校庆对于每所大学都是非常重要的一件事情，它不仅标志着大学的诞生，也可以反映学校的发展历史。今年正值新加坡国立大学成立 115 周年，也是笔者毕业五周年之际，于是提笔写下此文回忆当年参加毕业典礼的往事。
新加坡国立大学（National University of Singapore，简写为 NUS）是新加坡的第一所高等学府，是亚洲的顶级名校，其前身是一所成立于 1905 年的海峡殖民地医学学校，距今已经有 115 年的历史。目前，新加坡国立大学已经是拥有多个学院的综合性研究大学，其中包括理学院，工学院，商学院，医学院等重要学院。在泰晤士报的排名和 QS 世界大学排名上，NUS 一直在亚洲甚至世界上名列前茅。
在 NUS 的官网上，也换成了 115 校庆的页面，同时右上角的标语 “SHAPING THE FUTURE” 引入眼帘。其标语的中文含义大致就是形塑未来。并且可以在 NUS 的官网上看到校长寄语。
其大致的中文意思就是：
打开时光的卷轴，走进新加坡国立大学的峥嵘岁月，重温那些为国家为社会作出的卓著贡献。这一刻，我们深受鼓舞，我们倍感振奋！百余年间，新国大的优秀传统和创新精神一路引领着我们，追求卓越，精益求精，不断取得新的成就。
近年来，我们不断提升教学质量，增强科研能力，探索创业培育，构建一流的教育管理体系。未来我们将继续提供高质量的变革性教育，开展前沿和具有实际应用的科学研究，发扬传承创新精神，建设一个更美好的新加坡和世界。
新加坡国立大学始终致力于塑造未来。欢迎你与我们一起秉承新国大的创新精神，共同实现新国大立志成为全球最具创新力大学之一的愿望。
——陈永财教授
新加坡国立大学校长
除此之外，在 NUS 的 115 周年的官网上也可以看到近些年来 NUS 在科学研究和教育领域所取得的成绩和里程碑。无论是 Technology（科技），Health（健康），Enterprise（企业），Community（社区），Sustainability（可持续发展）方面，新加坡国立大学都取得了非常不错的成绩。
由于今年的新冠病毒猖獗，3 月份的 NUS Open Day 也从线下转成了线上，开启了史上第一次的 NUS E-Open House。在 2020 年的 5 月份至 6 月份，新加坡也开启了封城的策略，鼓励民众戴口罩少出行。往年的 7 月份，都是学生们参加毕业典礼的时候。但是今年为了防控疫情的需要，毕业典礼将会从 2020 年 7 月份延期至 2021 年的 1 月份延期半年，具体时间等待通知。
当年笔者是在 2015 年从 新加坡国立大学毕业的，当年毕业的时候学生签证恰好到期，最终还是通过办理旅游签证的方式进入新加坡。当时的毕业典礼是在 University Cultural Centre 举行，时间跨度从 2015 年 7 月 6 日 – 24 日。每个学生参加毕业典礼的时间是根据学位，院系等情况来进行安排的，有可能被安排在早上，下午，或者晚上。
在参加毕业典礼的时候，学生们都必须穿上相应的衣服，然后上台领取学位证书和成绩单。在 Universal Cultural Centre 外面，会有各种各样的道具以供毕业生们拍照。
在 2005 年，恰好是新加坡国立大学百年校庆的日子，其百年校庆的口号是 Unleashing Minds, Transforming Lives（思想任驰骋，生活显姿彩）。为了庆祝百年校庆，新加坡国立大学和新加坡邮政（Singapore Post）在 2005 年 4 月 20 日联合发行了纪念邮票。不过，当年的笔者才刚刚进入大学，从来没有想过会在五年之后进入 NUS 攻读博士学位，有的时候人生总是那么的奇妙。
在参加毕业典礼的时候，学生需要穿着学位服，而学位服的获得方式有两种，一是通过购买的方式，二是通过租赁的方式。每个学生参加毕业典礼的时候，都会获得两张门票，也就是说可以带自己的亲戚朋友去参加毕业典礼。学校的礼堂分两层，毕业的学生需要穿着相应的毕业服坐在第一层，然后在轮到自己的时候上台领取毕业证，再返回自己的位置。而自己的亲戚朋友可以坐在二楼，从高处观看所有的学生参加毕业典礼。而颁发毕业证的时候，都是校长亲自颁发，一般情况下这是普通学生能够接触到校长的唯一机会。
笔者当年参加毕业典礼的时候是 2015 年 7 月 9 日，当天下午就与众多博士生一起在 NUS 的校园里面合影留念，毕竟博士毕业是人生的一件大事。上图都是当年在 NUS 已经取得博士学位的同学们，里面绝大部分已经走向了学术科研的道路，只有少数人投身进入了工业界。
当时笔者在参加毕业典礼的时候，拿到一张毕业典礼的日程安排，时间是在 2015 年 7 月 9 日的晚上 8:00 开始，其中包括领导讲话，优秀毕业生致辞，颁发学位证书等环节。到了颁发学位证书，需要学生们上台的时候，学生们要先在旁边按照既定顺序站好，然后逐一上台领取毕业证并且与校长合影。而在每个学生领取毕业证书的时候，屏幕上都会呈现出文凭的名称和学生的名字。
新加坡是一个降雨非常频繁的国家，第一年在 Science Library 学习的时候，大约在 11，12 月份雨季，每天到了下午 4:00 的时候，笔者在图书馆内就会发现外面在下雨，而且下雨的时间非常准时。总让人觉得新加坡是一个定时降雨的国家。正因为如此，在 NUS 总有各种各样的雨棚，可以让教职工和学生不需要淋雨从一栋楼走到另一栋楼，或者走向车站。从 NUS 数学系的 S17 到 S16 也是存在一条过街天桥的，当年每次思考数学遇到瓶颈的时候，就喜欢来这里透气，看一看外面的风景换一换思绪。
