Talk:Heaviside step function

message: The definition is self-contradictory: Who can correct it?
S.
:Dirac delta function also defines it, with the ≥ switched for a ≤ - but this still doesn't match the written definition, unless I was lied to by years of math teachers and zero is negative. --Brion 08:55 Oct 26, 2002 (UTC)
Looked it up, it's =1 at zero (your teachers were right, Brion, zero, is neither neg nor pos). Funnily enough, the book I have here, they give the positive part first, then the negative, which seems the wrong way round. If I remember correctly, the value at zero isn't crucial, and different definitions exist: can be H(0)=0, H(0)=1, and H(0)=0.5 and of course, H(0)=cream cheese. needs checking though. -- Tarquin 10:16 Oct 26, 2002 (UTC)
In Japan, it is thought that <BR>
 
: 
    \int^{0}_{-\epsilon} \delta(x) dx = \int^{+\epsilon}_{0} \delta(x) dx = 1/2 
       for all \epsilon > 0   <BR>
 
and every image  H(x)  of Heaviside step function   H : \mathbb{R} \ni x \longrightarrow H(x) \in H(\mathbb{R}) \subset \mathbb{R}   is 
  
: 
    H(x) = \int^{x}_{-\infin} \delta(\xi) d\xi = \left\{\begin{matrix} 0 & \left( x < 0 \right) \\ 1/2 & \left( x = 0 \right) \\ 1 & \left( x > 0 \right) \end{matrix}\right. 
  <B>.</B>
 
Koiki Sumi 00:00 15 Sep 2003 (UTC)
: I kind of doubt there's any agreement at all, even in Japan...
:: I absolutely agree with Japanese definition. Consider a broadened or continuous δ(x)-type function of finite width with any symmetric shape. Integrate this from minus infinity to zero, and you will see that the integral contains exactly half the δ(x) function, and so is equal to 1/2. Now take the limit as the width of the δ(x) function goes to zero, thus obtaining a true δ(x) function. The integral from minus infinity to zero remains equal to 1/2. This is exactly the value of the step function at 0. 71.32.46.85 (talk) 20:45, 5 February 2019 (UTC)
thread_title:  (pre-TOC comments) 

message: i can't find the fourier transform of the heaviside function anywhere...anyone willing to share their expertise :)
Comment at support.
Charles Matthews 13:59, 31 Jan 2004 (UTC)
replies: text: Reminds me: I once won five pints of beer on a bet that it was in the textbooks (and drank three of them). As for the comment below, what is in 'the books' is typically wrong or misleading. I suppose the article can try to explain why. Charles Matthews 14:55, 11 Dec 2004 (UTC)
thread_title:  fourier transform? 

message: I was prettying up that "integral representation of the step function" at the end, and upon looking at it, i don't think it's correct.  maybe it is for the Signum function , but i don't think it is for the step function.  BTW, if we define the step function strictly in terms of the  \sgn() , i think the Fourier Transform of it comes out nicely.  also, the step function should either be undefined for x=0 or be defined to be 1/2 at x=0, but not either 1 or 0.  r b-j 03:30, 11 Dec 2004 (UTC)
thread_title:  "integral representation of the step function" 

message: At the following address you will find the Fourier transform of the Heaviside Step Function
http://mathworld.wolfram.com/HeavisideStepFunction.html
:Hmmm - I'm not saying that's wrong. I would say that 1/x is not a locally integrable function. Therefore using it to represent a Schwartz distribution is not in itself a naive kind of definition. The formula therefore needs some commentary: the difference of the delta function and the reciprocal function is a combination that seems to require some discussion. Charles Matthews 12:48, 19 Apr 2005 (UTC)
thread_title:  Fourier transform of the Heaviside Step Function 

message: I removed this:
: There are some trials to put analytical functions to numerically calculate Unit Step Function. The study published on  has shown that it is possible to mimic the Unit Step Function. The results were verified using Mathematica software. 
 
