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

[Mathematica] Simple Problem with the plot function

  1. May 8, 2013 #1
    I have a formula for the fibonacci sequence (with 1 being the first) and I noticed that the 12th fibonacci number was 144. I thought that was a neat coincidence, so I I headed over to mathematica to see if this (and 1) were the only numbers that had this property. I was almost certain that it was. So I plotted the two graphs, but only the x^2 graphed showed up. Here was my input. My question is about what I did wrong.

    phi := (1 + Sqrt[5])/2

    phih := (1 - Sqrt[5])/2



    Plot[{f, y}, {n, -20, 20}]

    Now, this works when I remove the phih term, which works well enough that I can clearly see that only 1 and 12 are solutions. On the other hand, this is not plotting something that it should be plotting...

    Anyway, thank you in advance for the help.
  2. jcsd
  3. May 8, 2013 #2
    The problem is that plot tries to make a continuous plot. It then runs into the problem that phih^x is not a real-valued function for non-integer x, and refuses to plot it altogether. To just plot the values at integer values of x, you can use DiscretePlot:

    DiscretePlot[{f, y}, {n, -20, 20, 1}]
  4. May 8, 2013 #3
    That's really neat and useful. I was so accustomed to just glazing over (1-Sqrt[5])/2, that I forgot that it was negative. If you raise it to the power of 1/2 for example, then the answer is imaginary. That's why it is not a real-valued function, correct? Thank you, sir.
  5. May 8, 2013 #4
    You are welcome. And, yeah, if the function has a non-zero imaginary part for at least one value of x, then clearly it's not real-valued on the whole domain (i.e. the plot range). However, Mathematica is able to handle cases where the function is piecewise real-valued, it just doesn't plot the part where the values are comples, see eg.
    Plot[(1 + I*HeavisideTheta[x - 5]*HeavisideTheta[6 - x]), {x, 0, 10}]
    (I would prefer it to give an error message or a warning, but it doesn't seem to do that.)

    Your function is a bit worse though as it is only real-valued on integer values of x. The proof just uses the general definition of the power of a number a (negative, complex etc.), which makes use of the complex logarithm:
    [itex]a^b=e^{b log a}[/itex].
    If a is real and negative, then this simplifies to
    [itex]a^b=e^{bLog(|a|)+ib\pi}=|a|^b e^{ib\pi}[/itex],
    which is only real for integer values of b.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook