文章目录

• • • 1 Probability and Statistics
    • 2 Linear Algebra
    • 3.Calculus
    • 一览图
    • hw0题目链接

1 Probability and Statistics

不妨假设C(n,K)=C(n,K)，0≤K≤NC(n, K)=C(n, K) ，0 \le K\le NC(n,K)=C(n,K)，0≤K≤N成立，只需证明C(n+1,K)=(n+1)!K!(n−K+1)!，0≤K≤NC(n+1, K)=\frac {(n+1)!}{K!(n-K+1)!} ，0 \le K\le NC(n+1,K)=K!(n−K+1)!(n+1)!​，0≤K≤N即可

1. K=0，C(n+1,K)=1C(n+1, K)=1C(n+1,K)=1，符合假设
2. 1≤K≤n1\le K\le n1≤K≤n，如下

C(n+1,K)=C(n,K)+C(n,K−1)=n!K!(n−K)!+n!(K−1)!(n−K+1)!=n!(n−K+1+K)K!(n−K+1)!=n!(n+1)K!(n−K+1)!=(n+1)!K!(n−K+1)!\begin{aligned}C(n+1, K)&=C(n, K)+C(n,K-1)\\&=\frac {n!}{K!(n-K)!}+ \frac {n!}{(K-1)!(n-K+1)!}\\&=\frac {n!(n-K+1 + K)}{K!(n-K+1)!}\\&=\frac {n!(n+1)}{K!(n-K+1)!}\\&=\frac {(n+1)!}{K!(n-K+1)!}\end{aligned}C(n+1,K)​=C(n,K)+C(n,K−1)=K!(n−K)!n!​+(K−1)!(n−K+1)!n!​=K!(n−K+1)!n!(n−K+1+K)​=K!(n−K+1)!n!(n+1)​=K!(n−K+1)!(n+1)!​​

• 10次硬币4次正面C104210=105512\frac{C_{10}^{4}}{2^{10}}=\frac {105}{512}210C104​​=512105​
• 13组，选2组进行组合13×12×C43×C42C525=64165\frac {13\times 12 \times C_4^3 \times C_4^2}{C_{52}^5}=\frac {6}{4165}C525​13×12×C43​×C42​​=41656​
  
• P(三次投掷有一次正面）=1−18=78P(三次投掷有一次正面）=1-\frac{1}{8}=\frac{7}{8}P(三次投掷有一次正面）=1−81​=87​
• P(三次投掷有三次次正面）=18P(三次投掷有三次次正面）=\frac{1}{8}P(三次投掷有三次次正面）=81​
  p=1878=17\mathbb p=\frac{\frac{1}{8}}{\frac{7}{8}}=\frac{1}{7}p=87​81​​=71​

• 随机位为0应该题目写错了，理解为大于小于0即可
  令 A={∣X∣=1}B={X<0}\begin{aligned}A&=\{|X| =1 \}\\B&=\{X<0 \}\end{aligned}AB​={∣X∣=1}={X<0}​

则p=P(B∣A)=P(AB)P(A)=P(X=−1)P(∣X∣=1)=12×1412×18+12×14=23\begin{aligned}p&= P(B |A)\\&= \frac{ P(AB)}{P(A)}\\&=\frac{ P(X=-1)}{ P(|X| =1 )}\\&=\frac{\frac 12 \times \frac 1 4}{\frac 12 \times \frac 1 8 +\frac 12 \times \frac 1 4}\\&=\frac 2 3 \end{aligned}p​=P(B∣A)=P(A)P(AB)​=P(∣X∣=1)P(X=−1)​=21​×81​+21​×41​21​×41​​=32​​

