|
Muhahahahaz
|
read my profile
sign my guestbook
Name: Spencer
Country: United States
State: California
Metro: Davis
Birthday: 12/30/1986
Gender: Male
Interests: eating, sleeping, doing homework... oh yeah, and doing anything else I possibly can (which is like never).
Expertise: Calculating the probability of me ever being completely happy, and then plugging that into the improbability drive.
Edit: Hmmm... it seems that it's not quite so improbable after all, too bad for traveling through space, though.
Occupation: Computer related
Industry: Engineering
Message: message me
AIM:  Muhahahahas
Member Since:
11/9/2003
|
|
| Today, I died in an xkcd accident!
| | |
| "Hi Muhahahahaz! It's been 1399 (wow, that's a big number) days since you joined Xanga..."
NOOOOOOOOOOOOOO!!!!
I missed the 1337 days mark! 
| | |
| Number Theory:
Grade Statistics:
Final Grade
Your Grade: 94.04% = A+
Median: 64.05%
Average: 68.37%
Standard Deviation: 16.12
and the Final Exam...
Grade Statistics:
Final
Your Grade: 98/100 = 98%
Median: 64.00%
Average: 67.58%
Standard Deviation: 17.71
A surprise here. Or maybe I should say A+ surprise, lol. I did as I expected on the final, but who knew I was the top student? Well, #1 out of 12 anyways... but I'll freaking take it, haha. Check out the graphs below. The graph for the Final Exam is pretty normal, but look at the graph for the Final Grades. The 70%-79% column is completely empty! Seems pretty strange to me. It was a small but fun class, and I loved learning about crazy Number Theory stuff. Now if only I can find out what I got on my Data Structures Final... unless they email our grades to us, I doubt we'll ever find out our Final Exam grades.
Final Grades:
Final Exam Grades:
 | | |
| OK, so finals today were fun, but I wantz mai graids now plz. Lol, sorry... too much "Caturday."
First, I had Data Structures at 10:30am. There were twelve questions pretty similar to the practice Final, so having done the practice Final made it pretty easy. The only trouble was this one question where it asks for an "external" sort algorithm (one where the data is too large to hold in memory). Apparently, this is one of the main uses for B-Trees, but I didn't make the connection. There's no reason to worry about it though - if I get full credit except for a zero on that question (very possible, except I'll probably get some points for it), then I will still end up with around 97.5% in the class, since I had a perfect score before the Final, lol. Come to think of it, the midterm grades were never accessible online in any way... I wouldn't be surprised if they didn't even tell us our Exam scores and just made us wait for the official grades. That would suck, but at least it's not a class I'm worried about.
Then at 1:30pm, I had Number Theory. There was a nice break for lunch (which ended up being about an hour and a half for me, since I finished Data Structures early), so I was plenty happy after my Taco Bell and Pepsi. There were 7 questions consisting of 4 proofs, 1 computation, 1 series of 3 definitions, and 1 series of 10 T/F questions. The computation question was kinda strange, since you would need a calculator to get the true continued fraction for sqrt(3). It should be an infinite and periodic continued fraction, but without a calculator I just estimated the first few terms. The definitions and the T/F questions were simple enough. There was just one T/F question that I wasn't quite sure about, but it turns out that I got it right. Then there were the 4 proofs.
The Euler pseudoprime proof was easy. Once you wrote down the definition of an Euler pseudoprime, it was pretty obvious. The proof dealing with continued fractions and the fibonacci numbers was also pretty easy, IF you could remember the recursive formula for continued fractions, which I couldn't, but more on that later. The other two proofs were pretty fun and interesting.
