What is the derivative of this function

>↑
What did he mean by this?

just integrate my shit up f.a.m.

I hope you're trolling user

>"""""""function"""""""
rofl its a function you moron.

en.wikipedia.org/wiki/Dirac_delta_function

[math] /delta' [/math] is the linear map on the space of differentiable functions given by
[math] \delta'(f)=-f'(0)[/math]

>Dirac Delta "function"

that's not a function

>From a purely mathematical viewpoint, the Dirac delta is not strictly a function, because any extended-real function that is equal to zero everywhere but a single point must have total integral zero

But what's wrong with that? f(x)=0 is a function with an integral of 0

The Dirac distribution doesn't have a null integral ya git

>Integral of zero means it's not a function

Shut the fuck up

3

en.m.wikipedia.org/wiki/Unit_doublet

It's called the unit doublet, and like the Dirac delta, it's not a proper function but a distribution. Intuitively it's like the point-dipole in physics.

>he doesn't know the integral is 1
wew

>The Dirac distribution doesn't have a null integral ya git
? the picture of the OP shows a function thats 1 at 0 and 0 everywhere else.
its not the dirac distribution

>Unit_doublet

The delta function is commonly represented as an arrow, where the arrow's height is equal to the value of the integral (in this case 1).

That would be the Kronecker Delta as used in signal processing :^)

>? the picture of the OP shows a function thats 1 at 0 and 0 everywhere else.
How is freshman year?

I was in a workshop where we were learning about the Dirac Delta and one of the postgrad assistants kept insisting to me that its value was 1 at the origin.

It's 1/p where p is the width of the function, the area is 1

>How is freshman year?
do you think the image shows the delta function?

>What is the derivative of this function

The distribution that makes the product rule under the integral true.

This.

Goodman, Introduction to Fourier optics 3rd edition bitch

bitch

why does nobody here /ref/

It is the common representation for it, yes.

>giving reference for basic math
wew

bitch i could reference everything I post. information is useless unless you can verify it.

>referring a fucking physics book for distribution theory

ZIPPY ZAP THAT AINT CRAP

>It is the common representation for it, yes.
no? show me 3 textbooks using that representation!

Not that guy but how else would you represent it

any other representation would have width which is unacceptable.

why would I do that?

are you one of those "muh rect function limit" fags

>how else would you represent it
suggestive: as a line at 0 heading towards infinity.
for the purists: not at all.

>for the purists: not at all.
but purists understand that all functional representations are just as conventional you dip

>"From a purely mathematical viewpoint, the Dirac delta is not strictly a function,"

The limit at 0 is clearly 0, but the value obviously is not. Even if it were a function, or if we stretched what we meant by "function," it wouldn't be continunous, and thus wouldn't be differentiable. Asking for this """"function""""'s derivative makes absolutely [math]0[/math] sense.

THANK YOU

Now how about the second derivative? Just a function [math]u_2(t)[/math] s.t. [math](x * u_2)(t) = \dfrac{\text{d}^2x}{\text{d}t^2}[/math]? How would such a function behave around zero? Is there a general way of telling how all derivatives of the Dirac delta function behave around zero?