You are amazing! Most profs these days just read off slides or mumble to themselves as they write down things without explaining anything. As opposed to you, who speaks clearly and actually explains every single aspect =D Thanks!
saved my life. studying for my Diff Eq final tomorrow at UFlorida. big portion will be on the dirac delta, step, periodic functions' laplace transforms and convolation.
Because t means time in the context of the Dirac Delta function, you can evaluate the improper integral from 0 to (infinity), instead of (- infinity) to (infinity), since there is no such thing as "negative time"* *As I'm saying this in 2018 I really hope some future physicists don't come up with negative time.
Dear colleague, where have you been back in the days when I had to crunch through my quantum mechanics lessons? I do not know if you get much feedback, but let me assure you I enjoy your lessons tremendously, and I see a great usefulness in what you do. First anyone can run and re-run - which you cannot do with an actual professor. Second, your way of explaining things makes it clear what`s going on even if a rock listens. And to think that I always dreamed of a source like this ...
@runninriot15 I think it's not so much an issue with negative time as with the definition of Laplace transform. The dirac delta function could very well be applied to positions or (as you said) relative time frames, but it is mathematically meaningless to consider the Laplace transform of a function for t < 0, because the Laplace transform is defined only from t = 0 to t = infinity. Thank you Sal for this wonderful intuitive proof of a difficult concept. I owe you lots!
One torus-shaped mug , two hypercubes of sugar and a nice Fourier wave caused by the ripples as they splash in make any cup of coffee worth drinking in the wee-early/late/crazy hours of any day(:
... in the dark hours between three o clock in the morning and dawn, living on coffee and equations alone. If that sounded a bit enthusiastic, it is because I am. Very good vids. Thanks so much for posting. Have an especially nice day silk ;-))
I understand how you did it visually but I was just wondering if there was any way to solve that integral analytically? As in by just numbers, without drawing te functions?
Isn't the integral of the Dirac delta function equal to 1 when evaluated from -infinity to +infinity. So wouldn't the integral be roughly 1/2 when evaluated from 0 to infinity? +KhanAcademy
+ramseycdavid Integral of Dirac delta(x-c) dx is 1 as long as the integral range includes the point c - anywhere else is zero by definition so doesn't matter whether you include it or not!
He said that we draw an arrow over the Dirac Delta Function with HEIGHT = 1 to show that its area is 1. How does it makes sense? Isn't the base very small (or an epsilon), so how will multiplying by a height of 1 will give an area of 1?
Thankyou Sal from University of Surrey. I will donate to your website If I manage to get a decent grade in Maths this year. You saved me from failing last year!
This was so long-winded after about 2:00 that it almost got confused. Should have demonstrated the property of integrating Dirac delta with any function. The result with Laplace fn then follows easily.
i know it's not intuitively true but: what if your c is negative, then your integral from 0 to inf would not cover it. but i guess negative time doesn't make sense either, unless you're talking in different reference frames
i know your comment is from 11 years ago, but i'll answer anyway haha. your question is very great!! if c was negative (which it can be), then the integral from 0 to infinity would be 0. the area is 1 (scaled however) only when c is within the limits of integration
if c = -1 for delta(t-c) then the laplace transform isn't e^-sc but 0 right? or am i wrong here? I thought of that because delta(t-c) would be 0 on the entire positive side, so you basically calculate just the integral of 0dt
Take a look at the last video where he introduces the Dirac delta function, there he makes an intuitive proof of that area of 1 ( ua-cam.com/video/4qfdCwys2ew/v-deo.html )
May I ask a question? At the poin c, delta function is not one, but infiniti. Then function times delta function at c is equal to inviniti, not that function itself. Why?
Aren't you mixing value of a function with the area of the function. Value of Dirac Delta is infinity but it's area is 1. Why are you multiplying the area of dirac delta with the value of f(t)?
consider the f(t) is a well behaved function and it could be consider as small as the width of dirac impulse at any point. so the product of f(t) and dirac function in that region woulb be equal to the f(a).... a is the point
It's an incorrect assumption to assume that infinity actually exists. Because it doesn't. You'd be better illustrating these functions under constraints. They are still valid, but make no sense using impossible "infinity" constraints.
Ugly scar was a good jelly Siri set the one right around the break up of that to be honest honestly lotta fun yeah lol classic break your femur Monday is a Mary habanero I hear not happening hey guys so don't be alarmed but like half of my legs are about to go into surgery I've been screaming pain right now love you say hi to grandma and grandpa for me hi yeah how do you sir Peyton passing out love you oh no on Montagna your volcano yeah that's a cone volcano and then there's also shield volcanoes what's the last type of volcano come on people con shield. Cc. Cccc c
You are amazing! Most profs these days just read off slides or mumble to themselves as they write down things without explaining anything.
