I am exhausted now...It is almost midnight but I have no idea about when I can go to sleep.
About Week 10, all I remember is the quiz about big-Oh expression. That is fairly simple I would say. And about Week 11, totally nothing.
The reason is simple: I went to USA for a short trip during the weekend of week 10, which is connected to the Fall Break on Nov 17 and 18. I was supposed to arrive in Toront on Tuesday by bus, but I was stuck by the blizzard around Buffalo.
That was terrible. We were stuck five days. I just arrived in Toronto yesterday afternoon. And I have been busy on everything I missed this week: lectures, assignments, and even a test.
Hopefully, I can write this blog as usual next week, or more than usual, given that next week will be terrifically harsh for me. I have one assignment due tonight which I have just finished, one for tomorrow afternoon, one quiz for tomorrow night, another assignment due on Tuesday, a term test on Wednesday and another two assignments due on Thursday and Friday. God bless me.
2014年11月23日星期日
2014年11月10日星期一
Week 9: Term Test 2
-I feel like that there is correct calculation in my solution of Quiz 7, but I am afraid of the format. I did not use the Sigma notation. Hope that won't hurt too much.
-I missed two lectures. It results in that something in the lecture notes are hard to understand by myself. I guess I have to ask Adam after the next tutorial.
-Test 2 was not difficult, nor easy. I made the same mistake: writing too slow. This time I finished question 1 and 3 first, and left little time for question 2. I was nervous because of that situation and this disturbed me. I ended up with not finishing that question. Ironically, after I breathed the fresh air out of the test room for a couple of minutes, I worked that question out. I was so frustrated. Nevertheless, I know that I must look forward and do better works in order to make my mistake up.
-I missed two lectures. It results in that something in the lecture notes are hard to understand by myself. I guess I have to ask Adam after the next tutorial.
-Test 2 was not difficult, nor easy. I made the same mistake: writing too slow. This time I finished question 1 and 3 first, and left little time for question 2. I was nervous because of that situation and this disturbed me. I ended up with not finishing that question. Ironically, after I breathed the fresh air out of the test room for a couple of minutes, I worked that question out. I was so frustrated. Nevertheless, I know that I must look forward and do better works in order to make my mistake up.
2014年11月4日星期二
Week 8: Assignment 2 and Big-Oh
-The tutorial goes on well as usual. Adam does a really good job as a successor of Yana. He explains things in an easy and efficient way.
-It takes me a while to understand the entire meaning of the Big-Oh expression. Besides, I still have a little doubt about the part "produce non-negative real numbers as output". What I wonder is, certain functions can produce negative numbers as output with natural numbers as input, such as n^2-3n or n^2+n-5, if we take n=1. I should really talk over this with Prof Heap or Adam sometimes.
-Assignment 2 is not difficult, but it definitely requires a whole bunch of carefulness to finish it flawlessly. Although I have checked it several times, I am still afraid about some mistaks in format (e.g. indentation) and in detailed proof.
-It takes me a while to understand the entire meaning of the Big-Oh expression. Besides, I still have a little doubt about the part "produce non-negative real numbers as output". What I wonder is, certain functions can produce negative numbers as output with natural numbers as input, such as n^2-3n or n^2+n-5, if we take n=1. I should really talk over this with Prof Heap or Adam sometimes.
-Assignment 2 is not difficult, but it definitely requires a whole bunch of carefulness to finish it flawlessly. Although I have checked it several times, I am still afraid about some mistaks in format (e.g. indentation) and in detailed proof.
2014年10月26日星期日
Week 7: The Provlem-solving Episode about Determining the Value of π Using Basic Tools Only
In this SLOG I am going to explore the proble about how to calculate the value of π using a ruler, a compass, a protractor, a pencil and paper only.
-Before that, I must say I feel so good that my last quiz result is quite good. It proves that I am following the course well. However, it seems like that since this week, Adam will be our new TA. Well, I must say people will miss Yana's class. But of course I don't mean that Adam is not a good TA. In fact, he is so opposite to a bad TA.
-And I have nothing much about the lectures (or I am just too eager to write about the problem-solving part lol.)
-The problem is, suppose that we all know there is a certain relationship between the radius of a circle and its circumference, which is c = 2πr in formula (c represents the circumference and r represents the radius). The problem is, if we don't know the value of π, how can we measure it (as approximate as possible) with nothing but a ruler, a compass, a pencil and some pieces of blank paper?
1. Understanding the problem
The unknown is simply the value of π. In other words, we are going to find the value of π (or an approximate number, as precise as possible). The data we know includes the formular c = 2πr. The condition should be that there is a circle (the formula applies to circles), and we know the circumference and the radius of it.
Once we reach the condition, we must be able to determine the unknown. Let's separate the condition into 1) drawing a circle, 2) get the radius and 3) get the circumference. As we have a compass and a ruler, we can easily make a circle with any radius we want, or just simply draw one and then measure the radius. Therefore, we get the radius. However, it is impossible to measure the circumference of the circle with the tools we have. In conclusion, 1) and 2) are directly possible to satisfy but we meet problem with 3).
2. Devising a plan
The connection between the data and the unknow is obvious: π = c/(2r). But we are in a smaller problem now: How can we get c as precise as possible?
I recall my experience during the first time I learn about circles. We were asked about and could intuitively answer that:
Both the area and perimeter of the inscribed square of a circle are smaller than the area and the perimeter (circumference) of the circle. When it comes to circumscribed square, the area and the perimeter are all bigger than the ones of the circle.
And this applies to ALL regular polygons.
Somehow I realize that the more sides the regular polygons have, the more its area and perimeter are approaching the one of the circle. I can prove it with square and regular hexagon. Once I prove it, I can draw a regular polygon with relatively more sides and calculate the approximate c, thus get the approximate value of π.
3. Carrying out the plan
First, I draw a circle with radius of 10 cm with my compass and ruler, on a Cartesian coordinate system in the plane. Then I draw an inscribed square and an inscribed regular hexagon under the help from the coordinate system and the protractor.
After that I calculate the perimeter (represented by p below) of the square first. By dividing the square into four isosceles right triangles and using the Pythagorean Theorem, I get the side (s) of the square as 10 x 2^0.5, so p = 40 x 2^0.5. By measuring the sides with my ruler and adding them together I get 56 cm.
Then I divide the regular hexagon into six regular triangles, so that each s of the hexagon equals to the radius of the circle, which is 10 cm. Therefore, the p of the hexagon is 60 cm. As 60 > 56, it proves that p of the regular hexagon is closer to c than p of the square (because both p are still smaller than c by previous proofs.)
Similarly, I calculate, measure and prove that same thing happens on the circumscribed figures (p of the circumsribed hexagon < p of the circumscribed square, which means p of the hexagon is closer to c).
As p of the inscribed and the circumscirbed regular polygons are approaching c, the mean of them should be even closer. Therefore, I draw an inscribed regular octagon and an circumscribed one, mearsure and calculate the mean of both p of them, which give me an approximate number 64 cm.
I decide to use this number as the approximate value of c. so I try the π = c/(2r) = 64/(2 x 10) = 3.2.
4. Looking back
I can check and make sure it is approaching the real value of π (as I know it is 3.1415926...)! I know that if I test regular decagon or something with more sides, the number will be even more accurate, but I don't need to bother doing that! All I have to do is celebrate that I solve the problem! I have now found a way to calculate the value of π with basic tools only!
-Before that, I must say I feel so good that my last quiz result is quite good. It proves that I am following the course well. However, it seems like that since this week, Adam will be our new TA. Well, I must say people will miss Yana's class. But of course I don't mean that Adam is not a good TA. In fact, he is so opposite to a bad TA.
-And I have nothing much about the lectures (or I am just too eager to write about the problem-solving part lol.)
-The problem is, suppose that we all know there is a certain relationship between the radius of a circle and its circumference, which is c = 2πr in formula (c represents the circumference and r represents the radius). The problem is, if we don't know the value of π, how can we measure it (as approximate as possible) with nothing but a ruler, a compass, a pencil and some pieces of blank paper?
1. Understanding the problem
The unknown is simply the value of π. In other words, we are going to find the value of π (or an approximate number, as precise as possible). The data we know includes the formular c = 2πr. The condition should be that there is a circle (the formula applies to circles), and we know the circumference and the radius of it.
Once we reach the condition, we must be able to determine the unknown. Let's separate the condition into 1) drawing a circle, 2) get the radius and 3) get the circumference. As we have a compass and a ruler, we can easily make a circle with any radius we want, or just simply draw one and then measure the radius. Therefore, we get the radius. However, it is impossible to measure the circumference of the circle with the tools we have. In conclusion, 1) and 2) are directly possible to satisfy but we meet problem with 3).
2. Devising a plan
The connection between the data and the unknow is obvious: π = c/(2r). But we are in a smaller problem now: How can we get c as precise as possible?
I recall my experience during the first time I learn about circles. We were asked about and could intuitively answer that:
Both the area and perimeter of the inscribed square of a circle are smaller than the area and the perimeter (circumference) of the circle. When it comes to circumscribed square, the area and the perimeter are all bigger than the ones of the circle.
And this applies to ALL regular polygons.
Somehow I realize that the more sides the regular polygons have, the more its area and perimeter are approaching the one of the circle. I can prove it with square and regular hexagon. Once I prove it, I can draw a regular polygon with relatively more sides and calculate the approximate c, thus get the approximate value of π.
