Lambert W Taylor Series Expansion [ Lagrange Inversion Theorem ]
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- Опубликовано: 9 сен 2024
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Let us do something special today! =) We are going to use the agrange Inversion Theorem to find out a Power Series Representation for the Lambert W Function! I hope you are going to enjoy this treat! :)
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Doing maths: yes papa
Proving theorem: yes papa
Lying?: no papa
don't kick me papa flammy
XD
XD
Yay!
Prove the Lagrange Inversion Theorem
Funny seeing you here, not
Now prove the Lagrange inversion theorem :D
Yes, would be very useful.
Yea, I really need this I'm searching this proof for days and i haven't found it
Maybe late, but here's a proof! You need to do a little work to get the expression used in the video, but gives the basic ideas.
users.math.msu.edu/users/magyarp/Math880/Lagrange.pdf
Agreed with some other comments, briefly discussing the radius of convergence would've been nice. Since n^n grows much faster than n!, this expansion only works for small z. I think it's z < 1/e if I recall correctly
I’d be curious to know how this Taylor series deals with the fact that W(z) has two branches. Does the same polynomial somehow describe both branches, or just one of them?
Two boards. Shlt is about to get serious
"Welcome back to anow video" I love your accent and I love you. Now that my differential equations class started, I hope you keep the differential equations videos coming!
But why is it called the Lambert W function? A guy named Lambert just liked the letter W?
Should pronounce it Lambert Vvvvv function, cuz that's much more deutlich, obviously.
Maybe because "L" is often used for Laplace transform and Lagrangian:)
W for weed
It's a Kamen Rider W reference
W for Wednesday my dude aaAaAAAaaAaaAAaaaAaaAaAaAAAHhhHhHHhHH
""It's bloody messy" 😂😂 thanks papa
ˋˋit falls a bit from the skyˋˋ
We might need to find its radius and interval of convergence. Maybe calculus stuffs, bprp can help!
Lambert W function is the best function in the universe :V!!!!!!!
Prove the Lagrange Inversion Theorem.
0:54 Mission failed. We'll get'em next time.
this function is a big W.
-- Labert
A really nice video - just one slight typo. For the series coefficients you define “g_n(z)” which should not be the case since you are taking the limit of the variable z (though you use x).
THE BOARD UNDER THE BOARD🤣🤣
Omega-Tau-Phi indeed, and well deserved. One of few physicists that can do proofs, is our Papa Flammy. Remember that those physicists are the ones that were the greatest.
So the mathematicians are the greatest physicists?
I come for the math, I stay cause papa is one sexy boi.
MAKE MATHS GREAT AGAIN!
Automatic captioning at 8:10 - WTF I am gonna call BND you sneaky boi
i got confused by the D^(n -1), thanks for showing the steps!
#wewantmore ummm.. i don't know I have no criticism but I want more.
San Samman
Probably Papa Flammy's source was using this notation. That's usually a sign of advanced material. However, there is no need for operator notation here.
Flammy lamby!
How does this work with the fact that W(z) has infinitely many branches. This definition only gives the principal branch.
Could you make a video on Lagrange multipliers to find functions of several variables extrema?
I wish he would slow down more for high school students watching this.
You seem like a chill guy... subbed
TheUnorthodoxGears
Hm, for me he looks like an arrogant smart ass, but in a funny and likeable way :D
Yaaay papa finally uploaded it ..
many thanks pappa
what's the radius of the convergence of this series ?
It wpuld be great if someone sent you a featured video where the proof is explained *wink wink wonk
Beautiful :)
Hey papa flammy, not directly related to the video but I had a question. I know sometimes when you're solving your diff eqs and you have dy/dx = 2 or something, you integrate both sides with respect to x. You always say you can cancel out the dx's on the right if ur a physicist or you can introduce a proper substitution. What do you mean when you say a proper substitution?
When you use separation of variables you end up with and integral, lets call it I, that looks like
I = integral f(y) dy/dx dx.
Let F be an anti-derivative of f. Then
F = integral f(y) dy ... dF/dy = f ... dF/dy dy/dx = f(y) dy/dx,
and by the chain rule of differentiation
dF/dy dy/dx
is nothing but
dF/dx.
Therefore,
I = integral dF/dx dx = F = integral f(y) dy
Akash Narayanan
In a calculus book I learned from, it was proved that first order derivative, for ex. dy/dx, can be treated like a fraction, so canceling out dx is not an improper procedure.
Higher order derivatives cannot be treated like fractions.
great video.
Fire Steinmeier! Flammy for President!
Please try to derive Lagrange Inversion Theorem in similar way.
It is very interesting tool.
Best regards
From which lecture in the university can I learn Lagrange inversion formula?
You have me hit mathematical climax with these series, papa.
I still want to see a proof of the Lagrange inversion theorem so that I don't have to keep using series reversions:/
Is 46 & pi the next version of that song?
