Category Archives: National University of Singapore

在新加坡的这五年—生活篇(六)

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

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。

NUS 的官网

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。

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

NUS 的里程碑

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。

NUS 2020 年的毕业典礼

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。

NUS 2015 年的毕业典礼

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。

2015 年的小狮子

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。

NUS 百年校庆纪念邮票

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。

毕业典礼结束

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。

博士毕业照(从左到右:赵博士,王博士,郦博士,蔡博士,徐博士,张博士)

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。

毕业典礼日程表
与校长近距离接触

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。

白天的 S17 天桥
晚上的 S17 天桥

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。

S17 的 4 楼办公室

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)

在新加坡的这五年—学术篇(三)

上一篇文章写了博士生毕业答辩的一些事情,这篇文章来写一写其他的内容。

无论在哪个高校,不少的博士生在准备博士答辩之前,就已经在导师的安排下早早找好了下家。有的同学是去美国,有的同学是去日本,有的同学是去欧洲。如果实在暂时没有外出的计划和打算,同学们也就留在新加坡继续从事科研工作,例如可以留在本校做研究员(Research Fellow)继续从事之前尚未完成的工作,也可以选择去 A*Star 等科研机构跟随其他教授。对于部分能力较强的博士,则可以直接找到国内外的教职,正式拿到学术圈的入场券。

刚进师门的时候,师兄们尚未毕业,等到师兄们毕业的时候,其实也有不少的选择。有的师兄去英国做了短期博士后,有的师兄则是去了智利等南美国家从事了好几年的博士后工作。虽然相对新加坡而言,智利确实也算不上特别好的出路,但是为了科研和论文,其实去智利从事几年科研工作也是非常不错的选择。翻阅了一下 Google,看了各个国家的人均 GDP,新加坡还是属于名列前茅的,智利算是南美洲还不错的国家之一。

新加坡人均GDP
新加坡人均 GDP

智利人均GDP
智利人均 GDP

中国人均GDP
中国人均 GDP

相比智利而言,巴西实在是一个不那么靠谱的国家。虽然巴西这个国家不那么靠谱,但是他们的数学好像还是可以的,尤其是动力系统方向。在 IMPA 数学研究所和圣保罗大学(University of Sao Paulo)大学都有不少杰出的数学家,不仅有菲尔兹奖得主,还有各种各样的科研工作者。对于在新加坡生活过的学生而言,虽然去巴西能够认识一些优秀的数学家,但是这对自身的科研发展而言并不是最优的选择。毕竟未来找教职的时候,学校除了看候选人的论文之外,也会看候选人的出身。而博士出身就包括导师的人脉,学校的排名等诸多因素。

IMPA
IMPA

在 Google 上搜索了一下新加坡的博士后工资和巴西的博士后工资。根据汇率兑换工具可知:5854 巴西雷亚尔 = 7534 人民币,不如 2019 年深圳的平均工资。5000 新加坡币 = 25000 人民币,这说明新加坡的博士后收入其实是远高于巴西的博士后水平的。而且在新加坡工作的博士生,通常待遇都还不错,一般情况下也能够找到 4000 – 5000 新币甚至更高收入的工作。不到万不得已,其实没有必要去做一份 7000 人民币的工作。

巴西博士后工资
巴西博士后的工资

新加坡博士后工资
新加坡博士后的工资

巴西作为距离中国最远的国家之一,从中国香港飞到圣保罗大约需要 26 个小时的时间,至少需要在欧洲转机一次才能够继续前往巴西,而且有不少航班都是转机两次。如果在南美洲从事科学研究的话,估计回家也是一件困难的事情,一年最多只有一次机会,极有可能三年的青春都在巴西度过了。

世界地图
世界地图

在豆瓣上,有一部可以排进 Top250 的电影,叫做《上帝之城》。讲述的是巴西黑帮在里约热内卢的故事,“上帝之城”是巴西的里约热内卢这座城市西南部的一片贫民窟。这部影片所聚焦的,正是这片贫民窟 60,70,80 三个年代中发生的一件件黑帮兴衰史。

上帝之城
上帝之城

2016 年的里约奥运会,有些运动员经历了抢劫,不仅财物被抢,还有各种证件和工作物品。可见巴西在里约奥运会期间也没有把治安问题解决好,让游客和运动员都蒙受了一定的损失。

里约奥运会_1
奥运代表团

在 2018 年 9 月 2 日,巴西国家博物馆发生火灾,近九成物品被烧毁。自 2013 年以来,巴西政府一直减少博物馆的预算。国家博物馆每年维护成本需要的 50 万巴西雷亚尔,但是获得的经费只有 5.4 万雷亚尔,博物馆出现了明显的欠缺维护迹象,如剥落的墙壁和裸露的电线。2018 年 6 月,巴西国家博物馆庆祝成立 200 周年,结果却在 2018 年 9 月份被烧毁。

巴西国家博物馆
巴西博物馆

在 2020 年的疫情面前,不少国家也有很多神操作,巴西自然也不例外,其感染人数也一直在持续增加,估计短期内并不能够解决这个问题。

新型冠状病毒20200503_巴西1新型冠状病毒20200503_巴西2

新型冠状病毒20200503_巴西3
巴西的疫情

在毕业答辩的当天,曾经有一个巴西博士后的岗位放在我的面前,我没有珍惜,等我失去的时候我并没有后悔,人生最开心的事情莫过于此。因为有无数的经验告诉我,选择比努力更重要。(未完待续)

在新加坡的这五年—学术篇(二)

本文写于毕业答辩五周年之际。对于每个博士生而言,毕业答辩都是一件非常重要的事情,这不仅关系着博士生能否顺利拿到学位,也会影响未来的职业生涯。

在博士答辩的时候,学生不仅需要向答辩委员会展示这几年在学校的研究成果,还需要得到答辩委员会的一致认可才能够拿到博士学位。每个学校对于博士生的毕业要求不一样,不同的学校有着不太一样的制度,但是整体流程却是大同小异的。下图是 NUS 数学系对于 Graduate Programme 的学生的一些基本要求。包括在 24 个月之内通过 Qualify Exam,学习 MA5198 这门课,完成英语课程的学习,CAP 达到 3.5 以上,最后就是写完博士毕业论文。

NUS_PHD_requirement
博士生的毕业要求

毕业论文作为 Step 6 就代表着这是博士生从学校毕业的最后一步了。当时为了记录自己搞科研写论文的时候,专门使用了 Google Calendar,然后把自己的工作时长记录下来。每周搞数学科研的时长平均是 20 个小时左右,毕竟进行创新性工作是比体力劳动累很多的。一天看书八小时没啥问题,但是一天思考数学难题八个小时就是因人而异了。

工作时长
2014 年的每周工作时长

从 NUS 提交论文的流程来看,其实博士生想提交一篇论文并不是一件非常容易的事情。先不说撰写毕业论文的路程之艰辛,就说说这提交论文的流程就足以花费很长时间。根据 NUS 的学术日历,博士生们需要在一些特定的时候提交论文才会比较“划算”。为什么用“划算”这个词语呢?因为 NUS 的博士生在前四年是不需要交学费的,但是在博士第五年开始延期的时候就需要缴纳一定的学费,而学费的多少完全取决于在博士生什么时候提交论文。如果在每个学期的前两周提交论文,那么这个学期是不需要交学费的。如果在 Recess Week 之前提交论文,那么这个学期只需要缴纳一半的学费。如果提交论文的时间过了 Recess Week,那么博士生就要缴纳本学期全部的学费。所以,对于想要交论文的博士生,通常都会在 Vacation 的期间尽量把论文全部写好,然后赶在开学的前两周提交论文,这样的话一来有充足的时间撰写论文,二来不用缴纳下一个学期的学费。

NUS学校日历
NUS 的学术日历

NUS 的博士生在前四年是免学费的,但是到了第五年的时候就要开始交学费了。而学费的涨幅也是十分惊人的,记得在 2010 年的时候一年的学费大约是 13000 SGD,而到了 2020 年的时候一年的学费已经涨到了 40000 SGD。虽然学费涨幅惊人,但是只要博士生能够按时毕业,就可以避免缴纳高额的学费。这可能也是督促博士生按时毕业的一个好方法。

NUS学费
NUS 的学费

言归正传,在进行答辩之前需要在学校和院系提交博士生的毕业论文。第一次提交论文初稿的时候需要提交四份打印稿,并且装订成册。方便答辩委员会的成员阅读和翻阅。当时打印论文的时候,是由大峰哥带路去 Queenstown 打印,价格比学校的 YIH 稍微便宜一点,时间也会相对快很多。