As opposed to you, who speaks clearly and actually explains every single aspect =D Thanks!
"profs these days" written 13 years ago, I'm dead.
"Don't judge me by the straightness of my axes" made my day !
fourth time getting the differantial class, first time understanding. thanks!
saved my life. studying for my Diff Eq final tomorrow at UFlorida. big portion will be on the dirac delta, step, periodic functions' laplace transforms and convolation.
I spent 2 hours trying to figure out how to find the Laplace Transform with the sifting property of the delta function. Thank you!
There's a tiny error in the last yellow expression where you wrote delta{ } instead of L{ }
Khan academy is great...👍👍👍
Just like wow😅... amazing teaching.. you were really ahead of time sir
Because t means time in the context of the Dirac Delta function, you can evaluate the improper integral from 0 to (infinity), instead of (- infinity) to (infinity), since there is no such thing as "negative time"*
*As I'm saying this in 2018 I really hope some future physicists don't come up with negative time.
my diff eq teacher gave us an equation with a delta function and refused to explain what it is. this saved my life
Best explanation of the sifting property of the delta function I've ever seen!
Just when i'd given up hope of understanding these things, I came across this video. You're a godsend. Thank you. :)
Wow, I love the way math can do anything. And I love the way you explain things. your use of colors really helps.
Its like sal learned to teach at khan academy. He nails it!
Best Demonstration till now....!
Hats Off!
Thank you so much for this wonderful video. Now, I know what Dirac delta function is.
It was great.
I have a non-mathematical question! Could you please let me know what is the brand and model of the optical pen you are using?
god bless your intuition Khan. Don't memorize, understand.
One of the most useful explanations I've seen. Thank you
very nice video, I like the way how you break down the integral and gave intuition.
excellent teaching !!! i don't think it can be clearer than that!
i can hear you in my head when im doing math questions lol
thank you for your videos
Khan is really Khan.
Many thanks khan.
This was very enlightening. Thank you!
thanks a lot, now Dirac delta function is not a mistery anymore !!
Dear colleague, where have you been back in the days when I had to crunch through my quantum mechanics lessons?
I do not know if you get much feedback, but let me assure you I enjoy your lessons tremendously, and I see a great usefulness in what you do. First anyone can run and re-run - which you cannot do with an actual professor. Second, your way of explaining things makes it clear what`s going on even if a rock listens.
And to think that I always dreamed of a source like this ...
So funny and very easy to understand, thanks!
@runninriot15 I think it's not so much an issue with negative time as with the definition of Laplace transform. The dirac delta function could very well be applied to positions or (as you said) relative time frames, but it is mathematically meaningless to consider the Laplace transform of a function for t < 0, because the Laplace transform is defined only from t = 0 to t = infinity.
Thank you Sal for this wonderful intuitive proof of a difficult concept. I owe you lots!
I am highly thankful to you Sir.
Excellent demonstration👍
Superb explanation!
At 11:27 he writes a delta symbol where it should be a Laplace symbol. Good video though.
This is the most beautiful thing I've seen today
Absolutely brilliant
tomorrow is my exam and this is so helpful
Thank you so much Sal...
thank you this is easy to understand
Thank you!
One torus-shaped mug , two hypercubes of sugar and a nice Fourier wave caused by the ripples as they splash in make any cup of coffee worth drinking in the wee-early/late/crazy hours of any day(:
kinda sad you didn't make these videos last year, might have saved my failing grade
... in the dark hours between three o clock in the morning and dawn, living on coffee and equations alone.
If that sounded a bit enthusiastic, it is because I am.
Very good vids. Thanks so much for posting.
Have an especially nice day
silk
;-))
here! have a comment after 8 years plus a like too.
wao another cracking good video. thanks alot
God bless you!
i hope this helps in my tomorrow's exam
VERY WELL STATED
I understand how you did it visually but I was just wondering if there was any way to solve that integral analytically? As in by just numbers, without drawing te functions?
thank you sir
thank you so much
"the plain, vanilla delta function"
do not judge me by the straightness of my axis
Isn't the integral of the Dirac delta function equal to 1 when evaluated from -infinity to +infinity. So wouldn't the integral be roughly 1/2 when evaluated from 0 to infinity? +KhanAcademy
ramseycdavid no, because it only has "area" under where t=c, and he did this for c>0, so if t
+ramseycdavid Integral of Dirac delta(x-c) dx is 1 as long as the integral range includes the point c - anywhere else is zero by definition so doesn't matter whether you include it or not!
I was thinking the same thing. =)
No, because c is positive it it could be negative then we can say this
awesome!
thanks
Master❤
I was doing some googling on this function and friend that it's integral is the function sign(x). Could someone explain this?
