This is what happens when a teacher actually knows what he is teaching. Amazing.

Whenever I learn something new from the book, I always polish it with your lectures

Watching lectures with this guy is like watching a Netflix series, you’re eager to see what’s going on in the next episode

These lecture series are truly precious. MIT and all the professors involved in this monumental effort are to be commended for their superb contribution to the advancement and propagation of human knowledge. The same also goes for the anonymous financial contributors whose generous donations have made it possible for these lectures to be made available to the general public, free of charge. 👏🏻👏🏻👏🏻

the moment he talks about the derivative of step function is zero except the one point everything happens and my mind was like 🤯🤯🤯 all the questions in my mind were explained. OMG, thank you so much professor

His course on Linear algebra is fantastic and has helped me a lot. I wish to thank him personally one day. Thank You Prof. Strang

Thank you Professor Gilbert! May you have students as enthusiastic as your good explanations!

Happy retirement Professor Strang. Your commentaries are very clear.

Coolest chalkboard I've ever seen. MIT does things right. No doing an equation and going across the entire room until it's too long to even focus

I can't thank you enough for sharing. i love the intuitive approach.

The way he explain is amazing .

Such a pleasure to watch the explanation of the derivative of the Step function and your explanation of the Sifting Property of the Delta Function. It's very valuable when doing Laplace and Fourier Transforms.

11:41 The intuition behind that equations is brilliant ❤️❤️❤️

what an excellent way of teaching such complex terms. I just wanna give him a hug

7:26 this is how long it took for me to understand something my lecturer tried and failed to explain over the course of two hours. Wtf. I wish they could clone Gilbert Strang and have him teach in every institution.

Superb explanation of the Heavyside and Delta functions. No textbook I've read mentions the Heavyside function's relation to the Delta function. The common "explanation" is to state that the integral of the Delta function is defined as one!!!

Awesome video...1 year on KZfaq and no dislikes.exceptional.

I can't believe that in just 1 minute i understood step function! Best prof ever!! 🌹🌹🌹

The Dirac distribution is the Fourier transform of unity and a special case of convolution, where A*f=g, g(x)=d(x-y). f(y)dy , if we imagine the gravitational interaction as a function of g(x) and the electromagnetic interaction as a function of f(y), then these forces (i.e. the lines of force) only interact when x is equal to y ( the Dirac impulse).

-god I love this guy. I feel like he was a better teacher then all my real life teachers haha

5:47 "If we take derivatives, we get crazyness". I feel ya 100% bruh :D

Professor Gilbert strang painstakingly explains the problem with sincerity thank you sir.

Such a gifted teacher, unbounded mind.

The Heaviside Step Function and the Dirac Delta Function are both extreme limits of their smoother, slightly more well-behaved counterparts, the Error Function and the Gaussian. The Gaussian is a smooth bell curve, exp(-(x squared)), and has finite area under it (it integrates to the square root of pi, famously). The Error Function is the name given to the function whose derivative is the Gaussian, though it has no real formulaic representation in x outside of its series expansion. The Error Function looks like a smooth version of the step function, one with a somewhat rounded off, curved step. In the limit as the full width at half maximum of the Gaussian goes infinitely narrow, it converges to the Dirac Delta Function, and in the limit of infinitely-square step shape, the Error Function converges to the Heaviside Step Function. Since the Gaussian is the derivative of the Error Function, (which can be shown by looking at the series expansions of both) it stands to reason that the Dirac Delta Function should be the derivative of the Heaviside Step Function.

The differential equation y'(t) = ay(t) + delta(t - T) can only hold when t != T since y'(t) is undefined at T. I guess delta functions are a shorthand for that idea? Theorem: pick any x in [a, b]. The integral of f over [a, b] is independent of f(x). Proof: consider any partition P of [a, b]. Now add x - 1/n and x + 1/n to that partition to get Q. Since Q is a finer partition than P (formally P is a subset of Q), then undersum-oversum gap of Q is smaller than that of P. But the oversum and undersum of Q equal those of P on every subinterval except in a small neighborhood around x. The contribution of that neighborhood can be made as small as we like by picking n large enough. Let g(y) = f(y) whenever y != x and arbitrary when y = x. Then the difference between the integral of g and the integral of f can be made smaller than any epsilon = 1/n, implying that those two integrals are equal. [The existence of such a g is exactly what I mean by "[...] independent of f(x)".] But then if the delta function is 0 everywhere except when x = 0 it must have the same integral as the zero function (g(x) = 0 everywhere), which is zero. Hence the delta "function" cannot be a function. But integrals and derivatives are only defined for functions. So what goes on here is that we're adding some abstract symbol whose behavior is given by its definition [similar to the way a sequence is not a number, but arithmetic on its limit functions like arithmetic on the sequence so it's kinda'-sorta'-like a number]. But we're not told the meaning of all expressions containing the delta. We are not shown the rules of its algebra, and they are not justified. Note that in the final example, y(t) is discontinuous at T (limit 0 from below and 1 from above) and thus not differentiable at T, i.e. y'(t) is not defined when t = T. So the only sensible meaning I can make of the delta function is my initial statement: it's some abstract token which we use to pretend that y is differentiable everywhere. In that way I guess it's like the infinity symbol: if two series diverge to positive infinity their sum also diverges to positive infinity, and in that sense (+oo) + (+oo) = (+oo). But note that this doesn't extend neatly to differences: the harmonic series minus itself converges to 0, but the harmonic series minus twice itself goes to (-oo). So is (+oo) - (+oo) equal to 0 or (-oo)? The question has no answer. To understand algebra with the delta "function" and exactly what is permitted, we would need theorems characterizing it.

You are a great person in teaching

The best explanation of Delta function ever

Excellent presentation of the topics. Thanks DrRahul Rohtak India

OMG, the integral sign that he draw, it's perfect!!!

