The course is over, and the final exam looms over the horizon of my every thought, like an inescapable eventuality. So it's a good as any to reflect back on CSC236H1F as a whole.
Lectures were fun and easy-enough to follow; perhaps the only CS lectures I didn't mind attending. The tutorial/quiz system was very fair. I had the weekend to familiarize myself with the topic, and then I got to see a similar question to the quiz being worked through. The assignments were posted far-enough from their due dates, and so I had a pretty decent time-frame to work through them. Their level of difficulty (and sometimes the type of questions) were unlike anything seen in lectures or tutorials, but that is usually the case with problem sets. The mid-terms were somewhere between the assignments and the quizzes in terms of difficulty and length. The second mid-term was a little too focused on a single concept, and the two questions were very closely related. It just felt like if you were not good at that one concept, you could forget about getting a good mark on that test (which I did!).
Now, in my head, I understand every concept in this course, and I was on top of my game pretty much for the whole semester. But I certainly feel like my mark doesn't reflect that. So what happened? I honestly don't know. Even though it really doesn't feel like it, this course might end up being my lowest mark so far in university.
There is still hope though. I'm hoping that the final won't be too challenging, and that I can improve my mark by doing good on that.
And that's that. Thanks for reading.
Wednesday, 5 December 2012
A Series Convergence Problem - Cont'd
This is a continuation of the last post.
4. Looking back
I went and slept on the problem, and I'm not exaggerating, woke up with an epiphany about the problem. All night I dreamt about series and floating sigma signs. Other than the sheer weirdness of waking up with a solution to a math problem, especially one that wasn’t that hard, I did realize that there was a much simpler and more intuitive solution to the problem! The approach that I wrote up in the last post is perhaps a little brute-forcing the way to a solution. This new solution uses an evident truth about series to prove the condition in one step. Here:
Assuming a_n converges.
Since (SIGMA)a_n converges, then ((SIGMA)a_n)^2 must also converge.
Since a_n => 0, then ((SIGMA)a_n)^2 >= (SIGMA)a_n^2 >= 0.
Then by the pinching theorem, (SIGMA) a_n^2 converges.
Then, given that a_n converges, then (a_n)^2 must converge.
Solutions are either valid or invalid, but I somehow feel this is a better solution. If only you had the time to sleep on every problem in the world…
4. Looking back
I went and slept on the problem, and I'm not exaggerating, woke up with an epiphany about the problem. All night I dreamt about series and floating sigma signs. Other than the sheer weirdness of waking up with a solution to a math problem, especially one that wasn’t that hard, I did realize that there was a much simpler and more intuitive solution to the problem! The approach that I wrote up in the last post is perhaps a little brute-forcing the way to a solution. This new solution uses an evident truth about series to prove the condition in one step. Here:
Assuming a_n converges.
Since (SIGMA)a_n converges, then ((SIGMA)a_n)^2 must also converge.
Since a_n => 0, then ((SIGMA)a_n)^2 >= (SIGMA)a_n^2 >= 0.
Then by the pinching theorem, (SIGMA) a_n^2 converges.
Then, given that a_n converges, then (a_n)^2 must converge.
Solutions are either valid or invalid, but I somehow feel this is a better solution. If only you had the time to sleep on every problem in the world…
Tuesday, 4 December 2012
A Series Convergance Problem
I don't know why, but I have just overcome with the urge to write up a SLOG
post about problem-solving, specifically a problem of mathematical
nature. So let's waste no time, and get to it!
If you remember from my first post, one of my favorite courses from last year happened to be MAT137. And one of my favorite topics in that course had to do with proofs regarding the convergence/divergence of various series. As such, I decided to dig through my stuff from last year, and find a question that I didn't get around to attempting to solve last year. A lemma from the textbook caught my eye, whose proof was left to the students as an exercise:
Instead of chicken scratches on a piece of paper, I decided to take the organized approach Prof. Heap referred to in the SLOG handout.
1. Understanding the Problem
Let's think about what it means for a series to be convergent. Mathematically, this means: For all ε > 0, there exists there exists some N’ in N(natural numbers), so that for all n >= N’, |a_n| < ε. So knowing nothing about sequence “a_n,” I would have to somehow show that the same thing is true about sequence “b_n = (a_n)^2”. This condition certainly doesn’t seem impossible.