First, there was one where we had to prove that the only positive integer solution to 3^n - 2^m = -1 is m = 2 and n = 1 (3^1 - 2^2 = 3 - 4 = -1). I had fun with this one. You basically have to consider the values of 3^n and 2^m mod 6. 3^n is always 3 mod 6, and 2^m is either 2 or 4 mod 6 (2 if m is odd, 4 if m is even). Then you break it down into cases of m being odd or even. If m is odd, then 3^n - 2^m is equivalent to 3 - 2 mod 6 (which is 1, rather than -1), thus m cannot be odd. If m is even, then we let m = 2k, so that we have that 3^n - 2^m = 3^n - 2^2k is equivalent to 3 - 4 = -1. Thus it is possible for m to be even, but then we notice that 3^n = 2^m - 1 = 2^2k - 1 = (2^k + 1)(2^k - 1), which ends up leading to a contradiction. 2^k + 1 and 2^k - 1 must both be powers of 3 (including 3^0 = 1) since their product is 3^m. Obviously, 2^k + 1 cannot be 1 since 2^k > 0. Also, 2^k - 1 = 1 implies that k = 1, so that m = 2 and n = 1, the solution we already know. Hence we now let 2^k + 1 = 3^i and 2^k - 1 = 3^j for i, j > 0. If k is even, then 2^k + 1 is equivalent to 4 + 1 = 5, but 3^i is always equivalent to 3 (mod 6). Similarly, if k is odd, 2^k - 1 is equivalent to 2 - 1 = 1, which is not 3 (mod 6). Hence we see that there is only one possible solution. I can't believe I actually wrote out my proof, but there it is. I saw the powers of 2 and 3 and figured I should consider them mod 6. From there, it was just an exploration of cases, and I had been hoping to get an even m so as to factor 2^m - 1, so I knew what to do once I did.
Second, we had to prove that no positive integer of the form 4^m*(8k + 7) could be represented as the sum of three squares, for m, k >= 0 (every positive integer can be represented as the sum of four squares). First, it says that we may assume that the statement is true for m = 0, since we proved this in the homework. It turns out that this can be proved by infinite descent, which I realized as I was working toward the solution. Since 4 divides 4^m*(8k + 7) for m > 0, we see that 4 must divide the sum in question, x1^2 + x2^2 + x3^2. Also, since x^2 is equivalent to 0 or 1 mod 4 for all x, we see that each of these squares must be equivalent to 0 mod 4 (so that 0 + 0 + 0 = 0), which means that each of the squares is divisible by 4, or that each of the numbers must be even. Thus dividing by 4 on both sides gives:
(x1/2)^2 + (x2/2)^2 + (x3/2)^2 = 4^(m-1)*(8k + 7)
This is the key equation. It shows that if we have a solution for any m, there must also be a solution for m - 1, so that by infinite descent, there must be a solution for m = 0. But this contradicts the fact that no such solution exists, thus the statement is proven.
Lastly, there was the proof involving continued fractions and the fibonacci numbers. At this point, I had completed all the other questions and I had about half an hour left. The only problem was that I hadn't studied the recursive formula for continued fractions. If I could remember it, then the proof was a simple proof by induction. I had already shown the base case, so now I just needed the formula so that I could show the inductive step, which seemed like it should be relatively simple. I ended up spending most of the remaining time trying to figure out the recursive formula through various examples, since it is difficult to actually derive. There were about 5 or 10 minutes left, and I finally figured it out. Part of the problem was that I had made some arithmetic mistakes, but when I checked over everything I realized my error. Formula in hand, the proof was very straightforward and I was able to breathe a sigh of relief. I was pretty worried that I wouldn't be able to figure out the formula. It's a good thing that I had plenty of time.
After my final, all was well. I did a few things that I needed to do, and then I came home and slept. For four hours. Life is good. 
Well, not quite yet. I still await my official grades. I'm confident that I did well on these Exams, but it will be much more reassuring to see the results. I'm also anxious to see if I got any A+'s, lol. For Data Structures, that's actually what I expect, but it's also possible for Number Theory. There are only 12 students in Number Theory, so it all just depends on where I stand with respect to maybe 1 or 2 top students. With such a small class, there can only be one A+, if there even is one, so I can only hope to be #1, lol. I'm obsessive, I know, but in a healthy way. I always try to do better (up to a point, called insanity).
Ok, well I should probably sleep now. Tomorrow awaits me with various obligations, and hopefully with my official grades as well. | | |
| Probability Theory:
Grade Statistics:
Final Grade
Your Grade: A
Median: 66.40%
Average: 66.00%
Standard Deviation: 18.68
and the Final Exam...
Grade Statistics:
Final
Your Grade: 192/200
Median: 60.00%
Average: 61.11%
Standard Deviation: 26.36
Another relief. It turns out that I got 192/200 on the final (where the high was 198), so that's like 96%, or possibly more like 97% with the curve. I'm kinda surprised that I got such a high score since I totally BSed that "hard" problem, but just like the TA told us to, I decided to "write something down," which appears to be enough. The Final Grades graph is pretty interesting... it's very evenly distributed. My straight percentage was 90.3%, so I'm probably the third one above 90%. The Final Exam graph, however, is fairly sporadic. Isn't it fun to look at graphs? Lol.
Final Grades:
Final Exam Grades:
 | | |
|
|
