Chapter 5 - Section 5.3 - Recursive Definitions and Structural Induction - Exercises - Page 357: 2

Part A: $f(1) = -6$, $f(2) = 12$, $f(3) = -24$, $f(4) = 48$. Part B: $f(1) = 16$, $f(2) = 55$, $f(3) = 172$, $f(4) = 523$. Part C: $f(1) = 1$, $f(2) = -3$, $f(3) = 13$, $f(4) = 141$. Part D: $f(1) = 3$, $f(2) = 3$, $f(3) = 3$, $f(4) = 3$.

For parts A,B,C and D, we know that for each recursive formula that $f(0) = 3$. In part A we are told that $f(n+1) = -2f(n)$. If we plug in $n=0$, we see that $f(0+1) = -2f(0)$, so $f(1) = -6$. To find $f(2)$ we can plug in $n=1$ to find that $f(1+1) = -2f(1)$, so $f(2) = 12$. This process can be continued to find $f(3) = -24$ and $f(4) = 48$. We can use this same approach for the remaining parts of this question. For Part B, if we let $n=0$, then it follows that $f(1) = 3(f(0)) +7 = 16$. Continuing we see that $f(2) = 55$, $f(3) = 172$, and $f(4) = 523$. For Part C if we let $n=0$ then we find that $f(1) = f(0)^2 -2f(0)-2 = 1$. Plug in ascending values of $n$ to find $f(2) = -3$, $f(3)=13$, and $f(4) = 141$. Solve Part D exactly the same as before, plugging $n=0$ into the recursion to find that $f(1) = 3^{\frac{f(0)}{3}} = 3$. Continue to find $f(2) = 3$, $f(3) = 3$, and $f(4)=3$.