对于博士生的答辩,除了导师会在现场之外,也需要从院系里面选择一两位教授,还需要从外校选择一位比较精通该领域的学者,共同组成该学生的答辩委员会。答辩委员会不仅要在学生答辩的时候出席,还要对学生提交的博士论文进行审核,查看该学生的学术成果是否能够达到毕业的要求。如果答辩委员会在审核期间对学生的论文有所疑问的话,也可以通过邮件的方式来向院系的秘书和学生发邮件,要求学生提供更加详细的资料和证明方法。不过发邮件提问的时候答辩委员会的教授都是匿名的,学生不能知道是哪位审稿人所提出的问题。一般审稿的期限为 3 个月左右,一旦达到了这个期限,并且审稿没有问题的话,就可以让院系安排博士生的毕业答辩了。

毕业答辩的邮件通知_full
答辩邮件

2015 年 4 月 24 日,是笔者在新加坡国立大学(National University of Singapore) 的答辩日子。每次有博士生答辩的时候,数学系的行政人员们就会提前发送一封邮件给大家,通知大家系里面有人去答辩了,请感兴趣的老师和同学自行前往。一般情况下,答辩的时长是一个小时左右。除此本校的教授无需专程移动之外,答辩委员会的成员也需要从各个国家飞往当地进行毕业答辩。在答辩的时候,学生需要准备一份 PPT 来讲解之前几年在学校做的科研工作,还需要提前准备一些可能面临的挑战点以应对大家的提问。听讲座的人背景不一样,所提出的问题也有可能是五花八门的,所以做好充足的准备是非常有必要的。在答辩的前两天,有的导师也会将答辩委员会的教授们介绍给博士生认识,毕竟学术圈也是一个圈子,混个脸熟还是十分有必要的。不过学术圈的人通知消息的方式都比较特殊,记得当时导师通知开会和吃饭都是通过邮件的方式,和现在通过微信和电话的方式有明显差异。

毕业聚餐的地点
聚餐地点

在博士生进行答辩陈述结束之后,答辩委员会的成员会让博士生暂时离开教室一段时间,然后他们在教室里面填写必要的资料,决定博士生是否能够通过这次答辩。如果顺利通过的话,教授们会走出来通知学生已经通过了答辩。如果不幸没有通过的话,就只能让学生持续修改论文和再接再厉了。除了参加博士生的答辩之外,学术圈的教授们聚集到一个学校,总会开设一两个讲座或者进行必要的聚餐。这次自然也不例外,笔者还有幸在答辩的当天下午听了一个教授的讲座。到了吃饭时间,下午 5:00 左右,导师就把学生和教授喊在一起,去外面聚餐吃饭。不过当时吃饭的地点就是在学校附近的食阁,大家会点上不少下酒菜,然后点上很多啤酒(PS:新加坡的酒真的是贵),在食阁那里谈天说地。当时留在师门的人也不多了,大师兄和二师兄已经顺利毕业离开了新加坡,还剩下三师兄和我在一起搞科研和答辩。不过在我答辩完了之后,三师兄也顺利提交了论文,在 2015 年的后半年也顺利通过了博士答辩。

大师兄的毕业证
大师兄的毕业证

记得在 2015 年 4 月 13 日的时候,大清早起来帮大师兄去拿 University Hall 拿学位证。NUS 有固定的发放学位的时间,及时大师兄是 2014 年交的论文,拿到学位证和成绩单也需要等到次年的 4 月份。因此,算好交论文的时间也是一门有趣的事情。不过导师从国内过来的时候,也是把三位师兄带到新加坡,因此三位师兄的年纪略比我大一些。师兄们也是经历了国内的硕士教育和国外的博士教育最终拿到学位,可见拿到博士学位之路真的不好走。

NUS_YIH
NUS 的 YIH

答辩完了之后其实还需要向学校提交一些材料和流程,提交材料的地点就是 NUS 的 YIH。记得当时提交论文不仅需要在网站上提交论文的 PDF 文件,并且把论文刻在一个光盘内才行(PS:这年头用光盘的时候真的是不多了)。当年在南京买的电脑 Y530 还带着光驱,但是从来没有使用过,在 2008 年买的时候绝对没有想到这台电脑刻的唯一一张光盘就是自己的博士毕业论文。(未完待续)

NUS E-Open House

新加坡(Singapore)

新加坡被誉为“花园城市”,又称为“狮城”,它位于马来半岛的南端,地处在马六甲海峡最南端的位置。属于热带雨林气候的新加坡,一年只有雨季与旱季两个季节,从而导致狮城降雨频繁,气温长期处于 25-34 摄氏度左右。新加坡是一个多民族融合的国家,英语作为其官方语言。

Singapore_Map_1
新加坡地图

Singapore_Map_2
新加坡夜景

新加坡国立大学(National University of Singapore)

位于新加坡西海岸的新加坡国立大学(National University of Singapore)是一所综合性大学,近十年在泰晤士报(Times)世界大学排名和 QS 世界大学排名上,NUS 都长期名列前茅,属于亚洲的顶级学府。NUS 的前身成立于 1905 年,今年恰好是 NUS 115 校庆的年份。发展至今,NUS 已经是拥有多个学院的综合性大学,其中包括理学院(Faculty of Science),工学院(Faculty of Engineering),商学院(Business School)等诸多学院。

NUS-Map-1
NUS 的地图

NUS Open Day(新加坡国立大学开放日)

学校的排名除了全体教职工的努力之外,优秀的生源也是保障学校能够持续运营的必要条件。因此,在每个学年的第二个学期,大约在 2,3月份的时候,NUS 都会开展 Open Day,目的之一就是对新加坡的本地学生和国际学生宣传 NUS。除了宣传自己的硬件设施,专业特色,教授质量之外,还会宣传校园生活(Campus Life),基础设施和历年优秀学生的就业情况。

当年笔者还在 NUS 就读的时候,就有幸参加过 NUS 的 Open Day,不过当年只是为了凑热闹而去观看了一些活动,并没有在申请学校之前就享受到这些福利。每年到了 Open Day 之际,都会有大批中学生或者理工院校的学生前来 NUS,通过参观 NUS 的情况,来判断 NUS 是否适合自身的发展。

NUS-Open_Day-2015-1
NUS Open Day 2015

NUS E-Open House

由于 2020 年的开局实在是不利,世界上的诸多国家都受到了新型冠状病毒的影响,自然新加坡也不例外。可能是因为这个原因,2020 年的 NUS Open Day 就从线下(Offline)搬到了线上(Online),通过网络这一个重要的媒介来开展 Open Day。今年的 NUS E-Open House 应该是历史上首次在网上举办的校园开放日活动,同时校方在 NUS 的官网上也进行了大力的宣传。

NUS-E-OpenDay-Facebook-3
NUS E-Open House 2020

除此之外,NUS 也通过 Facebook 账号也进行了推广,并且学生们可以通过 Facebook,YouTube,Instagram,Zoom 等诸多社交网络工具来全方位的了解 NUS。

NUS-E-OpenDay-Facebook-1

NUS-E-OpenDay-Facebook-2

由于本次的 NUS E-Open House 为期九天,从 2020-02-26 到 2020-03-05,个人感觉 Open Day 不如称之为 Open Week。在这段时间,NUS 的所有学院将会通过网络向全世界展示 NUS 的魅力。在这段时间里,学生和家长,包括想要继续深造的职场人士都可以通过社交网络了解到 NUS 的方方面面。

NUS-E-OpenDay-Web-1
2020 年 NUS E-Open House 的日程安排

除了学院之外,每个学院都会基于自身的条件来设计相关的项目。因此,向外部人士展示各个学院的项目细节也是一个非常重要的环节。

NUS-E-OpenDay-Web-2
项目列表

当然,学习是大学生活的一部分,除了学习之外,校园生活也是许多家长和学生关心的话题之一。于是,丰富多彩的校园生活则是 NUS 的一大特色。

NUS-E-OpenDay-Web-3
校园生活

NUS 的院系

Faculty of Arts and Social Sciences(艺术与社会科学学院)

前几年,随着直播(Live)这门技术的蓬勃发展,直播已经融入了人们生活的方方面面。FASS 的学生则是通过直播的方式向大家介绍了学院的地理位置,食堂,校园生活,学习等诸多内容,让观众感受到了在校学生的热情与活力。

NUS-E-OpenDay-FASS-5

当然,只有学生的宣传是远远不够的,此时需要整个院系的努力才能够把宣传力度加到最大。在 NUS e-Open House 的主页上,可以找到 FASS 的活动时间和宣传安排,使用各种各样的宣传渠道,对这个专业感兴趣的学生就会早早订阅并且关注这一消息。

NUS-E-OpenDay-FASS-2
FASS 的宣传

NUS-E-OpenDay-FASS-4
FASS 的专业

在 FASS 中,学生们可以选择的专业也是非常多的,不仅包括中文,日语,哲学,历史等常见课程,还包括心理学,社会科学等方向。

Business School(商学院)

在商学院,不仅邀请了院系的教授与未来的学生们直接对话,还邀请了学生代表进行发言,通过亲身经历来向大家展示这几年在商学院得到的知识与心得。

NUS-E-OpenDay-Business-3NUS-E-OpenDay-Business-1NUS-E-OpenDay-Business-2

School of Continuing and Lifelong Education(持续与终身教育学院)

