Science:Math Exam Resources/Courses/MATH307/April 2009/Question 07 (a)

MATH307 April 2009

Work in progress: this question page is incomplete, there might be mistakes in the material you are seeing here.

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q3 (d)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q6 (d)  •  Q7 (a)  •  Q7 (b)  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  Q8 (d)  •  

   Other MATH307 Exams

  •  December 2012  •  April 2013  •  April 2012  •  April 2006  •  December 2008  •  December 2010  •  

Question 07 (a)
Given the recurrence relation $x_{n+1}=3x_{n}-2x_{n-1}$ with the initial condition $x_{0}=a$ and $x_{1}=b$, (a) Solve the recurrence relation. (Give a general scalar expression for $x_{n}$ in terms of n, a, and b).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We can start by finding the first few values of this relation. $x_{2}=3b-2a$ $x_{3}=7b-6a$ $x_{4}=15b-14a$ $x_{5}=31b-30a$ With these, we can guess at a scalar expression for $x_{n}$ and then use induction to prove that it is correct. By inspection, we come up with the following: $x_{n}=(2^{n}-1)b-(2^{n}-2)a$. Proof: $n=0$: $x_{0}=(2^{0}-1)b-(2^{0}-2)a=0b-(-1)a=a$ $n=1$: $x_{1}=(2^{1}-1)b-(2^{1}-2)a=1b-0a=b$ $n=2$: $x_{2}=(2^{2}-1)b-(2^{2}-2)a=(4-1)b-(4-2)a=3b-2a$, which is what we found using the recurrence. Assume that for $n\leq k$, the expression is true. We now want to prove that the expression is true for $n=k+1$. Using the recurrence relation and our induction hypothesis, we find that $x_{k+1}=3x_{k}-2x_{k-1}$ $=3((2^{k}-1)b-(2^{k}-2)a)-2((2^{k-1}-1)b-(2^{k-1}-2)a)$ $=b(3(2^{k}-1)-2(2^{k-1}-1))-a(3(2^{k}-2)-2(2^{k-1}-2))$ $=b(2^{k}(3-1)-3+2)-a(2^{k}(3-1)-2)$ $=b(2^{k+1}-1)-a(2^{k+1}-2)$, which is the expression we had come up with. So for all integers n, $x_{n}=(2^{n}-1)b-(2^{n}-2)a$.