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I Function in this Equation

  1. Mar 3, 2013 #1
    (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.
  2. jcsd
  3. Mar 4, 2013 #2


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    [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]
  4. Mar 4, 2013 #3


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    I think the OP was looking for i if y is odd and 1 if y is even
  5. Mar 4, 2013 #4
    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.
  6. Mar 4, 2013 #5


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    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).
  7. Mar 4, 2013 #6


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    This can be written as

    ##i^\sigma## with ##\sigma=2y-1+(-1)^y## is possible, too.
  8. Mar 4, 2013 #7
    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?
  9. Mar 5, 2013 #8


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    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.
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