# In terms of n 1, 1, -1, -1, 1, 1, -1, -1,

1. May 12, 2010

### ziggie125

1. The problem statement, all variables and given/known data

Put in general terms 1, 1, -1, -1, 1, 1, -1, -1, ...

2. Relevant equations

3. The attempt at a solution

obviously (-1)^n, alternates 1, -1, 1, -1...

I have no idea how to figure this out. I thought it might have a sin function in it possibly.
Thanks a lot for your help.

2. May 12, 2010

### jbunniii

If you defined a function f(n) such that

f(0) = 0
f(1) = 0
f(2) = 1
f(3) = 1
f(4) = 2
f(5) = 2
...

then

$$(-1)^{f(n)}$$

would work, right? So try to find a simple formula for f(n).

3. May 12, 2010

### Martin Rattigan

The $i^{th}$ term (starting with $i=1$) could be $\sqrt{2}sin\{(2i-1)\frac{\pi}{4}\}$, but don't use that. You could probably use something similar to generate the $f(i-1)$ that jbunniii has suggested.

Of course, if the first number after the dots start is 42 you're in trouble.

4. May 12, 2010

### ziggie125

Hey thx a lot for the help. Either method works fine, as long as you can figure out f(n) for the first one.