Taylor Theorem Proof

  Рет қаралды 11,759

Dr Peyam

Dr Peyam

Күн бұрын

Пікірлер: 54
@jetstreamsam3737
@jetstreamsam3737 Жыл бұрын
Dude,i just wanted to tank you,You are saving people from wasting hours searching for a simple proof.❤
@112BALAGE112
@112BALAGE112 5 жыл бұрын
A more sophisticated name for "Junk" would be Lagrange remainder.
@jacoboribilik3253
@jacoboribilik3253 5 жыл бұрын
"For the sake of brevity we will always refer to this remainder as junk" quoting Euler
@lacnetopanky6512
@lacnetopanky6512 5 жыл бұрын
Lagrange remainder I just way of expressing the difference between function and its approximation. There exists also Cauchy remainder, integral remainder and maybe more of them. I think junk is better term for this kind of junk.
@pauselab5569
@pauselab5569 9 ай бұрын
probably something like O(f^(n)) since it might not have a n+1 derivative
@chirayu_jain
@chirayu_jain 5 жыл бұрын
I always wanted this proof
@AhmedIsam
@AhmedIsam 5 жыл бұрын
I swear I've been looking for this since the moment my eyes fell on Taylor series in Calculus book in my high school.
@jacoboribilik3253
@jacoboribilik3253 5 жыл бұрын
This is the best proof of Taylor's theorem, in my opinion. The rest is ok, but none of them is so simple and yet so powerful to delve into the question of what functions can and can't be represented by an infinite polynomial. Dr.Peyam, I would recommend that you make a video on taylor polynomial uniqueness, it's a beautiful fact in math.
@marcotanzilli4810
@marcotanzilli4810 5 жыл бұрын
Love this proof Dr Peyam! And it's even much easier than the one they usually teach in my uni
@ugursoydan8187
@ugursoydan8187 4 жыл бұрын
this is one of the most ingenious proof that I have seen in my life. thanks you very much. and also we can continue to taking integrals infinite times. doesn't we?
@maxsch.6555
@maxsch.6555 5 жыл бұрын
Wow I've never seen this proof. Thanks for sharing! :)
@benjamincolson
@benjamincolson 5 жыл бұрын
Could you make a part 2 that addresses convergence?
@LuisBorja1981
@LuisBorja1981 5 жыл бұрын
What a swift Taylor Theorem proof!
@davidmillerdrums
@davidmillerdrums 5 жыл бұрын
Wow you are so right about that proof. Soooooo nice. Sincere thanks.
@fedefex1
@fedefex1 5 жыл бұрын
Great! Multivariable Taylor proof now!
@drpeyam
@drpeyam 5 жыл бұрын
To get the multi variable version with f(x), apply this video to g(t) = f(tx), where t is real, and set t = 1
@fedefex1
@fedefex1 5 жыл бұрын
@@drpeyam thanks!
@s00s77
@s00s77 5 жыл бұрын
do we lack anything besides induction on n and the observation that f is at least C(n-1)?
@pauselab5569
@pauselab5569 9 ай бұрын
the historical proof was about newton.lagrange interpolations done at nearly the same point.
@tgx3529
@tgx3529 5 жыл бұрын
Dr Peyam,. Is here contemplated some assumed, that the function f is of class C k, in connection with the JUNK?
@dgrandlapinblanc
@dgrandlapinblanc 5 жыл бұрын
Excellent. Thank you very much.
@Handelsbilanzdefizit
@Handelsbilanzdefizit 5 жыл бұрын
instead of a powerseries: f(x) = a0 + a1 x + a2 x² + a3 x³ + ... You can develop many functions as a kind of taylor product: f(x) = a0 * a1^x * a2^x² * a3^x³ * ...
@ElizaberthUndEugen
@ElizaberthUndEugen 5 жыл бұрын
I don't follow the step where $\int_a^x f'(a)$ becomes $f'(a) (x-a)$. Why can we do that?
