1 problem found
The value \(V_N\) of a bond after \(N\) days is determined by the equation $$ V_{N+1} = (1+c) V_{N} -d \qquad (c>0, \ d>0), $$ where \(c\) and \(d\) are given constants. By looking for solutions of the form \(V_T= A k^T + B\) for some constants \(A,B\) and \(k\), or otherwise, find \(V_N\) in terms of \(V_0\). What is the solution for \(c=0\)? Show that this is the limit (for fixed \(N\)) as \(c\rightarrow 0\) of your solution for \(c>0\).
Solution: Suppose \(V_T = Ak^T + B\), then \begin{align*} && Ak^{T+1}+B &= (1+c)(Ak^T+B) - d \\ \Rightarrow && (k-1-c)Ak^T &= cB -d \\ \Rightarrow && k &= 1+c \\ && B &= \frac{d}{c} \\ && A &= V_0 - B \\ \Rightarrow && V_N &= (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c} \end{align*} When \(c = 0\), \(V_{N+1} = V_N - d \Rightarrow V_N = V_0 - Nd\). \begin{align*} \lim_{c \to 0} \left ( (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c} \right) &= \lim_{c \to 0} \left ( \frac{(V_0 c-d)(1+c)^N + d}{c} \right ) \\ &= \lim_{c \to 0} \left ( \frac{V_0c - d-Ncd+NV_0c^2 + o(c^2) + d}{c} \right ) \\ &= \lim_{c \to 0} \left ( V_0 - Nd + o(c) \right ) \\ &= V_0 - Nd \end{align*}