1. maxP(A∩B)=min{P(A),P(B)}=0.3maxP(A \cap B)=min\{P(A),P(B)\}=0.3maxP(A∩B)=min{P(A),P(B)}=0.3
2. minP(A∩B)=0minP(A \cap B)=0minP(A∩B)=0
3. maxP(A∪B)=P(A)+P(B)−minP(AB)=0.7maxP(A \cup B)=P(A)+P(B)-minP(AB)=0.7maxP(A∪B)=P(A)+P(B)−minP(AB)=0.7
4. minP(A∪B)=P(A)+P(B)−maxP(AB)=0.4minP(A \cup B)=P(A)+P(B)-maxP(AB)=0.4minP(A∪B)=P(A)+P(B)−maxP(AB)=0.4

2 Linear Algebra

初等行变换即可，r为2
(121103112)⟶(2)−(1)(1210−22112)⟶(3)−(1)(1210−220−11)⟶(2)/2(1210−110−11)⟶(3)+(2)(1210−11000)\left( \begin{matrix} 1 & 2 & 1 \\ 1 & 0& 3 \\ 1 & 1& 2 \end{matrix} \right) \overset{(2)-(1)} {\longrightarrow} \left( \begin{matrix} 1 & 2 & 1 \\ 0 & -2& 2 \\ 1 & 1& 2 \end{matrix} \right)\overset{(3)-(1)} {\longrightarrow} \\ \left( \begin{matrix} 1 & 2 & 1 \\ 0 & -2& 2 \\ 0 & -1& 1 \end{matrix} \right)\overset{(2)/2} {\longrightarrow}\left( \begin{matrix} 1 & 2 & 1 \\ 0 & -1& 1 \\ 0 & -1& 1 \end{matrix} \right)\overset{(3)+(2)} {\longrightarrow} \left( \begin{matrix} 1 & 2 & 1 \\ 0 & -1& 1 \\ 0 & 0& 0 \end{matrix} \right) ​111​201​132​​⟶(2)−(1)​​101​2−21​122​​⟶(3)−(1)​​100​2−2−1​121​​⟶(2)/2​​100​2−1−1​111​​⟶(3)+(2)​​100​2−10​110​​

(024242331)−1=(024242331)∗∣(024242331)∣=(−210−124−128−66−4)−16\left( \begin{matrix} 0 & 2 & 4 \\ 2 & 4& 2 \\ 3 & 3& 1 \end{matrix} \right)^{-1}=\frac{\left( \begin{matrix} 0 & 2 & 4 \\ 2 & 4& 2 \\ 3 & 3& 1 \end{matrix} \right)^*}{|\left( \begin{matrix} 0 & 2 & 4 \\ 2 & 4& 2 \\ 3 & 3& 1 \end{matrix} \right)|}=\frac{\left( \begin{matrix} -2 & 10 & -12 \\ 4 & -12& 8 \\ -6 & 6& -4 \end{matrix} \right)}{-16}​023​243​421​​−1=∣​023​243​421​​∣​023​243​421​​∗​=−16​−24−6​10−126​−128−4​​​

(A−λE)=(3−λ1124−λ2−1−11−λ)⟶(4−λ4−λ4−λ02−λ0002−λ)=(4−λ)(2−λ)2(A-\lambda E)=\left( \begin{matrix} 3-\lambda & 1 & 1 \\ 2 & 4-\lambda & 2 \\ -1 & -1 & 1-\lambda \end{matrix} \right) \longrightarrow \left( \begin{matrix} 4-\lambda & 4-\lambda & 4-\lambda \\ 0 & 2-\lambda & 0 \\ 0 & 0 & 2-\lambda \end{matrix} \right)\\ =(4-\lambda)(2-\lambda)^2 \\ (A−λE)=​3−λ2−1​14−λ−1​121−λ​​⟶​4−λ00​4−λ2−λ0​4−λ02−λ​​=(4−λ)(2−λ)2

• 当特征值为4时，有：
  A−4E=(−111202−1−1−3)⟶(101012000)A-4E=\left( \begin{matrix} -1 & 1 & 1 \\ 2 & 0 & 2 \\ -1 & -1 & -3 \end{matrix} \right)\longrightarrow \left( \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix} \right)A−4E=​−12−1​10−1​12−3​​⟶​100​010​120​​

