Chapter 5 - Section 5.3 - Recursive Definitions and Structural Induction - Exercises - Page 357: 4

Part A: $f(2) = 0$, $f(3) = -1$, $f(4) = -1$, $f(5) = 0$. Part B: $f(2) = 1$, $f(3) = 1$, $f(4) = 1$, $f(5) = 1$. Part C: $f(2) = 2$, $f(3) = 5$, $f(4) = 33$, $f(5) = 1214$. Part D: $f(2) = 1$, $f(3) = 1$, $f(4) = 1$, $f(5) = 1$.

For parts A, B, C and D, we know that for each recursive formula that $f(0) = 1$ and that $f(1)=1$. In part A we are told that $f(n+1) = f(n) - f(n-1)$. If we plug in $n=1$, we see that $f(1+1) = f(1) - f(0)$, so $f(2) = 0$. To find $f(3)$ we can plug in $n=2$ to find that $f(2+1) = f(2) - f(1)$, so $f(3) = -1$. This process can be continued to find $f(4) = -1$ and $f(5) = 0$. 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 $f(2) = f(1)f(0)$ = 1. Continuing we see that $f(3) = 1$, $f(4) = 1$, and $f(5) = 1$. For Part C if we let $n=1$ then we find that $f(2) = f(1)^2 + f(0)^3 = 2$. Plug in ascending values of $n$ to find $f(3) = 5$, $f(4)=33$, and $f(5) = 1214$. Solve Part D exactly the same as before, plugging $n=1$ into the recursion to find that $f(2) = f(n)/f(n-1) = 1$. Continue to find $f(3) = 1$, $f(4) = 1$, and $f(5)=1$.