Beautiful. Nobody has explained the derivation of taylors polynomial remainder on youtube as concise as you did. 11 years later and its still benifiting us calculus students...... ❤
Thankyou very much, keep it up Khan Academy. I love the Pure Maths videos and the financial ones. In particular I never use anything else in the world to research economics or finance and i'm now the most enlightened person in my circle of friends on the subject :). Don't give it up.
At 8:23, how is the integral of E^n+1 dx == E^n, when the integral power rule calls for adding 1 to the exponent and dividing by that number? Any explanation would help! Thank you
That tripped me up too. I think it's because the E^(n+1) refers to the n+1'th derivative of E, so it is not an exponent. Each time you integrate a derivative, the derivative goes down a "level" you could say. It's just an annoying notation thing I guess.
I had the same question. He forgot this constant, the integral should be |E'[n] (x) + a|, where a is a constant. However, |E'[n] (x) + a | >= |E'[n] (x)| - b, where b is a sufficiently big number to make this statement true, so: |E'[n] (x)| - b
when there is an equation, it is okay to use one constant for the whole equation as there is two constants, one for each side of the equation, which both can be moved to one side of the equation and make a new constant. So, you can say that he just used a shortcut.
For me, it's better that you use M instead of max abs(f^(n+1)(x)) in [a,b] like in the book I have, because it clearly shows that it is a constant and not a function which got me very confused when I first learned this stuff.
I just have an issue with interpreting what happened in the second integration of your error term towards the end of the video, do you mind clarifying my interpretation of what happened Khan? If you are treating 'a' and 'M' as a constant, then the integral of 'M(x-a)' with respect to x, where x is the variable, then the expression won't be 'M/2 * (x-a)^2'. This is evident when you expand the expression before integrating, 'Mx - Ma'. '-Ma' is a constant so that integrated becomes '-aMx' (I just rearranged the order of the constants, because '-Max' might be confusing). Then 'Mx' integrated becomes 'M/2 * x^2'. So when you simplify the expression now: 'Mx(x-2a)/2'. I do however understand that it should be intuitive that the result of the Error Term would follow the general trend of the following polynomial term; but I am however stuck with interpreting that second integral. Thank you for in advance Khan.
Actually, both produce the same result. Take your result and don't factor out the Mx. This is your result integral = (Mx^2)/2 - Mxa + constant Take Sal's result and expand it out integral = (M(x-a)^2)/2 + constant = (Mx^2 - 2Mxa + Ma^2)/2 + constant2 integral = (Mx^2)/2 -Mxa + (Ma^2)/2 + constant2 But (Ma^2)/2 is a constant so his integral is integral = (Mx^2)/2 - Mxa + anotherconstant idk how helpful this is to you as your comment is 3 years old but I figured I would write an answer anyways.
Around minute 8, he says that the integral of the (n+1)th derivative of E is the nth derivative of E. But when you take an integral/antiderivative, the exponent gets BIGGER, not smaller. Confused about this part!
Oh, so we are not only finding the error for a Taylor Polynomial modeling a function, like sinx. We are also finding the errors for the derivatives (or slope of the tangent lines) a Taylor Polynomial models for a function. Ah, I see why this error bouding is so important to the function as a whole.
Even you was confused with E(x) as expected value of probability. I think that you should stay with R[n](x) as this is less confusing, especially for students that in the same time as Calculus 2 have Introduction to Statistics, straight after introduction to Probability, where all letter of different alphabet got special meaning. Never mind, thanks for the video.
There's something I don't understand. From 6:22 to 7:33, he claims that the absolute value of the integral of some function is LESS THAN OR EQUAL TO the integral of some function in absolute value. How is this possible? We are trying to find the area between a curve and the x-axis. If you are taking the absolute value of the integral, you will ALWAYS get a positive value, because the negative would be negated. If you are taking the absolute value first, the integral can end up negative, and then we would have negative area. Take for example, f(x) = 2x. The integral of 4x^(3) is x^4, and bounded between x = -3 and x = -1, the area bounded between there and the x-axis is -80... the absolute value is then taken, to give you 80. Meanwhile, if you took the absolute value of 4x^(3), you would get 4x^(3)... then the integral of that would give you a negative value. Therefore, it seems to me as if the absolute value on the outside is LARGER than inside.
