Discussing the Near Step Function for γ=0.4823241136337762

AntonVrba
Messages
92
Reaction score
0
I would like to invite comment to the near step function below.

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

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
 

Attachments

  • step.gif
    step.gif
    2.6 KB · Views: 502
Last edited:
Physics news on Phys.org
Indeed. If you change all the x^2s to x^4s (or x^{1000}s), then it'll get even closer to a step function.
 
I just noticed that the step function can be simplified even more
<br /> \Mvariable{step}(x)=<br /> \frac{{e^{-\frac{\gamma }{{x^2}}}}}{{\sqrt{1-\frac{{e^{-\frac{\gamma }{{x^2}}}}}{{x^2}}}}}<br /> <br /> =\frac{{{\alpha}^{\frac{1}{{x^2}}}}}{{\sqrt{1-\frac{{{\alpha}^{\frac{1}{{x^2}}}}}{{x^2}}}}}\\<br />

for above gamma= 0.4823241136337762 , alpha =0.61734693877551, hence

<br /> \Mvariable{step}(x)=<br /> \frac{{{0.61734693877551}^{\frac{1}{{x^2}}}}}{{\sqrt{1-\frac{{{0.61734693877551}^{\frac{1}{{x^2}}}}}{{x^2}}}}}\\<br />
 
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 \gamma 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:
 
Data said:
Well, it's not really "simpler." You have the same number of arbitrary constants, QUOTE]


:eek: arbitrary :eek:
just play with the function and you will see that gamma or alpha has a rather very limited range for the function not to become complex for any real number:biggrin:

By proper choice of the Alpha you can make the function overshoot (become slightly larger than one) as in the example given, or the function always remaining slightly less than one i.e approach one at infinity, and that in a only very limitted range of values.
 
Last edited:
arbitrary within a certain domain is what I should have said, of course. In this case, you need \gamma &gt; 1/e.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top