Proof of Taylor's Theorem from Real Analysis

[@pairadeau]

Is it okay to love someone you've never met? I love you, Sir.

[@ozzyfromspace]

Great proof! I always understood it in terms of the cauchy mean value theorem so this new perspective really helps. Kudos, Scott. :)

[@ivangarcia30a]

This is a wonderful proof! Thank you

[@sonofjameson4752]

Great job,well explained with details that can easily be missed

[@WithASideOfFries]

Bro. This shit is ridiculous. You are so ridiculously well spoken and enthusiastic about this content and its truly blowing my mind, the ease with which you prove such a seemingly complicated theorem. Thanks so much, man, this is such an underrated video.

[@NobodiesOfficial]

It is indeed a complicated theorem. The fact that the functions he defined are given to you immediately does not make it less complicated.

[@ericgilkey3549]

@@NobodiesOfficial That's why it's typically one of the last theorems taught in a first analysis class.

[@andrewoceallaigh3789]

Excellent Video! I have to learn this theorem and a proof for Analysis class and this is by far the best explanation of the proof I have seen. I generally don't like analysis, this video made me appreciate the subject much more.

[@danielnarcisozuglianello8281]

I am really thrilled by how well organised your work is... Besides, you're a great teacher. Thanks

[@DavidDelli]

Thank you very much for your help. It was an easy way to understand Taylor's Theorem. Thanks!

[@amengioio]

Great video! Thank you for sharing this.

[@bornabiljan1294]

Excellent video, thank you very much!

[@nissypaul173]

Am really thankuful to u for this proof..... Thanku sir It's our exam time Due to corona virus issues, our class are suspended... N our teacher was not able to cover the syllabus... When I tried to refer the text... I wasn't able to understand anything... Later I started searching in KZbin for lectures n thus I found urs.... It is really useful for meeee... Thankuuu sir... May God bless u Abundantly.... Love you sir❤❤❤❤❤

[@sweetmintkiss]

I wish our professors would've explained things so well....i didn't understand a thing during 2 hour calc lecture...and here all clear within 15 minutes.....

[@adityam2407]

Ирина Шур , I think they did. we just didn't pay attention

[@arunyadav3329]

Ирина Шур you are 👉 right

[@Big_Sploosh]

Exactly what i needed. Thank you very much!

[@fatmanurmolla5573]

Thank Rolle's Theorem saving us from long proofs I have seen on Taylor's Theorem. 🙏🏻

[@mihaidumitrescu1325]

Such an elegant proof and very nicely explained! Thank you :)

[@vai_-cn9br]

I see....the joy of teaching. Good job sir👍

[@punditgi]

Scott does it again! 👍

[@abbasbookwala]

I thought more deeply about it and found there was no issue is defining M the way it was..as the error was merely scaled by a factor of (x-x0)^(n+1) this itself being a fixed number.

[@evanwonisch788]

Thanks! That was great!

[@VikeingBlade]

That surprisingly made sense.... first real analysis proof i've ever seen (I believe.) I always thought RA'd be super confusing. (Although, I don't know Rolle's theorem, so I had to just take his word on that.) Although, when he graphs g(t)... it had g^(k)(x0) = 0, not g(x0)...? Dont get that. Seemed like an error to me (but id doubt that) Very interesting! Thank you very much.

[@Germanyduck]

Long time since this but maybe what's around 8:50 is what you're looking for to explain g^k(x0)=0

[@eamon_concannon]

That was great explanation. Thanks

[@최고은-v2o]

Wow, It's clear and easy explanation. My english is not good, but I could understand everything what you say. Thank you!

[@postnetworkacademy]

Really very good teaching and explaination.

[@abbasbookwala]

I get the feel that in defining M, (x-xo)^(n+1) has been contrived (without proving) in its denominator to facilitate this proof by then merely proving M=f(c)^(n+1)/(n+1)! Does this leave any deficiency in the overall proof of Taylor's theorem?

[@advaita9230]

Sir how can you apply rolles to g^k (x0) and g (x) because they are not the same function which gives zero value. . Plss let me know. ...

[@CengizhanSen]

How are we able to apply Rolle's theorem for g(x)=0 and the k-th derivative of g at x_0 is 0?

[@estebanlopez1344]

thanks a lot, please keep doing this kind of videos

[@jonathanweigl]

Thank you very much for the explanation! I have a question, hope someone reading this can help me. Why is it so obvious that a function f : [a,b] ------ R can be approximated by a polynomial expression, and why the expression is specifically as it is presented in the theorem? The proof was wonderfull, but as I understand it the professor assumes here that Pn(x) has to exist and moreover it has to exist in Taylor's specific form. In other words, he prooved why Rn(x) is as shown in the theorem assuming Pn(x) exists. How do I explain/proof Pn(x)? Thanks so much.