在离开新加坡的前一个晚上（2015年7月17日晚上），笔者独自一人坐在 S17 的 4 楼会议室里，回想着这几年来新加坡生活的点点滴滴。从刚下飞机到进入 NUS 学习，从 Block 602 到 PGPR 再到 UTown，从通过博士资格考试到最终拿到博士学位。虽然经历了五年时间，但是每次回想起来却感觉是昨天的故事。正如歌词里唱的“有过多少往事，仿佛就在昨天；有过多少朋友，仿佛还在身边”。（未完待续）
infinite paraboloid
hyperbolic paraboloid
sphere with radius 
and center 
cylinder
Plane
Parabola
: 1 question. Especially, 
and 
(Integration by parts).
: 1 question, where 
is a positive real number. Especially, 
and 
.
and reverse the order of dx and dy).
and the domain 
plane. If the domain 
, where 
. Pay attention to the orientation of the curve on the boundary, i.e. the right hand rule).
).
is called a critical point of f(x), if 
on some interval I, then f(x) is increasing on the interval I. Similarly, if 
on some interval I, then f(x) is decreasing on the interval I.
is 
where 
is the slope of the tangent line.
is 
because the Chain Rule of derivatives.
at the point (16,16).
at the both sides, we get
That means the tangent line of the curve at the point (16,16) is y-16=-(x-16). i.e. y=-x+32.
, then calculating the derivative directly. i.e.
then (16,16) corresponds to 
From the derivative of the parameter functions, we know
then 
and 
Find dy/dx and express your answer in terms of 
,
is a function with one variable 
with ordinary differential equation 
The solution is
with some constant 
.
where 
is a function with one variable 
and 
is a function with one variable 
with some constant 
The integrating factor is 
That means
and take the integration under x at the both sides,
for some constant 
or 
, since the following formulas:
for the first order ordinary differential equation 
then we must make use of the initial condition to calculate the constant C after we solved the equation.
infinite paraboloid
cylinder
is a function, 
is a surface S. Then the surface integral is
denotes the cross product between 
and 
,
denotes the length of the vector 
for all 
, and 
, since 
and 
the cross product 
, such that 
determines the velocity of the fluid at 
. The flux is defined as the quantity of fluid flowing through 
does not just flow along 
has both a tangential and a normal component, then only the normal component contributes to the flux. Based on this reasoning, to find the flux, we need to take the dot product of 
is a vector field, 
is a scalar function. That means,
is a bounded domain and 
is a vector field.
is a vector field. 
is a compact surface and 
is the boundary of 
The curve 
has the positive orientation, that means following the right hand rule.
is a smooth function,
is the domain of 
.
