Category Archives: National University of Singapore

新加坡国立大学的数据科学与机器学习项目介绍

新加坡国立大学(National University of Singapore)是一所综合性的大学,根据泰晤士报和世界大学排名来看,NUS 在整个亚洲的排名是非常靠前的。同时,数学系(Department of Mathematics)则是在理学院(Science)下的一个院系。

NUS数学系的介绍

新加坡国立大学数学系的前身可以追溯到 1929 年的 Raffles College。当时理学院开设了数学,化学,物理三门课程,不过总共也就只有十个学生和三位教师,其中有一位是数学教师。第一届数学系的领导(从 1931 年到 1959 年)是Alexander Oppenheim教授,他是在美国芝加哥大学获得的博士学位。从 1929 年开始,在新加坡的教育系统中,数学教育事业得到了巨大的发展,对现在的新加坡国立大学和南洋理工大学的建立起到了至关重要的作用。

随着 NUS 的建立,数学系就进入了一个新的时代。新的校区在 Kent Ridge,1986 年理学院和数学系就在这里成立。这个时候,数学系就有了巨大的发展,不仅在本科生的招生规模方面有了巨大提高,在研究生项目规模上也有了一定的深度的提升。

NUS数学系首页

新加坡国立大学的数学系与国内的数学系有所不同。一般情况下,国内的数学系能够提供的专业包括数学与应用数学(Mathematics and Applied Mathematics),信息与计算科学(Information and Computing Science)与统计学(Statistics),有的时候会加上金融数学(Financial Mathematics)这一方向。而新加坡国立大学的数学系(Department of Mathematics)与统计系(Statistics)是分开的两个院系,虽然学生可以互相之间选择对方的课程,但是两者却是分属不同的院系。

NUS数学系本科专业

从数学系的首页来看,对于本科生而言,通常都有机会进行双学位的选择,例如:

  1. 计算机科学与数学;
  2. 经济学与数学;

因此,有不少的本科生都会有两个专业的学位证书。目标是为了让学生能够在未来从事数学研究和工业界的工作都打下坚实的基础。

NUS数学系硕士专业

NUS 的项目分成两块,coursework 项目和 Research 项目。第一个主要是以授课型的研究生为主,后者主要是为博士生或者准备攻读博士的人而准备的。对于硕士生而言,一般一两年就可以硕士毕业,只要修课的学分满了即可。

对于众多硕士生而言,进入 NUS 之前就要根据自身情况来选择一个合适的方向进行申请。对于硕士生的项目,数学系可以提供数学,金融数学等方向的课程,并且近期也提供了数据科学与机器学习专业(Data Science and Machine Learning)的课程。

项目介绍

申请条件

数据科学与机器学习项目是数学系,统计系,计算机系联合举办的为期一至两年的硕士生项目。期望学生的本科背景是数学方向,应用数学方向,统计与物理方向等。同时对学生的英语能力有一定的要求,希望对于母语不是英语的学生能够达到托福 85 分以上或者雅思 6.0 分以上的成绩。

项目课程要求

在课程安排方面,这个项目会有 20 个学分的课程,5 门核心课程,包括:

  1. Introduction to Big Data for Industry;
  2. Optimisation for Large-Scale Data-Driven Inference;
  3. Foundations of Machine Learning/Theory and Algorithms for Machine Learning;
  4. Cloud Computing;
  5. DSML Industry Consulting and Applications Project.

而选修课方面包括机器学习,数据挖掘,大数据,计算机视觉,金融数学等方向的专业课程。同样也会要求选择 20 个学分的选修课程即可。

学费

在学费方面,该项目是没有奖学金的,而一年的学费是 45000 新币,大约为 23 万人民币。

NUS申请流程

如果对这个项目感兴趣的同学,需要在 2020 年的 3 月 15 日之前提交申请,才有机会在 2020 年 8 月入学。

NUS数学系联系方式

如果有任何问题的话,可以考虑发邮件给 askmathpg@nus.edu.sg 或者拨打上面的电话号码。

 

Prediction of Final Exam 2014-2015 Semester I

Module:                 MA 1505 Mathematics I

Time:                      2 hours ( 120 minutes ), Saturday, 22-Nov-2014 (Morning)

Questions:             8 questions, each question contains two questions. i.e. 16 questions.

Average speed:     7.5 minutes per question.

