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

[@angelmendez-rivera351]

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]

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__]

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

[@drpeyam]

You’re back!!!!!!! 😍😍😍

[@blackpenredpen]

I am back from the 🏖

[@vladimirkhazinski3725]

Kiss already!

[@banana6108]

@@vladimirkhazinski3725 😑

[@ffggddss]

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

[@abhishekprasad6350]

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

[@colt4667]

BPRP uses a fish. Professor Julio uses a tomato.

[@chin6796]

MCU is for math creatures universe

[@damianbla4469]

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 :)

[@drekkerscythe4723]

The longer the beard, the higher the wisdom

[@glegle1016]

Dude needs to shave. That "beard" looks disgusting

[@muhammadusmonyusupov2556]

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

[@Kitulous]

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

[@just_a_dustpan]

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

[@agarykane2127]

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

[@stevenglowacki8576]

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]

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

[@adi8oii]

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]

Me too for x^x =2^64

[@spinecho609]

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

[@nesto9889]

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

[@wristdisabledwriter2893]

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

[@M-F-H]

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]

yea i am lol

[@abdomohamed4962]

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

[@jesseolsson1697]

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

[@d3vilman69]

Among the several Lambert W function videos I have watched, I think yours has the most clarity as you actually listed out the equations that defined the W function. Other videos just spit out W(some number) and call it a day without elaborating what W really is. Now I know W of a number actually produces a real number. Thank you.

[@JohnSmith-vd6fc]

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]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@helo3827]

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

[@danielvictoria3814]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@girlgaming1993]

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]

πi/2

[@SebastienPatriote]

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!

[@aprendiendoC]

To the original comment: That is not correct, since -pi/2 is not equal to ln(i)*e^ln(i). This is because ln(i)*e^ln(i) = pi/2 * i, which is not -pi/2

[@sharpnova2]

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

[@marianopatino939]

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]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@prohacker1373]

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

[@kepler4192]

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

[@jodikirsh]

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

[@a1175779]

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

[@tudor5555]

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 !

[@nosnibor800]

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]

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

[@eng.giacomodonato8514]

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

[@abdomohamed4962]

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

[@ameer_er2181]

@@abdomohamed4962 من ایرانیم

[@alberteinstein3612]

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

[@theimmux3034]

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]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@Zero-tg4dc]

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]

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

[@herculesmachado3008]

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

[@flamitique7819]

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 !

[@simonharris3797]

Cannot find this in as much detail elsewhere. Thank you

[@alexyanci7974]

41:53 - It's a + Great video

[@legendarynoob6732]

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*

[@darkahmedp]

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

[@61rmd1]

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

[@that_one_guy934]

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

[@kdmq]

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]

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

[@danielvictoria3814]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@محمدالنجفي-ظ1ه]

Oh god how much i love computing without using calculator thanks so much for this video please make more videos just like this is Just so useful ❤🌹

[@gtweak7]

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

[@RomainPuech]

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

[@anurag5565]

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.

[@MaximQuantum]

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

[@ffggddss]

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

[@abeldiaz1539]

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]

@@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

[@assassin01620]

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

[@blackpenredpen]

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

[@KjartanAndersen]

The beard of wisdom :) Absolutely great explanation.

[@ikocheratcr]

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.

[@gouharmaquboolnitp]

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

[@helo3827]

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

[@umeshprajapati7381]

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

[@umeshprajapati7381]

Plz sir give solution

[@asparkdeity8717]

@@umeshprajapati7381 0

[@lolaharwood619]

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!!

[@sueyibaslanli3519]

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😁

[@mista5796]

This dude is literally Mr Maths 👌

[@pragalbhawasthi1618]

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

[@farhansadik5423]

If there is someone who got a bit confused at 9:55 because he said why we had to take the domain of f to start from one, it's because if the domain was from -inf, then before reaching x=-1 it would have completed every x value till -1/e and after reaching 1 it would again start rising and complete each x value again, meaing the function is not one to one from -inf to 0(0 since the function approaches 0 to left of the graph and we have consider each y value from 0 to -1/e and the to 0 again) and one to one functions are not invertible

[@JMCoster]

Very, very good video ! 48 minutes top level

[@girlgaming1993]

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]

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

[@78rera]

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

[@panadrame3928]

39:28 nooooo it's xe^x - e^x !! thereby you have a false ∫W(x)dx... It'd rather be : xW(x) - (W(x) - 1)e^W(x). Then because W(x)e^W(x) = x we would have : xW(x) - x + e^W(x) , which we can also rewrite as x(W(x) - 1 + 1/W(x)) Pretty cool to see rational fraction X - 1 + 1/X appears here (invariant by X = 1/X)

[@panadrame3928]

Nvm he corrected the 2 mistakes 😂

[@АртемДараган-л1п]

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

[@teo9376]

This was one of the most interesting videos I’ve seen in a while. I will have to be back for more. I didn’t really get why we are able to look at only the principle angle for e^x=i for the last problem. I imagine it’s because it could equal any of those angles, so we just pick one that works nicely. So if we had W(-((pi+4)/2)) we would change the negative 1 to ie^((pi+4)/2)i). That is, assuming I didn’t misunderstand things.

[@cuboid2630]

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]

Thank you!

[@MrAnonymousfan1]

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?

[@madhavjuneja4333]

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

[@hunterkorble9134]

Bro ur always dishing free knowledge

[@theimmux3034]

W(x) is such a cool piece of math

[@einsteingonzalez4336]

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]

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

[@nathanaelmoses7977]

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

[@einsteingonzalez4336]

@@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]

@@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]

@@angelmendez-rivera351 Interesting

[@albertogarcia4177]

In 46:25 W(-π/2) has not sense cause the domain of this W branch (the main branch) is [-1/e, infinite) and we have -π/2

[@ausaramun]

I think that ln(i)=iπ/2 and W(-π/2)=iπ/2. For ln(i), we can rewrite the i in terms of e by using Euler’s formula, which will be e^{iθ}=cos(θ)+isin(θ), and we want it to be i. So the angle θ that corresponds to that is π/2, giving us cos(π/2)+isin(π/2), which equals i. Therefore, we can rewrite ln(i) as ln(e^i(π/2)) which equals iπ/2. For W(-π/2), we can again use the fact the e^i(π/2)=i so we can define that (iπ/2)·e^{iπ/2}=i^{2}·π/2, and i^{2} equals -1, thus giving us -π/2. So it would make sense that W(-π/2)=iπ/2.

[@angelmendez-rivera351]

Yes.

[@joaidane]

also -iπ/2 since -iπ/2·e^{-iπ/2} = (-iπ/2)/e^{iπ/2} = (-iπ/2)/i > -π/2 (cancelling i's directly or multiplying -iπ/2)/i top & bottom by i ) weird that i·i = -1 but i/i = 1 ! Late to the party, but interesting none the less :)

[@raviolivii2719]

Bit late, but here are my solutions to finding ln(i) and W(-π/2) ln(i): i = e^(iπ/2), ln(e^(iπ/2)) -> iπ/2 W(-π/2): Recall that ln(i) = iπ/2, we can rewrite -π/2 as ln(i)*e^ln(i), cancels to get (iπ/2)*i = -π/2 so W(ln(i)*e^ln(i)) is just ln(i)

[@joseluisrosales4104]

Ln (x/h) = Sqrt (x/h)-Sqrt (h/x) when x~h. Now consider x e^x=2. Ln (x)+x=Ln (2) and h=1 to approx Sqrt (x)- Sqrt (1/x)+×=Ln (2) or {(x-1)/(Ln (2)-x)}^2=x. And solve it as a qubic equation.

[@axelgiovanelli8401]

Legendary!!! Salute you from Argentina

[@MathNotationsVids]

Outstanding content and presentation. I really enjoy your videos!

[@chriswinchell1570]

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]

Good point!!!

[@archivewarrior8535]

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

[@blackpenredpen]

Thanks lol. You should see Professor Walter Lewin tho!

[@MrHK1636]

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]

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]

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.

[@SidneiMV]

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

[@UpstartRain]

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]

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).

[@hsh7677]

Thank you so much. I really enjoyed this!!

[@danielvictoria3814]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@gorilaylagorila2540]

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

[@danielvictoria3814]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@pratham_4355]

18:42 ln(i) =(π/2)•i W(-π/2) = (π/2)•i

[@ДмитрийОрлов-б9г]

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

[@NonTwinBrothers]

This one's for the books. The bprp epic

[@Casey-Jones]

wow ...... extreme hard core stuff

[@ARKGAMING]

42:03 I'm sure other people noticed it but it should be ... +e^w(x) +c and not ... -e^w(x) +c

[@absxn]

yup lol

[@saumyakathuria5594]

A lecture on use of dummy variable in Combinatorics pls

[@Hjerpower]

22:40 x+e^x=2 e^x=2-x e^-x=1/(2-x) e^2*e^-x=e^2/(2-x) e^(2-x)=e^2/(2-x) (2-x)e^(2-x)=e^2 2-x=W(e^2) x=2-W(e^2) Numerically they are the same answer

[@danielvictoria3814]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@_Blazing_Inferno_]

This is the answer I got as well. I find it quite interesting that W(e^2) has the property such that ln(W(e^2)) = 2 - W(e^2).

[@jiaweigong3411]

Every engaging; excellent work!

[@pierre7770]

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

[@cubicinfinity2]

I think BPRP's success has something to do with the fact that he's younger and more relatable than your calculus instructor. He also does a lot of responses to "how do you integrate this thing, lol" (it's youtube). And because it's youtube, it's not him talking to "students" it's "ladies and gentlemen". He's also more relaxed and it's not always "with respect to 'u'", but "in the 'u' world".

[@cubicinfinity2]

also that cheeky smile.

[@Xnoob545]

19:35 brb

[@Reallycoolguy1369]

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]

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

[@brucefrizzell4221]

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

[@aspectator3680]

There is a mistake in 39:42 F(x)=xe^x-e^x

[@jschnei3]

I am in love with this video

[@hasanjakir360]

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

[@yibozhao4510]

Use Euler's formula e^(i*pi) = -1 can also solve 'ln(i)' and I think it's easier.

[@lietpi]

Loved every second of the video!

[@Mr_Mundee]

for number 7, you could've just realized that e^x - e^-x was just 2*sinh(x) so the equation would've become 2sinh(x)=1 and so x would be arcsinh(1/2)

[@jackhandma1011]

36:42 he put plus C like it's a natural reflex. Nice.

[@joseph_newt0n]

15:39 ln(i) = iπ/2

[@ginopaperino2608]

in the 8th problem i think we could divide everithing by e^x so we had xe^(x)+1=2*e^(-x) then we can move the 2 to the left side and the 1 to the right side so (x-2)*e^(-x)=-1 then we can multiply by -e^2 now we have that (2-x)*e^(2-x)=e^2 from that we can find that x=2-W(e^2). I hope it's right and pls forgive my bad english!

[@MuhammadyusufK]

of course 13:07

[@SimonPetrikovv]

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?

[@shikshokio1]

Great explanation, thank you!

[@curtiswfranks]

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).