the inequalities:  $U(f, P_n) - \epsilon < R_n \leq U(f, P_n$  and  $L(f, P_n) \leq R_n < L(f, P_n) + \epsilon$ .  This gives us:
+ $U(f) \leq \sup_n(U(f, P_n)) = lim R_n = \inf_n(L(f, P_n)) \leq L(f)$
+Since we also have  $L(f) = U(f)$  by integrability,  $lim R_n = \int_0^1 f(x)dx$ , as desired.
 === (b) ===
@@ Line 383: / Line 396: @@
 ===== ඞ ඞ ඞ ඞ ඞ =====
-{{ :math104:s:amongus.mp4 |}}
+{{ math104-s21:s:amongus.mp4 |}}

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