联系我们： 手动添加方式: 微信>添加朋友>企业微信联系人>13262280223 或者 QQ: 1483266981

The course covers theoretical foundations necessary for the in-depth understanding of modern optimisation methods for data science.?The optimisation methods are presented in the context of relevant applications, such as the training of artificial neural networks and data classification. The methods discussed in the course include (stochastic) gradient descent, projection and splitting techniques. The course prepares students for confident application of modern numerical methods to problems in data science and helps them build sufficient mastery of optimisation tools and techniques for designing and implementing tailored methods for solving new problems.

这是一份unsw新南威尔士大学MATH3201 的成功案例

动态系统和混沌|MATH3201 Dynamical Systems and Chaos

问题 1.

For any n×nn times n matrices AA and BB, we have:∑n=0∞(∑j+k=nAjj!Bkk!)=(∑j=0∞Ajj!)(∑k=0∞Bkk!). sum_{n=0}^{ infty} left( sum_{j+k=n} frac{A^{j}}{j !} frac{B^{k}}{k !} right)= left( sum_{j=0}^{ infty} frac{A^{j}}{j !} right) left( sum_{k=0}^{ infty} frac{B^{k}}{k !} right) .

证明 .

Proof: We know that each of these infinite series of matrices converges. We just have to check that they converge to each other. To do this, consider the partial sumsγ2m=∑n=02m(∑j+k=nAjj!Bkk!) gamma_{2 m}= sum_{n=0}^{2 m} left( sum_{j+k=n} frac{A^{j}}{j !} frac{B^{k}}{k !} right)

andαm=(∑j=0mAjj!)?and?βm=(∑k=0mBkk!) alpha_{m}= left( sum_{j=0}^{m} frac{A^{j}}{j !} right) text { and } beta_{m}= left( sum_{k=0}^{m} frac{B^{k}}{k !} right) We need to show that the matrices γ2m?αmβm gamma_{2 m}- alpha_{m} beta_{m} tend to the zero matrix as m→∞m rightarrow infty. Toward that end, for a matrix M=[mij]M= left[m_{i j} right], we let |M|=max|mij||M|= max left|m_{i j} right|. We will show that |γ2m?αmβm|→0 left| gamma_{2 m}- alpha_{m} beta_{m} right| rightarrow 0 as m→∞m rightarrow infty.A computation shows thatγ2m?αmβm=∑′Ajj!Bkk!+∑′′Ajj!Bkk! gamma_{2 m}- alpha_{m} beta_{m}= sum^{ prime} frac{A^{j}}{j !} frac{B^{k}}{k !}+ sum^{ prime prime} frac{A^{j}}{j !} frac{B^{k}}{k !} where ∑′ sum^{ prime} denotes the sum over terms with indices satisfyingj+k≤2m,0≤j≤m,m+1≤k≤2m j+k leq 2 m, quad 0 leq j leq m, quad m+1 leq k leq 2 m while ∑′′ sum^{ prime prime} is the sum corresponding toj+k≤2m,m+1≤j≤2m,0≤k≤m j+k leq 2 m, quad m+1 leq j leq 2 m, quad 0 leq k leq m Therefore|γ2m?αmβm|≤∑′|Ajj!|?|Bkk!|+∑′′|Ajj!|?|Bkk!| left| gamma_{2 m}- alpha_{m} beta_{m} right| leq sum^{ prime} left| frac{A^{j}}{j !} right| cdot left| frac{B^{k}}{k !} right|+ sum^{ prime prime} left| frac{A^{j}}{j !} right| cdot left| frac{B^{k}}{k !} right| Now∑′|Ajj!|?|Bkk!|≤(∑j=0m|Ajj!|)(∑k=m+12m|Bkk!|) sum^{ prime} left| frac{A^{j}}{j !} right| cdot left| frac{B^{k}}{k !} right| leq left( sum_{j=0}^{m} left| frac{A^{j}}{j !} right| right) left( sum_{k=m+1}^{2 m} left| frac{B^{k}}{k !} right| right) This tends to 0 as m→∞m rightarrow infty since, as we saw above,∑j=0∞|Ajj!|≤exp?(n|A|)<∞ sum_{j=0}^{ infty} left| frac{A^{j}}{j !} right| leq exp (n|A|)< infty Similarly,∑′′|Ajj!|?|Bkk!|→0 sum^{ prime prime} left| frac{A^{j}}{j !} right| cdot left| frac{B^{k}}{k !} right| rightarrow 0 as m→∞m rightarrow infty. Therefore limm→∞(γ2m?αmβm)=0 lim {m rightarrow infty} left( gamma{2 m}- alpha_{m} beta_{m} right)=0, proving the lemma. 问题 2. Let AA be an n×nn times n matrix. Then the solution of the initial value problem X′=AXX^{ prime}=A X with X(0)=X0X(0)=X_{0} is X(t)=exp?(tA)X0X(t)= exp (t A) X_{0}. Moreover, this is the only such solution. 证明 . Proof: The preceding proposition shows thatddt(exp?(tA)X0)=(ddtexp?(tA))X0=Aexp?(tA)X0 frac{d}{d t} left( exp (t A) X_{0} right)= left( frac{d}{d t} exp (t A) right) X_{0}=A exp (t A) X_{0} Moreover, since exp?(0A)X0=X0 exp (0 A) X_{0}=X_{0}, it follows that this is a solution of the initial value problem. To see that there are no other solutions, let Y(t)Y(t) be another solution satisfying Y(0)=X0Y(0)=X_{0} and setZ(t)=exp?(?tA)Y(t) Z(t)= exp (-t A) Y(t) ThenZ′(t)=(ddtexp?(?tA))Y(t)+exp?(?tA)Y′(t)?=?Aexp?(?tA)Y(t)+exp?(?tA)AY(t)?=exp?(?tA)(?A+A)Y(t)?≡0. begin{aligned} Z^{ prime}(t) &= left( frac{d}{d t} exp (-t A) right) Y(t)+ exp (-t A) Y^{ prime}(t) &=-A exp (-t A) Y(t)+ exp (-t A) A Y(t) &= exp (-t A)(-A+A) Y(t) & equiv 0 . end{aligned} Therefore Z(t)Z(t) is a constant. Setting t=0t=0 shows Z(t)=X0Z(t)=X_{0}, so that Y(t)=Y(t)= exp?(tA)X0 exp (t A) X_{0}. This completes the proof of the theorem. Note that this proof is identical to that given way back in Section 1.1. Only the meaning of the letter AA has changed. 该课程涵盖了深入理解数据科学的现代优化方法所需的理论基础。优化方法是在相关应用的背景下提出的，如人工神经网络的训练和数据分类。课程中讨论的方法包括（随机）梯度下降，投影和分割技术。该课程为学生自信地将现代数值方法应用于数据科学中的问题做好准备，并帮助他们建立对优化工具和技术的充分掌握，以设计和实施解决新问题的定制方法。