Ahhh this was a great introductory lecture but I really need to get behind the in-depth math of delta-functions without the whole thing diverging into a graduation paper about distributions...

Thank you, Professor Strang!

"Dr. Strange" is pure fiction but "Dr. Strang" is REAL; here he is. He's a real-life SUPER HERO to me! I have his book on Linear Algebra and have been following him now for decades. Educators are the actual "super heroes", but never get the recognition they deserve. South Korea treats (pays) their educators like "Rock Stars", so should we.

clearly explained, brilliant, what a professor

best prof of all time

Love u mr Strang

only a minute in and he's great

This is one of those math things that make my brain happy

thank you very much for your lecture, now I am writing for ethnomathematics on onion farming using step function. God bless

Thank you Dr. Strang.

Sir H is not differenciable at t=0, then why are we considering that point?, if H was a step function which jumped to say 2 then the integral would have equal to 2?

I had the absolute worst professor for this class and didn’t learn anything from him. I learned differential equations via KZfaq University.

a very heart rending lecture.Inadequate explanation and unsatisfactory.

this is awsome!!!! thanks Proffesor!!

I hope I'm like this when I get old.

Thank you so much

Amazing teacher

for the last differential equation. I can`t get why when t is smaller than T y=0. if I substitute y=exp(a(t-T)) to the equation I still get equality when t < T. and 0 only when t=T.

11:01 me when I try to explain anything

Quick question: so in part 9:09 since delta function is 0 everywhere but in x=0, could we then put integral_{-a}^{a} diracdelta(t)f(t)dt=f(0) for all a real number? Or does the area of integration need to be from -infty to +infty?

This is really great.

Great video

6:41 Lets us not consider integral of dirac delta from -inf to inf as heaviside function. here you will see the uncertainty. 1. dirac delta is not 1 at t=0( its 1 in case of discrete time or kronecker delta, which is a discrete analog of dirac delta) 2.I can not understand why d/dt(H(t))|t=t0 = dirac delta(t0), how? when we donot know the value of dirac delta at t0. 3. integral -inf to inf dirac delta is 1 how? dirac delta is defined as a large value at strictly zero. how do you integrate such a quantity. If i consider splitting the integral from lower limit to 0 and 0 to upper limit, the answer of total integral is 0. I really donot know how to do it, please explain.

I am looking for the elementary properties of del function from which every other properties can be deduced. The visualisation of del function does not make sense technically.

Thank you, professor!

Great explanation!

I've never known this relation between step function and delta function since my college time, to be honest.

Imagine him going to the bakery and he lectures you about "Imagine I deposit one dollar..." and then continues with the Heavyside function and Dirac Delta...

this is amazing!!

Your description that the integral of Delta function equals to H(t) is not correct. H'(t)=Delta function(t) = 0 at except x=0, then the integral of it is Zero by Lebesgue integral theory and does not equal to H(t). Do you know that the derivative of Cantor function takes Zero at almost everywhere, and that the integral of it is also Zero ? But Cantor function is an increasing continuous function and not Zero. How do you consider the relation between H(t) and Delta function. The distribution theory is needed.

Thanks alot! God bless you!

But the derivative does not exist at t=0!!! Why would you say it equals infinity?

Thank you so much sir

Awesome Video like others

thank youuuu.

8:42 is what i wanted... I was sleepy all the moment, and then I was turned ON at this moment........... WOAAAHHHHHHHHH

Excellent!

I watched this simply because it was free. I have no idea what he was saying.

Great video! 🙏

6:11 let's just appreciate the perfect integration symbol

For the example starting on 11:21, why can't, up to t = T, y(t) = [e^(at) - 1]? That also satisfies the initial condition, does it not? Isn't 0, what Strang says is the value for the y up to t = T, just the case where a=0 for the general solution I gave? If so, wouldn't it be better to be consistent, since he doesn't explicitly specify any value of a in the second condition of the solution, and wishes to be general with respect to the variable a?

Just out of curiosity, if the step function is defined as 1 for t>=0 why do we say the slope at t=zero is infinity? The slope should be 0 at t=0. Actually, if the function is defined for all values of "t" than there should be no value of "t" where it's derivative is anything other than 0. Does my reasoning make any sense?

I have a question for the community/teacher that I can't understand. At time @8:57, he has the integral of [the Dirac Delta Function {d(t)} multiplied by some function {f(t)}]; so the integral [d(t) * f(t) * dt], which he concludes is just f(0) because the integral of d(t) is just H(t) which when evaluated from -inf to +inf, which equals 1. And I get 1 * anything = anything, so only at input 0 can there ever be an output. However, what I'm not getting is this, shouldn't the answer, instead of f(0), actually be F(0), where F(x) = integral [f(t) * dt]? If not, I don't see why the function f(t) doesn't get integrated.

Is there anyone on the internet that can explain how to evaluate a delta function as an indefinite integral? I need it for my differential equations class but I can’t find any videos.

big fan

At minute 13:00 i think y(t=0)=0 not y(t=T)

Anyone else watching this during finals week cause their prof uses strang,a textbook

Anyone thinks that the chalk marks on the right hand side of the board @8:23 looks like a guy in agony lol

Se parece al vaquero de la pelicula toy

super!!!!!

control systems.... i almost failed this course

100 % interest rate lol

many a great lecturer has failed to adequately illuminate the delta function....but not the illustrious Mr. Strang.

wow!!

Click!

A blackboard in 2015? Retro.

Sir ..plz send solutions for Ur linear algebra book by Gilbert strong

Clearly another lecture for engineers🙄.

bizarre concept

I'm pretty sure this is not true.

no entendí del todo gnte xd

Great explanation!