Chapter 5 - Section 5.3 - Recursive Definitions and Structural Induction - Exercises - Page 357: 3

Part A: $f(2) = -1$, $f(3) = 5$, $f(4) = 2$, $f(5) = 17$. Part B: $f(2) = -4$, $f(3) = 32$, $f(4) = -4096$, $f(5) = 536870912$. Part C: $f(2) = 8$, $f(3) = 176$, $f(4) = 92672$, $f(5) = 25764174848$. Part D: $f(2) = -\frac{1}{2}$, $f(3) = -4$, $f(4) = \frac{1}{8}$, $f(5) = -32$.

These questions are different from the previous 2 exercises, in that each recursion relies on two previous values instead of just one. For parts A, B, C and D, we know that for each recursive formula that $f(0) = -1$ and that $f(1)=2$. In part A we are told that $f(n+1) = f(n) + 3f(n-1)$. If we plug in $n=1$, we see that $f(1+1) = f(1) + 3f(0)$, so $f(2) = -1$. To find $f(3)$ we can plug in $n=2$ to find that $f(2+1) = f(2) + 3f(1)$, so $f(3) = 5$. This process can be continued to find $f(4) = 2$ and $f(5) = 17$. We can use this same approach for the remaining parts of this question. For Part B, if we let $n=1$, then it follows that $f(2) = f(1)^2 f(0) = -4$. Continuing we see that $f(3) = 32$, $f(4) = -4096$, and $f(5) = 536870912$. For Part C if we let $n=1$ then we find that $f(2) = 3f(1)^2 -4f(0)^2 = 8$. Plug in ascending values of $n$ to find $f(3) = 176$, $f(4)=92672$, and $f(5) = 25764174848$. Solve Part D exactly the same as before, plugging $n=1$ into the recursion to find that $f(2) = \frac{f(0)}{f(1)} = -\frac{1}{2}$. Continue to find $f(3) = -4$, $f(4) = \frac{1}{8}$, and $f(5)=-32$.