3. Carrying out the plan
First, I draw a circle with radius of 10 cm with my compass and ruler, on a Cartesian coordinate system in the plane. Then I draw an inscribed square and an inscribed regular hexagon under the help from the coordinate system and the protractor.
After that I calculate the perimeter (represented by p below) of the square first. By dividing the square into four isosceles right triangles and using the Pythagorean Theorem, I get the side (s) of the square as 10 x 2^0.5, so p = 40 x 2^0.5. By measuring the sides with my ruler and adding them together I get 56 cm.
Then I divide the regular hexagon into six regular triangles, so that each s of the hexagon equals to the radius of the circle, which is 10 cm. Therefore, the p of the hexagon is 60 cm. As 60 > 56, it proves that p of the regular hexagon is closer to c than p of the square (because both p are still smaller than c by previous proofs.)
Similarly, I calculate, measure and prove that same thing happens on the circumscribed figures (p of the circumsribed hexagon < p of the circumscribed square, which means p of the hexagon is closer to c).
As p of the inscribed and the circumscirbed regular polygons are approaching c, the mean of them should be even closer. Therefore, I draw an inscribed regular octagon and an circumscribed one, mearsure and calculate the mean of both p of them, which give me an approximate number 64 cm.
I decide to use this number as the approximate value of c. so I try the π = c/(2r) = 64/(2 x 10) = 3.2.
4. Looking back
I can check and make sure it is approaching the real value of π (as I know it is 3.1415926...)! I know that if I test regular decagon or something with more sides, the number will be even more accurate, but I don't need to bother doing that! All I have to do is celebrate that I solve the problem! I have now found a way to calculate the value of π with basic tools only!
2014年10月21日星期二
Week 6: Break and work hard again
It seems that I am getting used to write the SLOG late. I definitely have to correct this at this weekend.
-No tutorial on Monday (Thknasgiving Holiday). However, I kind of understand the proof structure. I think I worte the rights steps during the quiz.
-I missed Wednesday's lecture and I found something unclear about "floor" and "ceiling" on Friday. Fortunately, I figured them out with the course notes. We came to actually prove somthing and it was quite fascinating.
-No tutorial on Monday (Thknasgiving Holiday). However, I kind of understand the proof structure. I think I worte the rights steps during the quiz.
-I missed Wednesday's lecture and I found something unclear about "floor" and "ceiling" on Friday. Fortunately, I figured them out with the course notes. We came to actually prove somthing and it was quite fascinating.
2014年10月15日星期三
Week 5: The Midterm
First I would like to apologize for that I completely forgot about writing this for last week. As a result, it is hard for me to recall what happened then so this one might be a little bit shorter.
-The fourth tutorial moved to practice on proof structures. To be honest, I have not quite got them yet. For example, I am still not sure about what is the perfect answer for the quiz #4. However, we do not have tutorial on Thanksgiving, so I will just wait until the tutorial in the next week.
-Fortunately, the midterm does not contain things about proof structure. Nevertheless, I didn't really finish the last question when the examiner said ''time's up''. It is plainly because I write too slowly and I definitely have to increse my writing speed.
-I have to say I don't remember anything difficult from last week's lectures.
-The fourth tutorial moved to practice on proof structures. To be honest, I have not quite got them yet. For example, I am still not sure about what is the perfect answer for the quiz #4. However, we do not have tutorial on Thanksgiving, so I will just wait until the tutorial in the next week.
-Fortunately, the midterm does not contain things about proof structure. Nevertheless, I didn't really finish the last question when the examiner said ''time's up''. It is plainly because I write too slowly and I definitely have to increse my writing speed.
-I have to say I don't remember anything difficult from last week's lectures.
2014年10月5日星期日
Week 4: Assignment 1 is done
-Quiz #2 was perfect. Meanwhile, my classmates seem to feel the exercise problems for tutorial are getting harder, as they spent more time asking questions about them. Fortunately, based on understanding about course materials, I can find answers to those questions so they are still not difficult to me.
-Nothing hard to comprehend during the lecture yet. Also, Prof Heap helped us on Assignment 1 during the last lecture hour and that benefited me a lot. Through that I found some mistaks and missing staff, and correct them in time.
-Assignment 1 was not difficult. On the other hand, it do took my group member and me quite a while to think about the problems, as well as to type the symbols and formulas. Some proofs also cost time as they were not short. Overall, it contains more complexity compared with difficulty. Hope we can get a good result from that work.
-Nothing hard to comprehend during the lecture yet. Also, Prof Heap helped us on Assignment 1 during the last lecture hour and that benefited me a lot. Through that I found some mistaks and missing staff, and correct them in time.
-Assignment 1 was not difficult. On the other hand, it do took my group member and me quite a while to think about the problems, as well as to type the symbols and formulas. Some proofs also cost time as they were not short. Overall, it contains more complexity compared with difficulty. Hope we can get a good result from that work.
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