1. ## Recursively defined functions

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.

2. Hello
Originally Posted by sjenkins
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.
We're given $\begin{cases}f(0)=3\\f(n+1)=-2f(n) \end{cases}$. To find $f(1)$ choose $n$ such that $n+1=1$. This gives $n=0$. Then use the relation $f(n+1)=-2f(n)$ with $n=0$ : $f(0+1)=-2f(0)\implies f(1)=-2\times 3=-6$. Does it help ?

3. 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????

4. Originally Posted by sjenkins
Solution:
f(1)=2f(0)+3=2X3+3=9
f(2)=2f(1)+3=2X9+3=21

Where does that 9 come from????
That's the value of $f(1)$. $f(2)$ is given by $f(2)=2{\color{red}f(1)}+3$ and has $f(1)$ has been computed just before, we know that it equals 9.

5. Ohhhhh....that makes complete sense. I knew I was just missing something simple.

Thanks again for the help!!

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### 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

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