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：

注意到x^\underline{m} = x\cdot(x-1)\cdot\cdot\cdot(x-m+1),x^\underline{-m}=\frac{1}{(x+1)(x+2)\cdot\cdot\cdot(x+m)},m\geq0.

题目等价于2032^\underline{10} = (x+10)(x+9)\cdot\cdot\cdot(x+1) = (x+10)^\underline{10},

记f(x) = (x+10)^\underline{10},对于x\geq0,f'(x) > 0，故f(x)在(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.

另一方面，考虑如下的积分，显然有H_n = \sum_{k=1}^{n}\frac{1}{k} < 1 + \int_1^n\frac{1}{x}dx = ln(n) + 1.

综上,ln(n) < H_n < ln(n) + 1.

(2)可以利用分布和分的方式计算，分布和分的公式如下：

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

注意和分布积分的联系与区别:

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

同时注意在有限和分中\sum_0^n f(k) \delta k = \sum\limits_{0 \leq k < n}f(k)

本题中令u(k) = H_{2k+1},\Delta u(k) = u(k+1)-u(k) = \frac{1}{2k+2} + \frac{1}{2k+3},\Delta v(k) = 1,v(k) = k,

故有:

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

主要求解\sum\limits_{0 \leq k < n}\frac{4k+5}{2k+3} \delta k,有：

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

综上：

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

类似的可以计算\sum\limits_{0 \leq k < n}k\cdot H_{k},\sum\limits_{0 \leq k < n}k^2\cdot H_{k},\sum\limits_{0 \leq k < n}\frac{H_{k}}{k(k+1)}等的封闭形式，参考思路为分布和分，取u(k) = H_k即可.

3

题目(18分)：

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

解答:

这是一道取整函数求和的题目，我们可以按照以下思路进行求解：

设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项对应的分量都是m.有：

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

而计算\sum\limits_{j=1}^mj\cdot3^{j-1}只需要利用高中的错位相减法即可，不再赘述，其封闭形式为\frac{2m-1}{4}3^m + \frac{1}{4}.

故:

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

带入n = 20221201,m=16,f(n)=302015856.

 

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：

对于随机变量X的概率生成函数G_X(z) = \sum\limits_{k\geq 0}Pr(X = k)z^k，我们有：

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

有了这2个公式，证明就轻而易举了。

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

根据乘法求导规则，H'(z) = F'(z)G(z) + F(z)G'(z),而E(W) = H'(z)|_{z=1} = (F'(z)G(z) + F(z)G'(z))|_{z=1},故E(W) = F'(1)\cdot 1 + 1\cdot G'(1) = E(X) + E(Y).

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

而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).

因此:

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

 

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：

先令K=10000,这样固定了上界，更好计算(方便理解)：

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

现在我们对K进行推广，设上界为N,令K = \lfloor\sqrt[4]{N}\rfloor,有：

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

右半部分和为\lfloor\frac{N}{K}\rfloor - \lceil K^3\rceil + 1 = \lfloor\frac{N}{K}\rfloor - K^3 + 1.

综上：

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

带入N = 20221124,w = 406725.

 

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THTHT和TTHTT的任意模式，S_A为A胜利的模式，S_B为B胜利的模式,有：

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

而定义A:B = \sum\limits_{k=1}^{min(l,m)}2^{k-1}[A^{(k)} = B_{(k)}],l = |A|,m=|B|.

综上有：

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

当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：

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

在此题中，A = HTTH,l = 4,故E(X) = p^{-1} + p^{-2}q^{-2} = 18.(fair\ coin\ that\ p = q = \frac{1}{2})

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：

考虑P_n和P_{n-1}的递推关系，将前n-1个平面投影到第n个平面上，那么第n个平面被划分为L_{n-1}个区域，所以

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

所以很容易将此推广到高维空间。

对于k维空间，被n个k-1维超平面最多的划分数量为P_n^{k}，那么

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

特别的

$$
P_n^1 =1+ n
$$

下面证明

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

利用数学归纳法证明该结论。

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，

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

假设结论对于n'\le n-1, k' \le k-1时结论成立，那么

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

所以结论对n'=n,k'=k也成立，从而结论得证。这里利用了如下恒等式：

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

对于n = 3时，有P_3^k = P_3^{k-1} + P_2^{k-1} = P_3^{k-1} + \frac{(n-1)n}{2} + 1

根据累加法，求出和为：

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