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. 👏🏻👏🏻👏🏻

@NazriB

2 жыл бұрын

Lies again? Hello DDF

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.

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

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

The way he explain is amazing .

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

Happy retirement Professor Strang. Your commentaries are very clear.

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

Such a gifted teacher, unbounded mind.

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

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

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

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.

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

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

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

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

@XxToXicVaGxX

5 жыл бұрын

He has 7 on this video. I fully believe that it is your fault.......... wow

The best explanation of Delta function ever

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!!!

clearly explained, brilliant, what a professor

Thank you, Professor Strang!

Thank you Dr. Strang.

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

@XxToXicVaGxX

5 жыл бұрын

well it is MIT. The best for the best man

this is awsome!!!! thanks Proffesor!!

Great explanation!

You are a great person in teaching

Thank you, professor!

Amazing teacher

Love u mr Strang

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).

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

Great video! 🙏

this is amazing!!

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.

This is really great.

Great video

Thanks alot! God bless you!

Thank you so much

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...

Excellent!

Excellent presentation of the topics. Thanks DrRahul Rohtak India

best prof of all time

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

Thank you so much sir

only a minute in and he's great

"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.

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?

thank youuuu.

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

Awesome Video like others

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.

@fahrenheit2101

6 ай бұрын

and they exist, but are just wayy more complicated. it's all part of distribution theory

@jonaskoelker

6 ай бұрын

Actually my theorem is bogus, because if g(x) is arbitrary, it is not guaranteed that the oversum-undersum delta for g can be made to shrink to 0, specifically around x. For example, take a constant function f(x) = 0 and let g(x) = f(x) + 1 exactly when x is rational. By repeated application of my "theorem" g and f integrate equally, but we cannot even integrate g.

big fan

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

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

super!!!!!

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.

@materiasacra

7 жыл бұрын

Old question, but cannot resist answering ;-) Yes, you do get equality upon substitution of the exponential into the differential equation, but you violate the initial condition y(0)=0. (Strang is assuming that the time of the 'kick' t=T actually occurs after t=0, so T>0.) For t

@serhiiaif3959

7 жыл бұрын

thank you

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.

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.

a very heart rending lecture.Inadequate explanation and unsatisfactory.

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?

@ThePharphis

6 жыл бұрын

I don't see how that satisfies the range properly. It would only work precisely at t = 0 (the initial condition), but it would be wrong for every other value until t = T, unless i'm completely misunderstanding.

@alirezalanjani7055

5 жыл бұрын

If you consider y(t)=[e^(at)-1] and substitute in the equation for any amount of 't', it doesn't satisfy the equation, it will be a=delta(t).

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...

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.

@lokmanehamdani945

2 жыл бұрын

I hope this will help u kzread.info/dash/bejne/k4SipLyAYtzFhbA.html

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.

@GauravSingh-ob5ok

6 жыл бұрын

That would have been the case if Int(f*g)=int(f)*int(g) but this is not the case. One way to look this problem is numerical integration so i ll sum f(i)*g(i) for i=-inf to +inf .now g is special , its derac delta so its value is zero except at i=0. Hence i can replace f(i) by f(0). Now take f(0) common so sum of f(i)*g(i)=f(0)*{g(1)+g(2)+...} =f(0)*1=f0. Hope this helps.

@sabarikrishnam1485

5 жыл бұрын

f(t)*d(t)=0 at all points except at 0. So it only needs to be integrated at 0, but f(t)=f(0)(ie constant) at t=0, and can be taken out of integral giving f(0)*integral[d(t)dt].

@chaitanyakandwal7827

5 ай бұрын

thank you so much for that intuitive explanation!! @@sabarikrishnam1485

wow!!

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.

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

11:01 me when I try to explain anything

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

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?

@axelnils

9 ай бұрын

If there’s no slope, how did the function go from 0 to 1?

@MrPabloguida

8 ай бұрын

@@axelnils Well, interesting question, but isn't this why it is called the jump function?

Click!

Se parece al vaquero de la pelicula toy

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

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

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

100 % interest rate lol

@mbaye501

5 жыл бұрын

Remember when Finance PHDs were saying :" negative interest rates lol" They are not losing so much now

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

A blackboard in 2015? Retro.

@seansethi196

5 жыл бұрын

You say retro, I say timeless. Consider your computer and phone will be totally obsolete in about 20 years.

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

@mitocw

4 жыл бұрын

A possible alternative is linear algebra courses that have problem sets with solutions. You can filter for them on this list: ocw.mit.edu/courses/find-by-topic/#cat=mathematics&subcat=linearalgebra. We hope this helps!

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

control systems.... i almost failed this course

bizarre concept

@MicroageHD

4 жыл бұрын

It's awesome!

Clearly another lecture for engineers🙄.

@Demoneric25

3 жыл бұрын

Yes, and I´m grateful for it

I'm pretty sure this is not true.

no entendí del todo gnte xd

@physicalgraffitti

2 ай бұрын

pero lo voy a estudiar bn, gracias por su atención

Great explanation!