求得 x⃗1=(12−1)\vec x_1=\left( \begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right)x1​=​12−1​​

• 特征值为2时，同理可以求得x⃗2=(10−1)\vec x_2=\left( \begin{matrix} 1 \\ 0 \\ -1 \end{matrix} \right)x2​=​10−1​​和x⃗3=(1−10)\vec x_3=\left( \begin{matrix} 1 \\ -1 \\ 0 \end{matrix} \right)x3​=​1−10​​

(a)
不妨认为M∈Rm×n,U∈Rm×m,Σ∈Rm×n,V∈Rn×nM\in\mathbb R^{m\times n},U\in \mathbb R^{m\times m},\Sigma\in \mathbb R^{m\times n}, V\in \mathbb R^{n\times n}M∈Rm×n,U∈Rm×m,Σ∈Rm×n,V∈Rn×n易得Σ†∈Rn×m\Sigma^{\dagger}\in \mathbb R^{n\times m}Σ†∈Rn×m
所以有MM†M=UΣVTVΣ†UTUΣVT=U(ΣΣ†)ΣVT=UImΣVT=UΣVT\begin{aligned} MM^{\dagger}M &=U\Sigma V^TV\Sigma^{\dagger} U^TU\Sigma V^T\\ &=U(\Sigma\Sigma^{\dagger})\Sigma V^T\\ &=UI_m\Sigma V^T\\ &=U\Sigma V^T \end{aligned}MM†M​=UΣVTVΣ†UTUΣVT=U(ΣΣ†)ΣVT=UIm​ΣVT=UΣVT​
(b)
M†=VΣ†UT(M可逆，则m=n，且Σ†=Σ−1)=VΣ−1UT=(UΣVT)−1=M−1\begin{aligned} M^{\dagger} &= V\Sigma^{\dagger} U^T(M 可逆，则m=n，且\Sigma^{\dagger}=\Sigma^{-1}) \\ &=V\Sigma^{-1} U^T \\ &=(U\Sigma V^T)^{-1}\\ &=M^{-1} \end{aligned}M†​=VΣ†UT(M可逆，则m=n，且Σ†=Σ−1)=VΣ−1UT=(UΣVT)−1=M−1​

(a)
∀x，xTZZTx=(ZTx)T(ZTx)≥0\forall x，x^T ZZ^T x= (Z^Tx)^T(Z^Tx)\ge 0∀x，xTZZTx=(ZTx)T(ZTx)≥0，得证
(b)

• 必要性：由实对称矩阵可知，存在正交矩阵Q和对角矩阵Λ\LambdaΛ使面等式成立
  QTAQ=Λ,Λ=diag{λ1,...,λn}Q^TAQ=\Lambda,\Lambda =\text{diag} \{\lambda_1,...,\lambda_n\}QTAQ=Λ,Λ=diag{λ1​,...,λn​}

由已知易知，Λ\LambdaΛ>0，则必要性证明成功

• 充分性：取x=Qei≠0,i=1,...,nx= Qe_i\neq 0,i=1,...,nx=Qei​=0,i=1,...,n，且ei∈Rn,(ei)j=1{i=j}e_i \in \mathbb R^n ,(e_i)_j = 1\{i=j\}ei​∈Rn,(ei​)j​=1{i=j}

则有xTΛx=eiTQTAQei=eiTΛei=λi>0\begin{aligned} x^T \Lambda x &= e_i^T Q^TA Qe_i \\ &=e_i^T \Lambda e_i \\ &=\lambda_i\\ &>0 \end{aligned}xTΛx​=eiT​QTAQei​=eiT​Λei​=λi​>0​充分性证明成功

1. max⁡utx=1\max u^tx=1maxutx=1，u和x同向
2. min⁡utx=−1\min u^tx=-1minutx=−1，u和x反向
3. min⁡∣utx∣=0\min |u^tx|=0min∣utx∣=0，u和x正交