[@eamon_concannon]

He defines the quantity M in terms of f(x) and Pn(x). This is done without assuming that f(x) can be approximated by Pn(x). The only relation we need use between f(x) and Pn(x) is that they and their first n derivatives are equal at x=x_0. He then shows what M has to be which leads to the fact that f(x) can be approximated better and better by Pn(x) as n approaches infinity.

[@josefsifuentes4618]

Consider the first order Taylor polynomial - it's the tangent line. It's a good approximation near the point of tangency (assuming sufficient smoothness), but its rate of change (slope) remains constant, while the rate of change of the function changes. So to capture the rate of change of the rate of change, we find a second order Taylor polynomial - a tangential quadratic whose slope changes at the same rate of change that the slope of f changes - but only at the point of tangency. Of course the rate of change of the slope of the quadratic is constant, while the rate of change of the rate of change of f likely changes. So in comes the cubic Taylor polynomial. And so on and so on.

[@Frostbitecgi]

it helped a lot thank you sir :D

[@oceansofmath4732]

Thank you so much.

[@DouglasHPlumb]

Thanks, what confused me at first was your motivation in defining M. It looked like you pulled that out of nowhere.

[@samkirkiles6747]

This proof is the same as the proof in Rudin Principles of Mathematical Analysis on page 111 and if you're used to Rudin, he does things like this all the time :)). Basically, the intuition for defining that M is we want M to satisfy the equation f(B)=P(B)+M(B-a)^n, specifcially we want that M=f^n(x)/n! The definition written in this video is a rearranged version of this statement. In this comment I used the notation and indexing that Rudin uses so I would highly reccomend checking out that proof!

[@DouglasHPlumb]

@@samkirkiles6747 I bought the two books on Real Analysis by Tao and have Bartlett. I'm not going to study them but I have them for when I have questions about proofs. Do you have an opinion on Tao's Real Analysis 1 & 2?

[@zitianwang1205]

Very clearly explained!

[@algorithmo134]

Do you have a series of real analysis lectures?

[@madanbiswas3750]

Hello! sir, I am from Kolkata in India. A lot of thanks to you for this wonderful proof of a very important theorem you have shown here ,sir. My request you to , please, discuss on the L'Hospital theorem on limit to find Indeterminate forms. I am searching it for a long. If , already discussed, then, please give me that link to the reply of this very comment.

[@raidenstannate6373]

Great! Thanks for the proof

[@dheerajdheeraj6960]

Wonderful sir 💖

[@Rsingh1]

Is partial differentiation being used?

[@constructofeternity]

Its a really great proof scott, but how do you think maths of this proof has evolved?

[@maxfruman9552]

This is amazing👏👏👏

[@sindhud9298]

Sir what is the general formula for taylor's series using 2 variables?

[@raheelgill6267]

Sir will u plz explain why we take all terms at point x° but last term at c???

[@cameronspalding9792]

Taylor Series were meant to be an improvement of the tan line approximation

[@lexs7218]

“saying that the proof is not the fun part really hurt my feelings”- proof (also proofs are always the interesting part)

[@parvezpatel9809]

Wonderful explanation, just one doubt how does this prove that the remainder goes to zero when the polynomial terms go to infinity

[@dadyasuo8151]

the (n+1)! in the denominator goes to infinity as n->infinity, meaning the remainder goes to 0. He explained this in the first part of the video.

[@devlu1957]

Good sir

[@ChechoColombia1]

sir, thank you, for real

[@yumingjiang7819]

Very nice!

[@eruiluvatar6688]

Why is g(x0) = 0? g(x0)=f(x0)-Pn(x0)-M(x0-x0)^(n+1) Only gk(x0) is = 0

[@vineetphysics4074]

It's really very nice proof. But what's difference in interval (c,x)and (x,c) ? I think you might have taken in One interval as x_0 and in the other as x

[@sandunliyanaarachchi3800]

nice explane sr thank you

[@adamvictorio566]

thanks a lot sir

[@smitagrawal8689]

Awesome

[@aadishjain2378]

Why in g(x) M is taken as constant It is a function of x as well

[@giannispolychronopoulos2680]

aadish jain he is not differentiating with respect to “x” but rather, to “t”, so M is considered a constant.

[@rev10034]

Super...

[@erick.gudino]

gracias :3

[@edwardsteen3058]

At 8:31, you say that g(x_0) is zero. Given the definition of M, isn't there a problem with division by zero?

[@kamehamehaDdragon]

M denominator is never zero since the theorem says that X != Xo.

[@BTCIVVOTTESAIKUMAR]

Super sir

[@samarthsai9530]

Sir can you please explain me what happens at 9:38 when he says that g(x_0)=0 but on the board it is written that the k derivative of g(x_0)=0. Please help me.

[@cameronspalding9792]

Dude: is there a similar theorem for multi variable functions or even complex functions

[@julesk2294]

Omggg i hope there is. Multivariable taylor series

[@tsunningwah3471]

zhin

[@qualquan]

Get a better board write larger