The biggest difficulty with the Lambert W function is that, like the logarithmic function, it becomes multivalued when continued to the complex plane, but unlike with the complex logarithm, the branches complex W multifunction cannot be obtained by simply adding integral multiples of 2·π·i. In fact, the other complex branches of the Lambert W multifunction cannot be computed analytically from the values of the principal branch, unlike with the complex logarithm.

@sabinrawr

3 жыл бұрын

I was initially saddened by BPRP's polite decline to do all values for #15, but after your explanation, I'm actually pleased by this decision! Thanks!

@__hannibaalbarca__

2 жыл бұрын

I ll investigate this branches when I have free times it's interesting.

You’re back!!!!!!! 😍😍😍

@blackpenredpen

3 жыл бұрын

I am back from the 🏖

@vladimirkhazinski3725

3 жыл бұрын

Kiss already!

@banana6108

3 жыл бұрын

@@vladimirkhazinski3725 😑

@ffggddss

3 жыл бұрын

@@blackpenredpen Is that an umbrella in the sand that you're back from? (I.e., the beach?) Fred

3b1b has π creatures. BpRp:Here's my fish.

@colt4667

3 жыл бұрын

BPRP uses a fish. Professor Julio uses a tomato.

@chin6796

3 жыл бұрын

MCU is for math creatures universe

The longer the beard, the higher the wisdom

@glegle1016

3 жыл бұрын

Dude needs to shave. That "beard" looks disgusting

@muhammadusmonyusupov2556

3 жыл бұрын

@@glegle1016 common man. That's not your business

@Kitulous

3 жыл бұрын

i just broke 69 likes. sorry it's 70 now, couldn't resist

@just_a_dustpan

3 жыл бұрын

The beard doesn’t make the wisdom. The wisdom makes the beard

@agarykane2127

2 жыл бұрын

@@just_a_dustpan you must have a long beard if you say such wisdom

I have a master degree in mathematics (although I'm working as an accountant) and watch math youtube videos occasionally, but I never heard of this function before. Very strange. It reminds me of the guy who did the algebra portion of my master's oral exam saying that he had never seen one of the results that I worked on in the analysis portion of the exam (Stone-Weierstrass Theorem). Math is a big field, and there's plenty out there to learn. Sometimes you just never study something that's been studied by other people because you never needed to know it for what you worked on.

@kepler4192

2 жыл бұрын

Thanks to his videos, I’ve learnt about tetration and lambert W function

@adi8oii

Жыл бұрын

I am taking the Calc course at ug level rn, and I am having to learn this method because my calc textbook (Spivak) had a strange question in the very first chapter: solve the inequality x + 3^(x) < 4 (Spivak always asks weird questions lmao). So anyway, after putting it through an online inequality solver I learnt that it requires the Lambert (W) function and here I am...

@roccorossi5396

Жыл бұрын

Me too for x^x =2^64

@spinecho609

8 ай бұрын

im very surprised it hasnt come up as a physicist, especially since x+e^x kind of forms are so common!

@nesto9889

6 ай бұрын

you use eulers number in physics? im scared@@spinecho609

wow that was 48 mins ... it passed like 5 mins !!

@jesseolsson1697

3 жыл бұрын

it's amazing what learning feels like when you have a great teacher in a subject you love

Me during my high school Calc math class: *on my phone for the whole class* Me during a 48 min math video: *Fully engaged and even pause to do problems myself*

@danielvictoria3814

3 жыл бұрын

Just watch this impressive Maths channel... it’s very nice like this kzread.info/dron/ZDkxpcvd-T1uR65Feuj5Yg.html

@prohacker1373

2 жыл бұрын

are u allowed to use phone in the class?( i am high school student from india)

@kepler4192

2 жыл бұрын

@@prohacker1373 I’m pretty sure you shouldn’t be allowed

@jodikirsh

2 жыл бұрын

@@prohacker1373 We aren't allowed to, but most kids try to do it secretly.

Anyone still laughing at his joke “just buy another calculator”

@M-F-H

3 жыл бұрын

On that token, if your calculator doesn't have an "ln" button, then most likely it also doesn't have an e^x button (neither a ^ button) [any known counter example??] So the (1) is of limited practical beyond the first step of approximation where you can put e¹ = 2.7 ...

