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:

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

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：

$$\sum _0^{n}u(k)\mathrm{\Delta }v(k)\delta k=u(k)v(k){|}_0^n-\sum _0^{n}v(k+1)\mathrm{\Delta }u(k)\delta k.$$

注意和分布积分的联系与区别:

$${\int }_a^b{f}^{\prime }(x)g(x)dx=f(x)g(x){|}_a^b-{\int }_a^{b}f(x)g^{\prime }(x)dx.$$

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:

$$\begin{gathered} \sum _{0\le k<n}{H}_{2k+1}=k\cdot H_{2k+1}{|}_0^n-\sum _{0\le k<n}(k+1)\cdot (\frac{1}{2k+2}+\frac{1}{2k+3})\delta k \\ =n\cdot H_{2n+1}-\frac{1}{2}\sum _{0\le k<n}\frac{4k+5}{2k+3}\delta k \end{gathered}$$

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

$$\begin{gathered} \sum _{0\le k<n}\frac{4k+5}{2k+3}\delta k=\sum _{0\le k<n}(2-\frac{1}{2k+3})\delta k=2n-\sum _{0\le k<n}\frac{1}{2k+3}\delta k \\ =2n-\sum _{0\le k<n}[(\frac{1}{2k+2}+\frac{1}{2k+3}-\frac{1}{2k+2})]\delta k \\ =2n-\sum _{0\le k<n}(\frac{1}{2k+2}+\frac{1}{2k+3})\delta k+\frac{1}{2}\sum _{0\le k<n}\frac{1}{k+1}\delta k \\ =2n-\sum _{1\le k<2n+1}\frac{1}{k+1}\delta k+\frac{1}{2}\sum _{0\le k<n}\frac{1}{k+1}\delta k \\ =2n-(H_{2n+1}-1)+\frac{1}{2}{H}_n \end{gathered}$$

综上：

$$\sum _{0\le k<n}{H}_{2k+1}=(n+\frac{1}{2})H_{2n+1}-\frac{1}{4}{H}_n-n-\frac{1}{2}.$$

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：

$$\begin{gathered} f(3^m)=f(n)+(3^m-n)m=\sum _{k=1}^{3^m}\lceil lo{g}_{3}k\rceil =\sum _{j,k}j[j=\lceil lo{g}_{3}k\rceil ][1\le k\le 3^m] \\ \mathrm{由}\mathrm{于}j=\lceil lo{g}_{3}k\rceil ,\mathrm{可}\mathrm{知}j-1<lo{g}_{3}k\le j,\mathrm{即}{3}^{j-1}<k\le 3^j \\ \mathrm{故}\mathrm{原}\mathrm{式}=\sum _{j,k}j[3^{j-1}<k\le 3^j][1\le j\le m]=\sum _{j=1}^{m}j\cdot (3^j-3^{j-1}) \\ =2\cdot \sum _{j=1}^{m}j\cdot 3^{j-1} \end{gathered}$$

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:

$$f(n)=2\cdot \sum _{j=1}^{m}j\cdot 3^{j-1}-(3^m-n)m=n\cdot m-\frac{1}{2}{3}^m+\frac{1}{2},m=\lceil lo{g}_{3}n\rceil$$

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：

$$\begin{gathered} E(X)=G_X^{\prime }(z) \\ V(X)=G_X^{″}(z)+G_X^{\prime }(z)-G_X^{\prime }(z{)}^2 \end{gathered}$$

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:

$$\begin{gathered} E(W^2)=H^{″}(1)+H^{\prime }(1)=[F^{″}(z)G(z)+2F^{\prime }(z)G^{\prime }(z)+F(z)G^{″}(z)+F^{\prime }(z)G(z)+F(z)G^{\prime }(z)]{|}_{z=1} \\ =(F^{″}(1)+F^{\prime }(1))+(G^{″}(1)+G^{\prime }(1))+2F^{\prime }(1)G^{\prime }(1) \\ =E(X^2)+E(Y^2)+2E(X)E(Y) \\ \mathrm{故}V(W)=E(W^2)-E^2(W)=E(X^2)+E(Y^2)+2E(X)E(Y)-(E(X)+E(Y){)}^2 \\ =(E(X^2)-E^2(X))+(E(Y^2)-E^2(Y)) \\ =V(x)+V(Y) \end{gathered}$$

 

5

题目(18分)：

$$\begin{gathered} Wecallanintegerkwinnernumberifandonlyifkisdivisiblebythefloorofitsfourthroot, \\ thatis,if\lfloor \sqrt[4]{k}\rfloor |k.Findouthowmanywinnernumberstherearefrom1to20221124. \end{gathered}$$

