Heaviside Step Function Puns

(Personal note: I only play on Last Azlanti; with about a thousand hours on PFKM so far. This is not meant as a complaint. Just a musing.)

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Early Game: Peril! You misstep once, or merely hope that RNG will favor you—it won't. You're dead. Back to the drawing board.

Late Game: You are a god! You buff and you are golden, nothing can touch you. Unless you take on a threat far above your punch that you cannot eliminate with current party—in which case, refer to "Early Game".

Problem is that the transition from Early Game to Late Game happens like a speed bump. (Hence why it looks like Heaviside Step Function [#1].)

Specifically, at around lvl 7 or 8, your buffs transition you into god.

I remember the time when I was jumped by 10 ferocious manticores or a full flight of wyverns and thought "oh, shit!" — only to be surprised that they were all dead by turn 3 with no scratch to my lvl 7 or 8 party.

Issue with this:

1. You must buff your zillion buffs each new map—which takes ages with no in-game way to speed it up out of combat.
2. Short of encounter design gimmicks, of which there are only few, the difference between TPK and a no-scratch victory is hair thin. (Hence again the Heaviside Step Function.)

Basically, when your power scales cubically with your lvl [#2] threats against you will be either pitiful or hopelessly difficult.

(I know, I know, this is inherent to Pathfinder and not the fault of the developers. I wonder if D&D 5th edition (BG3?) will be any better on this issue...)

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[#1] c.f. Heaviside Step Function looks like... a stephttps://en.wikipedia.org/wiki/Heaviside_step_function

[#2] Think about it: your total damage & mitigation increases faster than quadratically with your level. Cubically is a reasonable estimate.

I am trying to find the Laplace transform of 4H(t+2)e^t, which I know to be 4e^(2s-2)/(s-1); however, using the In: LaplaceTransform[4HeavsideTheta[t+2}e^t,t,s], I am getting 4/(s-1). Does anyone know why this is happening? Thanks!

Hi guys, I have this unit step function here and I have to plot the graph:
f(t) = t + 3(t-2)u(t-2) + (3-t)u(t-3) + 3(5-t)u(t-5)

So my first intuition is to expand the entire function and group them into their different time domains as such:
f(t) = t + 3tu(t-2) - 6u(t-2) + 3u(t-3) - tu(t-3) + 15u(t-5) - 3tu(t-5)

I'm not sure if expanding is correct and I'm not sure how to continue from here.

I'm a neuroscientist (slightly on the computational side; I use generalized linear models to model poisson spike data) and enjoy reading/learning about various machine learning topics. One thing that I've been unable to find an answer for is why, besides the problematic dirac derivative, the heaviside (step) function is almost never used these days.

To me, the heaviside function has a nice neural interpretation: an action potential of a neuron. It either happens or it doesn't. Other activation functions could theoretically be thought of as the probability of firing, or perhaps rate of firing, but that's a less satisfying answer than I presume is buried out there in the literature.

I understand current activation functions have nice derivatives that make the actual training of the model computational tractable. But is there any other reason to use things like ReLU and tanh that a heaviside (step) function itself cannot?

In other words, if you take a deep neural net and replace ReLU's and tanh's with the heaviside (step) function, would performance be similar (assuming you could efficiently train the heaviside (step) function version)?

I hypothesize that you may need to add more neurons or layers to perhaps make up for some of the complexity ReLU's and tanh's confer. However, is there anything these nice activation functions do that heaviside itself cannot?

P.S. My search for multi-layer perceptrons have born little fruit, considering they too use ReLU's and tanh's.

(resolved)

We are directed to find the definite integral (by sketching a graph and using geometry methods; we just started learning integrals) of [H(t-2) + H(t)], where H(t) = 0 if t<0 and 1 if t>=0, integrating from t=[-2,4].

So far I have sketched H(t) and concluded that the definite integral of it (from -2 to 4) is 4 because it's only the rectangle area under the the graph for t=[0,4].

Similarly, I found the definite integral of H(t-2) (from -2 to 4) to be 2 because by sketching H(t-2) I observe that this is just H(t) shifted 2 units right, so the rectangle of area in question is now just under the graph of H for only t=[2,4].

Thus, by properties of integrals, I know that the integral of [H(t-2)+H(t)] from t=-2 to t=4 is the sum of the integrals, ie 2+4 which is 6.

Where I am completely confused and lost is how to add these two functions and sketch the graph of the result. I don't understand how to add two piecewise functions that have the "step" in a different place, but the value outputs of the functions are the same constants (0 or 1, specifically). Anything I can picture geometrically does not give me an integral result of 6, but usually 8 or more.

Thanks for any help

I keep seeing different definitions of the Heaviside step function. Some say it's undefined in x=0, some say it's 0 in x=0, and I've even seen some say that it's 1/2 in x=0...

Could anyone explain? Thanks in advance:P

I have to prove that δ(2t) = 1/2 δ(t) but I’m having trouble with it.

