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math104-s21:s:antonthan

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math104-s21:s:antonthan [2021/05/07 08:37]
45.48.153.5 [Practice Final Solution Sketches] 		math104-s21:s:antonthan [2026/02/21 14:41] (current)
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 === (a) === === (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) === === (b) ===
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 Therefore, the set of continuous functions is closed. 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) === === (c) ===
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 ==== Problem 14 ==== ==== 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 ==== ==== 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 ==== ==== Problem 16 ====
 False. False.
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 ==== Problem 17 ==== ==== 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 ==== ==== Problem 18 ====
  
 === (a) === === (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) === === (b) ===
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 ===== ඞ ඞ ඞ ඞ ඞ ===== ===== ඞ ඞ ඞ ඞ ඞ =====
-{{ :math104:s:amongus.mp4 |}}+{{ math104-s21:s:amongus.mp4 |}}
math104-s21/s/antonthan.1620376667.txt.gz · Last modified: 2026/02/21 14:44 (external edit)

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