Scores:                  20% mid-term exam, 80% final exam. i.e. Each question in the final exam is 5%.

Remark:                 Another Possibility: 5 Chapters, each chapter contains 1 big question, and each question contains three small questions, i.e. 15 questions. 8 minutes per question.

The contents in high school:

Trigonometric functions, some basic inequalities and identities.

The contents before mid-term exam: Please review the details of them.

Chapter 2: Differentiation

Derivatives of one variable functions, derivatives of parameter functions, Chain rule of derivatives, the tangent line of the curve, L.Hospital Rule, critical points of one variable, local maximum and local minimum of one variable function.

Chapter 3: Integration

Integration by parts, Newton-Leibniz Formula, the area of the domain in the plane, the volume of the solid which is generated by a curve rotated with an axis.

Chapter 4: Series

Taylor Series and Power Series, radius of convergence of power series, the convergence domain of power series, the sum of geometric series and arithmetic series.

Chapter 5: Three Dimensional Spaces

Cross Product and Dot Product of vectors, projection of vectors, the equation of the plane and the line in 3-dimensional space, Distance from a point to a plane, Distance from a point to a line, the distance between two lines in two or three dimensional spaces, the distance between two parallel planes. Intersection points of two different curves.

The contents after mid-term exam: Must prepare them.

By the way, 2-3 questions means at least 2 questions, at most 3 questions. 0-1 question means 0 question or 1 question.

Geometric Graphs in Three Dimensional Space:

http://www.wolframalpha.com

z=x^{2}+y^{2}, z=-(x^{2}+y^{2})              infinite paraboloid

z=x^{2}-y^{2}             hyperbolic paraboloid

(x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}=R^{2}  sphere with radius R>0 and center (x_{0},y_{0},z_{0})

x^{2}+y^{2}=R^{2},  y^{2}+z^{2}=R^{2}, z^{2}+x^{2}=R^{2}       cylinder

ax+by+cz=d, \text{ where } a,b,c,d \in \mathbb{R}             Plane

y=x^{2}+c \text{ and } x=y^{2}+c, \text{ where } c\in \mathbb{R}             Parabola

Chapter 6: Fourier Series:

Fourier Coefficients of functions with period 2\pi: 1 question. Especially, a_{2014} and b_{2014}  (Integration by parts).

Fourier Coefficients of functions with period 2L: 1 question, where L is a positive real number. Especially, a_{2014} and b_{2014} (Integration by parts).

Calculate the summation of Fourier coefficients: 0-1 question. Especially, \sum_{n=0}^{\infty} a_{n} and \sum_{n=1}^{\infty} a_{n}.

Cosine and sine expansion of function on the half domain: 1 question.

Chapter 7: Function of Several Real Variables

Directional derivatives, partial derivatives, gradient of functions with two or three variables, Chain Rule of partial derivatives: 1-2 questions. (Pay attention to whether the vector is a unit vector or not. If it is not a unit vector, you should change it to a unit vector first, and then calculate the directional derivatives).

Critical points of two variable functions (saddle point, local maximum, local minimum): 0-1 question. (Calculate the partial derivatives first, then evaluate the critical points, so we can decide the property of the critical points from some rules).

Lagrange’s method: 0-1 question. (Calculate the maximum value of functions under some special conditions. Construct the function first, evaluate partial derivatives secondly, and calculate the critical points of the new functions. In addition, if you use  inequality “arithmetic mean” is greater than “geometric mean”, then the question will become easier.)

Chapter 8: Multiple Integral

Double integral, polar coordinate: 1 question. (The formula of polar coordinate in the plane).

Reverse the order of integration of double integral: 1 question. (Draw the picture of domain R and reverse the order of dx and dy).

Volume of the solid: 1 question. (Double integral, find the function z=z(x,y) and the domain R on the xy-plane. If the domain R is a disk or a sector, then you can use the polar coordinate).

Area of the surface: 1 question. (Partial Derivatives of functions with two variables, the domain R on the xy-plane. If the domain R is a disk or a sector, then you can use the polar coordinate. The area of a surface is a special case of the surface integral of a scalar field).

Triple integral: 0-1 question. (The method to calculate the triple integral is similar to double integral).

Chapter 9: Line Integrals

Length of a curve: 0-1 question. (Parameter equation of the curves. Length of a curve is a special case of line integral of a scalar field).