学习不仅仅是学生时期的事情,在这个高速发展的时代,学习这件事情将会伴随我们每个人一生的时间。于 2016 年成立的 School of Continuing and Lifelong Education 会给每个成年人提供持续教育的机会。它主要是为了想学习新技能的成年人或者想拿到学位的成年人而设置的。在这个学院有本科项目和硕士项目,还有短期项目或者各种培训。如果想拿到某些证书或者学位,其实这里是个不错的选择。

当年在新加坡国立大学(NUS)读书的时候,身边就有一些攻读 Master 学位的同学,当时他们就已经在公司工作了,在工作之余会选择 part time 的硕士在进行攻读。只要达到了学分或者做完了相应的 Project 就可以拿到 Master 学位。

NUS-E-OpenDay-SCALE-7

NUS-E-OpenDay-SCALE-2

Department of Mathematics(数学系)

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。

在本次的 NUS E-Open House 中,Science 的每个院系都提供了相应的宣传资料,并且所有学科统一模板,由各个学科的教授来向大家介绍这些专业的背景,学习内容和相关优势。在 NUS 的数学系(Department of Mathematics),则是由 Prof Tan 来给大家介绍院系的相关内容。

NUS-E-OpenDay-Science-3

通过 Prof Tan 的PPT,我们可以得到选择数学系的五大原因分别是:

  1. Finding Good Jobs;
  2. Wide Range of Career Choices;
  3. Multiple Pathways;
  4. Myriads of Real World Applications;
  5. Life Long Learning Skills。

NUS-E-OpenDay-Science-Math-6
数学系学生的就业统计

从学校统计到的数学系学生毕业出路来看,其实数学系学生们的选择范围还是相对广泛的。不仅包括教育行业(Education)和科研行业(Research),还包括金融(Finance),科技(Technology),信息管理(Information Management)等诸多热门方向。就业的公司不仅包括星展银行(DBS),花旗银行(CitiBank),还包括 Google,Facebook 等科技公司。如果学生在读大学之前并不确定未来要做什么方向的话,并且对学习数学有一定的能力和兴趣的前提下,在本科期间选择数学专业其实是一个不错的选择。

NUS-E-OpenDay-Science-Math-8
数学系学生的就业方向

俗话说,数学是科学的基础。数学不仅仅是数学书上的一道道定理,而是可以解决现实生活问题的重要工具。在天气预测方向,动力系统(Dynamical System)有着独特的应用;在机器学习领域,微积分,线性代数与概率论则为这门学科提供了理论基础。在金融领域,Black-Scholes 方程,Monte Carlo 模拟则是其中的重要模型。从这些学科的发展来看,数学不仅可以为这些学科提供理论基础,也是解决这些学科难题的重要工具之一。

NUS-E-OpenDay-Science-Math-12
数学在各个学科中的应用

NUS-E-OpenDay-Science-Math-16
NUS 数学系的项目

数学系的课程以难著称,有人说:“没有比数学书更好的劝退材料”。要想劝退一个人学习数学,其实是非常简单的。在 NUS 数学系,数学系的课程难度就是课程 ID 的第一个数字。例如 MA1100,就表示 Level 1 的课程,属于数学系的入门课。MA4270 就表示 Level 4 的课程,属于数学系本科课程中较难的课程。学生们可以根据自身的情况和时间安排选择相关的课程进行学习。

NUS-E-OpenDay-Science-Math-18
数学系的课程

当然,每个专业都有着不同的细分方向,虽然刚入学的时候,同一院系的学生所学课程都是一样的,但是随着时间的迁移,不同的学生会选择不同的方向进一步的深造。在数学系,通常会有 Math(数学),Applied Mathematics(应用数学),Quantitative Finance(金融数学),数据科学与分析(Data Science & Analytics)这几个方向。学生们需要在学习的过程中根据自身的兴趣和需要选择最合适自己的方向。

NUS-E-OpenDay-Science-Math-17
NUS 数学系的项目

在本科的时候,通常都会有不少学有余力的学生,不满足于只学习一门课程,或者不满足于仅仅拿到一个学位。此时,NUS 数学系还可以提供双学位的项目(Double Degree Programme),或者选择双主修的项目(Double Major Programmes)。学生可以在其中选择经济学,管理学,商业分析,计算机科学,信息安全这几个方向进行辅修,从而把在数学系学到的知识进一步地应用在其他学科中。

NUS-E-OpenDay-Science-Math-22
数学系的双学位

Application(NUS 的申请)

在新加坡国立大学的官网上,可以找到其申请的入口,学生们可以根据自身的情况申请相应的项目,从而度过一个丰富多彩的大学生活。

NUS-E-OpenDay-Application-1
NUS 的申请

NUS-E-OpenDay-Application-3
NUS 的申请流程

结束语

通过新加坡国立大学的校园开放日(NUS Open Day),我们可以了解到 NUS 的院系特点,校园生活,学习项目等很多内容。而 NUS E-Open House 则是在这个特殊时期的一个创新,充分利用互联网的优势,把 NUS 的特色通过网络传递给大家,向大家充分展示了 NUS 的魅力之处。

参考资料:

  1. 新加坡国立大学的博士生项目介绍
  2. 新加坡的留学生活
  3. 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

Prediction of Final Exam 2013-2014 Semester I

Module:                 MA 1505 Mathematics I

Time:                      2 hours ( 120 minutes )

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 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}             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 and center (x_{0},y_{0},z_{0})

x^{2}+y^{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 series, Parseval’s identity: 2-3 questions. ( Integration by parts, calculate the sum of Fourier coefficients, period 2L functions ( where L is a positive real number), calculate the value of some special series from Fourier series, cosine expansion and sine expansion of function on the half domain).

Chapter 7: Multiple Variable Functions

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 Integration

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

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

Volume of the solid: 1 question. ( Double integrals).

Area of the surface: 1 question. ( Partial Derivatives of two variable functions, Polar Coordinate).

Chapter 9: Line Integrals

Length of the curve: 0-1 question. ( Parameter equation of the curves).

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).

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.

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, Cross product of vectors, the length of vectors).

Surface integrals of vector fields: 1 question. ( The equation of surface, Cross product and Dot product of vectors).

Stokes’ Theorem: 1 question. ( Pay attention to the orientation).

Divergence Theorem: 0-1 question. ( Triple integrals).

MA 1505 Tutorial 11: Surface Integral, Divergence Theorem and Stokes’ Theorem

Surface Integrals of Scalar Fields: Assume f: U \subseteq \mathbb{R}^{3} \rightarrow \mathbb{R} is a function, r: D\subseteq \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} is a surface S. Then the surface integral is

\iint_{S} f dS= \iint_{D} f(\textbf{r}(x,y)) || \textbf{r}_{x} \times \textbf{r}_{y} || dxdy

where the left hand side is the surface integral of the scalar field and the right hand side is the multiple integration. \textbf{r}_{x} \times \textbf{r}_{y} denotes the cross product between \textbf{r}_{x} and \textbf{r}_{y} ,

|| \textbf{r}_{x} \times \textbf{r}_{y} ||  denotes the length of the vector \textbf{r}_{x} \times \textbf{r}_{y}.

Remark.  If f(x,y,z)=1 for all (x,y,z) \in \mathbb{R}^{3} , and r: D\subseteq \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} is a surface, then

the left hand side is \iint_{S} dS = \text{ the surface area of } S.

the right hand side is \iint_{D} \sqrt{1+(f_{x})^{2}+ (f_{y})^{2} } dxdy , since \textbf{r}(x,y)=(x,y,f(x,y)), \text{ where } (x,y) \in D, \textbf{r}_{x}=(1,0,f_{x}) and \textbf{r}_{y}=(0,1,f_{y}), the cross product \textbf{r}_{x} \times \textbf{r}_{y}= (-f_{x}, -f_{y},1).

That means:

\text{ the surface area of } S= \iint_{D} \sqrt{1+(f_{x})^{2}+(f_{y})^{2} }dxdy.

Surface Integrals of Vector Fields:

Imagine that we have a fluid flowing through S, such that \bold{F}(x) determines the velocity of the fluid at \bold{x}. The flux is defined as the quantity of fluid flowing through S per unit time.

This illustration implies that if the vector field is tangent to S at each point, then the flux is zero, because the fluid just flows in parallel to S, and neither in nor out. This also implies that if \bold{F} does not just flow along S, that is, if F 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 \bold{F} with the unit normal vector to S at each point, which will give us a scalar field, and integrate the obtained field as above.

1280px-Surface_vectors

Assume \textbf{F} : U \subseteq \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} is a vector field, r: D\subseteq \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} is a surface S. Then the surface integrals of the vector field F is

\iint_{S} \textbf{F} \cdot d \textbf{S} = \iint_{S} \textbf{F} \cdot \textbf{n} dS

The left hand side is the surface integral of vector field and the right hand side is the surface integral of scalar function, since \textbf{F} \cdot \textbf{n} is a scalar function. That means,

\iint_{S} \textbf{F} \cdot d \textbf{S} = \iint_{S} \textbf{F} \cdot \textbf{n} dS = \iint_{D} \textbf{F}( \textbf{r}(x,y)) \cdot ( \textbf{r}_{x} \times \textbf{r}_{y}) dxdy

Divergence Theorem (Gauss’s theorem or Ostrogradsky’s theorem)

This theorem is a result that relates the flow (that is, flux) of a vector field through a surface to the behavior of the vector field inside the surface. More precisely, the divergence theorem states that the outward flux of a vector field through a closed surface is equal to the volume integral of the divergence over the region inside the surface. Intuitively, it states that the sum of all sources minus the sum of all sinks gives the net flow out of a region.

\iint_{S} \textbf{F} \cdot d \textbf{S} = \iiint_{V} \nabla \cdot \textbf{F} dV = \iiint_{V} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}) dxdydz

where V \subseteq \mathbb{R}^{3} is a bounded domain and \partial V=S, \textbf{F}=(P,Q,R) is a vector field.

800px-Divergence_theorem.svg

Stokes’ Theorem

\int_{\partial \Sigma} \textbf{F} \cdot d\textbf{r} = \iint_{\Sigma} ( \textbf{curl F} ) \cdot d \textbf{S}

where \textbf{curl F}= (\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}) \textbf{i} + (\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}) \textbf{j} + ( \frac{\partial Q}{\partial x}- \frac{\partial P}{\partial y}) \textbf{k} is a vector field. \Sigma is a compact surface and \partial \Sigma  is the boundary of \Sigma. The curve \partial\Sigma has the positive orientation, that means following the right hand rule.

429px-Stokes'_Theorem.svg

MA 1505 Tutorial 9 and 10: Line Integral and Green’s Formula

Line integral of a scalar field:

Assume f: U \subseteq \mathbb{R}^{n} \rightarrow \mathbb{R} is a smooth function,

where U is the domain of f.

\int_{C} f ds =\int_{a}^{b} f(r(t)) \cdot ||r^{'}(t)|| dt

where r:[a,b] \rightarrow C is a smooth curve.

Line integral of a vector field:

Assume \mathbf{F}: U \subseteq \mathbb{R}^{n} \rightarrow \mathbb{R}^{n} is a smooth vector function,

\int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{a}^{b} \mathbf{F}( \mathbf{r}(t)) \cdot \mathbf{r}^{'}(t) dt

where r:[a,b] \rightarrow C  is a smooth curve.

Green’s Formula: 

Assume \bold{F}=(P,Q) is a vector field,

\oint_{\partial D} \bold{F}\cdot d\bold{s}=\oint_{\partial D} Pdx + Qdy = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{ \partial P}{\partial y}) dxdy,

where D is the domain and \partial D denotes the boundary of D. The orientation of \partial D satisfies the left hand rule. That means if you walk along the boundary of D, the domain D must be on your left.

429px-Green's-theorem-simple-region.svg

D is a simply connected region with boundary consisting four boundaries C_{1}, C_{2}, C_{3}, C_{4}, the orientation is counterclockwise.

macroscopic_microscopic_circulationmacroscopic_microscopic_circulation_hole

In the first graph, \partial D which denotes the boundary of D has only one closed curve C and the orientation of C is counterclockwise. However, in the second graph, \partial D contains two curves, i.e. the blue one and the red one. The orientation on the blue one which is the outer boundary of D is counterclockwise, the orientation on the red one which in the inner boundary of D is clockwise. That means if you walk along the boundary of D, the domain D must be on your left. This is the left hand rule.

Corollary of Green’s Theorem

Assume D is a domain in the plane, Area(D) denotes the area of D, then the area can be calculated from the following formulas:

Area(D)=\iint_{D} dxdy

= \oint_{\partial D} xdy = \oint_{\partial D} -ydx =\oint_{\partial D} (-\frac{y}{2} dx +\frac{x}{2} dy)

Fundamental Theorem of Line Integral:

\int_{C}\nabla f \cdot d\bold{r}=f(\text{terminal point})-f(\text{initial point}),

where C denotes a curve from initial point to terminal point and f is a scalar field.

Conservative Vector Field:

1. \bold{F}=(P,Q,R) is called a conservative vector field, if there exists a scalar field f such that \nabla f=\bold{F}. It is equivalent to these conditions:

\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}, \frac{\partial R}{\partial x}=\frac{\partial P}{\partial z}.

2. \bold{F}=(P,Q) is called a conservative vector field, if  there exists a scalar field f such that \nabla f=\bold{F}. It is equivalent the condition \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}.

Path Independence:

A key property of a conservative vector field is that its integral along a path depends only on the endpoints of that path, not the particular route taken.

For example, if \bold{F}=(P,Q) or \bold{F}=(P,Q,R) is a conservative vector field, then the value of the line integral I=\int_{C} \bold{F}\cdot d\bold{r} depends only on the initial point and terminal point of the curve C. That means if \bold{F} is a conservative vector field, the curves C_{1} and C_{2} have the same initial points and terminal points, then these two line integrals are equal: \int_{C_{1}} \bold{F}\cdot d\bold{r}=\int_{C_{2}}\bold{F}\cdot d\bold{r}. For this reason, a line integral of a conservative vector field is called path independent.

Question 1. For each non-zero constant a>0, let C_{a} denote the curve y=a \sin x, where 0 \leq x \leq \pi. Let

I(a)= \int_{C_{a}} (1+y^{3})dx + (2x+y)dy

Find the minimum value of I(a) in the domain a>0.

Solution.

Method (i).  Use the definition of line integration.

Since y=a \sin x , 0 \leq x \leq \pi , dy= a \cos x dx ,

I(a)= \int_{0}^{\pi} (1+a^{3} \sin^{3} x) dx + (2x+ a \sin x) a \cos x dx

= \int_{0}^{\pi} ( 1+ a^{3} \sin^{3} x + 2a x \cos x + a^{2} \sin x \cos x) dx.

Since

\int_{0}^{\pi} (a^{3} \sin^{3} x) dx = \frac{4}{3} a^{3},

\int_{0}^{\pi} (2a x \cos x) dx =-4a,

\int_{0}^{\pi} (a^{2} \sin x \cos x) dx =0,

we get

I(a)= \pi + \frac{4}{3} a^{3} - 4a on a>0 .

I^{'}(a) = 4a^{2}-4 .

The minimum value is taken at a=1, the I(1)= \pi -\frac{8}{3}.

Method (ii). Use Green’s Formula.

Consider the domain D bounded by y= a\sin x and x=0, P(x,y)=1+y^{3} and Q(x,y)= 2x+y.

i.e.

D: 0\leq x \leq \pi, 0\leq y \leq a \sin x.

From Green’s Formula, pay attention to the orientation,

\iint_{D} ( 2-3y^{2}) dxdy = -I(a) + \int_{0}^{\pi} dx

Therefore,

I(a) = \pi - \iint_{D} ( 2- 3y^{2}) dxdy

= \pi - \int_{0}^{\pi} \int_{0}^{a \sin x} (2-3y^{2}) dy dx

= \pi - \int_{0}^{\pi} ( 2a \sin x - a^{3} \sin^{3} x) dx

= \pi - 4a + \frac{4}{3} a^{3}.

The derivative of I(a) is I^{'}(a) = 4 a^{2}-4, the minimum value is taken at a=1, and I(1)= \pi-\frac{8}{3}.

Question 2. Prove the area of the disc with radius R is \pi R^{2}.

Solution. 

Method (i). Definition of Integration.

Area=4 \int_{0}^{R} \sqrt{R^{2}-x^{2}} dx

= 4R^{2} \int_{0}^{\frac{\pi}{2}} cos^{2}\theta d \theta  (x=\sin \theta)

= 4R^{2} \int_{0}^{\frac{\pi}{2}} \frac{1+ \cos (2\theta)}{2} d\theta

= \pi R^{2}.

Method (ii). Green Formula

Area(D)=\iint_{D} dxdy

= \oint_{\partial D} xdy = \oint_{\partial D} -ydx =\oint_{\partial D} (-\frac{y}{2} dx +\frac{x}{2} dy)

For D=\{ x^{2}+y^{2}\leq R^{2} \}, on \partial D, x=\cos \theta, y=\sin\theta, where \theta \in [0, 2\pi).

Area(D)= \oint_{\partial D} x dy

= \int_{0}^{2\pi} R \cos \theta \cdot R \cos \theta d\theta

= \pi R^{2}.

Question 3. Prove the area of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1 is \pi ab.

Solution.  It is similar to Question 2, and x=a \cos \theta, y= b \cos \theta.

Question 4.  Calculate

\int_{C} 2xy dx + (x^{2}+z)dy +y dz

where C consists two line segments: C_{1} from (0,0,0) to (1,0,2), and C_{2} from (1,0,2) to (3,4,1).

