x+e^x=2 e^x=2-x x=ln(2-x) x=ln(2-ln(2-ln(2-ln(...))))

@blackpenredpen

5 жыл бұрын

I LOVE THIS SOLUTION!!!!!

@kwea123

4 жыл бұрын

No, how can this be correct? You need to prove that the sequence a(n+1)=ln(2-an) converge, otherwise this infinite thing has no sense. Like if I pose x=1-x, does x=1-(1-(1-...))??

@user-qq6si7zv3t

4 жыл бұрын

@@kwea123 It does converge

@kwea123

4 жыл бұрын

@@user-qq6si7zv3t I know, I didn't say it doesn't converge, but my point is that it needs proof, as it's not obvious at all!

@angelmendez-rivera351

4 жыл бұрын

AI葵 While it is true that, normally, one needs to prove the convergence of the sequence, it is not necessary in this case because this is actually already a well-known expression for the well-studied branches of the Lambert W map.

bprp: "okay lets do some math for fun" me: *I N S T A N T R A G I N G B O N E R*

@fredericchopin6445

3 жыл бұрын

lmfao

@plislegalineu3005

3 жыл бұрын

@@fredericchopin6445 It's written Fryderyk you anti-Polish pro-French guy (don't take it to ur heart)

I tried differently for example 2... my instinct was to raise e to both sides to get rid of the "+" e^(x+e^x) = e^2 (e^x)[e^(e^x)] = e^2 e^x = W(e^2) x = ln[W(e^2)] which is also around 0.44285... BTW nice video! I'm new to this concept as it wasn't taught (not even mentioned) in school! Good thing KZread recommended this for me!

@user-tv5qs2mb4p

Жыл бұрын

I thought same logic with you 😂

@meraldlag4336

Жыл бұрын

Same here lol, I was so amazed when it turned into the right format

@Tim3.14

11 ай бұрын

To prove these are equivalent: Note that W(e^2) = y where y satisfies y*e^y = e^2. Divide that by e^y to get y = e^(2-y), take the natural log to get ln(y) = 2-y. Substitute W(e^2) for y to get ln(W(e^2)) = 2-W(e^2)

@rainerzufall42

8 ай бұрын

Hmm... or this way: ln[W(x)] + W(x) = ln[W(x)] + ln[e^W(x)] = ln[W(x) * e^W(x)] = ln(x). Either way: ln[W(x)] = ln(x) - W(x).

《Who put the square. .... yeh that person was me》 《OMG , who put the + here.......yes he was me 》 😂legend

@blackpenredpen

5 жыл бұрын

XaXuser yup yup !!!

@jofx4051

4 жыл бұрын

When you ask question and you answer it yourself 😂

@leif1075

4 жыл бұрын

@@blackpenredpen isn't there a way to solve it without knowning and using the lambert function??

@moonlightcocktail

4 жыл бұрын

Me looking at the code I programmed at 2 am last night

@thecheesegenius3817

2 жыл бұрын

@@leif1075 nope

On the second one, let y=e^x, this gives shortly the solution : Ln(y)+y = 2 Ln(y)+y.Ln(e) = 2 Ln(y)+Ln(e^y)=2 y e^y=e^2 y=W(e^2) x=Ln(W(e^2))

@jatloe

4 жыл бұрын

Nice!

@mutesniper4729

4 жыл бұрын

I got this answer too

@artemanoha2266

4 жыл бұрын

MuteSniper , same. Is it correct?

@ThAlEdison

4 жыл бұрын

@@artemanoha2266 Wolfram alpha confirms that ln(W(e^2))=2-W(e^2)

@brandonlopez9474

4 жыл бұрын

@@ThAlEdison thanks for that broda

Even now I'm graduated in Math I'm still learning from you lmao

That's pretty cool, *isn't it?* :D

@anglaismoyen

Жыл бұрын

Yes

As I've read the comment, noone has been talking about that; the method used in equation 2 can be used to demonstrate the Wien's displacement law from the Planck law. You try to find the maximum of the Planck law as a single variable (lambda), with T being a fixed parameter; then you should end up with (x-5)e^x + 5 = 0. And it's exactly the same method as BPRP used in this video, but I am going to do it for the physics students if they want a detailed answer: (x-5)e^x + 5 = 0 e^5(x-5)e^(x-5) = - 5 (x-5)e^(x-5) = -5e^-5 Apply W in both sides: x-5 = W(-5e^-5) x = 5 + W(-5*e^-5)

@nikhilnagaria2672

2 жыл бұрын

Can you elaborate (about how the laws connect)?

@vinceheins

5 ай бұрын

thats smart😊

Great video. I never learned about the Lambert W function for my engineering degree and was always fascinated by the problems my friends and I considered "unsolvable". Your videos helped a lot!

Thank you blackpenredpen for making these videos about Lambert's W-function. Very nice explaining and creative solutions!

I got a similar answer for the second one. I used the substitution x = ln(y), and I plugged both sides into the exponential function, which got me the equation y*e^y = e^2. Applying the lambert function, and reversing the substitution, this got me ln(W(e^2)) = x.

