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make a difference. I updated the discrete section to say that. As you said, you can fix this approximation by taking the limit from the opposite side. However this approximation doesn't seem to have anything to do with the discrete case, so I removed it from that section. I could have moved it to
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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
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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
751:":): {\displaystyle \begin{align} H(x)&=\lim_{ \varepsilon \to 0^+} -\frac{1}{2\pi i}\int_{-\infty}^\infty \frac{1}{\tau+i\varepsilon} e^{-i x \tau} d\tau \\ &=\lim_{ \varepsilon \to 0^+} \frac{1}{2\pi i}\int_{-\infty}^\infty \frac{1}{\tau-i\varepsilon} e^{i x \tau} d\tau. \end{align}}
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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
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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 :) ....
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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
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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
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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"....
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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:
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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.
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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.
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used (outside of
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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.
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All of this is weird. The
Heaviside work has nothing to do with Fourier. Prof. Howie, when head of the Cavendish, told me that physical reality was composed of sine waves.
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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. --
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the "analytic approximations" section but for all x except zero, you don't need the limit. Basically I think this formula is not useful, so I've removed it entirely.
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It's uninteresting, of restricted applicability, (strictly speaking) incorrect (the inverse trigonometric functions do not have unique definitions) and constitutes
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817:. The mathematisisation of physics has been very damaging. Fourier is about a train of identical waveforms, not about a single step. Ivor Catt 11 Aug 2016
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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
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619:{\displaystyle H(x)=\int _{-\infty }^{x}\delta (\xi )d\xi =\left\{{\begin{matrix}0&\left(x<0\right)\\1/2&\left(x=0\right)\\1&\left(x: -->
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There appears to be an error with this section. It looks like it assumes there is some service (MathML?) running on the user's machine at port 6011:
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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.
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u is more familiar to me, of course, but if a majority of articles use h I guess that's ok. I wonder if u is used to distinguish it from H(f) =
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where the second representation is easy to deduce from the first, given that the step function is real and thus is its own complex conjugate.
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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?
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Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to
Restbase.") from server "
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is the most common definition is signal processing. I, too, would support a consistent definition throughout
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No, it's not just you. On google scholar there are 2500 "papers" using "heavyside function" and 35000 using "heaviside function".--
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a couple of meaningless phrases in it) and just objectively presented the reasons why each choice might be useful.
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i can't find the fourier transform of the heaviside function anywhere...anyone willing to share their expertise :)
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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. --
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uses a Θ(). To me, it looks like the H() notation should be used to be more consistant with the rest of
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it does not even make sense to talk of a value at zero, since such objects are only defined almost everywhere.
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has measure zero in the delta distribution, but its measure under each smooth approximation family becomes
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There is no way that the derivative of anything is simpler than an explicit piecewise-constant definition
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has shown that it is possible to mimic the Unit Step
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The simplest definition of the Heaviside function is as the derivative of the ramp function
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At the following address you will find the Fourier transform of the Heaviside Step Function
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I was taught with a u. Probably another one of those engineer/mathematcian differences. -
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Is the Heaviside step functin defined for x = 0? Looking at the alternartive definition
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While these approximations converge pointwise towards the step function, the implied
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I support that different notation should be mentioned. I noticed σ notation, also. --
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I think that H is the most commonly used notation in mathematics and θ in physics.
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I've seen H and θ. The different notations should be mentioned and referenced. --
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for a smooth probability distribution η. So I have also removed the following:
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I removed the following assertion from the text, because I think it is false:
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Often an integral representation of the Heaviside step function is useful:
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Further, the ramp function has no defined derivative at zero.--
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do not converge to the Heaviside step function in the sense of
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I kind of doubt there's any agreement at all, even in Japan...
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Hmmm - I'm not saying that's wrong. I would say that 1/x is
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Reminds me: I once won five pints of beer on a bet that it
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The definition is self-contradictory: Who can correct it?
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1620:distributions
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1611:
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1240:
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1069:
1068:
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1066:
1065:
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1051:
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1023:
1000:
994:
986:
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937:
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889:
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877:
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869:
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864:
863:
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816:
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799:
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791:
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780:
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1176:"Heaviside"
1140:Mathematica
928:Boothinator
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1767:Equation
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