Lambert W Function (domain, range, approximation, solving equations, derivative & integral)

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Comparing the natural log function ln(x) and the product log function W(x).
0:00 ln(x) vs. W(x)
1. Solve e^x=2 & ln(2), 0:16
2. Solve xe^x=2 & W(2), 3:01
Newton's Method: • Newton's method and Om...
3. Domain, range, & graph for ln(x), 6:53
4. Domain, range, & graph for W(x), 9:16
5. Nice values for ln(x), 13:30
6. Nice values for W(x), 15:47
7. Solve e^x-e^(-x)=1, 19:54
8. Solve x+e^x=2, 22:40
9. Solve x^x=2, 25:25 (the t-shirt teespring.com/cute-math-cat-6)
10. Derivative of ln(x) by implicit differentiation, 27:30
11. Derivative of W(x) by implicit differentiation, 28:35
Integral of an inverse function, 32:50
12. Integral of ln(x) by the formula, 36:52
13. Integral of W(x) by the formula, 38:42
14. What's ln(i)? 43:50
15. What's W(-pi/2)? 46:11
Read more about the Lambert W function on Wikipedia: en.wikipedia.org/wiki/Lambert...
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Опубликовано:

6 июл 2024

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Комментарии : 457

@angelmendez-rivera351 3 года назад

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 года назад

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 года назад

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

@abhishekprasad6350 3 года назад

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

@colt4667 3 года назад

BPRP uses a fish. Professor Julio uses a tomato.

@chin6796 3 года назад

MCU is for math creatures universe

@drekkerscythe4723 3 года назад

The longer the beard, the higher the wisdom

@glegle1016 3 года назад

Dude needs to shave. That "beard" looks disgusting

@muhammadusmonyusupov2556 3 года назад

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

@Kitulous 3 года назад

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

@just_a_dustpan 3 года назад

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

@agarykane2127 2 года назад

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

@drpeyam 3 года назад

You’re back!!!!!!! 😍😍😍

@blackpenredpen 3 года назад

I am back from the 🏖

@vladimirkhazinski3725 3 года назад

Kiss already!

@banana6108 3 года назад

@@vladimirkhazinski3725 😑

@ffggddss 3 года назад

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

@abdomohamed4962 3 года назад

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

@jesseolsson1697 3 года назад

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

@stevenglowacki8576 2 года назад

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 года назад

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

@adi8oii Год назад

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 Год назад

Me too for x^x =2^64

@spinecho609 7 месяцев назад

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

@nesto9889 6 месяцев назад

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

@marianopatino939 3 года назад

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 года назад

Just watch this impressive Maths channel... it’s very nice like this ru-vid.com/show-UCZDkxpcvd-T1uR65Feuj5Yg

@prohacker1373 2 года назад

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

@kepler4192 2 года назад

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

@jodikirsh 2 года назад

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

@wristdisabledwriter2893 3 года назад

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

@M-F-H 3 года назад

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 года назад

yea i am lol

@damianbla4469 3 года назад

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 :)

@girlgaming1993 3 года назад

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 года назад

πi/2

@SebastienPatriote 2 года назад

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!

@JohnSmith-vd6fc 3 года назад

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 года назад

Just watch this impressive Maths channel... it’s very nice like this ru-vid.com/show-UCZDkxpcvd-T1uR65Feuj5Yg

@helo3827 3 года назад

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

@danielvictoria3814 3 года назад

Just watch this impressive Maths channel... it’s very nice like this ru-vid.com/show-UCZDkxpcvd-T1uR65Feuj5Yg

@sharpnova2 3 года назад

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

@nosnibor800 3 года назад

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.

@VesaKo 3 года назад

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

@a1175779 3 года назад

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

@flamitique7819 3 года назад

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 !

@eng.giacomodonato8514 3 года назад

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

@abdomohamed4962 3 года назад

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

@ameer_er2181 2 года назад

@@abdomohamed4962 من ایرانیم

@tudor5555 3 года назад

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 !

@Zero-tg4dc 2 года назад

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.

@MrMatthewliver 3 года назад

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

@gtweak7 3 года назад

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

@ffggddss 3 года назад

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

@abeldiaz1539 3 года назад

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 года назад

@@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

@legendarynoob6732 3 года назад

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*

@simonharris3797 3 года назад

Cannot find this in as much detail elsewhere. Thank you

@alberteinstein3612 2 года назад

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

@MaximQuantum 2 года назад

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

@61rmd1 3 года назад

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

@assassin01620 3 года назад

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

@blackpenredpen 3 года назад

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

@ikocheratcr 3 года назад

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.

@theimmux3034 3 года назад

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 года назад

Just watch this impressive Maths channel... it’s very nice like this ru-vid.com/show-UCZDkxpcvd-T1uR65Feuj5Yg

@Kestrel2357 3 года назад

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

@danielvictoria3814 3 года назад

Just watch this impressive Maths channel... it’s very nice like this ru-vid.com/show-UCZDkxpcvd-T1uR65Feuj5Yg

@pierre7770 2 года назад

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

@RomainPuech 3 года назад

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

@MathNotationsVids 3 года назад

Outstanding content and presentation. I really enjoy your videos!

