# 2022北航具体数学期末试题

### 1

#### 题目(8分)：

Solve\ the\ equation\ with\ respect\ to\ x\ given\ that\ x >0:

$$2032^\underline{10} \cdot x^\underline{-10} = 1.$$

#### 解答：

f(x) = (x+10)^\underline{10},对于x\geq0,f'(x) > 0，故f(x)(0,+\infty)上递增，故该方程要有解则只有唯一解，显然唯一解为:x=2022.

### 2

#### 题目(12分)：

Donote\ H_n\ the\ n^{th}\ harmonic\ number.Please:

(1)Prove\ that\ for\ n > 1,we\ have\ ln(n) < H_n < ln(n) + 1(4分)

(2)for\ a\ given\ n > 0,find\ the\ closed\ form\ of\ \sum\limits_{0 \leq k < n}H_{2k+1}(8分)

#### 解答

(1)一方面，考虑如下的积分，有ln(n) = \int_1^n\frac{1}{x}dx < \sum_{k=1}^{n}\frac{1}{k} = H_n.

(2)可以利用分布和分的方式计算，分布和分的公式如下：

$$\sum_0^n u(k)\Delta v(k) \delta k = u(k)v(k)|_0^n - \sum_0^n v(k+1)\Delta u(k)\delta k.$$

$$\int_a^b f'(x)g(x)dx = f(x)g(x)|_a^b - \int_a^b f(x)g'(x)dx.$$

$$\sum\limits_{0 \leq k < n}H_{2k+1} = k\cdot H_{2k+1}|_0^n - \sum\limits_{0\leq k<n}(k+1)\cdot (\frac{1}{2k+2} + \frac{1}{2k+3})\delta k \\ =n \cdot H_{2n+1} - \frac{1}{2}\sum\limits_{0 \leq k < n}\frac{4k+5}{2k+3} \delta k$$

$$\sum\limits_{0 \leq k < n}\frac{4k+5}{2k+3} \delta k = \sum\limits_{0 \leq k < n}(2 - \frac{1}{2k+3}) \delta k = 2n - \sum\limits_{0 \leq k < n}\frac{1}{2k+3}\delta k\\ =2n - \sum\limits_{0 \leq k < n}[(\frac{1}{2k+2} + \frac{1}{2k+3} - \frac{1}{2k+2})] \delta k\\ = 2n - \sum\limits_{0 \leq k < n}(\frac{1}{2k+2} + \frac{1}{2k+3})\delta k + \frac{1}{2}\sum\limits_{0 \leq k < n}\frac{1}{k+1}\delta k\\ = 2n - \sum\limits_{1 \leq k < 2n+1}\frac{1}{k+1} \delta k + \frac{1}{2}\sum\limits_{0 \leq k < n}\frac{1}{k+1} \delta k\\ = 2n - (H_{2n+1} -1) + \frac{1}{2}H_n$$

$$\sum\limits_{0 \leq k < n}H_{2k+1} = (n+\frac{1}{2})H_{2n+1} - \frac{1}{4}H_n - n - \frac{1}{2}.$$

### 3

#### 题目(18分)：

Calculate\ f(20221201)\ for\ f(n)=\sum_{k=1}^n\lceil log_3k\rceil.

#### 解答:

m = \lceil log_3n \rceil,为了计算方便，我们对原和式增加3^m - n项,这3^m - n对应的分量都是m.有：

$$f(3^m) = f(n) + (3^m - n)m = \sum\limits_{k=1}^{3^m}\lceil log_3k \rceil = \sum\limits_{j,k}j[j = \lceil log_3k \rceil][1 \leq k \leq 3^m]\\ 由于j = \lceil log_3k \rceil,可知j - 1 < log_3k \leq j,即3^{j-1} < k \leq 3^j\\ 故原式=\sum\limits_{j,k}j[3^{j-1} < k \leq 3^j][1 \leq j \leq m] = \sum\limits_{j=1}^{m}j\cdot(3^j - 3^{j-1})\\ = 2\cdot\sum\limits_{j=1}^mj\cdot3^{j-1}$$

$$f(n) = 2\cdot\sum\limits_{j=1}^mj\cdot3^{j-1} - (3^m - n)m = n\cdot m - \frac{1}{2}3^m + \frac{1}{2},m = \lceil log_3n\rceil$$

