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

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Homework Statement



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

Homework Equations





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.
 
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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).
 
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.
 
Hey thanks a lot for the help. Either method works fine, as long as you can figure out f(n) for the first one.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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