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--- math104-s21:s:antonthan [2021/05/07 07:15]
45.48.153.5 [Headline]
+++ math104-s21:s:antonthan [2026/02/21 14:41] (current)
@@ 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 112: / Line 115: @@
 Take the limit, by fixing $x=y$.
 ==== Problem 10 ====
@@ Line 132: / Line 136: @@
 $$\lim_{n \to \infty} \sup_x |\sum_{k=n+1}^\infty (-1)^kf_k(x)| = 0$$
 This is what we wanted, so the series of functions converges uniformly.
 ==== Problem 13 ====
-We set $\epsilon > 0$.  By continuity, there is a $\delta > 0$ such that $|x-y| < \delta \implies |f(x)-f(y)| < \epsilon$.  We set $n$ large enough such that $1/n < \delta$, implying that $|f((k+1)/n) - f(k/n)| < \epsilon$.  This makes it for even $n$:
+We set $\epsilon > 0$.  By continuity, there is a $\delta > 0$ such that $|x-y| < \delta \implies |f(x)-f(y)| < \epsilon$.  We set $n$ large enough such that $1/n < \delta$, implying that $|f( (k+1)/n) - f(k/n)| < \epsilon$.  This makes it for even $n$:
 $$|1/n \sum_{k=1}^n (-1)^k f(k/n)| < \epsilon / 2 < \epsilon$$
 For odd $n$:
-$$|1/n \sum_{k=1}^n (-1)^k f(k/n)| < \epsilon (n-1)/(2n) + |f(1)|/n$$
+$$|1/n \sum_{k=1}^n (-1)^k f(k/n)| < \epsilon (n-1)/(2n) + |f(1 )|/n$$
 For the odd case, we can set an $n$ even larger than the previous one so that it is less than $\epsilon$.  This means that for large enough $n$, $|1/n \sum_{k=1}^n (-1)^k f(k/n)| < \epsilon$.  By def. of a limit, the limit is equal to 0, as desired.
-==== Headline ====
-==== Headline ====
+==== 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.
+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 $lim f(y_n) = 1$, 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 $r \neq 0$):
+$$y_a = r/m, y_b = r/n$$
+$$y_a/y_b = n/m$$
+$$\sqrt{p_b/p_a} = n/m$$
+LHS is irrational, and RHS is rational, so contradiction.  Since $x_n$ contains only 1 element in A, $lim f(x_n) = 0$ for all real $r$.  We are done.
+==== 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) ===
+Indicator function for rationals.
 ===== Question List =====
 **1. How do you use the rational root theorem to prove that something is irrational?**
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 ===== ඞ ඞ ඞ ඞ ඞ =====
-{{ :math104:s:amongus.mp4 |}}
+{{ math104-s21:s:amongus.mp4 |}}

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