![r:[a,b] \rightarrow C](https://s0.wp.com/latex.php?latex=r%3A%5Ba%2Cb%5D+%5Crightarrow+C+&bg=ffffff&fg=2b2b2b&s=1&c=20201002)
is a smooth curve.
is a smooth vector function,
is a vector field,
is the domain and 
denotes the boundary of 
The orientation of 
, the orientation is counterclockwise.
denotes the area of 
is called a conservative vector field, if there exists a scalar field 
. It is equivalent to these conditions:
.
depends only on the initial point and terminal point of the curve 
and 
have the same initial points and terminal points, then these two line integrals are equal: 
. For this reason, a line integral of a conservative vector field is called path independent.
, let 
denote the curve 
, where 
Let
in the domain 
, 
, 
,
on 
.
.
the 
bounded by 
and 
and 
is 
the minimum value is taken at 
.
on 
where 
is 
.
is a conservative vector field. Since
is a conservative vector field, and we can assume 
, i.e. 
Hence,
for some constant 
and 
with 
it is counterclockwise, 
is from 0 to 
it is clockwise, 
to 0.
,
, the domain 
and 
are partial derivatives of 
and 
respectively.
and the projection of it on 
the function 
is 
the volume of the ellipsoid is
and 
the determinant of Jacobian matrix is 
or 0/0, then we can use the L.Hospital Rule to calculate the limit. Precisely, the L.Hospital Rule is 
is a differentiable function and its derivative 
.
.
.
.
.
where y=1/x.
where a>0.
is
. Calculate the value of S.
to show that S=1.
is 
. Take the integration of the function on the interval [0,1], we get
.
.
is 
. Differentiate f(x) and get 
. Moreover, 
and 
.
This implies f(0)=0. Assume
we get 
That means f(1)=1.
Therefore, f(1)=1.
Answer is 
.
.
.
,
and 
.
, 
, where 
and 
.
.
.
.
, we get
denotes the partial derivative of 
with respect to the variable 
.
and 
. It is same as 
and 
.
.
is non-negative. That means the region R is 
. i.e. 
. Therefore, we should calculate the integration
is a real interval, calculate the maximum value of
![I=[-1,1].](https://s0.wp.com/latex.php?latex=I%3D%5B-1%2C1%5D.&bg=ffffff&fg=2b2b2b&s=0&c=20201002)
Therefore, the maximum value of the integration is
, then 
and 
and 
is a two variable function, then we use the notations to describe the partial derivatives of 
denotes the partial derivative of f under the variable x.
denotes the partial derivative of f under the variable y.
,
.
is a unit vector, i.e. its length is 1. If 
is 
at the point p, then we can define the directional derivative of 
,
,
, 
. That means
is 
and 
, we know
for all 
, for all 
is a 
function and 
. 
is a function of y.
is a function of x.
for some function 
and 
.
. Second, take the partial derivative of the function with respect to the variable x. That means,
from the partial differential equation. Therefore, the function 
is independent of the variable x. i.e. the function is a function of variable y.
, 
, then we take the integration of x at the both sides,
, the left hand side is 
, the right hand side is 
for some function 
That means, 
. and 
is a function of 
is a function of y. Assume 
. Take the integration of y at the both sides, we have
for some function 
for some functions 
and 
are non-negative. Find the maximum value of 
.
and 
, therefore the maximum value is taken at 
and 
The maximal value of g(L) is taken at 
.
i.e. 
and 
are non-negative variables. 
. Find the maximum value of 
the critical point is taken at 
i.e. the maximum value of f(x,y,z) is taken at 
, i.e. 
and 
i.e. 
and 
where 
and 
are constants.
ZHANG RONG 