@waynewang5773

3 жыл бұрын

yea i am lol

32:45 This is why we all love your teaching. This is the method nobody teaches in the universities and nobody else shows on youtube etc. Thank you very much :)

W(-pi/2)=W(ln(i)*e^ln(i))=[ln(i)]. Fun math man, thank you for the problem. My friend and I had a lot of fun taking it on.

@user-Loki-young0515

2 жыл бұрын

πi/2

@SebastienPatriote

2 жыл бұрын

I feel so dumb. I thought the question was W(pi/2), not W(-pi/2). So I found +/- ipi/2 for W(-pi/2) but kept searching. I like your solution too!

Your exposition on the Lambert W function has been quite illuminating. I would say it generated at least 100 foot-Lamberts of Luminance. Thanks.

@danielvictoria3814

3 жыл бұрын

Just watch this impressive Maths channel... it’s very nice like this kzread.info/dron/ZDkxpcvd-T1uR65Feuj5Yg.html

YES!!! FINALLY!!!! I am waiting for this for so long!!! Thank you!!

@danielvictoria3814

3 жыл бұрын

Just watch this impressive Maths channel... it’s very nice like this kzread.info/dron/ZDkxpcvd-T1uR65Feuj5Yg.html

i was literally thinking about coding a W function calculator (with Newton's method) the other day i really love your and peyam and penns content so much. makes me want to start a channel myself

I've been looking for videos about the subject for weeks since you talked about it in your videos, and now I have the perfect to understand the w function perfectly! Thank you ao much and keep up the good work, you're the best !

Dude you got a talent of teaching ! not only this is explained really well but it's also entertaining. It's a pleasure to learn math from you. Thanks to you, at my calc 2 exam I got 18/20. So many thanks and don't stop because your are saving grades over here !

This was really helpful in understanding Lambert's motives for creating such a function. Thank you!

Thanks for this. I came across this x.e^x function several years ago when modelling multi-access protocols (ALOHA in particular) - and discovered, it is an example of a "one way function" i.e. it has no inverse. I did not know about the W function (I am an Engineer, not a Mathematician). I managed using MathCad to plot the inverse graph, and it demonstrated beautifully the limiting traffic intensity of ALOHA (due to access collisions), which then bends back on itself, from the limit. It is a poor, early, multi-access, protocol developed at the University of Hawaii. So it is not one way. The W function is the inverse. By the way blackpen like the beard.

Used wolframalpha to simplify a complex equation and it returned with a product log function.... Having no idea what a “product log function” is, this video has been very helpful

Videos like these are a treasure, please keep these coming.

It's amazing!!!! I'm studying Newton's method now in the course of Numerical Methods for engineering!!!😆😆😆

@abdomohamed4962

3 жыл бұрын

where are you from ?? .. Im studying it too

@ameer_er2181

2 жыл бұрын

@@abdomohamed4962 من ایرانیم

Thank you so *freaking* much!!!!This was one of the best lectures on your channel.Need more lectures like this. Ah also I know it's late but *Happy New Year*

Fantastic, now I get what W(x) really does and how it works. Very nice explanations. I did not saw this one my subscriptions 5 weeks ago :( , but it is never too late.

THE BEST VIDEO OF YOUR CHANNEL Thank you for providing to learners bigger video like this one that don t necessarily make as much views as the other oned because it is less "sexy" yet more complete and useful

Thank you for the formula for integrating inverse functions :-)

Great video. At 25:00 I ended up getting 2-W(e^2) instead of ln(W(e^2)) and thought I had done something wrong, but it turns out they are the same thing.

Bravo Mr Bprp, nice and clear video, very understandable...thank you!

Cannot find this in as much detail elsewhere. Thank you

So at the end, #s 14 & 15 show us that ln(i) = W(-½π) Fred

@abeldiaz1539

3 жыл бұрын

Couldn't we solve it more easily by using Euler's Formula? I haven't finished watching the video but... e^(i*pi) = -1 ln both sides i*pi = ln(-1) -1 can be expressed as i^2 so the above is the same as i*pi=ln(i^2) Drop the exponent down to multiply with the natural log i*pi=2*ln(i) Flip and divide both sides by 2 and ln(i)=(pi/2)i Similarly using Euler's formula, e^(ipi) =-1 sqrt both sides and express sqrt as a 1/2 exponent and sqrt of -1 as i You have i = e^(ipi/2) Knowing -pi/2 is the same as (pi/2)*(-1) which is the same as (pi/2)*i*i and plugging in that other expression for i gives (pi/2)i*e^(ipi/2), which we see we have the same expression multiplying our e as to what the exponent of e is, so the Lambert W of that expression is just (pi/2)i, which was the same as the ln(i) we saw earlier. There might be an easier way, but that's how I solved it. 😅 EDIT: Just finished watching, and I see my method doesn't account for the additional solutions to ln(i), but my question is for #15) why is it okay for him to not include the "+2npi" to his exponent for e?... Could you have added the expression to the inside of the original W(-pi/2) part or no? 🤔