I have to use the approximation of the unit step function u_Δ(t) to prove it. In this function, the step is not instantaneous but is a straight line with slope 1/Δ from t=0 to t=1. The derivative of this function is δ_Δ(t), which is constant at 1/Δ between t=0 and t=Δ.

So:

u_Δ(t) = { t/Δ for 0 < t < Δ

δ_Δ(t) = { 1/Δ for 0 < t < Δ

Then I would guess

u_Δ(2t) = { 2t/Δ for 0 < 2t < Δ

δ_Δ(2t) = { 2/Δ for 0 < 2t < Δ

Both δ_Δ(t) and δ_Δ(2t) in this case can be integrated and shown to be 1, but the one with 2t is higher and narrower. I don’t see how this should result in the thing I should prove.

This solution on Slader uses a different method, where the 2t is inserted in the range, but not in the value. This is probably the correct way but I don’t understand why.

I've got the following initial value problem and the Heaviside function is giving me some confusion. I know I need to take the Laplace transform of the problem to solve for x(t) but I'm stuck with the (2t_0-t) instead of (t-t_0).

t>0, t0>0, x(0)=0, x'(0)=0   

cx' + kx = f_p*𝛿(t-t_0) + f_h*Heaviside(2t_0-t) 

and the Heaviside is defined as

Heaviside(t-t_0) = 1 for t>t_0 and 0 for t<t_0

Can someone please explain me to me how you distribute that first line to get the second line? I have been staring at this thing for an hour.

https://imgur.com/a/VDFNwR3

also unrelated question, does anybody know the 'general' method to convert any piecewise function to a Heaviside function?

https://preview.redd.it/97di3l5s5hn71.png?width=1101&format=png&auto=webp&s=acb28a2e5ad862c00cd537d58557bc5b75e651ad

Hello,

The following initial value problem is to be solved using methods of Laplace transforms, however it's forcing term is a Heaviside Function (Unit Step FUnctions)

I know I will have to take the laplace of both sides of the given equation dx/dt.

Step 1 i thought was to re-write the given equation accounrding to t H(t)=h, 0<=t<=Lambda H(t)=h, 1<=t<=(1+lambda) 0 Otherwise

We know that H(t) is the heat generated from people machines etc and is always positive (increaing) [0,infinity)

Is my first step to find x(t) by using the initial conditions from 0<=t<=(7/4)hr in order to get a general solution to the DE. After substituting, I would solve it using laplace transforms

Since x(0)=T* then at 0<=t<=(7/4), x(0)=18

I was able to find that x(t)=18 ? Please verify.

If that is indeed the first step how do I incorporate the use of the unit step fuction/ window functions. When do i express in unit step functions or do it?

Also, should i be evaluating using laplace the DE as giving (without putting in initial conditions)

Please help and thank you so much in advanced!

https://preview.redd.it/l9gl2xmq5hn71.jpg?width=2248&format=pjpg&auto=webp&s=c6bb49be2a4d5ff67c8f7917dc8ca208fadb15cc

https://preview.redd.it/6g6c6wmq5hn71.jpg?width=2216&format=pjpg&auto=webp&s=16c808e84b32ca434c77fcad74204529e3bb4394

https://preview.redd.it/dg7v5ymq5hn71.jpg?width=2064&format=pjpg&auto=webp&s=5c8d6829c2b8b69afffed71d7d2157b60270e5ee

Hi friends,

I have a question. How do you integrate a Heaviside Theta function that is dependant on a logarithm. I mean, H(a·log(x))dx

Thanks in advance

Having trouble with the question and the answer I got is 1/2 . I am not sure if it's right. So asking for help.

If we consider the distribution f(x; y; z) described by f(x; y; z) =H(x) where H is the Heaviside step function. How do we compute (in the sense of distributions) gradf (in vector analysis formalism) and df (differential geometry formalism).

Hello Everyone.

Equivalent visosity:

"((mus)*(T<Tm)) +1000*(flc2hs(T-Tm, 20))+(mul)*(T>Tl)"

It turns to yellow text>> unexpected unit of input.

&gt;> (mul and mus are in [Pa.s]<<

&gt;>Tm is melting temp =1798[K]<<

When I use flc2hs(T==Tm): It works but I am not sure about it.

Can anyone please help me out ??

Thank you.

https://preview.redd.it/jyuqc9zbw7v41.png?width=439&format=png&auto=webp&s=f9376b0a805860088a202139326dffe40a7caafa

Hey guys, say I've got a heaviside step function and I need to graph it out, how can I convert the heaviside function into piecewise equations?

For example (I use H, others use step or U I believe) - f(t) = H(t−1) + 2(t−2)H(t−2)−5H(t−3)−4(t−5)H(t−5)

How would I go about graphing this out on a piecewise graph? I tried getting the equations by doing like, (2t-2) - (1) and (-5 - (2t-3)) and (-4t+20 - (-2t-2)) but it doesn't seem to really be working.