Solution. 

Method (i).  From the definition of line integral,

C_{1}: r_{1}(t)=(t,0,2t), C_{2}: r_{2}(t)=(1+2t, 4t, 2-t).

\int_{C_{1}} 2xy dx + (x^{2}+z)dy +y dz=0

\int_{C_{2}} 2xy dx + (x^{2}+z)dy +y dz

=\int_{0}^{1}( 48 t^{2} + 24 t +12) dt

=40.

Method (ii). Check the vector field \mathbf{F}=(2xy, x^{2}+z, y) is a conservative vector field. Since

\frac{\partial}{\partial y} (2xy)= \frac{\partial}{\partial x} (x^{2}+z)=2x

\frac{\partial}{\partial x}(y)=\frac{\partial}{\partial z}(2xy)=0

\frac{\partial}{\partial y}(y)=\frac{\partial}{\partial z}(x^{2}+z)=1

Therefore, \mathbf{F} is a conservative vector field, and we can assume \nabla f=\mathbf{F} , i.e. f_{x}=2xy, f_{y}=x^{2}+z, f_{z}=y. Hence,

f(x,y,z)=x^{2}y+yz+K for some constant K.

Therefore, the answer is f(3,4,1)-f(0,0,0)=40.

Question 5. Evaluate

\oint_{C} (x^{5}-y^{5}) dx + (x^{5}+y^{5}) dy,

where C denotes the boundary with positive orientation of the region between the circles x^{2}+y^{2}=a^{2} and x^{2}+y^{2}=b^{2} with 0<a<b.

Solution.

Method (i). The definition of the line integral.

On circle x^{2}+y^{2}=b^{2}, it is counterclockwise, x= b \cos \theta \text{ and } y= b \sin \theta,  \theta is from 0 to 2\pi.

On circle x^{2}+y^{2}=a^{2}, it is clockwise, x=a \cos \theta \text{ and } y= a \sin \theta, \theta is from 2\pi to 0.

On the circle C_{b}: x^{2}+y^{2}=b^{2} ,

\oint_{C_{b}} (x^{5}-y^{5})dx + ( x^{5}+y^{5}) dy

= b^{6} \int_{0}^{2\pi} ( \cos^{5} \theta - \sin^{5} \theta) \cdot ( - \sin \theta) + ( \cos^{5} \theta + \sin^{5} \theta) \cdot \cos \theta ) d \theta

= b^{6} \int_{0}^{2\pi} ( \cos^{6} \theta + \sin^{6} \theta) - \sin \theta \cos \theta ( \cos^{4} \theta - \sin^{4}\theta ) d\theta

= b^{6} \int_{0}^{2\pi} ( 1- 3\sin^{2} \theta \cos^{2}\theta - \frac{1}{2} \sin 2\theta \cos 2\theta ) d\theta

= b^{6} \int_{0}^{2\pi} ( \frac{5}{8} -\frac{3}{8} \cos 4\theta -\frac{1}{4} \sin 4 \theta ) d\theta

= b^{6} \cdot \frac{5}{4} \cdot \pi.

Pay attention to the orientation, we get the answer is

\frac{5 \pi}{4} ( b^{6}-a^{6}).

Method (ii). Green’s Theorem.

\oint_{C} (x^{5}-y^{5}) dx + (x^{5}+y^{5}) dy,

= \iint_{D}( 5 x^{4}+ 5 y^{4}) dxdy

= 5 \int_{0}^{2\pi} \int_{a}^{b} r^{5}( \cos^{4} \theta + \sin^{4} \theta) dr d \theta

= 5 \int_{a}^{b} r^{5} dr \cdot \int_{0}^{2 \pi} ( \cos^{4} \theta + \sin^{4}\theta) d\theta

= \frac{5( b^{6}-a^{6})}{6} \int_{0}^{2\pi} (1-2\sin^{2}\theta \cos^{2} \theta) d\theta

= \frac{5(b^{6}-a^{6})}{6} \cdot \frac{3 \pi}{2}

= \frac{5}{4} \pi ( b^{6}-a^{6}).

MA 1505 Tutorial 8: Surface Area and Volume

Assume z=z(x,y) is a surface on \mathbb{R}^{3}, the domain R is the projection of the surface z=z(x,y)  on xy-plane. Then the area of the surface is

\iint_{R} \sqrt{1+z_{x}^{2}+z_{y}^{2}} dxdy,

where z_{x} and z_{y} are partial derivatives of z=z(x,y) with respect to the variable x and y respectively.

If a surface is z=z(x,y)\geq 0 and the projection of it on xy-plane is R, then the volume bounded by xy-plane and the surface z=z(x,y) is

\iint_{R} z(x,y) dxdy.

Theorem 1. 

The surface area of the sphere with radius R is 4\pi R^{2}.

The volume of the sphere with radius R is \frac{4}{3} \pi R^{3}.

Proof.  The equation of the sphere with radius R is x^{2}+y^{2}+z^{2}=R^{2}.

First we calculate the surface area of sphere.

Assume R=\{ (x,y): x^{2}+y^{2} \leq R^{2} \}, the function z=z(x,y)=\sqrt{ R^{2}-x^{2}-y^{2}}.

Then

z_{x}=(-x)/ \sqrt{R^{2}-x^{2}-y^{2}},

z_{y}=(-y)/ \sqrt{R^{2}-x^{2}-y^{2}}.

Therefore

\sqrt{1+z_{x}^{2}+z_{y}^{2}} =R/ \sqrt{R^{2}-x^{2}-y^{2}}.

The surface area of half-sphere is

\iint_{x^{2}+y^{2}\leq R^{2}} \frac{R}{\sqrt{R^{2}-x^{2}-y^{2}}} dxdy

= \int_{0}^{2\pi} \int_{0}^{R} \frac{Rr}{\sqrt{ R^{2}-r^{2}}} dr d\theta

= 2 \pi R \int_{0}^{R} \frac{r}{\sqrt{R^{2}-r^{2}}} dr

= 2 \pi R^{2}

Hence, the total surface area of the sphere with radius R is 4\pi R^{2}.

Second, we calculate the volume of the sphere with radius R.

V= 2\iint_{x^{2}+y^{2} \leq R^{2}} \sqrt{R^{2}-x^{2}-y^{2}} dx dy

= 2 \int_{0}^{2\pi} \int_{0}^{R} \sqrt{R^{2}-r^{2}}\cdot r dr d\theta

= 2 \cdot 2\pi \cdot \int_{0}^{R} \sqrt{R^{2}-r^{2}}\cdot r dr

= \frac{4}{3} \pi R^{3}.

Theorem 2. The volume of the ellipsoid \frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}}=1 is \frac{4}{3}\pi abc.

Proof.

The upper bound of the volume is

z= c\cdot \sqrt{1-\frac{x^{2}}{a^{2}}- \frac{y^{2}}{b^{2}}}.

The lower bound of the volume is

z=- c\cdot \sqrt{1-\frac{x^{2}}{a^{2}}- \frac{y^{2}}{b^{2}}}.

Assume R=\{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1 \}, the volume of the ellipsoid is

V= 2 \iint_{R} c \cdot \sqrt{1-\frac{x^{2}}{a^{2}}- \frac{y^{2}}{b^{2}}} dxdy

= 2 c \int_{0}^{2\pi} \int_{0}^{1} \sqrt{1-r^{2}} \cdot (r \cdot a \cdot b) dr d\theta

where we use the substitution x=a \cdot r \cos \theta and y=b\cdot r \sin \theta, the determinant of Jacobian matrix is a\cdot b\cdot r.

Therefore, the value equals to

2abc \cdot (2\pi) \int_{0}^{1} \sqrt{1-r^{2}} r dr = \frac{4}{3} \pi abc.

MA 1505 Tutorial 2: Integration

L.Hospital Rule: if the ratio is \infty/\infty or 0/0, then we can use the L.Hospital Rule to calculate the limit. Precisely, the L.Hospital Rule is 

\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \frac{f^{'}(x)}{g^{'}(x)},

where a is a finite real number or infinity.

If f(x) is a continuous function, then F(x)= \int_{a}^{x} f(t) dt is a differentiable function and its derivative F^{'}(x)=f(x).

General Leibniz Integration Rule.

\frac{d}{d\theta} ( \int_{a(\theta)}^{b(\theta)} f(x,\theta)dx)

= \int_{a(\theta)}^{b(\theta)} f_{\theta}(x,\theta) dx + f(b(\theta), \theta)\cdot b^{'}(\theta) - f(a(\theta),\theta) \cdot a^{'}(\theta)

Question 1. Calculate the value S=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x+\cos x}dx .

Solution. 

