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Math 112 Solutions for 5.1, p. 359

1.
(a) $ \{1,\frac{5}{4}, \frac{3}{2},\frac{7}{4},2\}$. (b) $ LHS(4)=\frac{11}{4}, RHS(4)=\frac{13}{4}$. (c) $ LHS(8)=\frac{23}{8},
RHS(8)=\frac{25}{8}$. (d) $ \{1,1+\frac{1}{n},1+\frac{2}{n},\cdots,1+\frac{n-1}{n}, 2\}$. $ LHS(n)=2+\dfrac{2}{n^{2}}[1+2+3+\cdots +(n-1)],
RHS(n)=2+\dfrac{2}{n^{2}}[1+2+\cdots +n]$. (e) $ LHS(n)=3-\frac{1}{n},
RHS(n)=3+\frac{1}{n}$. (f) 3 (g) The trapezoid has bases 2 and 4 and altitude 1, so has area 3.
2.
(a) $ \pi$. (b) $ LHS(n)$ will be an overstimate, $ RHS(n)$, an underestimate, because function is decreasing. (c) 3.58422; 2.25089. (d) 3.39532; 2.72865. (e) Yes; $ LHS(n)$ should decrease as $ n$ increases, and $ RHS(n)$ should increase.
3.
(a) Because the function is increasing, $ LHS(n)$ is an underestimate and $ RHS(n)$ is an overestimate. (b) $ LHS(3)=$ 0.715249, $ RHS(3)=$ 1.238848. (c) $ LHS(6)=$ 0.863382, $ RHS(6)=$ 1.125182. (d) As $ n$ increases, underestimates should increase and overestimates should decrease. That is what we see here. (e) about 1.
4.
$ A\approx \frac{32}{3}$.
5.
(a) There is no inscribed polygon, so the inner area is zero. (b) The smallest circumscribed polygon is a square with area 1, so the outer area is 1.
7.
Proof: Suppose that $ \mathbb{N}-S$ is non-empty. Then $ \mathbb{N}-S$ has a least element $ m$. Now $ 1\in S$, so $ m\neq 1$. Hence $ m-1\in S$, since $ m$ is the least element not in $ S$, so $ m=(m-1)+1\in S$, a contradiction. Therefore $ S=\mathbb{N}$.

8.
(a) Let $ S=\{n\in\mathbb{N}\;\vert\;1+2+\cdots +n=\frac{n(n+1)}{2}\}$. $ 1\in S$ since $ 1=\frac{1(1+1)}{2}$. Suppose $ k\in S$. Then $ 1+2+\cdots
+k=\frac{k(k+1)}{2}$. Hence $ 1+2+\cdots
+k+(k+1)=\frac{k(k+1)}{2\vert}+(k+1)=\frac{(k+1)(k+1+1)}{2}$, and $ k+1\in S$. Hence $ S=\mathbb{N}$, and the given statement is true for all positive integers $ n$.

(b) Let $ S=\{n\in\mathbb{N}\;\vert\;
1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{n(n+1)(2n+1)}{6}\}$. $ 1\in S$ because $ 1=\frac{(1)(1+1)(2\cdot 1+1)}{6}$. Suppose $ k\in S$. Then $ 1^{2}+2^{2}+\cdots+k^{2}+(k+1)^{2}=\frac{k(k+1)(2k+1)}{6}+(k+1)^{2}=
\frac{k(2...
...g(k+1)}{6}\cdot(k+1) =\frac{(k+1)(2k^{2}+7K+6)}{6} =
\frac{(k+1)(k+2)(2k+3)}{6}$, and $ k+1\in S$. Hence $ S=\mathbb{N}$, and the statement is true for all positive integers $ n$.

(e) Let $ S=\{n\in \mathbb{N}\;\vert\;8^{n}-3^{n}$    is divisible by $ 5\}$. Then $ 8^{1}-3^{1}=5$ is divisible by 5, so $ 1\in S$. Suppose $ k\in S$. Then $ 8^{k}-3^{k}$ is divisible by 5. Hence $ 8^{k+1}-3^{k+1}=8^{k+1}-3\cdot
8^{k}+3\cdot 8^{k} -3^{k+1}=8^{k}(8-3)+3(8^{k}-3^{k})$ is divisible by 5 because each term is. Hence $ k+1\in S$, so $ S=\mathbb{N}$ and $ 8^{n}-3^{n}$ is divisible by 5 for every positive integer $ n$.

10.
If $ \delta =\frac{\epsilon}{4}$, then $ \vert x_{1}-x_{2}\vert<\delta
\Rightarrow
\vert x_{1}^{2}-x_{2}^{2}\vert=\vert x_{1}-x_{2}\vert\vert x_{1}+x_{2}\vert<\frac{\epsilon}{4}\cdot
4=\epsilon$, because $ x_{1},x_{2}<2$.

11.
(a) If $ f$ is uniformly continuous, then given $ \epsilon>0$ there exists $ \delta>0$ such that if $ p_{1},p_{2}\in[a,b]$ and $ \vert p_{1}-p_{2}\vert<\delta$ then $ \vert f(p_{1})-f(p_{2})\vert<\frac{\epsilon}{b-a}$, by the definition in Problem 10.

(b) If $ \frac{b-a}{n}<\delta$, then let $ M_{i}$ and $ m_{i}$ be the points in the $ i$th subinterval at which $ f$ has its maximum and minimum, respectively. Then $ f(M_{i})-f(m_{i})<\frac{\epsilon}{b-a}$, by part (a), so that

$\displaystyle US(n)-LS(n)$ $\displaystyle = [f(M_{1})+\cdots +f(M_{n})]\frac{b-a}{n}-[f(m_{1})+\cdots +f(m_{n})]\frac{b-a}{n}$    
  $\displaystyle = [(f(M_{1})-f(m_{1}))+\cdots (f(M_{n})-f(m_{n})]\frac{b-a}{n}$    
  $\displaystyle < [\frac{\epsilon}{b-a}+\cdots \frac{\epsilon}{b-a}]\frac{b-a}{n}$    
  $\displaystyle = n\cdot\frac{\epsilon}{b-a}\cdot\frac{b-a}{n}$    
  $\displaystyle = \epsilon.$    

(c) If $ US(n)-LS(n)<\epsilon$, then as $ n\to\infty$, $ US(n)-LS(n)\to 0$. That means $ US(n)$ and $ LS(n)$ have the same limit, so $ R$ has area.




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Jason Grout 2003-03-28