Line integrals of scalar fields: 1 question. (The equation of line segment, the equation of the circle with radius R, the length of vectors). Geometric meaning: the area of the wall along the curve.

Line integrals of vector fields: 1 question. (The equation of line segments, the equation of the circle with radius R, Dot product of vectors). Physical meaning: Work done.

Conservative vector fields and Newton-Leibniz formula of gradient vector fields: 0-1 question. (Definition of conservative vector field and its equivalent condition. When the value of a line integral of vector field is independent to the curve C, where C has the fixed initial point and the terminal point?).

Green’s Theorem: 1 question. (Two cases: the boundary is open; the boundary is closed. If the curve is open, you should close it by yourself.) Pay attention to the orientation, i.e. anticlockwise and left hand rule.

Chapter 10: Surface Integrals

Tangent plain of a surface: 0-1 question. (Partial derivatives, Cross product of two vectors, Normal vector of a plane)

Surface integrals of scalar fields: 1 question. (The equation of surface z=z(x,y) and the projection of the surface on the xy-plane, Cross product of vectors, the length of vectors. Change the surface integrals of scalar fields to double integrals).

Surface integrals of vector fields: 1 question. (The equation of surface z=z(x,y) and the projection of the surface on the xy-plane, Cross product and Dot product of vectors).

Stokes’ Theorem: 1 question. (This is a rule on line integrals of vector fields and surface integrals of vector fields. Remember the operator curl. Pay attention to the orientation of the curve on the boundary, i.e. the right hand rule).

Divergence Theorem: 0-1 question. (This is a rule on surface integrals of vector fields and triple integrals. Remember the operator div).

Prediction of Middle Term Test

Module:                 MA 1505 Mathematics I

Time:                     1 hours ( 60 minutes )

Questions:            10 Multiple Choice Questions.

Average speed:     6 minutes per question.

Scores:                  20% in final score.

The contents in high school:

Trigonometric functions, some basic inequalities and identities.

Questions in middle term test:

Question 1. Derivatives, Tangent line of a function, Intersection point of tangent line and x-axis, y-axis. Basic Rules of differentiation, Chain Rule.

Question 2. Critical points of a function, how to calculate the maximum and minimum value of a function.

Question 3. Integration by parts, integrate trigonometric functions.

Question 4. Fundamental theorem of calculus.

Question 5. Find the area which is bounded by some curves.

Question 6. Mathematical models. ( e.g. light and ball drop, ship and so on).

Question 7. Radius of convergence of a power series, the interval of convergence of a power series.

Question 8. Calculate the Taylor series of functions, Calculate the coefficients of Taylor series.

Question 9. How  to use Taylor series to calculate the solution of an equation.

Question 10. How to use Taylor series to calculate the summation of some series. ( Integration and differentiation).

Question 11. The length of a curve, the tangent line of a curve.

Question 12. Dot product and cross product of two vectors, equation of planes, normal vector of a plane, distance between a point and a plane.

MA 1505 Tutorial 1: Derivative

Definition of Derivative:

f^{'}(x)=\lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}

Rule: Assume f(x) and g(x) are two differentiable functions, the basic rules of derivative are

(f\pm g)^{'}(x)=f^{'}(x)\pm g^{'}(x)

(f\cdot g)^{'}(x)= f^{'}(x) g(x) + f(x)g^{'}(x)

(f/g)^{'}(x)=(f^{'}(x)g(x)-f(x)g^{'}(x))/(g(x))^{2}

(f\circ g)^{'}(x)=f^{'}(g(x))g^{'}(x)

Definition of Critical Point: x_{0} is called a critical point of f(x), if f^{'}(x_{0})=0.

If f^{'}(x)>0 on some interval I, then f(x) is increasing on the interval I. Similarly, if f^{'}(x)<0 on some interval I, then f(x) is decreasing on the interval I.

Tangent Line: Assume f(x) is a differentiable function on the interval I, then the tangent line of f(x) at the point x_{0}\in I is y-f(x_{0})=f^{'}(x_{0})(x-x_{0}), where f^{'}(x_{0}) is the slope of the tangent line.

Derivative of Parameter Functions: Assume y=y(t) and x=x(t), the derivative y^{'}(x) is y^{'}(t)/x^{'}(t), because the Chain Rule of derivatives.

Question 1. Calculate the tangent line of the curve x^{\frac{1}{4}} + y^{\frac{1}{4}}=4 at the point (16,16).