It's uninteresting, of restricted applicability, (strictly speaking) incorrect (the inverse trigonometric functions do not have unique definitions) and constitutes original research. EdC 15:14, 4 June 2006 (UTC)
thread_title:  Analytic Exact Form of Unit Step Function 

message: It feels wrong to define H(0) as 1/2 and then immediately say the H(0) seldom matters and can be defined in various ways. However, it would also feel wrong to show a definition by case analysis which considered only x<0 and x>0. Would it be a horrible idea simply to remove the first displayed formula and just rely on the prose in the first sentence, plus the graph? Henning Makholm 01:37, 26 November 2006 (UTC)
thread_title:  H(0) in the intro paragraph 

message: Is it just me, or is it an incredible coinicdence that the "Heaviside" function looks like y = 1, with the negative region "heavier" (i.e. "pushed" down to 0)? The "heavier side", or "heavyside", sounds a lot like "Heaviside".... Timeroot (talk) 04:38, 27 January 2009 (UTC)
thread_title:  "Heaviside" 

message: It is so painful to label a maths article as needing references,thats why i won't do it. guys, may you add more in-text references? try to get references from several sources. any one with advanced engineering text book may do better...smile :) .... Freshymail-user_talk:fngosa--the-knowledge-defender 18:16, 27 August 2009 (UTC)
thread_title: references

message: Is the Heaviside step functin defined for x = 0? Looking at the alternartive definition
:H(x) = \lim_{z \rightarrow x^-} ((|z| / z + 1) / 2)
H(0) would be 0. Is is sure it shouldn't be z\rightarrow x^+ or just z\rightarrow x? --Kri (talk) 15:30, 15 October 2009 (UTC)
replies: text: Oh sorry, found the section H(0) now. Still wonder about the limit thing though. --Kri (talk) 15:34, 15 October 2009 (UTC)
thread_title:  Value in 0? 

message: I removed the following assertion from the text, because I think it is false:
If one differentiates the elements of the approximating family, then one does indeed get weak convergence to the delta measure.  In fact, these all do converge as distributions as well, since their derivatives are all of the form k \eta(k x) for a smooth probability distribution η.  So I have also removed the following:
--173.75.156.204 (talk) 11:55, 22 October 2009 (UTC)
thread_title:  Non-convergence? 

message: As the lead section rightly says, the value at H(0) is mostly irrelevant, and its value is usually just chosen for concreteness (if at all).  I updated the H(0) section to reflect this.  I toned down support for H(0)=1/2 (which seemed almost to be promotion of some author's favourite choice, and had a couple of meaningless phrases in it) and just objectively presented the reasons why each choice might be useful.  
I took out the bit about using a subscript to denote the value at zero, which the article said "may be used".  Well of course it may be used; a number painted in red on your forehead "may be used".  Unless it's also true that it actually is used (outside of Wikipedia) its mention here is pointless.  What's worse this notation is sometimes used but means something completely different: it means H(x) translated by the subscript. Quietbritishjim (talk) 01:30, 11 February 2010 (UTC)
thread_title:  H(0) (again) 

message: From Heaviside_step_function#Zero_argument:
: [...] it does not even make sense to talk of a value at zero, since such objects are only defined almost everywhere.
Should this not be "almost nowhere" (complement of almost everywhere)?
If my point is valid, could we use the term "almost nowhere" linking this to almost everywhere?
Hulten (talk) 13:42, 14 November 2016 (UTC)
thread_title:  Almost nowhere 

message: > The simplest definition of the Heaviside function is as the derivative of the ramp function
There is no way that the derivative of anything is simpler than an explicit piecewise-constant definition Quietbritishjim (talk) 11:38, 21 December 2017 (UTC)
thread_title:  Confusing lead section 

message: What is the use of the fact: "This function is the cumulative summation of the Kronecker delta"? It seems to me to be without any sensible meaning. Madyno (talk) 08:46, 3 April 2019 (UTC)
thread_title: Kronecker

message: I’ve made an equation for x≠0: \frac{x+|x|}{2x} Meckersnapper (talk) 18:09, 14 September 2020 (UTC)
thread_title: Equation

message: Define unit step function 2409:4042:D98:A190:0:0:8308:650E (talk) 20:19, 6 December 2021 (UTC)
thread_title:  M3