honestly, this guy may not show difficult examples that I may be tested on, however his theory lessons are what makes hime so great! Thank you very much Sal :)
"...IN A NEW COLOUR!" *Fails to click on a colour* "That is not 'a new colour'!" 😂😂😂 PS. This was the best explanation of MacLaurin's Series I've seen till date. When we did this in college, our prof seemed to use Aladin's magic lamp to bring this series into existence. Thank you for this.
Hey Sal, I think I am in love with you. This is me shooting my shot. I have a winning personality, fun hobbies, and I will spend every moment with you like it is my last. Much love xoxo
I was really bad at Maclauren and Taylor series when I was in Calc 1, and now I'm learning about Laurent series in Complex Analysis. This video helped me tremendously to understand it because I'm too lazy/exhausted to read about it. Thanks again, Mr. Khan.
this is indeed around first year uni math in the usa, but they do start teaching about polynomials in algebra 2 which was taught in 10th or 11th grade i forget
Amazing! When my college professor explained this he didn't even bother to tell us what we were actually doing/finding. He just basically gave us the formula. Thanks so much!
@Yakushii not really, because you wouldn't be able to find the second derivative etc, what it does allow you to do is turn a function into a polynomial ie one just involving x to certain powers, so you can make sin(x) into a function of lots of x's of different powers added together.
@dickie4thepeople i think of it as an approximation. First we write an equation, then we add an equation for the gradient, then we add an equation for the rate of how the gradient changes (using differentiation) etc... so our approximation for the rest of the line is becoming better.... sorry i dont completely grasp this either. but i hope this kinda helps!
You can evaluate at any number, but a Maclaurin series is evaluated at zero. That's the only distinction. Zero can be replaced with 'a' or any variable holding the place for any number.
isnt this series restricted to functions that exist at f(0) for example log sin x cant be expressed as a polynomial in this way am i correct? or do we take limits in those cases and express the function as a summaion of limiting cases
Basically the binomial expansion method that we are told to cram is also a case of the Taylor series just like the Maclaurin series is but not on basis terms. One is centered around avalue and one just aims at expanding a function.
Thank you so much!! I have finals later this week, and I was sick when this concept was introduced, and now I understand it!! Thank you thank you thank you!!
They are both awesome. Sal tends to focuses a little more on theory/understanding while Patrick is usually more direct and demonstrates how to apply the concepts.
Where does it come in 11th?? I am also an Indian and I studied this only while doing my B-Tech. Yes, you have some specific functions whose Taylor's expansions you have to learn in 11th and 12th, if you are preparing for the JEE exam, but you still don't learn the derivation and that it comes from Taylor's and Mclaurin series....You just memorize the expansions of specific functions
Awesome! The best interpretation of Maclaurin series! BUT i would really appreciate if you would show How we get Taaylor from Maclaurin. And why we use (x-a) in Taylor.
Woah man, who knew that it is THIS intuitive. Well, I guess most things are... but if it wasn't for you I would've gone my whole life thinking that this is some kind of elusive abstract thing that some guy thought of and it's just the way it is, and we should accept it and use it. Man, why does no one tell us this, insted of just giving us the formula. Stupid. I was first 'taught' Taylor series without drawing any graphs whatsoever :/
at that particular point (doesn't have to be zero) we want our function and the actual function to have the same value, then the same derivative, then the same second derivative, then the same 3rd derivative etc.. and the more you do that, the more likely it will look like the actual function Basically we are saying you can approximate a function if you keep taking the derivative of a certain point
When I first saw this, I was puzzled by the fact that the sequence of derivatives at ONE point only can approximate the WHOLE function when keep adding terms. I still don’t get the intuition here.
Khan academy as I know it in the past was made so that people can watch the videos and understand the concepts without getting bored and without getting confused and before they were doing well in staying consistent with the curriculum they are tackling... but as of now, when I watch Khan academy videos and I easily get bored, I easily get confused, and there is no consistency with any curriculum that I know of. I don't know if it has to do anything with that I am in college or if my understanding of the topic is low, but if I know something, it is that Khan academy has fallen in my opinion
@qwobify If you differentiate f^4(x)*x^4/24. You get f^3(x) * 4x^3/24= f^3(x) * x^3/6. Differentiating that gives f^2(x) * 3x^2/6 = f^2(x) * x^2/2. And finally you get f(0) at the end, just as before.
