1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Step function

  1. Apr 29, 2005 #1
    I would like to invite comment to the near step function below.

    \Mvariable{step(x)}=\frac{{e^{-\frac{\gamma }{{x^2}}}}}{{\sqrt{1-\frac{{e^{-\frac{\gamma }{{x^2}}}}}{{x^2}}}}}

    the above function evaluates to nearly 1 for |x|>1 and nearly zero for |x|<0.3

    for gamma = 0.4823241136337762 I have attached the plots of step and 1-step

    here are some spot values for
    x and step(x)
    0.05 1.15E-83
    0.1 1.84E-21
    0.2 6.55E-6
    0.3 5.11E-3
    0.4 6.11E-2
    0.5 2.32E-1
    0.6 0.518
    0.8 0.932
    1.0 1.0068
    2.0 1.006
    5.0 1.00092
    10 1.00022
    50 1.0000090

    Attached Files:

    Last edited: Apr 29, 2005
  2. jcsd
  3. Apr 29, 2005 #2
    Indeed. If you change all the [itex]x^2[/itex]s to [itex]x^4[/itex]s (or [itex]x^{1000}[/itex]s), then it'll get even closer to a step function.
  4. Apr 30, 2005 #3
    I just noticed that the step function can be simplified even more
    \frac{{e^{-\frac{\gamma }{{x^2}}}}}{{\sqrt{1-\frac{{e^{-\frac{\gamma }{{x^2}}}}}{{x^2}}}}}


    for above gamma= 0.4823241136337762 , alpha =0.61734693877551, hence

  5. Apr 30, 2005 #4
    Well, it's not really "simpler." You have the same number of arbitrary constants, though I guess you have a couple fewer symbols in general (you got rid of two minus signs - but you could do that just by specifying that [itex]\gamma[/itex] must be negative). I'd usually just leave it in the exponential form, but that's a subjective choice based on the fact that I like the letter e :wink:
  6. Apr 30, 2005 #5
    Last edited: Apr 30, 2005
  7. Apr 30, 2005 #6
    arbitrary within a certain domain is what I should have said, of course. In this case, you need [itex]\gamma > 1/e[/itex].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?