### 4

#### 题目(20分)：

Suppose\ there\ are\ three\ random\ variable\ X, Y\ and\ W,\\ and\ their\ probabilistic\ generating\ functions\ are\ F(z), G(z)\ and\ H(z)\ respectively. \\Given\ that\ H(z) = F(z)\cdot G(z), prove\ that

(1)E(W) = E(X) + E(Y)(10分)

(2)V(W) = V(X) + V(Y)(10分)

#### 解答：

$$E(X) = G_X'(z)\\V(X) = G_X''(z) + G_X'(z) - G_X'(z)^2$$

(1).F(1) = G(1) = H(1) = \sum\limits_{k\geq 0}Pr(k) = 1

(2).类似的，H''(z) = F''(z)G(z) + 2F'(z)G'(z) + F(z)G''(z)

E(W^2) = \sum\limits_{k\geq 0}Pr(W=k)k^2 = \sum\limits_{k\geq 0}Pr(W=k)[k(k-1)z^{k-2} + kz{k-1}]|_{z=1} = H''(1) + H'(1).

$$E(W^2)=H''(1)+H'(1) = [F''(z)G(z) + 2F'(z)G'(z) + F(z)G''(z) + F'(z)G(z)+F(z)G'(z)]|_{z=1}\\ = (F''(1) + F'(1)) + (G''(1)+G'(1)) + 2F'(1)G'(1)\\ =E(X^2) + E(Y^2) + 2E(X)E(Y)\\ 故V(W) = E(W^2) - E^2(W) = E(X^2)+E(Y^2)+2E(X)E(Y) - (E(X)+E(Y))^2\\ = (E(X^2) - E^2(X)) + (E(Y^2)-E^2(Y))\\ =V(x)+V(Y)$$

### 5

#### 题目(18分)：

$$We\ call\ an\ integer\ k\ winner\ number if\ and\ only\ if\ k\ is\ divisible\ by\ the\ floor\ of\ its\ fourth\ root,\\ that\ is, if\ \lfloor\sqrt[4]{k}\rfloor | k.Find\ out \ how\ many\ winner\ numbers\ there\ are\ from\ 1\ to\ 20221124.$$

#### 解答：

$$\sum\limits_{1\leq k \leq 10000}[\lfloor\sqrt[4]{k}\rfloor | k] = \sum\limits_{k,n}[k = \lfloor\sqrt[4]{k}\rfloor][k | n][1 \leq n \leq 10000]\\ = \sum\limits_{k,m,n}[k^4 \leq n \leq (k+1)^4][n = km][1 \leq n \leq 10000]\\ = 1[注释：上界] + \sum\limits_{k,m}[k^4 \leq n \leq (k+1)^4][1 \leq k < 10][注释：其余部分]\\ = 1 + \sum\limits_{1\leq k < 10}(\lceil\frac{(k+1)^4}{k}\rceil - \lceil k^3\rceil)\\ = 1 + \sum\limits_{1\leq k < 10}(4k^2 + 6k + 5)$$

$$W = \sum\limits_{1\leq k < K}(4k^2 + 6k + 5) + \sum\limits_m[K^4 \leq Km \leq N]\\ = 4\cdot \frac{(K-1)K(2K-1)}{6} + \frac{1}{2}(11+6K-1)(K-1) + \sum\limits_m[m \in [K^3,\frac{N}{K}]]$$

$$\sum\limits_{1\leq k \leq N}[\lfloor\sqrt[4]{k}\rfloor | k] = \frac{2}{3}(K-1)K(2K-1) + \frac{1}{2}(K-1)(6K+10) +\lfloor\frac{N}{K}\rfloor - K^3 + 1,K = \lfloor\sqrt[4]{n}\rfloor.$$