Quick question: in the general notation for the Maclaurin series, why is x^n divided by n! and not just n? If you had x^4, it's derivative would be 4x^3, so wouldn't you only need to divide by 4, not 4! ? Any enlightenment would be appreciated. :)
Ahhh this would've been helpful about 2 months ago! I eventually got it but this would've made the process much faster. I take Calc 3 in the fall... Hint hint... But thanks for all you do your vids are an amazing help!
@khanacademy Sal, when a function is approximately written as a polynomial, does the approximation depend on the number of coincident points or even the shape? For example, in the first try, the approximated polynomial might be touching the function at more than one point in the neighborhood of x=0. But in the 2nd try, it cannot touch at more than one point around zero as it is a tangent now. So the number of coincident points have decreased though we have generated a better shape. How?
I still don't understand i get lost at 3:06 when he makes p(x)=f(0)+f'(0)x. Why did he add f'(0)x?? how does adding f'(0)x make p(x) and f(x) have he same first derivative. Also how does adding all these extra derivative terms give us a better approximation?? Thank you for any clarification!
Take the derivative of p(x)=f(0)+f'(0)x and notice that it is precisely the first derivative of f(x). This means that we have improved our approximation slightly. Therefore it can be improved further by matching higher order derivatives of the approximation to that of the original function.
+purplefire5 "how does adding f'(0)x make p(x) and f(x) have he same first derivative?" -> just derive it man, you'll see they do have the same derivative (remember that f(0) is a constant)!
Well I don't think that the line he draws at minute 5:00 should be linear. Because p'(x) is a linear line and p(x) should be something like y=x^2. Am I wrong?
Thanks so much for this video Sal... is there a video for proof of Taylor Series Expansion for function with 2 variables? Thanks again for all the videos.
Google Taylor or Maclaurin proofs. The full definitions of these series expansions have them going to an infinite number of terms anyway. Sal just stopped after a few derivations when in reality, these keep going forever.
I would like to give some of my intuition that I understanded from Khan maybe it would help someone. When we have p(x) = f(0) + f`(0)x + 1/2f``(0)x^2 + 1/6f````(0) + 1/2 * 3 * + 1/1 * 2 * 3 * 4 f````(0)x^4. Each time we take the derivative we keep decreasing the power term by one and dividing term too. For example if we take 4th derivative. the first would be f`(x) = f`(x) + xf``(x) + (3 * 1/(1 * 2 * 3))f```(x)) + (4/1 * 2 * 3 * 4)(f````(x))x^3. Each time we delete one of the factorials.
Sometimes you need these tiny details, these tiny bits of intuition to get the picture. You really help with this. Thanks.
honestly, this guy may not show difficult examples that I may be tested on, however his theory lessons are what makes hime so great!
Thank you very much Sal :)
I hope I'm not the only one who burst out laughing at "well Sal that's a hoarrible approximation".
I chuckled too
YOU try doing better with a straight line 😂😂
I love the "aha!" moment you get in the middle of watching the video. Nothing quite like it!
"...IN A NEW COLOUR!"
*Fails to click on a colour*
"That is not 'a new colour'!"
😂😂😂
PS. This was the best explanation of MacLaurin's Series I've seen till date. When we did this in college, our prof seemed to use Aladin's magic lamp to bring this series into existence. Thank you for this.
Aladdin's magic lamp 😂😂😂
Khan Academy just has this magic that touches parts of everyone's brains and gets it working like no one else could and would ...
That sounds weirdly romantic.
Hahahah man you just gotta ruin it!