@ivanzivkovic7572

2 жыл бұрын

@@abeldiaz1539 the line of reasoning you used from the Euler identity doesn't work, logarithm identities that work for real numbers don't work for the complex valued logarithm, you could say -1 is (-1)^2 and get the same result for ln(-1), which would be incorrect bc ln(-1) is (3pi/2 + 2k*pi)*i, k integer Anyway, you can't do the same as he does in 14 in 15 bc the branches of the Lambert W function are not 2k*pi*i apart, if you look at the definition of Lambert W you'll see that it is an inverse of x*e^x, and if you were to add 2k*pi*i to the exponent of e, even though that factor won't change you'd have to add it to the x that multiplies it as well, which means you would get a different result

Again, you are explaining something what recently grabbed my attention when i was wondering through world of wikipedia math!

@danielvictoria3814

3 жыл бұрын

Just watch this impressive Maths channel... it’s very nice like this kzread.info/dron/ZDkxpcvd-T1uR65Feuj5Yg.html

I needed this, because I’ve never truly known what Lambert W was

Really good video, beautifully put together. Thank you !!

Thank you! The format of comparing natural logs with Lambert functions is very helpful. Could you prepare a similar video with comparing circular trig functions and how they relate to analogous Jacobi elliptic functions as well as inverse trig functions and integrals and how they relate to analogous elliptic integrals?

Outstanding content and presentation. I really enjoy your videos!

I love this kind of long videos a lot... And especially when it's by bprp...

Very, very good video ! 48 minutes top level

Excellent idea: work with the inverse of the function and observe properties: W (f (x)) = x.

I did the integral of the Lambert W function by integrating both sides of W(x) = x/e^W(x). Glad to discover I got the right answer this way.

@danielvictoria3814

3 жыл бұрын

Just watch this impressive Maths channel... it’s very nice like this kzread.info/dron/ZDkxpcvd-T1uR65Feuj5Yg.html

The beard of wisdom :) Absolutely great explanation.

I've reached the point in High School where I can actually follow what's going on :D

In Azerbaijan, it is 8:00 and I wake up for my IELTS exercise; however I am watching you albeit all of them are known by me😁

Thanks for your hard work and good videos ))) o love this function ahahha ))) you are the best math teacher ))

You cheeky little blighter!) Love all ur content, especially about imaginary equations like cos(x)=2 etc. Peace!!!!!

20:04 e^0 plus e^0 definitely equals one.

@blackpenredpen

3 жыл бұрын

Lol! I was thinking I had 2 on the right hand side, just like my next question.

Legendary!!! Salute you from Argentina

Every engaging; excellent work!

When I watch a blackpenredpen video: I don't understand but I feel like I got smarter.

@umeshprajapati7381

3 жыл бұрын

Is this function f(x)=(x)^3/2 is differential at x=0

@umeshprajapati7381

3 жыл бұрын

Plz sir give solution

@asparkdeity8717

3 жыл бұрын

@@umeshprajapati7381 0

Loved every second of the video!

Thank you so much. I really enjoyed this!!

@danielvictoria3814

3 жыл бұрын

Just watch this impressive Maths channel... it’s very nice like this kzread.info/dron/ZDkxpcvd-T1uR65Feuj5Yg.html

Problem 7 alternative: e^x-e^(-x)=1 1/2(e^x-e^(-x))=1/2 sinh x = 1/2 x = arcsinh 1/2 x = 0.481

Wow great video! I learned a lot, thank you!

@danielvictoria3814

3 жыл бұрын

Just watch this impressive Maths channel... it’s very nice like this kzread.info/dron/ZDkxpcvd-T1uR65Feuj5Yg.html

Brilliant lecture! Thank you.

41:53 - It's a + Great video

So that's the product logarithm... but what's x+xe^x=2? What if we have (1/x + 1)e^x = 2?

@angelmendez-rivera351

3 жыл бұрын

For this, you need a generalization of the Lambert W concept.