@helo3827 3 года назад

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

@umeshprajapati7381 3 года назад

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

@umeshprajapati7381 3 года назад

Plz sir give solution

@asparkdeity8717 3 года назад

@@umeshprajapati7381 0

@MrAnonymousfan1 3 года назад

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?

@herculesmachado3008 3 года назад

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

@gouharmaquboolnitp 3 года назад

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

@JMCoster 4 месяца назад

Very, very good video ! 48 minutes top level

@axelgiovanelli8401 2 года назад

Legendary!!! Salute you from Argentina

@lietpi Год назад

Loved every second of the video!

@pragalbhawasthi1618 3 года назад

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

@KjartanAndersen 3 года назад

The beard of wisdom :) Absolutely great explanation.

@gorilaylagorila2540 3 года назад

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

@danielvictoria3814 3 года назад

Just watch this impressive Maths channel... it’s very nice like this ru-vid.com/show-UCZDkxpcvd-T1uR65Feuj5Yg

@jiaweigong3411 9 месяцев назад

Every engaging; excellent work!

@user-nr3yb3ki9p 2 года назад

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

@sawyerandrobbie 3 года назад

This is great! Thank you!!!

@ThePhysicsMathsWizard 3 года назад

Nice one, i love it!!

@hsh7677 3 года назад

Thank you so much. I really enjoyed this!!

@danielvictoria3814 3 года назад

Just watch this impressive Maths channel... it’s very nice like this ru-vid.com/show-UCZDkxpcvd-T1uR65Feuj5Yg

@alexyanci7974 3 года назад

41:53 - It's a + Great video

@anushrao882 3 года назад

Yess! This is so cool.

@garyewart9185 2 года назад

Brilliant lecture! Thank you.

@kdmq 3 года назад

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

@papanujian7758 3 года назад

waiting so long. finally

@SamiSalah92 2 года назад

LOVE YOU

@jschnei3 3 года назад

I am in love with this video

@sueyibaslanli3519 3 года назад

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😁

@cthzierp5830 6 месяцев назад

Happy new year to you as well :)

@hunterkorble9134 2 года назад

Bro ur always dishing free knowledge

@neilgerace355 3 года назад

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

@user-xk3en1tj2e Год назад

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

@chriswinchell1570 3 года назад

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 года назад

Good point!!!

@einsteingonzalez4336 3 года назад

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 года назад

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

@nathanaelmoses7977 3 года назад

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

@einsteingonzalez4336 3 года назад

@@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 года назад

@@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 года назад

@@angelmendez-rivera351 Interesting

@nicopb4240 3 года назад

Thank you very much!

@saumyakathuria5594 3 года назад

A lecture on use of dummy variable in Combinatorics pls

@78rera Год назад

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

@cuboid2630 3 года назад

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 года назад

Thank you!

@UpstartRain 2 года назад

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 Год назад

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).

@Reallycoolguy1369 2 года назад

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 года назад

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

@MrHK1636 3 года назад

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 года назад

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 года назад

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.

@w__a__l__e 3 года назад

thanks dude.

@Casey-Jones 3 года назад

wow ...... extreme hard core stuff

@girlgaming1993 3 года назад

15:36. Woohoo i solved it (pretty sure) Basically I guessed and got it right. It uses e^(pi*i)=-1. I thought that by square rooting both sides it would give me sqrt(-1) which is i, the goal. (e^(pi*i))^(1/2) multiply the exponents and get e^((pi*i)/2) (which equals i). So, ln(i)=(pi*i)/2

@violintegral 2 года назад

But it's also equal to i(π/2+2πn) for all integers n.

@hasanjakir360 6 месяцев назад

at 5:40 "it will work, because I did it already" had me laughing.

@seghirimoha9026 2 месяца назад

Thank you very much

@abdomohamed4962 3 года назад

nice video bro

@mista5796 4 месяца назад

This dude is literally Mr Maths 👌

@al-shaibynanong1237 3 года назад

I've been waiting for this. Thank you sir🥰

@theimmux3034 3 года назад

W(x) is such a cool piece of math

@Jamiree7 3 года назад

Very good lecture

@madhavjuneja4333 2 года назад

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

@anurag5565 3 года назад

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.

@darkahmed_codeforces_ Год назад

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

@SimonPetrikovv 3 года назад

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?

@Xnoob545 3 года назад

19:35 brb

@huzefa1991 2 года назад

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

@HariHari-wv6ht 3 года назад

happy new year belately

@archivewarrior8535 2 года назад

35:17 he really just makes a dashed line like it’s nothing. I’ve never seen that

@blackpenredpen 2 года назад

Thanks lol. You should see Professor Walter Lewin tho!

@wxadbpl 3 года назад

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?

@baptiste5216 3 года назад

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 ?

@miguelzabala 3 года назад

Awesome

@brucefrizzell4221 3 месяца назад

Retired computer geek learning NEW math for FUN . Just joined.

@NonTwinBrothers 2 года назад

This one's for the books. The bprp epic

@SidneiMV 3 месяца назад

*e^[W(x)] = x/W(x)* wow! *how useful* it is!!

@curtiswfranks 3 года назад

The W(x e^x) notation may seem weird, but that is because we do not do polynomial notation well. In this case, though, we do have some recourse: W = (id • exp)^(-1).

@joshmcdouglas1720 3 года назад

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

@indarajgochermaths5176 3 года назад

So nice video Great

@that_one_guy934 Год назад

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