Wonderful video, gives a lot of intuition on the matter, but how do you bind the C at 10:10? I got a bit confused over this part, because you have no control over C at the point where you choose it to be -Ma, how do you prove that it is? Really nice video, but kinda crucial flaw in the proof :/
When you have E^(n+1) you have the derivative value of E, and i can tell that because of the parenthesis, else it would be a exponential function. If you want to integrate a function, you are decreasing the derivative value by the value of the primitive. So, when you do one primitive of E^(n+1) you are subtracting the value one to the derivative level of E. If you wanted to do 2 primitives of E^(n+1) you would have E^(n-1) - and 2 constants - and so on.
I just thought of this two minutes ago so I might be wrong. When you complete the first integral and solve for the constant of integration, M is the maximum gradient of the E^(n)(x) while M is the gradient of the line Mx. For this reason in all cases excluding those where f(x) is a degree (n+1) polynomial, you can be even more restrictive. As Mx will always grow faster than E^(n)(x) you can say, E^(n)(x) < M(x-a). This aligns with the proof provided by the extended mean value theorem.
Amazing. Understood it. There is however another way (which is straightforward) to derive this error bound that I found in the following link: brilliant.org/wiki/taylor-series-error-bounds/ Is that really valid? There are statements that don't seem to be correct there. I can't understand it. For example, "Since the Taylor approximation becomes more accurate as more terms are included, the Pn+1(x) polynomial must be more accurate than Pn(x)" does not seem to be right.
That depends entirely on the nature of f(x). Here it may be assumed that f(x) is infinitely differentiable. Think about it, say f(x)=x^(n+1), then the (n+1)th derivative is clearly not zero. These are just for polynomial functions. Take e^x and sinx for instance. Any derivative of them is not going to be zero
![KSI - Dirty [Official Visualiser]](http://i.ytimg.com/vi/VoDh1h1dPUE/mqdefault.jpg)
Beautiful. Nobody has explained the derivation of taylors polynomial remainder on youtube as concise as you did. 11 years later and its still benifiting us calculus students...... ❤
Remember when math used to be numbers
Good times.
it's not the expected value function. Oh, did I mention it's not the expected value function? Oh btw it's not the expected value function
Smoothly sweep it under the carpet by saying it's the deviation from the expected value.
🤣🤣🤣
But... it looks like the expected value function..? Is it not the expected value function? Why would it be E(x) then?
Your explanation, accompanied with the drawings, about the two different integrals is stunningly clear! Good job AGAIN!! =)
You have no idea how much i appreciate this video...thank you so much!
Thankyou very much, keep it up Khan Academy. I love the Pure Maths videos and the financial ones. In particular I never use anything else in the world to research economics or finance and i'm now the most enlightened person in my circle of friends on the subject :). Don't give it up.
so... is that E the expected value function? it sure looks like it, i hoped anyone would make it more explicit if it was or not
@@candy_heart7191 that'sthejoke.jpg
I've only watched the two videos of this concept and it's already so much better than others on RUclips
This is life changing!
At 8:23, how is the integral of E^n+1 dx == E^n, when the integral power rule calls for adding 1 to the exponent and dividing by that number? Any explanation would help! Thank you
That tripped me up too. I think it's because the E^(n+1) refers to the n+1'th derivative of E, so it is not an exponent. Each time you integrate a derivative, the derivative goes down a "level" you could say. It's just an annoying notation thing I guess.
@@tonykorol5899 I had the same confusion as well - thanks for clearing it up!
It's not the power. It's the nth derivative
@@tonykorol5899thank you so much for your explanation ❤
At 08:10, why isn't a constand needed when taking indefinite integral of the (n+1)th derivative of error function?
