Proof of the Convolution Theorem

Proof of the Convolution Theorem,
The Laplace Transform of a convolution is the product of the Laplace Transforms, changing order of the double integral, proving the convolution theorem,
www.blackpenredpen.com

[@munashekaks7585]

this guy is a genius.we spent more than two weeks in lectures studying laplaces but i couldnt understand anything...the moment i watched videos from this guy hahahaha i bet im now a genius..lol...thanks lots

[@blackpenredpen]

Munashe Kaks great thanks

[@ingoclever1722]

So true !

[@univuniveral9713]

I know, right? The same genius brain brain that would have realized enlightenment if he had chosen a monastic path is being used to realise mathematical equations. He is also a goo teacher. I just subscribed, maybe he has more courses coming up.

[@aashsyed1277]

@@univuniveral9713 Brain brain u wrote 2 times

[@gerardogutierrez4911]

please do more differential equations. I took that class and learned a lot in a small amount of time, and now its all gone. we spent like two weeks on all of laplace, convolution, dirac's delta function, and a whole lot of stuff that required much more than two weeks to get into.

[@blackpenredpen]

Jerry Gutierrez will do!

[@gerardogutierrez4911]

Also, I finished watching your video, and I love the explanation. I was always catching up in my differential eq's class and never fully understood why the teacher did what she did, and after watching this video, I feel like the laplace of the convolution, actually makes sense, and like you, I can appreciate how something so nasty like laplace and convolution combine to make something succinct, and pretty neat to look at and think about.

[@nomvuseleloh]

I found it, Convolution, Thank you... you are a blessing

[@blackpenredpen]

this is the page with all the playlists.

[@blackpenredpen]

blackpenredpen.com/math/DiffEq.html

[@solomontadesse3443]

thanks mister you helps alot

[@satyamlal4461]

You are absolutely amazing!

[@simdriver6797]

Wait... what happened to the e^(-st) (right after the "note")?

[@centreville4048]

Amazing....

[@UETLHRCED15]

1-u(tau-t)=u(t-tau) are both equal???

[@ahmadhalim260]

Thank you

[@brajesh457]

thanks

[@owenpino6893]

Is this part of calculus??

[@aneeshsrinivas9088]

convolution is the worst notation ever,change my mind. as someone who is almost always typing math, * meaning multiplication should be standard,its been this way since like prealgebra, so this is like changing the meaning of + to be subtraction to me. this notation is the biggest middle finger for anyone who types math all the time and honestly i came up with a better notation for that, f(t)*_{(a,b)}g(t)=∫_a^b f(u)g(t-u)du.

[@andirijal9033]

Are you M.Sc or Ph.D ?

[@sukursukur3617]

I expected you gave up

[@Eknoma]

1:25 did my man just say Calculus 3? How the fuck is the american system so slow? Everything he has ever said is "Calculus 2" is taught in calc 1 or before here

[@HeraldoS2]

Man, what a huge video... Great work! But I found it a lot simpler going "backwards", using the subtitution theorem for multivariable integration and changing the integrating region from (R+)^2 to {(x,y) | 0

[@novanecros9145]

I have been trying to prove this theorem myself for about a month now because I could honestly not find proof anywhere and just seeing this video in my recommended made me smile so widely I already smashed that subscribe button. I'll be watching your videos from now on my dude. :D

[@oybekkhakimjanov5944]

Thank you! You've explained this so clearly.

[@JordHaj]

2:55 the function graph looks like end brace you draw (using Laplace transform you should put input in braces) P.s. yea my English is very bad

[@randomguy2836-g3i]

you can actually do u=t-v, du=dt after changing variables, and the integral will work very nicely without any unit step functions.

[@AakashSingh-ew2kz]

Such an elegant proof of Convolution theorem using the unit step function.✨️

[@edwardalexis6801]

Are you going to explain the deconvolution too?

[@UETLHRCED15]

when u introduce at start u(tau-t) t acts as constant (t=a ) at start but at end in u(t-tau) you choose tau as constant (tau=a) and t as variable ,please explain this ambiguity ???

[@calvinjackson8110]

I cannot follow what he is doing. Will have to look at this video more than once. HE knows what he is doing, but I do not.

[@Alkis05]

11:03 Not all those functions are continuos in the interval. Namely the u(v-t) function is not. Which makes the whole thing not continuous at v=t. But I get it what you mean. The u function is not continuous, but can be integrated in that period

[@angelmendez-rivera351]

Alkis05 It being continuous is not relevant for the integration. You can integrate step functions over any interval.