2. Devising a Plan
This problem looks simple enough, but it is a bit different from the other series behaviour problems I've solved so far. From my recollection (which, I admit, is a bit wonky), either the sequence "a_n" on which the series is based on is usually explicitly given, or the formula for the series can either be: a) manipulated to a sum that has easily provable behaviour, or b) can be linked with a function whose end-behaviour can be proven using a variety of theorems. But in this case, I know nothing about the series or the sequence "a_n" on which the series is based on.
I have a few ideas about how to manipulate the relationship |a_n| < ε or - ε < a_n < ε, to get some facts about (a_n)^2, namely multiplying out the inequality by “a_n” throughout. Then, I believe, I could use the pinching theorem to justify the condition for “b_n = (a_n)^2”.
3. Carrying out the Plan
Assuming a_n converges.
Then |a_n| < ε.
Then - ε < a_n < ε.
Then – ε(a_n) < a_n(a_n) = ((a_n)^2) < ε(a_n).
Then it suffices to show the two series bounding (a_n)^2 converge for (a_n)^2 to
converge.
For – ε(a_n): – ε(– ε) > – εa_n > – ε( ε).
For ε(a_n): ε(– ε) < εa_n < ε( ε).
So (a_n)^2 is bound between – ε( ε) and ε( ε).
So –ε^2 < - εa_n < (a_n)^2 < εa_n < ε^2. And by the pinching theorem, (a_n)^2
converges.
Then, given that a_n converges, then (a_n)^2 must converge.
And there you go! It was that simple. I wish I did this last year, so I could totally feel like a mathematics prodigy.
If you remember from my first post, one of my favorite courses from last year happened to be MAT137. And one of my favorite topics in that course had to do with proofs regarding the convergence/divergence of various series. As such, I decided to dig through my stuff from last year, and find a question that I didn't get around to attempting to solve last year. A lemma from the textbook caught my eye, whose proof was left to the students as an exercise:
Assume that some series (SIGMA)a_n is convergent, and that an => 0 for all n in N(natural numbers). Prove that the series (SIGMA)(a_n)^2 also converges.
Instead of chicken scratches on a piece of paper, I decided to take the organized approach Prof. Heap referred to in the SLOG handout.
1. Understanding the Problem
Let's think about what it means for a series to be convergent. Mathematically, this means: For all ε > 0, there exists there exists some N’ in N(natural numbers), so that for all n >= N’, |a_n| < ε. So knowing nothing about sequence “a_n,” I would have to somehow show that the same thing is true about sequence “b_n = (a_n)^2”. This condition certainly doesn’t seem impossible.
2. Devising a Plan
This problem looks simple enough, but it is a bit different from the other series behaviour problems I've solved so far. From my recollection (which, I admit, is a bit wonky), either the sequence "a_n" on which the series is based on is usually explicitly given, or the formula for the series can either be: a) manipulated to a sum that has easily provable behaviour, or b) can be linked with a function whose end-behaviour can be proven using a variety of theorems. But in this case, I know nothing about the series or the sequence "a_n" on which the series is based on.
I have a few ideas about how to manipulate the relationship |a_n| < ε or - ε < a_n < ε, to get some facts about (a_n)^2, namely multiplying out the inequality by “a_n” throughout. Then, I believe, I could use the pinching theorem to justify the condition for “b_n = (a_n)^2”.
3. Carrying out the Plan
Assuming a_n converges.
Then |a_n| < ε.
Then - ε < a_n < ε.
Then – ε(a_n) < a_n(a_n) = ((a_n)^2) < ε(a_n).
Then it suffices to show the two series bounding (a_n)^2 converge for (a_n)^2 to
converge.
For – ε(a_n): – ε(– ε) > – εa_n > – ε( ε).
For ε(a_n): ε(– ε) < εa_n < ε( ε).
So (a_n)^2 is bound between – ε( ε) and ε( ε).
So –ε^2 < - εa_n < (a_n)^2 < εa_n < ε^2. And by the pinching theorem, (a_n)^2
converges.
Then, given that a_n converges, then (a_n)^2 must converge.
And there you go! It was that simple. I wish I did this last year, so I could totally feel like a mathematics prodigy.
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