Method (i). Take the derivative of the equation x^{\frac{1}{4}}+y^{\frac{1}{4}}=4 at the both sides, we get

\frac{1}{4}x^{-\frac{3}{4}} + \frac{1}{4}y^{-\frac{3}{4}} y^{'}=0.

Assume x=y=16, we have the derivative y^{'}(16)=-1. That means the tangent line of the curve at the point (16,16) is y-16=-(x-16). i.e. y=-x+32.

Method (ii). From the equation, we know y(x)=(4-x^{\frac{1}{4}})^{4} , then calculating the derivative directly. i.e.

y^{'}(x)=4(4-x^{\frac{1}{4}})^{3}\cdot (-1)\cdot \frac{1}{4}x^{-\frac{3}{4}}

Therefore, y^{'}(16)=-1.

Method (iii). Making the substitution x=4^{4}\cos^{8}\theta, y=4^{4}\sin^{8}\theta, then (16,16) corresponds to \theta=\pi/4. From the derivative of the parameter functions, we know

\frac{dy}{dx}= \frac{dy/d\theta}{dx/d\theta}=\frac{4^{4}\cdot 8\sin^{7}\theta\cdot \cos\theta}{4^{4}\cdot 8\cos^{7}\theta\cdot (-\sin\theta)}

If we assume \theta=\pi/4, then y^{'}(16)=-1.

Method (iv). Geometric Intuition. Since the equation x^{\frac{1}{4}}+y^{\frac{1}{4}}=4 is a symmetric graph with the line y=x, and (16,16) is also on the symmetric line. Therefore, the slope of the curve at the point (16,16) is -1. Hence, the tangent line is y=-x+32.

Question 2. Let y=(1+x^{2})^{-2} and x=\cot \theta. Find dy/dx and express your answer in terms of \theta.

Method (i). y=\frac{1}{1+x^{2}}= \sin^{2}\theta

\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta} = \frac{2\sin\theta \cos \theta}{-\sin^{-2}\theta}= - \sin^{2}\theta\sin2\theta.

Method (ii). \frac{dy}{dx}=-\frac{2x}{(1+x^{2})^{2}} = -\frac{2\cot \theta}{(1+\cot^{2}\theta)^{2}}=-\sin^{2}\theta\sin 2\theta.

Key Points of MA 1506

Recall Key Points of MA 1505:

1. Fundamental Theorem in Calculus:

(\int_{0}^{x} f(t)dt)^{'}=f(x)

2. Integration by Parts:

\int f(x) dg(x)= f(x)g(x) - \int g(x) df(x)

3. Derivatives and Integration:

( \sin x)^{'} = \cos x

(\cos x)^{'} = -\sin x

Hyperbolic Sine: \sinh x= \frac{e^{x}- e^{-x}}{2}

Hyperbolic Cosine: \cosh x=\frac{e^{x}+e^{-x}}{2}

(\sinh x)^{'}= \cosh x

(\cosh x)^{'}= \sinh x

MA 1506 Tutorials:

Ordinary Differential Equations:

1. Newton-Leibniz Formula

y=y(x) is a function with one variable x with ordinary differential equation y^{'}=f(x). The solution is

y=\int f(x) dx + C with some constant C .

2. Separable Equations

y=y(x) is a function with one variable x with ordinary differential equation N(y)y^{'}=M(x), where N(y) is a function with one variable y and M(x) is a function with one variable x.

The solution is \int M(x) dx = \int N(y) dy + C with some constant C.

3. One Order Ordinary Differential Equations

y=y(x) is a function with one variable x with one order ordinary differential equation y^{'}+P(x)y=Q(x). The integrating factor is R(x)= exp ( \int P(x) dx). That means

d( R(x) y) = R(x) Q(x) and take the integration under x at the both sides,

R(x)y=\int R(x)Q(x) dx + C for some constant C.

Sometimes, we need to make some substitution as z=y^{2} or z=\frac{1}{y} , since the following formulas:

2 y y^{'} = (y^{2})^{'},

-\frac{y^{'}}{y^{2}} = (\frac{1}{y})^{'}.

If there is an initial condition y(0)=A for the first order ordinary differential equation y^{'} + P(x)y=Q(x), then we must make use of the initial condition to calculate the constant C after we solved the equation.

4. Second Order Ordinary Differential Equations