👍👍👍
He said that we draw an arrow over the Dirac Delta Function with HEIGHT = 1 to show that its area is 1. How does it makes sense? Isn't the base very small (or an epsilon), so how will multiplying by a height of 1 will give an area of 1?
"Dont judge me by the straightness of my axes" - Sal Khan
Thankyou Sal from University of Surrey.
I will donate to your website If I manage to get a decent grade in Maths this year. You saved me from failing last year!
This was so long-winded after about 2:00 that it almost got confused.
Should have demonstrated the property of integrating Dirac delta with any function. The result with Laplace fn then follows easily.
Chaa gia hai jaani Ly
i know it's not intuitively true but:
what if your c is negative, then your integral from 0 to inf would not cover it.
but i guess negative time doesn't make sense either, unless you're talking in different reference frames
i know your comment is from 11 years ago, but i'll answer anyway haha. your question is very great!! if c was negative (which it can be), then the integral from 0 to infinity would be 0. the area is 1 (scaled however) only when c is within the limits of integration
@Macranius By the way, this was really usefull for me
S is variable how it can be taken out from integration?
god creating me : 5:37
i found that much more than "reasonable useful" ^_^
I think all mathematicians do.
A mathematician is a device for turning coffee into theorems. -Paul Erdos
if c = -1 for delta(t-c) then the laplace transform isn't e^-sc but 0 right? or am i wrong here? I thought of that because delta(t-c) would be 0 on the entire positive side, so you basically calculate just the integral of 0dt
Very . good ! - Thank . you . so . much ! :-)
@shrikant96 I think he studied electrical engineering and something else at MIT
At 11.16 how is f(c) at c = 0 , i.e. f(0) = 1?
I know the feeling... (:
부디 한글 자막좀 ...ㅠㅠ
How did you get 1 in 1:22
Take a look at the last video where he introduces the Dirac delta function, there he makes an intuitive proof of that area of 1 ( ua-cam.com/video/4qfdCwys2ew/v-deo.html )
@khan : all i wanna know is.....where did you study?!??
it says detla: T before L, watch carefully the video again
I am sorry, I still do not see what is it used for. We can use f(c) instead of inte(f(x)*delta(t-c)), which is much simpler! I am confused!
You are missing a factor of 1/2 as the integral is from 0 - infinity
My bad you are correct provided c > 0 , at c = 0 I think you need a limit from both sides even if infinitesimally small and then missing the 1/2
Wrong again: value is 1 if the point 0 is included in the integration
Why integral from zero to infinite is one
Is it just me or it says at the top "Detla" instead of "Delta"
is it just me or does it say "sais" in your comment instead of "says" XD hehehehe
@@thesufferingengineer6003 You are right, thank you for correcting my 8 years ago mistake
@@Macranius 😁♥️
Laplace transform of delta(t2-3t+2)=????
Use polynomial laplace formula
it says delta, umadson??
he spelled "the rock" function wrong
the guy who disliked this video is a mathematician and doesn't like generalized functions.
"Don't judge me by the straightness of my axes"
fallen angel, lost your way,
I thought you meant the title of the video, ambiguous statement is ambiguous.
May I ask a question?
At the poin c, delta function is not one, but infiniti. Then function times delta function at c is equal to inviniti, not that function itself.
Why?
Aren't you mixing value of a function with the area of the function. Value of Dirac Delta is infinity but it's area is 1. Why are you multiplying the area of dirac delta with the value of f(t)?
consider the f(t) is a well behaved function and it could be consider as small as the width of dirac impulse at any point. so the product of f(t) and dirac function in that region woulb be equal to the f(a).... a is the point
Sal saving my ass again :)
*judging you based on the straightness of your axes* (jk thank u for the video)
I don't hear anything. Why is that? :/
It's an incorrect assumption to assume that infinity actually exists. Because it doesn't. You'd be better illustrating these functions under constraints. They are still valid, but make no sense using impossible "infinity" constraints.
He said "pseudo infinity". Anyway, the existence of "infinity" is more tangible than the existence of numbers themselves!
Ugly scar was a good jelly Siri set the one right around the break up of that to be honest honestly lotta fun yeah lol classic break your femur Monday is a Mary habanero I hear not happening hey guys so don't be alarmed but like half of my legs are about to go into surgery I've been screaming pain right now love you say hi to grandma and grandpa for me hi yeah how do you sir Peyton passing out love you oh no on Montagna your volcano yeah that's a cone volcano and then there's also shield volcanoes what's the last type of volcano come on people con shield. Cc. Cccc c
If f(t) F(s) and g(t) G(s) then f(t)*g(t)++ F(s)G(s)e^-s
True
False
thank you!
"Don't judge me by the straightness of my axes"