S=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x +\cos x }dx

= \int_{0}^{\frac{\pi}{2}} \frac{\cos(\frac{\pi}{2}-t)}{\sin(\frac{\pi}{2}-t) + \cos( \frac{\pi}{2}-t)} dt

= \int_{0}^{\frac{\pi}{2}} \frac{\sin t}{\cos t+\sin t}dt

Therefore

2S=S+S

= \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x+\cos x} + \frac{\sin x}{\cos x+\sin x} dx

= \int_{0}^{\frac{\pi}{2}} 1 dx= \frac{\pi}{2}

Hence, S=\frac{\pi}{4} .

Question 2. Calculate the value S=\int_{0}^{\frac{\pi}{4}} \ln(1+\tan{x}) dx .

Solution. 

S=\int_{0}^{\frac{\pi}{4}} \ln(1+\tan x) dx

= \int_{0}^{\frac{\pi}{4}} \ln(1+\tan(\frac{\pi}{4}-x)) dx

= \int_{0}^{\frac{\pi}{4}} \ln(\frac{2}{1+\tan x})dx

= \frac{\pi \ln2}{4} - \int_{0}^{\frac{\pi}{4}} \ln(1+\tan x) dx

= \frac{\pi \ln2}{4} - S

Therefore, S=\frac{\pi \ln2}{8} .

Question 3. 

S=\int_{0}^{1} \frac{x^{4}}{x^{4}+(1-x)^{4}}dx

Solution.

S=\int_{0}^{1} \frac{x^{4}}{x^{4}+(1-x)^{4}}dx

= \int_{0}^{1} \frac{(1-x)^{4}}{x^{4}+(1-x)^{4}}dx

= 1- \int_{0}^{1} \frac{x^{4}}{x^{4}+(1-x)^{4}}dx

Therefore, S=0.5.

Question 4. 

\lim_{x\rightarrow 0} \frac{\sin x}{x}=1.

\lim_{x\rightarrow \infty} x\tan\frac{1}{x} = \lim_{y\rightarrow 0}\frac{\tan y}{y}=1, where y=1/x.

\lim_{x\rightarrow \infty} \frac{\ln x}{x^{a}}=0, where a>0.

MA 1505 Tutorial 3: Taylor Series

The Taylor Series of f(x) at the point x_{0} is

f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(x_{0})}{n!} (x-x_{0})^{n}.

e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n!}

\ln(1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{n}}{n}

\sin x= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n-1}}{(2n-1)!}

\cos x =\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!}

\frac{1}{1-x} =\sum_{n=0}^{\infty} x^{n}

Question 1. Let S=\sum_{n=0}^{\infty} \frac{1}{n!(n+2)} . Calculate the value of S.

Solution.

Method (i).

S=\sum_{n=0}^{\infty} \frac{1}{n!(n+2)}

= \sum_{n=0}^{\infty} \frac{n+1}{(n+2)!}

= \sum_{n=0}^{\infty} \frac{(n+2)-1}{(n+2)!}

= \sum_{n=0}^{\infty} (\frac{1}{(n+1)!}-\frac{1}{(n+2)!})

= 1

Method (ii). Integrate the Taylor series of xe^{x} to show that S=1.

The Taylor series of x e^{x} is \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} . Take the integration of the function on the interval [0,1], we get

\int_{0}^{1} xe^{x} dx

=\int_{0}^{1} \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} dx

= \sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{x^{n+1}}{n!} dx

= \sum_{n=0}^{\infty} \frac{1}{n!(n+2)}=S .

The left hand side equals to 1 from integration by parts.

Method (iii). Differentiate the Taylor series of (e^{x}-1)/x.

The Taylor series of f(x)= (e^{x}-1)/x is \sum_{n=1}^{\infty} \frac{x^{n-1}}{n!} . Differentiate f(x) and get f^{'}(x)= \sum_{n=2}^{\infty} \frac{x^{n-2}}{(n-2)!n} . Moreover, f^{'}(x)= \frac{e^{x}x-(e^{x}-1)}{x^{2}} and f^{'}(1)=1=S.

Method (iv). Assume the function f(x)=\sum_{n=0}^{\infty} x^{n}/(n!(n+2)). This implies f(0)=0. Assume

g(x)=\int_{0}^{x}f(t)dt= \sum_{n=0}^{\infty} \int_{0}^{x} \frac{t^{n}}{n!(n+2)}dt = \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+2)!} = \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+2)!} = \frac{1}{x}(e^{x}-1-x).

Since f(x)=g^{'}(x), we get f(x) = x^{-1}(e^{x}-1)-x^{-2}(e^{x}-1-x). That means f(1)=1.

Method (v). Assume the function f(x)=\sum_{n=0}^{\infty} x^{n}/(n!(n+2)).

f(x)= \sum_{n=0}^{\infty} \frac{x^{n}}{(n+1)!} - \frac{x^{n}}{(n+2)!} = x^{-1}\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!} - x^{-2}\sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+2)!} = x^{-1} (e^{x}-1) - x^{-2}(e^{x}-1-x). Therefore, f(1)=1.

Remark. There is a similar problem: calculate \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!(n+2)}. Answer is 1-2e^{-1}.

Question 2. Let n be a positive integer. Prove that

\frac{1}{2} \int_{0}^{1} t^{n-1}(1-t)^{2} dt= \frac{1}{n(n+1)(n+2)}

and calculate the value of the summation

S=\frac{1}{1\cdot 2 \cdot 3} + \frac{1}{3\cdot 4 \cdot 5} + \frac{1}{5\cdot 6\cdot 7} + \frac{1}{7\cdot 8 \cdot 9}+.... .

Solution. 

\frac{1}{2}\int_{0}^{1} t^{n-1}(1-t)^{2}dt

= \frac{1}{2} \int_{0}^{1} (t^{n+1}-2t^{n}+t^{n-1}) dt

= \frac{1}{2} (\frac{1}{n+2}-\frac{2}{n+1}+\frac{1}{n})

= \frac{1}{n(n+1)(n+2)} .

To calculate the value of S, there are two methods.

Method (i). The summation of S, n is only taken odd numbers. From the first step, we know the summation

S=\frac{1}{2} \int_{0}^{1} (1+t^{2}+t^{4}+t^{6}+...)(1-t)^{2}dt

= \frac{1}{2} \int_{0}^{1} \frac{1}{1-t^{2}} (1-t)^{2} dt

= \frac{1}{2} \int_{0}^{1} \frac{1-t}{1+t} dt

= \frac{1}{2} \int_{0}^{1} (\frac{2}{1+t}-1)dt

= \frac{1}{2}( 2\ln(1+t)-t)_{t=0}^{t=1}

= \ln 2 -\frac{1}{2} .

Method (ii).

Since \frac{1}{n(n+1)(n+2)}= \frac{1}{2}(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}) ,

S=\sum_{ odd} \frac{1}{n(n+1)(n+2)}

= \frac{1}{2} \sum_{odd} ( \frac{1}{n}- \frac{2}{n+1}+\frac{1}{n+2})

= \frac{1}{2} ( \frac{1}{1}-\frac{2}{2}+\frac{1}{3}+ \frac{1}{3}-\frac{2}{4}+\frac{1}{5}+ \frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{7}-\frac{2}{8}+\frac{1}{9}+...)

= \frac{1}{2} ( \frac{1}{1} + 2( -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-...))

= \frac{1}{2} ( 1 + 2 (\ln 2-1))

= \ln 2 -\frac{1}{2} .

Here we use the Taylor series of \ln(1+x)= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{n}}{n} and \ln 2= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....

Question 3. Assume \zeta(k)=1+\frac{1}{2^{k}} + \frac{1}{3^{k}} + ... = \sum_{m=1}^{\infty} \frac{1}{m^{k}}.

Prove

\sum_{k=2}^{\infty} (\zeta(k)-1)=1.

\sum_{k=1}^{\infty} (\zeta(2k)-1)=3/4.

Proof.

\sum_{k=2}^{\infty} (\zeta(k)-1)

= \sum_{k=2}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^{k}}

= \sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{m^{k}}

= \sum_{m=2}^{\infty} \frac{1}{(m-1)m}

= \sum_{m=2}^{\infty} ( \frac{1}{m-1} - \frac{1}{m})

= 1.

\sum_{k=1}^{\infty} ( \zeta(2k)-1)

= \sum_{k=1}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^{2k}}

= \sum_{m=2}^{\infty} \sum_{k=1}^{\infty} \frac{1}{m^{2k}}

= \sum_{m=2}^{\infty} \frac{1}{m^{2}-1}

= \sum_{m=2}^{\infty} \frac{1}{2} ( \frac{1}{m-1}-\frac{1}{m+1})

= \frac{1}{2}(1+\frac{1}{2})

= \frac{3}{4}.

Question 4. Calculate the summation S= \sum_{k=1}^{\infty} \frac{(-1)^{k} k}{4k^{2}-1}.

Solution. 