I had the same question. He forgot this constant, the integral should be |E'[n] (x) + a|, where a is a constant.
However, |E'[n] (x) + a | >= |E'[n] (x)| - b, where b is a sufficiently big number to make this statement true, so:
|E'[n] (x)| - b
when there is an equation, it is okay to use one constant for the whole equation as there is two constants, one for each side of the equation, which both can be moved to one side of the equation and make a new constant. So, you can say that he just used a shortcut.
I, uh, found my new favourite source where to check all this stuff :D Excellent presentation.
how is it that you can just "choose" the C you want?
thank you man you're really a genius!
For me, it's better that you use M instead of max abs(f^(n+1)(x)) in [a,b] like in the book I have, because it clearly shows that it is a constant and not a function which got me very confused when I first learned this stuff.
My book use f^(n+1)(t), with t in [a,b].
I just have an issue with interpreting what happened in the second integration of your error term towards the end of the video, do you mind clarifying my interpretation of what happened Khan? If you are treating 'a' and 'M' as a constant, then the integral of 'M(x-a)' with respect to x, where x is the variable, then the expression won't be 'M/2 * (x-a)^2'. This is evident when you expand the expression before integrating, 'Mx - Ma'. '-Ma' is a constant so that integrated becomes '-aMx' (I just rearranged the order of the constants, because '-Max' might be confusing). Then 'Mx' integrated becomes 'M/2 * x^2'. So when you simplify the expression now: 'Mx(x-2a)/2'.
I do however understand that it should be intuitive that the result of the Error Term would follow the general trend of the following polynomial term; but I am however stuck with interpreting that second integral. Thank you for in advance Khan.
Actually, both produce the same result. Take your result and don't factor out the Mx.
This is your result
integral = (Mx^2)/2 - Mxa + constant
Take Sal's result and expand it out
integral = (M(x-a)^2)/2 + constant = (Mx^2 - 2Mxa + Ma^2)/2 + constant2
integral = (Mx^2)/2 -Mxa + (Ma^2)/2 + constant2
But (Ma^2)/2 is a constant so his integral is
integral = (Mx^2)/2 - Mxa + anotherconstant
idk how helpful this is to you as your comment is 3 years old but I figured I would write an answer anyways.
@@johnprice233 thank you so much! old questions stay for new people :)
@@johnprice233 this helped me after after 3 years after you replied to someone who asked 6 years back
Around minute 8, he says that the integral of the (n+1)th derivative of E is the nth derivative of E. But when you take an integral/antiderivative, the exponent gets BIGGER, not smaller. Confused about this part!
It's not a exponent it's the derivative notation
Oh, so we are not only finding the error for a Taylor Polynomial modeling a function, like sinx. We are also finding the errors for the derivatives (or slope of the tangent lines) a Taylor Polynomial models for a function. Ah, I see why this error bouding is so important to the function as a whole.
You are really doing a great job!
Even you was confused with E(x) as expected value of probability. I think that you should stay with R[n](x) as this is less confusing, especially for students that in the same time as Calculus 2 have Introduction to Statistics, straight after introduction to Probability, where all letter of different alphabet got special meaning. Never mind, thanks for the video.
There's something I don't understand. From 6:22 to 7:33, he claims that the absolute value of the integral of some function is LESS THAN OR EQUAL TO the integral of some function in absolute value. How is this possible? We are trying to find the area between a curve and the x-axis. If you are taking the absolute value of the integral, you will ALWAYS get a positive value, because the negative would be negated. If you are taking the absolute value first, the integral can end up negative, and then we would have negative area. Take for example, f(x) = 2x. The integral of 4x^(3) is x^4, and bounded between x = -3 and x = -1, the area bounded between there and the x-axis is -80... the absolute value is then taken, to give you 80. Meanwhile, if you took the absolute value of 4x^(3), you would get 4x^(3)... then the integral of that would give you a negative value. Therefore, it seems to me as if the absolute value on the outside is LARGER than inside.
