Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    [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

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    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

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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.
     
  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

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: I Function in this Equation
  1. Functional Equation? (Replies: 4)

  2. Functional equations (Replies: 2)

  3. Functions and equations (Replies: 15)

Loading...