S=\sum_{k=1}^{\infty} \frac{(-1)^{k} k}{4k^{2}-1}

=\frac{1}{4} \sum_{k=1}^{\infty} (-1)^{k} ( \frac{1}{2k-1} +\frac{1}{2k+1})

= \frac{1}{4} \sum_{k=1}^{\infty} ( \frac{(-1)^{k}}{2k-1} - \frac{(-1)^{k+1}}{2k+1})

= \frac{1}{4} \cdot \frac{-1}{2-1} = - \frac{1}{4}.

MA 1505 Tutorial 7: Integration of Two Variables Functions

In the tutorial 7, we will learn to calculate the integration of two variables, reverse the order of integration and polar coordinate.

The formulas of polar coordinate are x=r \cos(\theta), y=r \sin(\theta), where r\in (0,\infty) and \theta \in [0, 2\pi).

\iint_{D} f(x,y) dxdy= \iint_{D^{'}} f(r \cos \theta, r \sin \theta) r dr d\theta

Question 1. The application of polar coordinate. Calculate the value of

I= \int_{-\infty}^{\infty} e^{-x^{2}}dx.

Solution.

Method (i).

I=\int_{-\infty}^{\infty} e^{-x^{2}} dx= \int_{-\infty}^{\infty} e^{-y^{2}}dy .

Therefore

I^{2}=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} dxdy

= \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^{2}} r dr d\theta

= 2\pi \int_{0}^{\infty} e^{-r^{2}}r dr

= 2\pi \frac{1}{2} e^{-r^{2}}|_{r=0}^{r=\infty}

= \pi .

Hence I=\sqrt{\pi} .

Method (ii).

Since I=\int_{-\infty}^{\infty} e^{-x^{2}}dx = 2\int_{0}^{\infty} e^{-x^{2}}dx = 2\int_{0}^{\infty} e^{-y^{2}}dy , we get

I^{2}=\int_{0}^{\infty} \int_{0}^{\infty} e^{-x^{2}-y^{2}} dy dx

Assume y=sx, we get

I^{2}=4 \int_{0}^{\infty} \int_{0}^{\infty} e^{-x^{2}(1+s^{2})} x ds dx

=4 \int_{0}^{\infty} \int_{0}^{\infty} e^{-(1+s^{2})x^{2}} x dx ds

=4 \int_{0}^{\infty} \frac{1}{2(1+s^{2})} ds

=4 \cdot \frac{1}{2} \arctan s|_{s=0}^{s= \infty}

= \pi

Therefore, I=\sqrt{\pi}

Question 2. Calculate the value of

\lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \int_{x}^{t} \sin{y^{2}} dy dx}{t^{4}}.

Solution.

Method (i). Leibniz Integration Rule.

\frac{d}{d\theta} ( \int_{a(\theta)}^{b(\theta)} f(x,\theta)dx)

= \int_{a(\theta)}^{b(\theta)} f_{\theta}(x,\theta) dx + f(b(\theta), \theta)\cdot b^{'}(\theta) - f(a(\theta),\theta) \cdot a^{'}(\theta)

Here f_{\theta}(x,\theta) denotes the partial derivative of f(x, \theta) with respect to the variable \theta.

In the question, assume G(x,t)=\int_{x}^{t} \sin{y^{2}} dy .

Making use of L’Hospital Rule, we have

\lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \int_{x}^{t} \sin{y^{2}} dy dx}{t^{4}}

= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} G(x,t)dx}{t^{4}}

= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} G_{t}(x,t)dx+ G(t,t)\cdot 1 - G(0,t)\cdot 0}{ 4 t^{3}}

= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \sin{t^{2}}dx}{4t^{3}}

= \lim_{t\rightarrow 0^{+}} \frac{ t \sin{t^{2}}}{4t^{3}}= \frac{1}{4}

Method (ii). Reverse the order of integration.

The integration domain is 0\leq x \leq t and x \leq y \leq t. It is same as 0\leq y \leq t and 0\leq x\leq y.

Answer= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \int_{0}^{y} \sin{y^{2}} dxdy}{t^{4}}

= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} y \sin{y^{2}}dy}{t^{4}}

= \lim_{t\rightarrow 0^{+}} \frac{ t \sin{t^{2}}}{4t^{3}}

=\frac{1}{4} .

Question 3. MA1505 2010-2011 Semester 2, Question 6(b).

Let R be a region of xy-plane, find the largest possible value of the integration

\iint_{R} (4-x^{2}-y^{2})dxdy.

Solution. 

Since we want to find the largest possible value, then we must guarantee that on the region R, the function f(x,y)=4-x^{2}-y^{2} is non-negative. That means the region R is 4-x^{2}-y^{2}\geq 0. i.e. x^{2}+y^{2}\leq 4. Therefore, we should calculate the integration

\iint_{x^{2}+y^{2}\leq 4} (4-x^{2}-y^{2}) dxdy

= \int_{0}^{2\pi} \int_{0}^{2} (4-r^{2})r dr d\theta

= 2\pi \int_{0}^{2} (4r-r^{3})dr

= 8\pi

Question 4. I \subseteq \mathbb{R} is a real interval, calculate the maximum value of

\int_{I} (1-x^{2}) dx.

Solution.

To calculate the maximum value of the integration, the maximal interval I=[-1,1]. Therefore, the maximum value of the integration is

\int_{-1}^{1} (1-x^{2}) dx = \frac{4}{3}.

Qustion 5. Calculate the multiple integration

\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} e^{x^{2}+y^{2}} dy dx.

Solution.

Method (i).  Use the polar coordinate.

\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} e^{x^{2}+y^{2}} dydx

= \int_{0}^{\pi/2} \int_{0}^{1} e^{r^{2}} r dr d\theta

= \frac{\pi}{2} \int_{0}^{1} e^{r^{2}} r dr

= \frac{\pi}{2} (\frac{e^{r^{2}}}{2}) |_{r=0}^{r=1}

= \frac{\pi}{4}(e-1).

Method (ii). Make the substitution y=sx, then dy=x ds.

The region is 0\leq x \leq 1 and 0\leq s \leq \sqrt{1-x^{2}}/x.

That is equivalent to 0 \leq s \leq \infty and 0 \leq x \leq 1/\sqrt{1+s^{2}}.

The integration is

\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}/x} e^{x^{2}+s^{2}x^{2}} xds dx

= \int_{0}^{\infty} \int_{0}^{1/\sqrt{1+s^{2}}} e^{(1+s^{2})x^{2}} x dx ds

= \int_{0}^{\infty} (\frac{1}{2(1+s^{2})} e^{(1+s^{2})x^{2}} |_{x=0}^{x=1/\sqrt{1+s^{2}}}) ds

= \int_{0}^{\infty} \frac{e-1}{2(1+s^{2})}ds

= \frac{e-1}{2} \arctan s|_{s=0}^{s=\infty}

= \frac{\pi}{4} (e-1).

MA 1505 Tutorial 6: Partial Derivatives and Directional Derivative

In the tutorial, we will learn the partial derivatives for multiple variable functions.

Assume z=f(x,y) is a two variable function, then we use the notations to describe the partial derivatives of f(x,y).

f_{x}=\frac{\partial f}{\partial x} denotes the partial derivative of f under the variable x.

f_{y}=\frac{\partial f}{\partial y} denotes the partial derivative of f under the variable y.

Similarly, we can also define the second derivative of f(x,y).

f_{xx}=\frac{\partial^{2} f}{\partial x^{2}} ,

f_{xy}=f_{yx}=\frac{\partial^{2}f}{\partial x\partial y}=\frac{\partial^{2}f}{\partial y\partial x}  \text{ if } f(x,y) \text{ is a } C^{2} \text{ function.}

f_{yy}=\frac{\partial^{2} f}{\partial y^{2}} .

Assume u=(a,b) is a unit vector, i.e. its length is 1. If f(x,y) is C^{1} at the point p, then we can define the directional derivative of f(x,y) at point p as

f_{x}(p) a + f_{y}(p) b

Theorem 1. Geometric mean is not larger than Arithmetic mean.

For n positive real numbers a_{1}, a_{2}, ..., a_{n} ,

(a_{1}...a_{n})^{\frac{1}{n}} \leq \frac{a_{1}+...+a_{n}}{n}

“=” if and only if a_{1}=a_{2}=...=a_{n}.

Theorem 2. Cauchy’s Inequality.

For 2n real numbers a_{1},..., a_{n}, b_{1},...,b_{n} ,

(a_{1}b_{1}+...+a_{n}b_{n})^{2}\leq (a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2})

“=” if and only if \frac{a_{1}}{b_{1}}=...=\frac{a_{n}}{b_{n}}.

Proof.

Method (i). Construct a non-negative function f(x) with respect to variable x

f(x)= \sum_{i=1}^{n}(a_{i}x-b_{i})^{2}= (\sum_{i=1}^{n} a_{i}^{2}) x^{2} - 2(\sum_{i=1}^{n} a_{i}b_{i}) x + (\sum_{i=1}^{n} b_{i}^{2}).

