i Function in this Equation

Big-Daddy

(i being the complex square root of -1 here.)

I have a function which is dependent on the term "y", where, if y is odd, the function is multiplied by i, whereas if y is even the function is multiplied by 1. (y is always a real integer greater than or equal to 0.)

How can I add an i^(some y function) term to the function to express this?

I have identified that i^(1+multiple of 4)=i, whereas i^(0+multiple of 4)=1.

I considered i^(4y-3), but not only does this break down for small values of y, but it yields wrong results for when I want the function to be multiplied by 1 as it never results in a multiple of 4 being the degree of i.

CompuChip

[itex](-1)^y[/itex] is 1 when y is even and -1 when y is odd.
So [itex]\tfrac12\left( (-1)^y + 1 \right)[/itex] is 1 when y is even and 0 when y is odd.
Does that help you proceed?

[edit]And of course i⁰ = 1.[/edit]

SteamKing

I think the OP was looking for i if y is odd and 1 if y is even

micromass

What about

[tex]\frac{1}{2}[(-1)^y + 1] + \frac{i}{2}[(-1)^{y+1} +1][/tex]

Edit: I guess CompuChip intended his post to be a hint so that the OP could guess this formula. But I don't see harm in giving the complete solution here.

CompuChip

Exactly micromass, mostly because I didn't feel like figuring out the details :)

Actually your solution is better than what I had in mind (which was something like [itex]i^\sigma[/itex] where [itex]\sigma[/itex] is based on what I wrote before).

mfb

This can be written as
$$\frac{1+i}{2}+\frac{1-i}{2}(-1)^y$$

##i^\sigma## with ##\sigma=2y-1+(-1)^y## is possible, too.

Big-Daddy

Thanks all! Your formulae check out perfectly.

I was wondering how you come up with something like this? Is it roughly trial and error or do you somehow work back from the final goal?

CompuChip

Using [itex](-1)^\epsilon[/itex] (where [itex]\epsilon = 0, 1[/itex]) is quite a common trick, especially in physics. From there it's just a bit of playing and shuffling around: you add one to shift the -1 to 0, and then you have to divide by 2 to scale the 2 back to 1.

Of course then you can repeat the trick with [itex](-1) \cdot (-1)^\epsilon = (-1)^{\epsilon + 1}[/itex]. This basically gives you two "indicator" functions: one which is zero if epsilon = 0 and one otherwise; and one which is zero if epsilon = 1 and zero otherwise. So you multiply one by the number you want for epsilon = 0 and the other one by the other value, leading to micromass' answer. mfb's answer is obtained by opening the brackets and rearranging so that instead of (...)1 + (...)i you get (...)1 + (...)(-1)^y.

You'll find that you pick up quite a lot of these "tricks" over time, and then it's often a matter of a bit of experience or luck to pick the right one and some creative shuffling around to adapt it to your needs.