Primaski |4x^3| = 4x^3 is true only for positive x
Take for example x=-1
|4(-1)^3| =4(-1)^3
|-4| = -4
4 = -4 which is obviously incorrect
Here absolute value of 4x^3 will be -4x^3 bcoz domain u r taking is -ve
Thank you 😊
Beautiful!
Does anyone knows where the video where he does examples of the remainder function is???
Wonderful video, gives a lot of intuition on the matter, but how do you bind the C at 10:10? I got a bit confused over this part, because you have no control over C at the point where you choose it to be -Ma, how do you prove that it is? Really nice video, but kinda crucial flaw in the proof :/
about 11 years late but you're right that is problematic. He should be taking a definite integral from a to x instead, you get the same result
Great stuff!
Mind blowing..
this is one of the most challenging concepts in calculus, why rush? honestly feels and sounds like sal is rushing through this
He spent half an hour on this concept though. Definitely is a lot of time spent on a concept.
where are the examples i cant find them
lol
eiiii i see you ryan !
It's a proof lol. It's EVERY example.
the best and the perfect
this is so hard man
and now it's easy
Thank you!
I want a graphical proof/demonstration
Why is the integral of E^(n+1) = E^n ?
Azad Kaya it isn't E^n+1 it's the n+1th derivative of E, I was confused too
When you have E^(n+1) you have the derivative value of E, and i can tell that because of the parenthesis, else it would be a exponential function.
If you want to integrate a function, you are decreasing the derivative value by the value of the primitive. So, when you do one primitive of E^(n+1) you are subtracting the value one to the derivative level of E. If you wanted to do 2 primitives of E^(n+1) you would have E^(n-1) - and 2 constants - and so on.
IThe function is not E^(n+1). The (n+1) is actually how many derivatives we've taken
What happened to the c (constant of integration)???
This video played at 2x speed is a godsend
I just thought of this two minutes ago so I might be wrong. When you complete the first integral and solve for the constant of integration, M is the maximum gradient of the E^(n)(x) while M is the gradient of the line Mx. For this reason in all cases excluding those where f(x) is a degree (n+1) polynomial, you can be even more restrictive. As Mx will always grow faster than E^(n)(x) you can say, E^(n)(x) < M(x-a). This aligns with the proof provided by the extended mean value theorem.
pura dimag lgaya hai pr chalo aa gya smjh mai
where is part 3?
In__
Your--
Mom--
Butt-
Holeee__-
Brilliant
Genius............
11:40 How?
You didnt explain why you replace M with f^(n+1)(x) at the very end.
Sir, is the order of error function related to this? Could we say the order of the error function is (n+1)?
squaring and dividing by the same power ..(I checked this for power of two)
it is not valid for the [0,2) region ...
thanksssssssssssssssss God please thank u so much
It might have been much easier and clearer if they used notation of Max instead of just M. M already is too well known for Middle value.
"max" notation is indeed used, just not in Khan's tutorial.
So now we see, that there was another assumption about "arbitrary function f" - it's n+1 derivative must be continuous.
Not really, we never used this assumption.
Amazing. Understood it. There is however another way (which is straightforward) to derive this error bound that I found in the following link:
brilliant.org/wiki/taylor-series-error-bounds/
Is that really valid? There are statements that don't seem to be correct there. I can't understand it.
For example, "Since the Taylor approximation becomes more accurate as more terms are included, the Pn+1(x) polynomial must be more accurate than Pn(x)" does not seem to be right.
Why isn't (n+1)th derivative of f(x) 0?
That depends entirely on the nature of f(x). Here it may be assumed that f(x) is infinitely differentiable. Think about it, say f(x)=x^(n+1), then the (n+1)th derivative is clearly not zero. These are just for polynomial functions. Take e^x and sinx for instance. Any derivative of them is not going to be zero
se x
Wow
none
zhian1
nifa
j
bilibili
zh
j j j j j j
bilibili