I dont know what I would do without you :')
Hey Sal, I think I am in love with you. This is me shooting my shot. I have a winning personality, fun hobbies, and I will spend every moment with you like it is my last. Much love xoxo
I was really bad at Maclauren and Taylor series when I was in Calc 1, and now I'm learning about Laurent series in Complex Analysis. This video helped me tremendously to understand it because I'm too lazy/exhausted to read about it. Thanks again, Mr. Khan.
this is indeed around first year uni math in the usa, but they do start teaching about polynomials in algebra 2 which was taught in 10th or 11th grade i forget
Amazing! When my college professor explained this he didn't even bother to tell us what we were actually doing/finding. He just basically gave us the formula. Thanks so much!
“TRY TO DO ANY BETTER USING A HORIZONTAL LINE THEN” 😂 loved that
okay but did i ask
@@andrewmontoya8511 Yes you did ask
@Yakushii not really, because you wouldn't be able to find the second derivative etc, what it does allow you to do is turn a function into a polynomial ie one just involving x to certain powers, so you can make sin(x) into a function of lots of x's of different powers added together.
Sal! You are just amazing. You know what? You are on the list of my most most most favourite teachers in my life.
You are that amazing.
@dickie4thepeople i think of it as an approximation. First we write an equation, then we add an equation for the gradient, then we add an equation for the rate of how the gradient changes (using differentiation) etc... so our approximation for the rest of the line is becoming better.... sorry i dont completely grasp this either. but i hope this kinda helps!
Got a vid on the radius of convergence?
You can evaluate at any number, but a Maclaurin series is evaluated at zero. That's the only distinction. Zero can be replaced with 'a' or any variable holding the place for any number.
that's pretty interesting, but how are you suppose to know the values of f'(0), f''(0), f'''(0), etc... if you aren't given the function f(x)???
Khan breaks everything down in understandable chunks yet without losing the generalization rigor of Maths - wonderful
isnt this series restricted to functions that exist at f(0)
for example log sin x cant be expressed as a polynomial in this way
am i correct?
or do we take limits in those cases and express the function as a summaion of limiting cases
I really, really like this.
Derivation of the Maclaurin Series from "CORE MATHS for A-level" by L. Bostock and S. Chandler, published by Stanley Thornes (Publishers) Ltd:
A power function of f(x) = (a + x)^n
Where, when considered with the general binomial theorem, gives:
a^n + a^(n-1) * x + a^(n-2) * x^2 + ...
Where a^n, a^(n-1), a^(n-2)... are all constants, to be reconsidered as:
a0 + a1 * x + a2 * x^2 + ...
f(x) = (a + x)^n = a0 + a1 * x + a2 * x^2 +...
f'(x) = a1 + (2)a2 * x +...
f''(x) = (2)a2 + ... (3)(2)a3 * x +...
f''(x) = (3)(2)a3 + (4)(3)(2)a4 * x +...
When x = 0:
f(0) = a0
f'(0) = a1
f''(0) = (2)a2 -> a2 = f''(x) / 2 = f''(x) / 2!
f'''(0) = (3)(2)a3 -> a3 = f'''(x) / (3)(2) = f'''(x) / 3!
...
f^n(0) = f^n(x) / n(n-1)(n-2)... = f^n(x) / n!
Therefore:
f(x) = (a + x)^n = a0 + a1 * x + a2 * x^2 +... = f(0) + f'(0) * x + [f''(0) / 2!] * x^2 + [f'''(0) / 3!] * x^3 + ... [f^n(0) / n!] * x^n
Bro what are u talking about
Basically the binomial expansion method that we are told to cram is also a case of the Taylor series just like the Maclaurin series is but not on basis terms. One is centered around avalue and one just aims at expanding a function.
i cant's take it in once,,,,, but watching again again & finally get it
Thank you so much!! I have finals later this week, and I was sick when this concept was introduced, and now I understand it!! Thank you thank you thank you!!
How are you doing bud
Colin Maclaurin is credited with the Maclaurin Series
They are both awesome. Sal tends to focuses a little more on theory/understanding while Patrick is usually more direct and demonstrates how to apply the concepts.
The best teachers aren't the smartest ones, it's the ones that doesn't require their students to be very smart to learn.