Consider the equation f(x)=0, there are only two possibilities: one is the equation f(x)=0 has only one root, the other one is the function has no real roots. Therefore,

\Delta=4(\sum_{i=1}^{n} a_{i}b_{i})^{2} - 4 ( \sum_{i=1}^{n} a_{i}^{2}) \cdot ( \sum_{i=1}^{n} b_{i}^{2}) \leq 0.

Hence, (a_{1}b_{1}+...+a_{n}b_{n})^{2}\leq (a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2}).

Moreover, if “=”, then f(x)=0 has only one root x_{0}, i.e. for all 1\leq i \leq n, a_{i}x_{0}-b_{i}=0. That means

\frac{a_{1}}{b_{1}} =...=\frac{a_{n}}{b_{n}}.

By the way, the solution of ax^{2}+bx+c=0 is \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} and \Delta=b^{2}-4ac.

Method (ii). Since a^{2}+b^{2}\geq 2 ab, we know

ab\leq \frac{1}{2}(\lambda^{2} a^{2}+ b^{2}/\lambda^{2}) for all \lambda \neq 0.

Assume \lambda^{2}=\sqrt{(\sum_{i=1}^{n}b_{i}^{2})/(\sum_{i=1}^{n} a_{i}^{2})} , for all 1\leq i \leq n,

a_{i}b_{i}\leq \frac{1}{2} (\lambda^{2}a_{i}^{2}+b_{i}^{2}/\lambda^{2})

Take the summation at the both sides,

\sum_{i=1}^{n} a_{i}b_{i} \leq \frac{1}{2}( \lambda^{2}\sum_{i=1}^{n}a_{i}^{2} + (\sum_{i=1}^{n} b_{i}^{2})/ \lambda^{2})= \sqrt{(\sum_{i=1}^{n} a_{i}^{2}) \cdot (\sum_{i=1}^{n}b_{i}^{2})}.

Question 1. Assume u(x,y) is a C^{2} function and u>0 . u(x,y) satisfies the partial differential equation u u_{xy}= u_{x}u_{y}.

Prove

(1) \frac{\partial \ln u}{\partial y} is a function of y.

(2) \frac{\partial \ln u}{\partial x} is a function of x.

(3) The solution of u(x,y) has the form u(x,y)=f(x) g(y) for some function f(x) and g(y) .

Proof.

(1) Method (i) Make use of derivative.

First, we know \frac{\partial \ln u}{\partial y}=\frac{u_{y}}{u} . Second, take the partial derivative of the function with respect to the variable x. That means,

\frac{\partial }{\partial x} (\frac{u_{y}}{u})= \frac{u_{xy}u- u_{x}u_{y}}{u^{2}}=0 from the partial differential equation. Therefore, the function \frac{u_{y}}{u} is independent of the variable x. i.e. the function is a function of variable y.

Method (ii) Make use of integration.

Since u u_{xy}=u_{x}u_{y} , \frac{u_{x}}{u}=\frac{u_{xy}}{u_{y}} , then we take the integration of x at the both sides,

\int \frac{u_{x}}{u} dx =\int \frac{u_{xy}}{u_{y}} dx , the left hand side is \ln u, the right hand side is \ln |u_{y}| + h_{1}(y) for some function h(y). That means, \frac{|u_{y}|}{u}=e^{-h_{1}(y)} . and \frac{u_{y}}{u} is a function of y .

(2) is similar to (1).

(3) From part (1), we know \frac{\partial \ln u }{\partial y} is a function of y. Assume \frac{\partial \ln u }{\partial y} = h_{2}(y). Take the integration of y at the both sides, we have

\ln u= \int h_{2}(y) dy + h_{3}(x) for some function h_{3}(x) . u = e^{\int h_{2}(y) dy} \cdot e^{h_{3}(x)} = g(y) \cdot f(x) for some functions f(x) and g(y).

Question 2. Assume L+K=150,  L and K are non-negative. Find the maximum value of f(L,K)=50 L^{0.4} K^{0.6} .

Solution.

Method (i). Langrange’s Method.

g(L,K,\lambda)=f(L,K)-\lambda(L+K-150)=50L^{\frac{2}{5}}K^{\frac{3}{5}}-\lambda(L+K-150).

Take three partial derivatives of g,

\frac{\partial g}{\partial \lambda} = -(L+K-150)=0

\frac{\partial g}{\partial L} = 50 \cdot \frac{2}{5} L^{-\frac{3}{5}}K^{\frac{3}{5}} - \lambda

\frac{\partial g}{\partial K}= 50 \cdot \frac{3}{5} L^{\frac{2}{5}} K^{-\frac{2}{5}} - \lambda=0

Solve these three equations, we get 2K=3L and L+K=150 , therefore the maximum value is taken at L=60 and K=90.

Method (ii). Change to one variable function.

Since L+K=150, we can define the one variable function

g(L)=f(L,150-L)=50 L^{\frac{2}{5}}(150-L)^{\frac{3}{5}}.

The derivative of g^{'}(L)=50 \cdot (\frac{2}{5} L^{-\frac{3}{5}}(150-L)^{\frac{3}{5}} - L^{\frac{2}{5}}\frac{3}{5}(150-L)^{-\frac{2}{5}}).

The critical point is L=60. The maximal value of g(L) is taken at L=60, K=90.

Method (iii). Mathematical Olympic Method.

Use the fact that the geometric mean is not larger than the arithmetic mean.

f(L,K)=50 L^{\frac{1}{5}}L^{\frac{1}{5}} K^{\frac{1}{5}} K^{\frac{1}{5}} K^{\frac{1}{5}}

= \frac{50}{3^{\frac{2}{5}} 2^{\frac{3}{5}}} (3L)^{\frac{1}{5}} (3L)^{\frac{1}{5}} (2K)^{\frac{1}{5}} (2K)^{\frac{1}{5}} (2K)^{\frac{1}{5}}

\leq \frac{50}{3^{\frac{3}{5}} 2^{\frac{2}{5}}} \frac{3L+3L+2K+2K+2K}{5}

= \frac{50}{3^{\frac{2}{5}} 2^{\frac{3}{5}}} \frac{6(L+K)}{5}

= \frac{50}{3^{\frac{2}{5}} 2^{\frac{3}{5}}} \frac{6\cdot 150}{5} .

The maximum value is taken at 3L=2K. i.e. L=60, K=90.

Question 3. Assume 24x+18y+12z=144 and x, y, z are non-negative variables. f(x,y,z)=18 x^{2} y z . Find the maximum value of f(x,y,z).

Solution.

Method (i). Langrange’s Method

g(x,y,z,\lambda)=f(x,y,z)-\lambda(24x+18y+12z-144) = 18 x^{2} y z-\lambda(24x+18y+12z-144)

Take four partial derivatives of g, the critical point is taken at x=z=1.5y. i.e. the maximum value of f(x,y,z) is taken at x=z=3, y=2.

Method (ii) Math Olympic Method

f(x,y,z)=18 x \cdot x \cdot y \cdot z

= \frac{18}{12*12*18*12} (12x) \cdot (12x) \cdot (18y) \cdot (12z)

\leq \frac{18}{12*12*18*12} (\frac{12x+12x+18y+12z}{4})^{4}

= \frac{18}{12*12*18*12} (\frac{144}{4})^{4}

The maximum value is taken at 12x=12x=18y=12z, i.e. x=z=3, y=2.

Question 4.  2012 Exam MA1505 Semester 1, Question 3(a)

Assume f(x,y) has continuous partial derivatives of all orders, if

\nabla f = (xy^{2}+kx^{2}y+x^{3}) \textbf{i} + (x^{3}+x^{2}y+y^{2}) \textbf{j},

Find the value of the constant k.

Solution.

Method (i) Use derivatives.

Since f has continuous partial derivative of all orders, f_{xy}= f_{yx}.

Since f_{x}= xy^{2}+kx^{2}y+x^{3} and f_{y}=x^{3}+x^{2}y+y^{2},

we have

\frac{\partial}{\partial y} (xy^{2}+kx^{2}y+x^{3})= \frac{\partial}{\partial x} (x^{3}+x^{2}y+y^{2})

This implies 2xy+kx^{2}=3x^{2}+2xy. i.e. k=3.

Method (ii). Use integration.

f_{x}=xy^{2}+kx^{2}y+x^{3} \Rightarrow f(x,y)=\frac{1}{2}x^{2}y^{2} + \frac{k}{3}x^{3}y + \frac{1}{4}x^{4} + h_{1}(y),

f_{y}=x^{3}+x^{2}y+y^{2} \Rightarrow f(x,y)=x^{3}y+\frac{1}{2}x^{2}y^{2}+\frac{1}{3}y^{3}+h_{2}(x).

Comparing them, we know k=3, h_{1}(y)=\frac{1}{3}y^{3}+C_{1} and h_{2}(x)=\frac{1}{4}x^{4}+C_{2}, where C_{1} and C_{2} are constants.

Therefore k=3.