How Can I Express This Function with a Complex Square Root Term?

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The discussion revolves around expressing a function that depends on an integer variable "y," where the function is multiplied by the complex unit "i" if "y" is odd and by 1 if "y" is even. Participants propose using the expression (1/2)[(-1)^y + 1] + (i/2)[(-1)^(y+1) + 1] to achieve this, which effectively distinguishes between odd and even values of "y." Additional insights include using the term i^σ, where σ is defined based on the parity of "y," to represent the desired function. The conversation highlights the use of common mathematical tricks, such as indicator functions, to simplify the expression. Overall, the community collaboratively arrives at a solution that meets the original requirement.
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(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.
 
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(-1)^y is 1 when y is even and -1 when y is odd.
So \tfrac12\left( (-1)^y + 1 \right) is 1 when y is even and 0 when y is odd.
Does that help you proceed?

[edit]And of course i0 = 1.[/edit]
 
I think the OP was looking for i if y is odd and 1 if y is even
 
What about

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

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.
 
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 i^\sigma where \sigma is based on what I wrote before).
 
micromass said:
\frac{1}{2}[(-1)^y + 1] + \frac{i}{2}[(-1)^{y+1} +1]
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.
 
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?
 
Using (-1)^\epsilon (where \epsilon = 0, 1) 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 (-1) \cdot (-1)^\epsilon = (-1)^{\epsilon + 1}. 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.
 
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