@drpeyam
@drpeyam 5 жыл бұрын
f’(a) is a constant (with respect to x), so just pull it out
@ElizaberthUndEugen
@ElizaberthUndEugen 5 жыл бұрын
@@drpeyam Ah, of course. Thanks!
@cach_dies
@cach_dies Жыл бұрын
Who's the first guy in the meme from the thumbnail?
@mathematicadeestremo6396
@mathematicadeestremo6396 5 жыл бұрын
Taylor ..... New-Zealand :)))
@redknight344
@redknight344 5 жыл бұрын
Awesome!!!!
@willnewman9783
@willnewman9783 5 жыл бұрын
But if you do it like this, then the constant M depends on how close x is to a. If you wanted it to hold for all x, M could go to infinity, right?
@Contradi
@Contradi 5 жыл бұрын
The constant M is bounded below by the maximum value of |f'''(x)| between x and a, you're correct. Choosing a more positive egative x value might require you to increase M. If f'''(x) is unbounded, then yes you could have it go for infinity for the "Junk" term to be valid for all x. This is in line with the expectation that using a lower order approximation for a function can result in an unbounded error as you move away from the point you use to make your approximation. e.g. my error will go to ∞ as x -> ∞ if I approximate a parabola with a line. (As a counter example, your error will not go to ∞ if you approximate, say, a sine function with a constant function like y = 0 or y=122)
@pranavsingh9284
@pranavsingh9284 5 жыл бұрын
enjoyed!
@AJ-fo3hp
@AJ-fo3hp 3 жыл бұрын
This is looks like wavelet analysis, sum and difference
@cmcatholic1798
@cmcatholic1798 2 жыл бұрын
Damn ross Taylor in the thumbnail 😂😂
@subhrajyotidutta4725
@subhrajyotidutta4725 5 жыл бұрын
Sir just a kind request. while writing on the board try not to cover up the board. I like you videos AF. Thank you : ) Love from india.😘
@RalphDratman
@RalphDratman 5 жыл бұрын
That is important. Also please stop and stand aside for a moment before erasing the board so we can pause the video at that point if necessary and review anything we have not yet understood.
@xsunshine99_45
@xsunshine99_45 5 жыл бұрын
Right no this video has 110 likes and *ZERO* dislikes You have to be darn good to pull that off in 2019
@drpeyam
@drpeyam 5 жыл бұрын
Yay!!! 😄
@6612770
@6612770 5 жыл бұрын
Simply delicious!
@slavdam2300
@slavdam2300 4 жыл бұрын
mind = blown
@蔺美云
@蔺美云 2 жыл бұрын
You made me laugh 😂 when you said the junk…..
@choungyoungjae8271
@choungyoungjae8271 3 жыл бұрын
cool
@shandyverdyo7688
@shandyverdyo7688 5 жыл бұрын
Someday, i'll understand. :):
@guitoo1918
@guitoo1918 5 жыл бұрын
Taylor Theorem never worked for me. My junk is probably just too big.
@gatitoconsueter
@gatitoconsueter 5 жыл бұрын
7:20 do you get it XD
@NamaSaya-wg9gn
@NamaSaya-wg9gn 5 жыл бұрын
Finally
@kmac5912
@kmac5912 5 жыл бұрын
Please solve the infinite product of x=1 to infinity of cos(pi/(x+2))
@pranavsingh9284
@pranavsingh9284 5 жыл бұрын
u dont pre assume something u just go tep by step and do it
@TIO540S1
@TIO540S1 Жыл бұрын
“Junk is very small…” hmmm…
@CaptainCalculus
@CaptainCalculus 5 жыл бұрын
Ross Taylor is always the 4th term
@billgrant7262
@billgrant7262 5 жыл бұрын
taytay series
@SR-kd4wi
@SR-kd4wi 5 жыл бұрын
Show Integration of 1/(x^x) from 0 to 1 =1.2....
@pranavsingh9284
@pranavsingh9284 5 жыл бұрын
u are don
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