Can someone help me get started with this...
Find f(1), f(2), f(3), f(4), and f(5) if f(n) is defined recursively by f(0)=3 and for n=0,1,2,...
f(n+1)=-2f(n)
I'm not looking for the answer, but for help in understanding.
Hello
We're given $\displaystyle \begin{cases}f(0)=3\\f(n+1)=-2f(n) \end{cases}$. To find $\displaystyle f(1)$ choose $\displaystyle n$ such that $\displaystyle n+1=1$. This gives $\displaystyle n=0$. Then use the relation $\displaystyle f(n+1)=-2f(n)$ with $\displaystyle n=0$ : $\displaystyle f(0+1)=-2f(0)\implies f(1)=-2\times 3=-6$. Does it help ?
I still feel so confused, but maybe if I give an example that is given in the book, I will understand what is going on.
Suppose that f is defined recursively by
f(0)=3
f(n+1)=2f(n)+3
Find f(1), f(2), f(3), f(4)
Solution:
f(1)=2f(0)+3=2X3+3=9
f(2)=2f(1)+3=2X9+3=21
Where does that 9 come from????