[@Alkis05]

@@angelmendez-rivera351 That is exactly what said... I was correcting him in regard to him saying every function was continuous.

[@angelmendez-rivera351]

Alkis05 I’m fairly certain he meant to say all of the functions are Riemann integrable.

[@morganrogers196]

Around 15.00, I'm confused because there were two e functions... and they merged into one function? The Laplace definition gives e^-vs(F(s)), so when you replace it back into the integral, wouldn't it give e^-s(t+v)?? Great videos! I love them and they've been so helpful for me!

[@chasecupp5550]

Morgan Rogers I was confused at first too and was thinking the same thing as you. That e^-st is part of the definition of the Laplace transform and our f(t) from the definition is f(t-v)u(t-v) in this case. We didn’t actually do anything with the integral and just replaced it as a whole with the known result of the shift theorem, because if you wrote out the formal definition for the Laplace Transform, substituting f(t-v)u(t-v) for f(t), it would be exactly that black integral.

[@ehmchris476]

I don't know why, but this is still easier than algebra

[@MuhammadWaleed-f6n]

Wow. Genius. Keep producing content like this

[@smittyflufferson1299]

Respect

[@thanveerahamed506]

Thank you so much bro!

[@yaleng4597]

我合理地懷疑1*1*1*1=t^3/3

[@blackpenredpen]

Just work it out. U don’t need to doubt 😆

[@mariuszgawron4997]

The Laplace Transform is itself an convolution. Moreover, convolution theorem works backward, it means that transform of convolution gives product of transforms!

[@yarooborkowski5999]

Could You derive invers Laplace transform formula simply from Laplace transform? F(s)=integral(e^-st f(t))dt -> f(t)=1/2πi integral(e^st F(s))ds

[@patipateeke]

Thank you for the excellent videos which taught me a great deal!! Your work is amazing and helps many people around the world to understand math better! I have a question about the converse of this theorem: does a multiplication in the t-domain correspond to a convolution in the s-domain? I think that for a Fourier transform, also the converse is true, but I'm not so sure whether this is true for the Laplace transform...

[@tanjimbinfaruk9145]

Brilliant proof!

[@blackpenredpen]

Yay!!

[@odongoabraham9464]

What a beauty.. mathematics genius

[@TawpeeBoyshuck]

I'm a little confused here at 6:33. Isn't u(3-v) the mirror over the vertical axis of u(v-3)?

[@smrtfasizmu6161]

Omg blackpenredpen has a video about this. This guy covered so many topics

[@AryanKumar-uu2jk]

2 plate momo laga do sir ji

[@david-yt4oo]

10:56 nice, new trick

[@ny6u]

convoluton in the time domain is equal to multiplication in the frequency domain.

[@SuHAibLOL]

Great video once again man

[@sujoy7471]

❤thanks sir

[@alvisarker9016]

Loved it ❤

[@brooax935]

probably the best channel to understand laplace thank u sir.

[@ferlywahyu342]

But how to solve laplace (f'(x)/g(x))

[@univuniveral9713]

definition of convolution is integrating from minus to plus inf.

[@ejercicioparatodos1886]

Nice

[@alvaro5704]

Finally i understand it! Mindblowing video dude

[@abatipatrick2417]

May God bless you for all you do, bro

[@K-Von]

hello, do you have any videos of Fourier series?

[@birdboat5647]

just make it phi bro

[@david-yt4oo]

9:00 those open intervals after you multiply by the unit step function (USF) throw me off because I think that the original area had that "slice" but when you multiply by the USF that "slice" is no lomger accounted for. maybe the way I said it was a bit confusing, but could you tell me why? (also, I think this is an ingenious way to approach the problem dude)

[@mattetor6726]

Thank you!

[@HomerBeeSimpson]

Is -(u(v-t) - 1) = u(t-v)?

[@warrengibson7898]

Yes

[@МихаилУжов-е2й]

The u is also known as " Heaviside step function" H(x)=d/dx(max{x,0})

[@mimithehotdog7836]

What if there are constants a,b, c? L{ c f(t)*g(t) }? L{ a f(t)*b g(t) }?

[@krejman]

make new functions f1(t)=c f(t)

[@mimithehotdog7836]

@@krejman , thanks

[@andykwan6215]

超美

[@kyang1305]

This is pure gold! This guy is a genius.

[@rollno4299]

Easy to understand..... 🌹🌹🌹🌹🌹🌹🌹🌹

[@gauravanand9478]

You are genius........ Really.....

[@balajip5030]

Thanks

[@bhavyaputta5002]

Simply awesome😃

[@mengzhenc3704]

A-W-E-S-O-M-E!