# Practice Paper 3 Question 7

Let $$a_n, b_n$$ be sequences of positive real numbers which satisfy $$a_0 = b_0$$ and the recursive matrix relation $$\big(\begin{smallmatrix}a_n \\b_n\end{smallmatrix}\big)=\big(\begin{smallmatrix} 1 & y \\0 & 1\end{smallmatrix}\big)\big(\begin{smallmatrix}a_{n-1} \\b_{n-1}\end{smallmatrix}\big).$$ What is the simplest non-recursive form of the ratio $$\frac{b_n}{a_n}?$$

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## Hints

• Hint 1
Evaluate $$\big(\begin{smallmatrix}1 & y \\0 & 1\end{smallmatrix}\big)\big(\begin{smallmatrix}u \\v\end{smallmatrix}\big).$$
• Hint 2
Try to express the relationship between $$\big(\begin{smallmatrix}a_n \\ b_n\end{smallmatrix}\big)$$ and $$\big(\begin{smallmatrix}a_0 \\ b_0\end{smallmatrix}\big)$$ as a non-recursive matrix equation.
• Hint 3
Simplify $${\big(\begin{smallmatrix}1 & y \\0 & 1\end{smallmatrix}\big)}^n,$$ by trying small values of $$n$$ or otherwise.

## Solution

We can unwind this matrix recurrence in the same way as a regular recurrence, $$\big(\begin{smallmatrix} a_n \\ b_n \end{smallmatrix}\big) = \big(\begin{smallmatrix} 1 & y \\ 0 & 1 \end{smallmatrix}\big) \big(\begin{smallmatrix} a_{n-1} \\ b_{n-1} \end{smallmatrix}\big)$$ $$={\big(\begin{smallmatrix} 1 & y \\ 0 & 1 \end{smallmatrix}\big)}^2 \big(\begin{smallmatrix} a_{n-2} \\ b_{n-2} \end{smallmatrix}\big)$$ $$=\ldots=$$ $${\big(\begin{smallmatrix} 1 & y \\ 0 & 1 \end{smallmatrix}\big)}^n \big(\begin{smallmatrix} a_{0} \\ b_{0} \end{smallmatrix}\big).$$ Using induction, we can show that $${\big(\begin{smallmatrix} 1 & y \\ 0 & 1 \end{smallmatrix}\big)}^n$$ $$={\big(\begin{smallmatrix} 1 & ny \\ 0 & 1 \end{smallmatrix}\big)}.$$ Hence, $$\big(\begin{smallmatrix} a_n \\ b_n \end{smallmatrix}\big)$$ $$={\big(\begin{smallmatrix} 1 & ny \\ 0 & 1 \end{smallmatrix}\big)} \big(\begin{smallmatrix} a_{0} \\ b_{0} \end{smallmatrix}\big)$$ $$=\big(\begin{smallmatrix} a_{0} + nyb_{0} \\ b_{0} \end{smallmatrix}\big).$$ Therefore the ratio $$\frac{b_n}{a_n}$$ is $$\frac{b_0}{a_0 + nyb_0} = \frac{1}{1+ny}.$$

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