How f(0) and its derivatives can have all the information of the entire curve? The curve could go in any direction after f(0).
Funny how Indian students do this in class 11 while in the US college students struggle in AP calc 😅
I live in England and im also doing this in highschool lol. Im also Indian btw :D
Where does it come in 11th?? I am also an Indian and I studied this only while doing my B-Tech. Yes, you have some specific functions whose Taylor's expansions you have to learn in 11th and 12th, if you are preparing for the JEE exam, but you still don't learn the derivation and that it comes from Taylor's and Mclaurin series....You just memorize the expansions of specific functions
Nice Video Dude
Awesome! The best interpretation of Maclaurin series! BUT i would really appreciate if you would show How we get Taaylor from Maclaurin. And why we use (x-a) in Taylor.
Woah man, who knew that it is THIS intuitive. Well, I guess most things are... but if it wasn't for you I would've gone my whole life thinking that this is some kind of elusive abstract thing that some guy thought of and it's just the way it is, and we should accept it and use it. Man, why does no one tell us this, insted of just giving us the formula. Stupid. I was first 'taught' Taylor series without drawing any graphs whatsoever :/
You're the world's best teacher, undoubtedly Mr. Sal Khan, Masha-Allah!
when he says "I don't have the computer power of my brain to draw it properly..." damn!! this guy has some imagination I really envy of him >.
After reading the comments I envy all of you that you all have mastered the Taylor theorem but I am still struggling!!😂😂
at that particular point (doesn't have to be zero) we want our function and the actual function to have the same value, then the same derivative, then the same second derivative, then the same 3rd derivative etc..
and the more you do that, the more likely it will look like the actual function
Basically we are saying you can approximate a function if you keep taking the derivative of a certain point
When I first saw this, I was puzzled by the fact that the sequence of derivatives at ONE point only can approximate the WHOLE function when keep adding terms. I still don’t get the intuition here.
I don't know what I'd do without you sal😭😭😭😘😘
@MilitaryMan006 This is first year uni math.
you are a star, and a hero, thank you for this
I have my final in two days and I absolutely love you
Hey are you alive
So is it possible to find the function of any given graph, just by reading the graph, using MacLaurin series?
Khan academy as I know it in the past was made so that people can watch the videos and understand the concepts without getting bored and without getting confused and before they were doing well in staying consistent with the curriculum they are tackling... but as of now, when I watch Khan academy videos and I easily get bored, I easily get confused, and there is no consistency with any curriculum that I know of. I don't know if it has to do anything with that I am in college or if my understanding of the topic is low, but if I know something, it is that Khan academy has fallen in my opinion
are there any Khan academy videos for convergence tests for series?
@qwobify If you differentiate f^4(x)*x^4/24. You get f^3(x) * 4x^3/24= f^3(x) * x^3/6. Differentiating that gives f^2(x) * 3x^2/6 = f^2(x) * x^2/2. And finally you get f(0) at the end, just as before.
Yes, more stuff on Series and Sequence. It was lacking in your Calculus Playlist.
Awesome teaching i love your teaching style
Quick question: in the general notation for the Maclaurin series, why is x^n divided by n! and not just n? If you had x^4, it's derivative would be 4x^3, so wouldn't you only need to divide by 4, not 4! ? Any enlightenment would be appreciated. :)
Thanks. This was very helpful in trying to understand my not so helpful textbook.
I Really Like The Video From Your Approximating a function at 0 using a polynomial
Wow. So well and simply explained. Thankyou.
Sal, you're so good at explaining. Thank you so much
So helpful thanks!
Could you do a video explaining time travel?
Well you use (x-a) in Taylor because you can choose any center point (a). So in the case of the Maclauren Series, a=0 so it' would just be (x).
"That's not a new color...."
Pretty much describes my understanding of Taylor series
I HAVE EXAM ON THIS ON MONDAYYYYY ahhhhhh THX SALLLL
Do I have to evaluate at zero? or can I evaluate at any number?
wow what concept! thanks do much for explaining it so simply i feel like a have a good understanding of it now
some people get so excited at things
Ahhh this would've been helpful about 2 months ago! I eventually got it but this would've made the process much faster. I take Calc 3 in the fall... Hint hint... But thanks for all you do your vids are an amazing help!
