Difference between revisions of "Midterm Page MA375Fall2008walther" - Rhea

Revision as of 16:05, 26 October 2008

The midterm is Tuesday, October 28th. Here is a list of example midterm questions.

Questions and Solutions to Sample Problems

Question: For the midterm we are allowed to complete one page with notes. Does anyone know if this must be hand written or can it be typed?

Answer: As mentioned today in class(10/21), we are allowed to complete one full page, 8.5x11", worth of cheat sheet. This can be typed, handwritten, crayon, etc. In addition, we may use any sort of calculator.

Question 21

Let f_i be the n-th Fibonacci number: f_0=0, f_1=1, f_2=1,... Prove that f_1 + f_3 + f_5 + ... + f_(2n+1) = f_(2n+2).

By the sequence, we initially see f_(2n+2) = f_(2n+1) + f_(2n). Since we are asked to prove something of the form f_(2n+2) = f_(2n+1) + [blah], we will assume the objective is true, and try to iron out f_(2n) to match. So, let's expand f_(2n): f_(2n) = f_(2n-1) + f_(2n-2). So, we have f_(2n+2) = f_(2n+1) + f_(2n-1) + f_(2n-2). We begin to see a pattern -- expanding an even-numbered subscript gives us an odd-numbered subscript and another even-numbered subscript to expand later. It would seem like we would end up with something like what we are trying to prove.

How can we prove this, though? The first thought that comes to mind would just be to write a sequence of expansions out as we were doing, add a couple of dots ("..."), and toss the solution at the end. However, that's not terribly rigorous. So, let's try to use induction instead:

Base case: Let n = 0. f_2 = f_1 + f_0 = f_1 (since f_0 = 0) by the definition of Fibonacci numbers, and f_2 = f_1 by the proposed recursion.

Inductive step: Assume that for a given integer n that f_1 + f_3 + f_5 + ... + f_(2n+1) = f_(2n+2). We want to show that f_1 + f_3 + f_5 + ... + f_(2n+1) + f_(2n+3) = f_(2(n+1)+2) = f_(2n+4). f_(2n+4) = f_(2n+3) + f_(2n+2) (by the Fibonacci sequence [Remember a couple of paragraphs ago? This was our initial plan.]) = f_(2n+3) + f_(2n+1) + ... + f_5 + f_3 + f_1 (by assumption [...and this is what makes the proof mathematically rigorous instead of "here's some '...'s, you get the idea"-based]).

Thus, for all integer n ≥ 0, f_1 + f_3 + f_5 + ... + f_(2n+1) = f_(2n+2).

Question 5

Can someone verify that I'm interpreting/doing this question right? I assume without repetitions means we can't have 11111, 22222, 11134, 23359 etc. The number of strings with 1 but not 2 is equal to the number of strings with 2 but not 1, so you just find one answer and double it. The number of strings with 1 but not 2 should be the number of ways to arrange the the digits 3-9 and 0 in the 4 remaining places. But since the 1 can go in any place, we multiply this by 5. So my answer is 2*5*(8*7*6*5) = 16800. Anyone else get this? --Mkorb 20:56, 26 October 2008 (UTC)

Is there going to be a review session for this midterm? --Podarcze 12:06, 26 October 2008 (UTC)

Wasn't the review session during thursday's class?

-I meant study session. I know people have been getting together to do homework on monday's and I was wondering if they are doing it this week. --Podarcze 21:05, 26 October 2008 (UTC)

Materials to Study

Would it be enough to study the HW and the sample questions? -Wooi-Chen Ng

I think we still have to read and study the notes that the professor gave us during class. I think it's helpful too since many examples were done during class time.--hussaini 19:48, 26 October 2008 (UTC)