3.Calculus

df(x)dx=−2e−2x1+e−2x=−21+e2x∂g(x,y)∂y=2e2y+e3xy26xy\begin{aligned} \frac {d f(x)}{dx} &= \frac {-2e^{-2x}} {1+e^{-2x}}=-\frac {2}{1+e^{2x}} \\ \frac {\partial g(x,y)}{\partial y} &= 2e^{2y}+e^{3xy^2} 6xy \end{aligned}dxdf(x)​∂y∂g(x,y)​​=1+e−2x−2e−2x​=−1+e2x2​=2e2y+e3xy26xy​

∂f∂v=∂f∂x∂x∂v+∂f∂y∂y∂v=−ysin⁡(u+v)−xcos⁡(u−v)\begin{aligned} \frac {\partial f}{\partial v} &=\frac {\partial f}{\partial x} \frac {\partial x}{\partial v}+ \frac {\partial f}{\partial y} \frac {\partial y}{\partial v}\\ &=-y\sin(u+v)-x\cos (u-v) \end{aligned}∂v∂f​​=∂x∂f​∂v∂x​+∂y∂f​∂v∂y​=−ysin(u+v)−xcos(u−v)​

一阶偏导：
∂E(u,v)∂u=2(uev−2ve−u)(ev+2ve−u)∂E(u,v)∂v=2(uev−2ve−u)(uev−2e−u)\begin{aligned} \frac{\partial E(u,v)}{\partial u} &=2(ue^v -2ve^{-u} )(e^v +2ve^{-u})\\ \frac{\partial E(u,v)}{\partial v} &=2(ue^v -2ve^{-u} )(ue^v -2e^{-u})\\ \end{aligned}∂u∂E(u,v)​∂v∂E(u,v)​​=2(uev−2ve−u)(ev+2ve−u)=2(uev−2ve−u)(uev−2e−u)​
二阶偏导：
∂2E(u,v)∂u2=2∂∂u(ue2v−2vev−u+2uvev−u−4v2e−2u)=2(e2v+2vev−u+2vev−u−2uvev−u+8v2e−2u)=2(e2v+4vev−u−2uvev−u+8v2e−2u)∂2E(u,v)∂v2=2∂∂v(u2e2v−2uvev−u−2uev−u+4ve−2u)=2(2u2e2v−2uev−u−2uvev−u−2uev−u+4e−2u)=2(2u2e2v−4uev−u−2uvev−u+4e−2u)∂2E(u,v)∂u∂v=2∂∂v(ue2v−2vev−u+2uvev−u−4v2e−2u)=2(2ue2v−2ev−u−2vev−u+2uev−u+2uvev−u−8ve−2u)\begin{aligned} \frac{\partial^2 E(u,v)}{\partial u^2} &=2\frac{\partial}{\partial u}(ue^{2v} -2v e^{v-u}+2uve^{v-u}-4v^2 e^{-2u})\\ &=2(e^{2v} +2v e^{v-u}+2ve^{v-u}-2uve^{v-u}+8v^2 e^{-2u}) \\ &=2(e^{2v} +4v e^{v-u}-2uve^{v-u}+8v^2 e^{-2u})\\ \frac{\partial^2 E(u,v)}{\partial v^2} &=2\frac{\partial}{\partial v}(u^2e^{2v} -2uv e^{v-u}-2ue^{v-u}+4ve^{-2u})\\ &=2(2u^2e^{2v} -2u e^{v-u}-2uv e^{v-u}-2ue^{v-u}+4e^{-2u})\\ &=2(2u^2e^{2v} -4u e^{v-u}-2uv e^{v-u}+4e^{-2u}) \\ \frac{\partial^2 E(u,v)}{\partial u \partial v} &=2\frac{\partial}{\partial v}(ue^{2v} -2v e^{v-u}+2uve^{v-u}-4v^2 e^{-2u})\\ &=2(2ue^{2v} -2 e^{v-u}-2ve^{v-u}+2ue^{v-u}+2uve^{v-u} -8v e^{-2u}) \\ \end{aligned}∂u2∂2E(u,v)​∂v2∂2E(u,v)​∂u∂v∂2E(u,v)​​=2∂u∂​(ue2v−2vev−u+2uvev−u−4v2e−2u)=2(e2v+2vev−u+2vev−u−2uvev−u+8v2e−2u)=2(e2v+4vev−u−2uvev−u+8v2e−2u)=2∂v∂​(u2e2v−2uvev−u−2uev−u+4ve−2u)=2(2u2e2v−2uev−u−2uvev−u−2uev−u+4e−2u)=2(2u2e2v−4uev−u−2uvev−u+4e−2u)=2∂v∂​(ue2v−2vev−u+2uvev−u−4v2e−2u)=2(2ue2v−2ev−u−2vev−u+2uev−u+2uvev−u−8ve−2u)​
代入题目条件u、v=1有
∂E(u,v)∂u∣u=1,v=1=2(e2−4e−2)∂E(u,v)∂v∣u=1,v=1=2(e2+4e−2−4)∂2E(u,v)∂u2∣u=1,v=1=2(e2+8e−2+2)∂2E(u,v)∂v2∣u=1,v=1=2(2e2+4e−2−6)∂2E(u,v)∂u∂v∣u=1,v=1=2(2e2−8e−2)\begin{aligned} \frac{\partial E(u,v)}{\partial u} \Big |_{u=1,v=1} &=2(e^2 -4e^{-2}) \\ \frac{\partial E(u,v)}{\partial v} \Big |_{u=1,v=1} &=2(e^2 +4e^{-2}-4) \\ \frac{\partial^2 E(u,v)}{\partial u^2}\Big |_{u=1,v=1}& =2(e^2+8e^{-2}+2) \\ \frac{\partial^2 E(u,v)}{\partial v^2}\Big |_{u=1,v=1}& =2(2e^2+4e^{-2}-6) \\ \frac{\partial^2 E(u,v)}{\partial u\partial v}\Big |_{u=1,v=1}& =2(2e^2 -8e^{-2}) \\ \end{aligned}∂u∂E(u,v)​​u=1,v=1​∂v∂E(u,v)​​u=1,v=1​∂u2∂2E(u,v)​​u=1,v=1​∂v2∂2E(u,v)​​u=1,v=1​∂u∂v∂2E(u,v)​​u=1,v=1​​=2(e2−4e−2)=2(e2+4e−2−4)=2(e2+8e−2+2)=2(2e2+4e−2−6)=2(2e2−8e−2)​
最后结果为：
∇E=(2(e2−4e−2)2(e2+4e−2−4)),∇2E=(2(e2+8e−2+2)2(2e2−8e−2)2(2e2−8e−2)2(2e2+4e−2−6))\begin{aligned} \nabla E= \left( \begin{matrix} 2(e^2 -4e^{-2}) \\ 2(e^2 +4e^{-2}-4) \end{matrix} \right) , \nabla^2 E = \left( \begin{matrix} 2(e^2+8e^{-2}+2) & 2(2e^2 -8e^{-2}) \\ 2(2e^2 -8e^{-2}) & 2(2e^2+4e^{-2}-6) \end{matrix} \right) \end{aligned}∇E=(2(e2−4e−2)2(e2+4e−2−4)​),∇2E=(2(e2+8e−2+2)2(2e2−8e−2)​2(2e2−8e−2)2(2e2+4e−2−6)​)​