### 6

#### 题目(16分)：

$$(1) Calculate\ the\ winning\ probabilities\ of\ two\ ending\ patterns\, THTHT\ and\ TTHTT.(8分)\\ (2)Calculate\ the\ expected\ amount\ of\ flips\ and\ its\ standard\ deviation\ when\ flipping\ fair\ coins\ and\ ending\ immediately\ after\ HTTH\ appears.(8分)$$

#### 解答：

(1)设N为没有包含模式THTHTTTHTT的任意模式，S_AA胜利的模式，S_BB胜利的模式,有：

$$1+N(T+H) = S_A + S_B\\ N = S_A\sum\limits_{k=1}^52^k[A^{(k)} = A_{(k)}] + S_B\sum\limits_{k=1}^5 2^k[B^{(k)} = A_{(k)}]\\ N = S_A\sum\limits_{k=1}^52^k[A^{(k)} = B_{(k)}] + S_B\sum\limits_{k=1}^5 2^k[B^{(k)} = B_{(k)}]\\$$

$$\frac{S_A}{S_B} = \frac{B:B - B:A}{A:A - A:B}$$

A= THTHT,B=TTHTT时，带入计算有B:B = 19,B:A = 1,A:A = 21,A:B = 1,故\frac{S_A}{S_B} = \frac{9}{10}.

(2)根据对应模式期望、方差公式：

$$E(X) = \sum\limits_{k=1}^l \widetilde {A_{(k)}}[A^{(k)} = A_{(k)}],\widetilde {A_{(k)}}为模式A中用p^{-1}替代H，用q^{-1}替代T的运算结果,l = |A|.\\ V(X) = (\sum\limits_{k=1}^l \widetilde {A_{(k)}}[A^{(k)} = A_{(k)}])^2 - \sum\limits_{k=1}^l (2k-1)\widetilde{A_{(k)}}[A^{(k)} = A_{(k)}]$$

V(X) = (p^{-1} + p^{-2}q^{-2})^2 - [(2*1 - 1)p^{-1} + (2*4 - 1)p^{-2}q^{-2}] = 210.

### 7

#### 题目(8分)：

$$How\ many\ isolate\ 3D\ parts\ we\ will\ get\ at\ most\ by\ separating\ a\ cube\\in\ the\ 3D\ Euclidean\ space\ with\ arbitrary\ n\ 2D\ planes?$$

#### 解答：

$$P_n = P_{n-1} + L_{n-1}$$

$$P_{n}^k = P_{n-1}^{k} + P^{k-1}_{n-1}$$

$$P_n^1 =1+ n$$

$$P_{n}^k =\sum_{i=0}^k \binom n i$$

k=1时，即为线段情形，显然有

$$P_{n}^k =\sum_{i=0}^k \binom n i =n +1$$

n=0时，

$$P_{n}^k =\sum_{i=0}^k \binom 0 i =1$$

\begin{aligned} P_{n}^k - P_{n-1}^{k} &= P^{k-1}_{n-1} \\ P_{n-1}^k - P_{n-2}^{k} &= P^{k-1}_{n-2} \\ \ldots \\ P_{1}^k - P_{0}^{k} &= P^{k-1}_{0} \\ P_n^k - 1&= \sum_{i=0}^{n-1}P^{k-1}_{i}\\ P_n^k & = 1+\sum_{i=0}^{n-1}P^{k-1}_{i} \\ &=1+\sum_{i=0}^{n-1} \sum_{j=0}^{k-1} \binom i j\\ &= 1+\sum_{j=0}^{k-1}\sum_{i=j}^{n-1} \binom i j\\ &= 1+\sum_{j=0}^{k-1} \binom {n} {j+1}\\ &= 1+\sum_{j=1}^{k} \binom {n} {j}\\ &= \sum_{j=0}^{k} \binom {n} {j} \end{aligned}

$$\sum_{j=i}^n\binom j i =\binom {n+1} {i+1}$$

$$P_3^n = 1 + \frac{1}{12}n(n+1)(2n+1) - \frac{n(n+1)}{4} + n$$

posted @ 2023-01-07 01:29  `Demon  阅读(190)  评论(0编辑  收藏  举报