@@WhattheHectogon Yeah it's gonna be on the new album
Hell yeah
Great video as always, I've been trying to find an inverse for the anti-derivative of the Maxwell-Boltzmann boi and couldn't find any worked examples of the Lagrange inversion theorem so this helps a lot
It would really be nice if you could provide an hemdsärmeligen proof of Lagrange’s inversion theorem. ;-)
Lagrange...
small people...
small...
i tried it with just tayloer seiries and got y(c)+lnu(x-c) u is constent
sorry ln 1+u(x-c)
this is wrong double checked again (the mistake was f''=-f'^2) working on this for the 2nd time i got a recursive formula for the Taylor series of the inverse of x^s*e^x but its a monster containing the sum of the n previous number in the seiris multiplied by ns
however it seems like it is diverging a lot so this makes this function weird never the less it is unalitic
Challenge: Solve a^n+b^n=c making n the subject using lambert W-function :)
Maximilian Koch
I assume you want a general solution for
a^x+b^x=c,
since n usually denotes natural numbers and this equation is not likely to have integer solutions.
It seems that this equation requires other function to solve it, since there are no x'es not in the exponent.
Let's see. We want the same base for exponential functions, say, a:
b^x =[a^(log_a b)]^x = a^(x*log_a b) .
Let's define a parameter p:
p=log_a b=ln b/ln a .
Our equation becomes
a^x + a^(x*p)=c ->
a^x + (a^x)^p=c .
Let's change a variable :
a^x=y -> y+y^p=c , y>0 .
So, we get a polynomial-ish equation in terms of y - not polynomial, because p may not be integer. If we get y, x is just
x=ln y/ln a, ln a=/=0, a=/=1.
a=1 is a trivial case.
What about p=ln b/ln a ? Let's explore the cases:
p=0 -> ln b=0 -> b=1, trivial.
p=1 -> ln b=ln a, b=a, trivial.
p=2 - quadratic equation.
p=3;4 - cubic and quartic equations. Algebraic solutions are known (wiki), but they are messy, especially for quartic.
We can also solve p=l/m, where l,m=1;2;3;4. For example, p=3/4 :
y^(3/4) + y=c
->[y^(1/4)]^3+[y^(1/4)]^4=c
y^(1/4)=t, y=t^4
->t^3 + t^4=c
If p
Maximilian Koch
It seems that equation y^p +y=c can be solved by Lagrange inversion theorem. Here is solution to equation y^p - y=c :
en.m.wikipedia.org/wiki/Lagrange_inversion_theorem
,Section "Example" . I guess that this equation can be turned to our form by using substitution y=(-1)^[1/(p-1)] *u, I extrapolate from article
en.m.wikipedia.org/wiki/Bring_radical
,section "Normal forms", paragraph "Bring-Jerrard normal form".
However, this substitution involves complex numbers in general, so... Dunno.
Give a proof of the error term in Simpsons Rule. I dare you!
AncientAncestor
Who likes calculating errors ? They tend to be dull, messy and tedious...
Now this theorem is something else ! I suspect that proving it requires some serious knowledge of theory of functions of complex variable...
@@Hexanitrobenzene Charity Livestream? Watching Papa Flammys slow descent into madness as the hours pass of him performing increasingly more tedious but absolutely essential calculations in order to prove an absolutely essesntial result in the field of numerical integration would probably be the best thing to ever happen to the internet.
do W(-π/2)
Hi, is it even possible to solve for x in 3^x+x^2-2=0 ?? Make a video plsss
Lucas Depetris
It seems that the constant term causes a lot of trouble. A similar equation without it is solved in wiki :
en.m.wikipedia.org/wiki/Lambert_W_function
,section "Solutions of equations", example 3. However, this method cannot be applied when constant term is present, because then square root cannot be extracted.
Help, Papa ! :D
I still can't understand Taylor Functions... did you do a video about this?
Also, what is that big D at 1:05?
That's mine.
Oh okay, thanks papa!
@@NoNTr1v1aL Lol
@@misotanniold787 okay, I got the idea. Now I have to elaborate on this argument! Thank you!
Proof of Lagrange Inversion Theorem:
May y(x,b)= x + b*f(y),
Near b = 0, we get the taylor expansion in b = 0:
x + sum from k = 1 to infinity x^k / k! (partial^k / partial x^k) (y(x,0))
(1)
y(x,b) = x - b*f(y(x,b))
partial y/partial x - 1 - b (partial f/partial y) (partial y/ partial b) = 0 means partial y/partial x (1-b*f'(y)) = 1
partial y/partial b - f(y(x,b)) - b (partial f/partial y) (partial y/ partial b) 0 means partial y/partial b (1-b*f'(y)) = f(y)
Therefore, partial y/partial b = f(y) partial y / partial x
(2)
Now we want to show that for all n >0, (partial^n y/partial ^n b) = (partial^n-1 / partial x^n-1) (f^n (y) partial y/partial x).
partial^2 y/ partial b^2 = (partial/partial b) (partial/partial b) (y) = (partial/partial b) (f(y) partial y/partial x)) = (partial/partial x) (f^2(y) partial y/partial x)
By recursion, we get the said 3rd step formula.
(3)
In (x,0), y=x.
partial y/partial x = 1
and partial^n/partial b^n (y(x,0)) = partial^n-1/partial x^n-1 (f^n(x)).
By (1), we get the forth step:
y = x + sum from k=1 to +infinity of b^k/k! (partial^k-1/partial x^k-1) f^k(x)
(4)
We have now proved the Lagrange Inversion theorem at x=0. A simple change of variable z=x+x_0 makes it in any real point.
:")
Where are you studying????
Flammable Maths okk ty. You are doing a grt job, keep it up. U r doing graduation or msc???
So are we having an affair with the Lambert W function? I see through the lies
:D
fiRsT11!!
Please no more lambert videos 😖 too much for me
More lambert, more lambert more lambert