@nathanaelmoses7977

3 жыл бұрын

Newton's method? Or you can somehow solve it with w(x)? Idk im terrible at math

@einsteingonzalez4336

3 жыл бұрын

@@nathanaelmoses7977 Newton's method isn't ideal because it gives a numerical answer, and such answers are mostly approximations. Finding the exact inverse helps us get the value quicker.

@angelmendez-rivera351

3 жыл бұрын

@@nathanaelmoses7977 It has been proven that you cannot solve equations of the form (a·x + b)·e^x = c·x using the Lambert function unless b = 0. This is why you need a generalization of the concept. There is one publication I once saw regarding a generalization that reminded me of the hypergeometric functions, essentially defining a function that is used to solve the equation e^(c·x) = a·[x - p(1)]·•••·[x - p(n)]/([x - q(1)]·•••·[x - q(m)]), although I do not remember if the publication was peer-reviewed. Once I find it again, I will link it here.

@nathanaelmoses7977

3 жыл бұрын

@@angelmendez-rivera351 Interesting

Bro ur always dishing free knowledge

15:40 answer is iπ/2 or to include all values of theta- i(2nπ+-π/2 ∪ nπ+(-1)^nπ/2)

Happy new year to you as well :)

can you please give an example where there are 2 solutions for lambert W, that is, there is a primary W0 solution and W-1 secondary solution as well and show how to use the newton-raphson on each branch?

In the equation x+e^x = 2, I went this way: 1 = (2-x)e^(-x) => e^2 = (2-x)e^(2-x), since xe^x is defined for [-1,+infty) (to use W), then we'd need 2-x >= 1 which means x1 won't have any solutions since x+e^x > 1 + e (since e^x is strictly increasing) and since e>1, then x+e^x > 2, right?

Could you help how to calc W-1 using a iterative method, as newton-raphson, for instance? :)

At the end, we finally know that a man who teach bicycle to that boy is a genius-man...

I haven't study yet this theory in my school but after watching it's seems like ... .

Hi Dr., Have you seen Dr. Peyam’s recent video on time shifted DE? He solved it for one particular shift but to do it in general requires the Lambert W function.

@drpeyam

3 жыл бұрын

Good point!!!

Thank you blackpenredpen!!! I really needed this lecture!! I just watched over and it's so concise (and better than other lessons lol) :D Thank you so much!!!!!!!!!

@blackpenredpen

3 жыл бұрын

Thank you!

But then do we need to also introduce a new function to solve equations with the form : x • w(x) = a If yes, do each time we introduce a new function to solve equations, does this new function also introduce new equations wich need a new function to be solved ?

pls tell sum of series n tends to infinity 1/(lnn)^2 is convegent or divergent?

ln(i) = i(pi/2) Solved using polar representation of i and Euler's formula. Assumed n = 0, but other values of n will also give other answers. i = 1e^{(i) (2n + 1) (pi/2)} because r = 1 and t = (2n+ 1) pi/2 taking ln(x) on both sides: ln(i) = i(2n+1)(pi/2) It gives us a family of equations.

Hello Mr BRRP. When I plug W(-pi/2) into my pari-GP calculator I get the message "domain error in W" Remember you showed that the domain of W is [-1/e, inf) and -pi/2 < -1/e. Could you please clarify.

Is there a way to get that derivatives for you poster?

Are ln(i) and W(-pi/2) both equal to i(pi/2) ? Got both of these using the polar form of i

This is great! Thank you!!!

Thanks! Can you please share what are the real life applications of Lambert W function??

32:50, I've never seen this method before ... thanks!

can you please make a follow-up video on different index values of the lambert function. and the expansion for W0 and W-1 (as they only provide the real solutions), my question arises when i tried to solve 2^x-x^8. I could calculate the 2 real roots but could not calculate the 3rd one as its value came from W-1 expansion Thx

Hello Mr hope you are doing well. l would like to ask you about the branches of the Lambert W function , Wn(X) . What does it differ from W(X).

is it possible to use the W function to solve x + lnx + (x^2+x)^(1/2) = 1? or just lnx + (x^2+x) = 0?

Hi. This 0.5x + e^-x = 2 is similar to #8 but I cannot solve it. Any clue? Thanks.

I am in love with this video

the first is (i(pi))/2.the second is I. Thank you professor

Is there a name for the inverse of (ln x)/x, or ⁿ√n (that is, the nth root of N, or xth root of x)?

In the Newton's Method, why do we necessarily start with X1 = 1 ? and why don't we get the appropriate answer if we start with X1 = 0 or some other value?

Okay, so you got a different final result using the Lambert W function. Please explain.

This was just recommended to me after I watched your (sinx)^sinx video. Perfect timing! Are there properties of the lambert W function that are analogous to addition and product rule for logs?

@hunterhare7647

Жыл бұрын

I think there's a "change of base" formula for the Lambert W function: e.g. x * n^x = y. In this case, the solution is W(y * ln(n))/ln(n).

An interesting question: does the equation x+x^x=3 for example need another different inverse function?

Yess! This is so cool.

I wonder if there is a real-life problem that requires to solve for an unknown, given its product with something raised to itself as power. Any idea?

I still confuse how can we find the value of Lambert W function W(ln2). In the case of the equation x^x = 2, I know the value x = 1.559 by wolfram alpha program from the other video. Meanwhile I dont have thats program. So how do we calculate the product log Lambert W functiom W(ln2) by hand manual calculation without thats program? (wolfram alpha ).Please explain me

I have one further question: qhat is the Taylor series expansion of W(x)?

I love everything you have done with the lambert W function and I have been teaching my students and colleagues how to use it! It has made it where I can solve almost any transcendental function equation... but what about something like (e^x)*(log(x,e))=15? Where x is both the exponent of the natural exponential and the BASE of the logarithm... now the the x's are 2 "levels" apart instead of 1 "level" like with the Lambert W function.

@walexandre9452

2 жыл бұрын

I think this exercise cannot be solved by the Lambert W function. Some exercises having 2 "levels" can be solved... but not this one.

A lecture on use of dummy variable in Combinatorics pls

Thank you very much!

Why do we ignore the bottom bit of the x(e^x) graph?

Can anyone solve: x + a * b * e^(x/c*d) = h, where a,b,c,d,h are positive real numbers. We saw how to solve x + e^x = a, but in my equation I have these weird constants messing up with me...

We define W(x) being the inverse of xe^x and ln(x) being the inverse of e^x. What if we also define W2(x) being the inverse of x×2^x as log2(x) is for 2^x We can extend this even further: W_y (x) is the inverse of x×y^x in 2021

@bernardovitiello

3 жыл бұрын

We definitely could, and that would probably be a good idea, but you can represent every other W_y using W on the base e (much like one can do with logarithms) W_y(x)*y^W_y(x)=x We can change the base of the first exponent using logarithms So W_y(x)*e^ln(y)*W_y(x)=x Now to apply W we must make sure the coefficient and the exponent are the same, which can be achieved by multiplying both sides by ln(y) ln(y)*W_y(x)*e^ln(y)*W_y(x)=ln(y)*x Finally W(x) is appliable so W(ln(y)*W_y(x)*e^ln(y)*W_y(x))=W(ln(y)*x) ln(y)*W_y(x)=W(ln(y)*x) W_y(x)=W(ln(y)*x)/ln(y)

@angelmendez-rivera351

3 жыл бұрын

You totally can do that, but doing this is relatively pointless, and there is no good incentive for it. The reason we even still talk about logarithms in other bases is because the binary logarithm has very important applications in computing, and the decimal logarithm in engineering, as well as the fact that logarithms with different bases are basically historical relics. These things do not hold for the W(x), as the study of this function in rigorous detail is a lot more modern, and there are virtually no applications to using an analogue of this in a different base.

Great! Why don't you do a lecture about zeta function?

Nice one, i love it!!

Hey, I was trying to solve the equation x + b sin(x) + a = 0, and eventually I got to the equation (c/t) exp(it) - (1/t) exp(-it) = d, which is similar to y exp(y), so I was wondering if anyone know if there's a way to solve it.

@walexandre9452

2 жыл бұрын

After an analysis of f(x) = x + bsin(x), we can see that f^-1(x) cannot exist in many intervals (but it can exist in anothers). So, a general solution for real x maybe doesn't exist as x = f^-1(y) . Complex numbers/functions or implicit solution are better candidates.

Hello Mr BPRP. Please do a video on L'Hopital's rule and prove why it works, that should be in Analysis 101.

This dude is literally Mr Maths 👌

Does W(x) exist for x in complex plane?

5d) Eulers identity (with tau since its more elegant): e^(i [tau] 1/4) = i (1/4 rotation of the complex plane = i) ln(e^x)=x so ln(e^(i [tau] 1/4))= i [tau] 1/4 or just ½pi*i