Thanks for any and all help!

Hello everyone. I want to apply the Heaviside function on the delta(phils) as shown in the figure. Can I apply the step function as a Heaviside function shown in the figure?

https://preview.redd.it/bodfkwq28xz41.png?width=804&format=png&auto=webp&s=472f5b93680e0a06f0d41c3395c5a28af440584f

I'm having trouble writing the coefficients in terms of (x-c).

Here are the two problems off of my test review: https://imgur.com/r8CuKGk

I believe I'm rewriting the functions in terms of u(x-c), u(x-2) in #18, and u(x-pi) in #19, correctly, but after that, I'm not sure how to continue the problems.

In #19, I also know that I'll have to use the Addition Formula to rewrite the sin and cos terms, but I also am not sure how to do that.

Find laplace transform of f(t)=u(t-a)*t^2

why do I get the wrong answer if i substitute t^2 with ((t-a)^2 +2ta -a^2), then transform for each "part". If it was +a^2 instead of -a^2 i would get the right solution. Use http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf as laplace table

Edit: So thanks to /u/Cerlancism for pointing out that you can actually hover over the number of upvotes the post has and it will tell you the % upvoted (on desktop). Still way more subtle and out of sight than it was before, idk if it works for mobile, and that is the sort of thing to be removed very easily with even less fanfare.

Edit: Ok so it's been several hours and it seems permanent (for me and a couple others who have commented). This post contains an observation which I editorialized by tying to GME (how it will make driving narratives easier). But this seems to be a new feature that will likely permanently replace the %upvoted stat (time will tell). If that happens, it will effect reddit as a whole. Furthermore, much of my observation of narratives is not isolated to GME but this is the community I am active on and which seems to be the most tuned in to this sort of corporate/PR interference with online discussion as GME has been the target of a "short and distort campaign" for months/years.

And if you are from all and think this doesn't happen on reddit, you probably weren't around to remember how well known gboob, who made a career out of explaining reddit to corporations, used to be. He isn't alone and the corporations don't pay reddit consultants for fun.

It must have worked for Youtube, so they are applying it here, though reddit has messed with obfuscating post karma in the past.

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This change plus that recent "REDDIT rEAlLY nEeds to gEt RiD of PoSt HisTorIES" posts are going to drive two massive changes that will make it way easier for shilling specific narratives or brands.

The fact that parade on wall street is banned from being linked sitewide, for supposedly "vote manipulating" while literal ads are sitting at 3k upvotes exactly almost permanently (which is obvious evidence of vote manipulation) just further my certainty that reddit is being primed for Wall street control (even more than they already are).

Edit: for those who don't see why this matters - upvote downvote percentage can be a good indicator of bot activity pumping or suppressing a topic.

Also, I'm on desktop and I use Brave with no reddit extensions and it says "x people here"

Others are reporting differently in the comments - this is just an FYI. Edit: I feel that this is likely due to a progressive rollout of a new feature, just like what happened on Youtube.

https://previe

I have some data that I suspect is based more on the Heaviside function than the Sigmoidal. However it is easier for me to fit Sigmoidal. If I fit a Sigmoidal curve to data with the function

y=a/(1+exp(-b(x+c))

Is (c) in the above equation equal to the step position of the corresponding Heaviside function? I tried looking at some research, but I feel like this proof is more classical...

Tried solving this but I'm not sure if I am right so far? Imgur

http://gyazo.com/63b7ed22424468ededfdbf93acbb8d7a.png?1367361430

hey there.

I took the laplace transform of both sides -- the only hard part is the right side, because its a step function.

it becomes the integral from 0 to 5 of e^(-st) * t dt, since after 5 it's just 0 forever.

integrating by parts i get te^(-st) +e^(-5s)/s -1/s

but this is wrong. where did I screw up?

I have a question about sine and cosine transform, and the question also includes a Heaviside function. I have tried to solve it but I think there is a problem with my solution. I suspect that it was caused by wrong integration of the Heaviside function.

Question: http://i.imgur.com/0R9kTtp.png My solution: http://i.imgur.com/gvCemN0.jpg

I would appreciate if anyone can help me on this question, thanks.

If we consider the distribution f(x; y; z) described by f(x; y; z) =H(x) where H is the Heaviside step function. How do we compute (in the sense of distributions) gradf (in vector analysis formalism) and df (differential geometry formalism).

If we consider the distribution f(x; y; z) described by f(x; y; z) =H(x) where H is the Heaviside step function. How do we compute (in the sense of distributions) gradf (in vector analysis formalism) and df (differential geometry formalism).

If we consider the distribution f(x; y; z) described by f(x; y; z) =H(x) where H is the Heaviside step function. How do we compute (in the sense of distributions) gradf (in vector analysis formalism) and df (differential geometry formalism).