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

MATH307 April 2009

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Question 07 (b)
Given the recurrence relation $x_{n+1}=3x_{n}-2x_{n-1}$ with the initial condition $x_{0}=a$ and $x_{1}=b$. (b) For what values of a and b will the sequence $x_{0},x_{1},x_{2},...$ converge to a finite limit?

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Hint

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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. Since every convergent sequence is Cauchy and every Cauchy sequence in $\mathbb {R}$ converges, we can write the sequence as a Cauchy sequence to find the values of a and b for which the sequence converges to a finite limit. So for every $\epsilon >0$ there is a natural number N such that for every n and m integers greater than or equal to N, $|x_{n}-x_{m}|<\epsilon$. Then $|(2^{n}-1)b-(2^{n}-2)a-(2^{m}-1)b+(2^{m}-2)a|<\epsilon$, and so $|b(2^{n}-2^{m})-(2^{n}-2^{m})a|<\epsilon$, that is, $|(b-a)(2^{n}-2^{m})|<\epsilon$. Since $|2^{n}-2^{m}|$ can be arbitrarily large and we want this to be true for all $\epsilon >0$, we propose that we need $b=a$. Suppose $b=a$. Then $|x_{n}-x_{m}|=|0(2^{n}-2^{m})|=0<\epsilon$, as required. Now suppose there is some value of a such that $a\not =b$ but $|(b-a)(2^{n}-2^{m})|<\epsilon$. Then let $k=|b-a|$ and take $\epsilon =k$. Then $|k(2^{n}-2^{m})|=|k(2^{n-m}(2^{m}-1))|$. Since we can take n and m to be any integers such that they are greater than or equal to N, take $m=N$ and $n=N+1$. Then $|k(2(2^{N}-1))|\geq |2k|>k$ since $N\geq 1$. This contradicts our assumption that there is a $k\not =0$ such that for any $\epsilon >0$, $|x_{n}-x_{m}|<\epsilon$. So the sequence converges to a finite limit when b = a.