解答：

先令K=10000,这样固定了上界，更好计算(方便理解)：

$$\begin{gathered} \sum _{1\le k\le 10000}[\lfloor \sqrt[4]{k}\rfloor |k]=\sum _{k,n}[k=\lfloor \sqrt[4]{k}\rfloor ][k|n][1\le n\le 10000] \\ =\sum _{k,m,n}[k^4\le n\le (k+1{)}^4][n=km][1\le n\le 10000] \\ =1[\mathrm{注}\mathrm{释}：\mathrm{上}\mathrm{界}]+\sum _{k,m}[k^4\le n\le (k+1{)}^4][1\le k<10][\mathrm{注}\mathrm{释}：\mathrm{其}\mathrm{余}\mathrm{部}\mathrm{分}] \\ =1+\sum _{1\le k<10}(\lceil \frac{(k+1{)}^4}{k}\rceil -\lceil k^3\rceil ) \\ =1+\sum _{1\le k<10}(4k^2+6k+5) \end{gathered}$$

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

$$\begin{gathered} W=\sum _{1\le k<K}(4k^2+6k+5)+\sum _m[K^4\le Km\le N] \\ =4\cdot \frac{(K-1)K(2K-1)}{6}+\frac{1}{2}(11+6K-1)(K-1)+\sum _m[m\in [K^3,\frac{N}{K}]] \end{gathered}$$

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：

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

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：

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

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：

$$\begin{gathered} E(X)=\sum _{k=1}^l\widetilde{A_{(k)}}[A^{(k)}=A_{(k)}],\widetilde{A_{(k)}}\mathrm{为}\mathrm{模}\mathrm{式}A\mathrm{中}\mathrm{用}{p}^{-1}\mathrm{替}\mathrm{代}H，\mathrm{用}{q}^{-1}\mathrm{替}\mathrm{代}T\mathrm{的}\mathrm{运}\mathrm{算}\mathrm{结}\mathrm{果},l=|A|. \\ V(X)=(\sum _{k=1}^l\widetilde{A_{(k)}}[A^{(k)}=A_{(k)}]{)}^2-\sum _{k=1}^l(2k-1)\widetilde{A_{(k)}}[A^{(k)}=A_{(k)}] \end{gathered}$$

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)：

$$\begin{gathered} Howmanyisolate3Dpartswewillgetatmostbyseparatingacube \\ inthe3DEuclideanspacewitharbitraryn2Dplanes? \end{gathered}$$

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

$$P_n^k=P_{n-1}^k+P_{n-1}^{k-1}$$

特别的

$$P_n^1=1+n$$

下面证明

$$P_n^k=\sum _{i=0}^k(\genfrac{}{}{0ex}{}{n}{i})$$

利用数学归纳法证明该结论。

k=1时，即为线段情形，显然有

$$P_n^k=\sum _{i=0}^k(\genfrac{}{}{0ex}{}{n}{i})=n+1$$

n=0时，

$$P_n^k=\sum _{i=0}^k(\genfrac{}{}{0ex}{}{0}{i})=1$$

假设结论对于n'\le n-1, k' \le k-1时结论成立，那么

$$\begin{array}{rl}{P}_n^k-P_{n-1}^k& =P_{n-1}^{k-1}\\ P_{n-1}^k-P_{n-2}^k& =P_{n-2}^{k-1}\\ \dots \\ P_1^k-P_0^k& =P_0^{k-1}\\ P_n^k-1& =\sum _{i=0}^{n-1}{P}_i^{k-1}\\ P_n^k& =1+\sum _{i=0}^{n-1}{P}_i^{k-1}\\ & =1+\sum _{i=0}^{n-1}\sum _{j=0}^{k-1}(\genfrac{}{}{0ex}{}{i}{j})\\ & =1+\sum _{j=0}^{k-1}\sum _{i=j}^{n-1}(\genfrac{}{}{0ex}{}{i}{j})\\ & =1+\sum _{j=0}^{k-1}(\genfrac{}{}{0ex}{}{n}{j+1})\\ & =1+\sum _{j=1}^k(\genfrac{}{}{0ex}{}{n}{j})\\ & =\sum _{j=0}^k(\genfrac{}{}{0ex}{}{n}{j})\end{array}$$

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：

$$\sum _{j=i}^n(\genfrac{}{}{0ex}{}{j}{i})=(\genfrac{}{}{0ex}{}{n+1}{i+1})$$

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：

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