@khanacademy Sal, when a function is approximately written as a polynomial, does the approximation depend on the number of coincident points or even the shape? For example, in the first try, the approximated polynomial might be touching the function at more than one point in the neighborhood of x=0. But in the 2nd try, it cannot touch at more than one point around zero as it is a tangent now. So the number of coincident points have decreased though we have generated a better shape. How?
Thanks for saving my grades!
Without Sal I wouldn't have passed high school
not very formal, but quite clear and intuitive. thank you!
Thanks Sal!
Yeah, thanks. I can never seem to freaking understand anything in calc2, I aced the first calc, but it's like there's a brick wall in calc 2.
Khan is king!
amazing explanation. thanks.
Well, I finally understand it... 3 years after finishing my calculus course.
Thank you Sal
from where term 1/2 f"(0) x^2 came in 3rd point?
This is pretty good. I got the idea about how Mclaurin Series generated. Thumbs up.
thank you khan
Good example, great explanation
This is gold. Thank you again. :)
i love you, Sal
I still don't understand i get lost at 3:06 when he makes p(x)=f(0)+f'(0)x. Why did he add f'(0)x?? how does adding f'(0)x make p(x) and f(x) have he same first derivative. Also how does adding all these extra derivative terms give us a better approximation?? Thank you for any clarification!
Take the derivative of p(x)=f(0)+f'(0)x and notice that it is precisely the first derivative of f(x). This means that we have improved our approximation slightly. Therefore it can be improved further by matching higher order derivatives of the approximation to that of the original function.
+TheNuclearpolitics Is it necessary to know the infinite higher order derivatives of the line?
+TheNuclearpolitics
Great explanation.
+purplefire5 "how does adding f'(0)x make p(x) and f(x) have he same first derivative?" -> just derive it man, you'll see they do have the same derivative (remember that f(0) is a constant)!
Luv u lot. Best explanation for Maclaurin series ever.U doing great work ...
"That's not a new color" 😂
Luv u guys
This guy know bio,chem,math and physics. damn
i wish i saw this video back when i was 20 -__- nice and easy explanation
i never tought taylor series would recieve a 1.1 million views
This was a really helpful video.
Well I don't think that the line he draws at minute 5:00 should be linear. Because p'(x) is a linear line and p(x) should be something like y=x^2. Am I wrong?
He is drawing P(x) = f(0) + f'(0) x which is eqn of line
Great explanation sir ❤❤
this guy, is the saviour!
.
Many,many thanks to khan academy!!
This video was Awesome
Thanks so much for this video Sal... is there a video for proof of Taylor Series Expansion for function with 2 variables?
Thanks again for all the videos.
saved my day
Does anybody know if there's a fast way to find the next video?
thanks for the proof!
Does anyone know where I can find the proof Sal is referring to at 12:20 ?
Google Taylor or Maclaurin proofs. The full definitions of these series expansions have them going to an infinite number of terms anyway. Sal just stopped after a few derivations when in reality, these keep going forever.
Nice one!
very helpful thanks alot
where do you take class?......
I think he differentiated the second video one time to many. ;O)
Thats actually genius.
Thank you so much.
I would like to give some of my intuition that I understanded from Khan maybe it would help someone. When we have p(x) = f(0) + f`(0)x + 1/2f``(0)x^2 + 1/6f````(0) + 1/2 * 3 * + 1/1 * 2 * 3 * 4 f````(0)x^4.
Each time we take the derivative we keep decreasing the power term by one and dividing term too. For example if we take 4th derivative.
the first would be f`(x) = f`(x) + xf``(x) + (3 * 1/(1 * 2 * 3))f```(x)) + (4/1 * 2 * 3 * 4)(f````(x))x^3.
Each time we delete one of the factorials.
Better Call Sal
ultimate, fantastic, mind blowing Boss.