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微积分 3 多元函数极限和连续性

$### Problem 7 Suppose $ R $ is the relation on $ \mathbb{Z} \times \mathbb{Z} $ where $ aRb $ means that $ a $ has the same lowest order (right-most) digit as $ b $. **Example:** $(a, b) \in R$ when $ a = 1234 $ and $ b = 64 $ because both numbers end in 4. Determine whether $ R $ is an equivalence relation. Show your reasoning. --- ### Solution Yes, $ R $ is an equivalence relation. - **Reflexive:** $ a $ ends with the same digit as itself, so $(a, a) \in R$. - **Symmetric:** If $ a $ ends with the same digit as $ b $, then $ b $ ends with the same digit as $ a $. Thus, if $(a, b) \in R$, then $(b, a) \in R$. - **Transitive:** If $ a $ ends with the same digit as $ b $, and $ b $ ends with the same digit as $ c $, then $ a $ ends with the same digit as $ c $. Hence, $(a, c) \in R$. If asked this on the final, you must show proof of all three required properties.$

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### Problem 7 Suppose $ R $ is the relation on $ \mathbb{Z} \times \mathbb{Z} $ where $ aRb $ means that $ a $ has the same lowest order (right-most) digit as $ b $. Example: $(a, b) \in R$ when $ a = 1234 $ and $ b = 64 $ because both numbers end in 4. Determine whether $ R $ is an equivalence relation. Show your reasoning. --- ### Solution Yes, $ R $ is an equivalence relation. - Reflexive: $ a $ ends with the same digit as itself, so $(a, a) \in R$. - Symmetric: If $ a $ ends with the same digit as $ b $, then $ b $ ends with the same digit as $ a $. Thus, if $(a, b) \in R$, then $(b, a) \in R$. - Transitive: If $ a $ ends with the same digit as $ b $, and $ b $ ends with the same digit as $ c $, then $ a $ ends with the same digit as $ c $. Hence, $(a, c) \in R$. If asked this on the final, you must show proof of all three required properties.

$### 6. Consider the Turing machine \[ T = (Q, \Lambda, q_0, \delta) \] where: \[ \begin{aligned} \delta(q_0, 0) &= (1, R, q_1) \\ \delta(q_0, 1) &= (1, R, q_1) \\ \delta(q_1, 0) &= (1, L, q_0) \\ \delta(q_1, 1) &= (0, R, q_0) \\ \delta(q_0, \#) &= (1, R, q_1) \\ \delta(q_1, \#) &= (\#, R, h) \end{aligned} \] For the following tape, determine the final tape when $T$ halts, assuming that $T$ begins in state $q_0$ at the position indicated by the arrow. Show all steps. ... # # 1 1 0 # # ... 起始箭头在第一个 1 上使用中文解题$

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### 6. Consider the Turing machine \[ T = (Q, \Lambda, q_0, \delta) \] where: \[ \begin{aligned} \delta(q_0, 0) &= (1, R, q_1) \\ \delta(q_0, 1) &= (1, R, q_1) \\ \delta(q_1, 0) &= (1, L, q_0) \\ \delta(q_1, 1) &= (0, R, q_0) \\ \delta(q_0, \#) &= (1, R, q_1) \\ \delta(q_1, \#) &= (\#, R, h) \end{aligned} \] For the following tape, determine the final tape when $T$ halts, assuming that $T$ begins in state $q_0$ at the position indicated by the arrow. Show all steps. ... # # 1 1 0 # # ... 起始箭头在第一个 1 上使用中文解题

$### 5. Consider the following program segment: ```text i := 1 total := 1 while i < n do i := i + 1 total := total + i ```` Let ( p ) be the proposition: [ p:\quad ( \text{total} = \frac{i(i+1)}{2} \ \text{and}\ i \le n ) ] Use mathematical induction to prove ( p ) is a loop invariant. --- ## Solution ### Basis Step Before the loop is entered, ( p ) is true since [ \text{total} = 1 = \frac{1(1+1)}{2} ] and [ i \le n ] --- ### Inductive Step Suppose ( p ) is true and ( i = k < n ) after the ( k - 1 )-th execution of the loop. Since ( p ) is true: [ \text{total} = \frac{k(k+1)}{2} ] Suppose that the while loop is executed again. Then ( i ) is incremented to [ i = k + 1 ] Total becomes: [ \begin{aligned} \text{total} &= \text{total}_{\text{prev}} + i \ &= \frac{k(k+1)}{2} + (k+1) \quad \text{(by inductive hypothesis)} \ &= \frac{k(k+1) + 2(k+1)}{2} \ &= \frac{(k+1)(k+2)}{2} \end{aligned} ] --- After the loop, ( i \le n ) and `total` is still of the required form. Therefore, ( p ) is a loop invariant. ∎ 使用中文解答$

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### 5. Consider the following program segment: ```text i := 1 total := 1 while i < n do i := i + 1 total := total + i ```` Let ( p ) be the proposition: [ p:\quad ( \text{total} = \frac{i(i+1)}{2} \ \text{and}\ i \le n ) ] Use mathematical induction to prove ( p ) is a loop invariant. --- ## Solution ### Basis Step Before the loop is entered, ( p ) is true since [ \text{total} = 1 = \frac{1(1+1)}{2} ] and [ i \le n ] --- ### Inductive Step Suppose ( p ) is true and ( i = k < n ) after the ( k - 1 )-th execution of the loop. Since ( p ) is true: [ \text{total} = \frac{k(k+1)}{2} ] Suppose that the while loop is executed again. Then ( i ) is incremented to [ i = k + 1 ] Total becomes: [ \begin{aligned} \text{total} &= \text{total}_{\text{prev}} + i \ &= \frac{k(k+1)}{2} + (k+1) \quad \text{(by inductive hypothesis)} \ &= \frac{k(k+1) + 2(k+1)}{2} \ &= \frac{(k+1)(k+2)}{2} \end{aligned} ] --- After the loop, ( i \le n ) and `total` is still of the required form. Therefore, ( p ) is a loop invariant. ∎ 使用中文解答

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Calculus 3: Multivariable Functions, Limits, and Continuity

Calculus 3: Multivariable Functions, Limits, and Continuity

Calculus 3: Multivariable Functions, Limits, and Continuity

Calculus 3: Multivariable Functions, Limits, and Continuity

Calculus 3: Multivariable Functions, Limits, and Continuity

Calculus 3: Multivariable Functions, Limits, and Continuity

Calculus 3: Multivariable Functions, Limits, and Continuity

Calculus 3: Multivariable Functions, Limits, and Continuity

微积分 3 多元函数 极限和连续性

微积分 3 多元函数极限和连续性