0
$\begingroup$

If we define the Heaviside step function H(x) in limit notation, as per below, this yields 1/2 at x=0. How might this be adjusted to give 0 or 1 at x=0?

$$ H(x)=\lim_{b \to \infty} \frac{1}{1+\frac{1}{b}^\frac{bx}{\ln(b)}} $$

Sorry, I don't like the title but couldn't think of anything better.

$\endgroup$
4
  • 1
    $\begingroup$ Why don’t you simply subtract/add 1/2? $\endgroup$ May 20, 2021 at 15:20
  • $\begingroup$ Adding or subtracting a half would cause the whole function to shift up or down, but I'm looking for an expression that shifts it up or down at just the point x=0. $\endgroup$
    – Noman
    May 20, 2021 at 18:03
  • 1
    $\begingroup$ Why do you need to define $H(x)$ as a limit? What's wrong with simply writing a piecewise definition? $\endgroup$ May 21, 2021 at 5:38
  • $\begingroup$ I'm working on a math that allows for defining differentiable step functions using novel numbers and was wondering if it has any utility in intuiting solutions that may otherwise not be obvious. All its expressions can be expressed as limits to zero or infinity, so I figured if it has any utility, it would be something about the notation itself. $\endgroup$
    – Noman
    May 21, 2021 at 11:21

1 Answer 1

0
+50
$\begingroup$

You can set

$$H(x):=\lim_{b\to+\infty}\frac{\frac{2}{\pi}\arctan(bx)+1-\exp(-bx^2)}{2}$$

and this is such that $H(0)=0$, but you can modify it to cover the other case.

$\endgroup$
2
  • 1
    $\begingroup$ Thanks for that, and could you possibly provide some insight as to how you derived this? $\endgroup$
    – Noman
    Apr 16, 2022 at 21:36
  • $\begingroup$ The idea is the same as in your example: let me talk about that. You have that your limit is $1/2$ when computed for $x=0$. It means you just need to subtract a functions which is $1/2$ in $x=0$ and $0$ everywhere else. Clearly $1/2\exp[-bx^2]$ in the limit does the job. This is to say that you can also adjust your example. I thought of using the arctangent because in the limit it becomes the step function (almost). Then I adjusted from there $\endgroup$
    – Gauge_name
    Apr 17, 2022 at 7:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .