Lambert W Function (domain, range, approximation, solving equations, derivative & integral)
Рет қаралды 224,819
Күн бұрынComparing 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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@angelmendez-rivera351 3 жыл бұрын
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.
@abhishekprasad6350 3 жыл бұрын
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
@drekkerscythe4723 3 жыл бұрын
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
@abdomohamed4962 3 жыл бұрын
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
@drpeyam 3 жыл бұрын
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
@marianopatino939 3 жыл бұрын
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 kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@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 Жыл бұрын
@@prohacker1373 We aren't allowed to, but most kids try to do it secretly.
@wristdisabledwriter2893 3 жыл бұрын
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
@stevenglowacki8576 2 жыл бұрын
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 6 ай бұрын
im very surprised it hasnt come up as a physicist, especially since x+e^x kind of forms are so common!
@nesto9889 5 ай бұрын
you use eulers number in physics? im scared@@spinecho609
@girlgaming1993 3 жыл бұрын
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 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!
@damianbla4469 3 жыл бұрын
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 :)
@JohnSmith-vd6fc 3 жыл бұрын
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 kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@sharpnova2 3 жыл бұрын
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
@helo3827 3 жыл бұрын
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 kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@a1175779 3 жыл бұрын
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
@nosnibor800 3 жыл бұрын
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.
@eng.giacomodonato8514 3 жыл бұрын
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
@Zero-tg4dc 2 жыл бұрын
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.
@flamitique7819 3 жыл бұрын
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 !
@MrMatthewliver 3 жыл бұрын
Thank you for the formula for integrating inverse functions :-)
@ffggddss 3 жыл бұрын
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
@VesaKo 3 жыл бұрын
This was really helpful in understanding Lambert's motives for creating such a function. Thank you!
@tudor5555 3 жыл бұрын
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 !
@alberteinstein3612 2 жыл бұрын
I needed this, because I’ve never truly known what Lambert W was
@gtweak7 3 жыл бұрын
Videos like these are a treasure, please keep these coming.
@legendarynoob6732 3 жыл бұрын
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*
@MaximQuantum 2 жыл бұрын
I've reached the point in High School where I can actually follow what's going on :D
@alexyanci7974 3 жыл бұрын
41:53 - It's a + Great video
@assassin01620 3 жыл бұрын
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.
@61rmd1 2 жыл бұрын
Bravo Mr Bprp, nice and clear video, very understandable...thank you!
@MathNotationsVids 3 жыл бұрын
Outstanding content and presentation. I really enjoy your videos!
@helo3827 3 жыл бұрын
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
@simonharris3797 3 жыл бұрын
Cannot find this in as much detail elsewhere. Thank you
@ikocheratcr 3 жыл бұрын
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.
@pierre7770 2 жыл бұрын
Really good video, beautifully put together. Thank you !!
@theimmux3034 3 жыл бұрын
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 kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@RomainPuech 3 жыл бұрын
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
@MrAnonymousfan1 3 жыл бұрын
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?
@pragalbhawasthi1618 3 жыл бұрын
I love this kind of long videos a lot... And especially when it's by bprp...
@JMCoster 3 ай бұрын
Very, very good video ! 48 minutes top level
@jiaweigong3411 8 ай бұрын
Every engaging; excellent work!
@herculesmachado3008 3 жыл бұрын
Excellent idea: work with the inverse of the function and observe properties: W (f (x)) = x.
@kdmq 2 жыл бұрын
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
@Kestrel2357 3 жыл бұрын
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 kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@lietpi Жыл бұрын
Loved every second of the video!
@axelgiovanelli8401 2 жыл бұрын
Legendary!!! Salute you from Argentina
@hunterkorble9134 2 жыл бұрын
Bro ur always dishing free knowledge
@garyewart9185 2 жыл бұрын
Brilliant lecture! Thank you.
@KjartanAndersen 3 жыл бұрын
The beard of wisdom :) Absolutely great explanation.
@darkahmed_codeforces_ Жыл бұрын
the first is (i(pi))/2.the second is I. Thank you professor
@gouharmaquboolnitp 3 жыл бұрын
I haven't study yet this theory in my school but after watching it's seems like ... .
@user-nr3yb3ki9p 2 жыл бұрын
Thanks for your hard work and good videos ))) o love this function ahahha ))) you are the best math teacher ))
@gorilaylagorila2540 3 жыл бұрын
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 kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@sawyerandrobbie 3 жыл бұрын
This is great! Thank you!!!
@hsh7677 3 жыл бұрын
Thank you so much. I really enjoyed this!!
@danielvictoria3814 3 жыл бұрын
Just watch this impressive Maths channel... it’s very nice like this kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@UpstartRain 2 жыл бұрын
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).
@cthzierp5830 4 ай бұрын
Happy new year to you as well :)
@jschnei3 3 жыл бұрын
I am in love with this video
@user-xk3en1tj2e Жыл бұрын
You cheeky little blighter!) Love all ur content, especially about imaginary equations like cos(x)=2 etc. Peace!!!!!
@einsteingonzalez4336 3 жыл бұрын
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
@78rera 10 ай бұрын
At the end, we finally know that a man who teach bicycle to that boy is a genius-man...
@neilgerace355 3 жыл бұрын
32:50, I've never seen this method before ... thanks!
@al-shaibynanong1237 3 жыл бұрын
I've been waiting for this. Thank you sir🥰
@that_one_guy934 Жыл бұрын
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
@madhavjuneja4333 2 жыл бұрын
15:40 answer is iπ/2 or to include all values of theta- i(2nπ+-π/2 ∪ nπ+(-1)^nπ/2)
@ThePhysicsMathsWizard 3 жыл бұрын
Nice one, i love it!!
@lolaharwood619 3 ай бұрын
I'm pretty sure you can use W function on a standard graphing calculator just by graphing y=xe^x and the value you want to take the w function of, i.e I.e 5=6xe^6x => W(5)= W(6xe^6x) W(5)=6x To find W(5): Function W maps y=xe^x Plot y=xe^x Plot y=5 Use (Ints) to find the x for which xe^x = 5 Let x value of intersection = c, c is a constant Then use W(5)= c 6x=c x=c/6 Please correct me if I'm wrong!!
@chriswinchell1570 3 жыл бұрын
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!!!
@mista5796 3 ай бұрын
This dude is literally Mr Maths 👌
@anushrao882 3 жыл бұрын
Yess! This is so cool.
@anurag5565 3 жыл бұрын
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.
@nicopb4240 2 жыл бұрын
Thank you very much!
@sueyibaslanli3519 3 жыл бұрын
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😁
@Xnoob545 3 жыл бұрын
19:35 brb
@girlgaming1993 3 жыл бұрын
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 жыл бұрын
But it's also equal to i(π/2+2πn) for all integers n.
@MrHK1636 3 жыл бұрын
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.
@Reallycoolguy1369 2 жыл бұрын
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.
@Jamiree7 3 жыл бұрын
Very good lecture
@Sesquipedalia Жыл бұрын
solving for : W(-π/2) **we know ln(i) = iπ/2 bec** a + bi can be written as rcosθ + risinθ. where "r" is the hypotenuse of triangle formed with sides "a" and "bi" = r(cosθ + isinθ) = r(e^iθ). in this case, r = 1. taking ln now since we need to find ln(i), we get ln(i) = iθ = iπ/2 (angle here is π/2 because we are adults now.) **now, W(-π/2) = W(ixiπ/2)** = W(ixln(i)) =W(e^ln(i)xln(i)) = ln(i) = iπ/2. dude. im 15 and was loosing interest in math, but you saved me and made me suddenly alot better than i was before, i dont have "log" in my book bec im only in grade 10, and i didnt understand anything about it. heck, i didnt understand alot of what is in my book aswell. but now im here knowing stuff way beyond log. all because of you. you made me addicted to math. thank you edit : im 16 now, today is my birthday
@nohaxjustxmod-sfs3984 Жыл бұрын
happy late birthday, we have somewhat of the same situation
@cuboid2630 3 жыл бұрын
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!
@huzefa1991 2 жыл бұрын
Thanks! Can you please share what are the real life applications of Lambert W function??
@seghirimoha9026 Ай бұрын
Thank you very much
@SimonPetrikovv 3 жыл бұрын
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?
@SidneiMV 2 ай бұрын
*e^[W(x)] = x/W(x)* wow! *how useful* it is!!
@saumyakathuria5594 3 жыл бұрын
A lecture on use of dummy variable in Combinatorics pls
@baptiste5216 3 жыл бұрын
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 ?
@SamiSalah92 Жыл бұрын
LOVE YOU
@wxadbpl 3 жыл бұрын
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?
@hasanjakir360 4 ай бұрын
at 5:40 "it will work, because I did it already" had me laughing.
@papanujian7758 3 жыл бұрын
waiting so long. finally
@w__a__l__e 3 жыл бұрын
thanks dude.
@brucefrizzell4221 2 ай бұрын
Retired computer geek learning NEW math for FUN . Just joined.
@dovidglass5445 3 жыл бұрын
I love all your videos, thanks so much! By the way, I think you made a small error at 20:07. You say that if e^x+e^(-x)=1 then we have the simple solution x=0, but this isn't true; if x=0 then we have 2=1 which obviously is impossible, so maybe the equation e^x+e^(-x)=1 would have been ok after all :)
@Jack_Callcott_AU 3 жыл бұрын
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.
@aurelosquino646 3 жыл бұрын
Great video! Is there an integral than can be computed in terms of W(x)? I mean, is there any function like e^(-x^2) for which we can now compute its primitive using W(x) and otherwise we can't?
@danielvictoria3814 3 жыл бұрын
Just watch this impressive Maths channel... it’s very nice like this kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@joshmcdouglas1720 3 жыл бұрын
Are ln(i) and W(-pi/2) both equal to i(pi/2) ? Got both of these using the polar form of i
@Casey-Jones 3 жыл бұрын
wow ...... extreme hard core stuff
@curtiswfranks 3 жыл бұрын
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).
@aayushrampal1524 3 жыл бұрын
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
@abdomohamed4962 3 жыл бұрын
nice video bro
@indarajgochermaths5176 3 жыл бұрын
So nice video Great
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