E(u,v)≈E(1,1)+∇ET(u−1v−1)+12!(u−1v−1)∇2E(u−1v−1)=(e−2e−1)2+2(e2−4e−2)(u−1)+2(e2+4e−2−4)(v−1)+(e2+8e−2+2)(u−1)2+(2e2+4e−2−6)(v−1)2+2(2e2−8e−2)(u−1)(v−1)\begin{aligned} E(u,v) &\approx E(1, 1)+ \nabla E^T \left( \begin{matrix} u- 1 \\ v- 1 \end{matrix} \right) + \frac 1 {2!} \left( \begin{matrix} u- 1 & v- 1 \end{matrix} \right) \nabla^2 E \left( \begin{matrix} u- 1 \\ v- 1 \end{matrix} \right) \\ &=(e-2e^{-1})^2 + 2(e^2 -4e^{-2}) (u-1)+ 2(e^2 +4e^{-2}-4)(v-1)\\ &+(e^2+8e^{-2}+2) (u-1)^2 + (2e^2+4e^{-2}-6)(v-1)^2+2(2e^2 -8e^{-2})(u-1)(v-1) \end{aligned}E(u,v)​≈E(1,1)+∇ET(u−1v−1​)+2!1​(u−1​v−1​)∇2E(u−1v−1​)=(e−2e−1)2+2(e2−4e−2)(u−1)+2(e2+4e−2−4)(v−1)+(e2+8e−2+2)(u−1)2+(2e2+4e−2−6)(v−1)2+2(2e2−8e−2)(u−1)(v−1)​

Aeα+Be−2α=A2eα+A2eα+Be−2α≥3A2eα×A2eα×Be−2α3=3A2B43\begin{aligned} &Ae^{\alpha} +Be^{-2\alpha}\\ &=\frac A 2 e^{\alpha}+\frac A 2 e^{\alpha} +Be^{-2\alpha}\\ &\ge 3 \sqrt[3] {\frac A 2 e^{\alpha}\times\frac A 2 e^{\alpha} \times Be^{-2\alpha}} \\ &=3\sqrt[3]{\frac {A^2B} 4} \end{aligned}​Aeα+Be−2α=2A​eα+2A​eα+Be−2α≥332A​eα×2A​eα×Be−2α​=334A2B​​​当且仅当
A2eα=A2eα=Be−2αα=13ln⁡2BA\begin{aligned}&\frac A 2 e^{\alpha}=\frac A 2 e^{\alpha} =Be^{-2\alpha}\\ &\alpha =\frac 1 3 \ln \frac {2B} A\end{aligned}​2A​eα=2A​eα=Be−2αα=31​lnA2B​​

原式展开有：
E(w)=12∑i=1d∑j=1dwiAijwj+∑i=1dwibi\begin{aligned} E(w) & =\frac 1 2 \sum_{i=1}^d \sum_{j=1}^d w_{i}A_{ij}w_j + \sum_{i=1}^d w_i b_i \end{aligned}E(w)​=21​i=1∑d​j=1∑d​wi​Aij​wj​+i=1∑d​wi​bi​​
由于A是对称矩阵：
∂E(w)∂wk=12∑j=1dAkjwj+12∑i=1dwiAik+bk=∑j=1dAkjwj+bk∂2E(w)∂wl∂wk=∂∂wl(∑j=1dAkjwj+bk)=Akl\begin{aligned} \frac{\partial E(w)}{\partial w_k} &=\frac 1 2 \sum_{j=1}^d A_{kj}w_j+\frac 1 2 \sum_{i=1}^d w_{i}A_{ik} + b_k \\ &=\sum_{j=1}^d A_{kj}w_j +b_k \\ \frac{\partial^2 E(w)}{\partial w_l\partial w_k} &= \frac{\partial}{\partial w_l} \Big(\sum_{j=1}^d A_{kj}w_j +b_k \Big)\\ &= A_{kl} \\ \end{aligned}∂wk​∂E(w)​∂wl​∂wk​∂2E(w)​​=21​j=1∑d​Akj​wj​+21​i=1∑d​wi​Aik​+bk​=j=1∑d​Akj​wj​+bk​=∂wl​∂​(j=1∑d​Akj​wj​+bk​)=Akl​​
合成为矩阵形式有：
∇E(w)=Aw+b∇E2(w)=A\begin{aligned} \nabla E(w)& =Aw+b\\ \nabla E^2(w)& =A \end{aligned}∇E(w)∇E2(w)​=Aw+b=A​

一览图

hw0题目链接