It'd be cool if there was a page of equations like these ones to solve.

Amazing ! I can improve myself in english and in maths thanks to your videos ! I love the way you use to solve problems, and it's always interesting. Maths for fun !

For the 2nd one : x + e^x = 2 exp(x + e^x) = exp(2) e^x • exp(e^x) = e^2 W(e^x • exp(e^x)) = W(e^2) e^x = W(e^2) x = ln(W(e^2)) and just to match your answer : W(e^x) = y e^x = y • e^y exp(x-y) = y y = exp(x - W(e^x)) So, ln(W(e^2)) = ln(exp(2 - W(e^2)) = 2 - W(e^2)

I have seen several videos on your channel, but this one is the best !

@blackpenredpen

5 жыл бұрын

Кирилл Бон thanks!!!!!

Good stuff! Thanks for posting!

For problem 2, wouldn't it be easier to eponentiate both sides. You get e^x e^(e^x)=e^2 => ye^y=e^2.=> x=ln[W(e^2)]. Where we are using y=e^x.

Awesome video! Would love to learn more about the exact definition of the lambert w function, is it defined through a taylor series just like the natural log function?

@angelmendez-rivera351

4 жыл бұрын

Being the inverse function to f(x) = xe^x IS the exact definition of W(x). If what you *meant* to say is that you want to know if an expression in terms of arithmetic functions or rational functions or trigonometric functions and their inverse maps, as well as exponentials and logarithms, then no, such an expression does not exist, at least no with finitely many operations. A power series for W(x) does exist, but this should not be surprising. Every function that is analytic at x = 0 has a power series.

I actually tried to do it!! Good video Also, do you think you could do a video on how to approximate the W function or at least how the computers do it? Thanks

@XanderGouws

5 жыл бұрын

You can use Newton's method to find the roots of we^w - x (where you're solving for w, and x is the value you want - treat it like a constant)

You have convinced me, I'll start using the Lambert function whenever I can 😉

@blackpenredpen

5 жыл бұрын

Dani Borrajo Gutiérrez yay!!!!!

Please do more of this stuff. Just great!

@blackpenredpen

5 жыл бұрын

SkyRider thank you!!! Yes more is coming

New subscriber ... believe me watch your videos every single day. Thank you so much

bruh ur literally my favorite human being

Try: x+log(x-1)-xlog(x-1)=0 It's a fun little exercise that involves utilising W(x).

You did it so fast... Btw, Could You recomend any calculator for Lambert W(x) function? Best regards

7:26 that makes my heart happy :)

I like these videos, but could you try doing things with complex numbers, such as W(i) or W(-2)?

@MichaelRothwell1

5 жыл бұрын

Yes, W with a complex argument arises when you solve x^x = i, which was requested in a comment to a recent BPRP video.

@General12th

5 жыл бұрын

Hell yeah!

Amazing!!

Interesting. Now I had learned how to solve these equations. Thank you👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻

maybe a video on the derivative(s) of the lambert W function would be cool

I love you so much for doing maths . that is magic how to process for resolve the second equation. Thanks a lot for increase my math perception 🥰

I learned of Newtons method for solving for roots in my computation for physics class. X_(n+1) = x_n - f(x_n)/f ' (x_n) You just try an initial x that is close to satisfying the answer. I tried x = 1 for x^2e^x -2 = 0 and get about 0.72 which is close enough. Trying this 3 times, I got 0.901201. Not bad for just pen and paper.

Hey!i just wanted to let you know that i liked the video :)

Nice! Thank you!

Can you please help me find the area enclosed by | |x| - |y| | less than or equal to 1 and x^2 +y^2 less than or equal to1..

I hope you’re going to make more videos like these!

@blackpenredpen

5 жыл бұрын

Bryan Franzoni ok!!!

is it possible to approximate the value of a W function using newton’s method

Mate you've become the Bob Ross of maths

@blackpenredpen

5 жыл бұрын

Sam Spedding what a complement! Thank you!!

@jofx4051

4 жыл бұрын

Wew KZread viewer has interesting comment that I don't expect I woild read this!

I have an problem can you Please solve Find the integral solution of the equation x^2-13y^2=-1

For second one I took exp() of both sides and ended up with x=ln(W(e^2))...are these the same or is something wrong

Fantastique.....lambert fonction Thanks.👍

what formula do we give the computer, to calculate W. not using the pre-defined function.

@ViktorKronvall

5 жыл бұрын

ekueh Wikipedia has a formula for a Taylor expansion around 0 for W: en.wikipedia.org/wiki/Lambert_W_function#Asymptotic_expansions

@Nickesponja

5 жыл бұрын

If you want to find out W(z), you can consider the equation x*e^x-z=0 and use something like Newton's method to approximate x=W(z). Actually bprp has a video doing just that by hand (kzread.info/dash/bejne/d56J0sFuiJnPn5c.html), but of course with a computer it's much faster

@XanderGouws

5 жыл бұрын

You can use "productlog(a)" in WolframAlpha, you could also write a script in python to do it with Newton's method, and if you're going to be using python you could also use a package called SciPy special functions

@General12th

5 жыл бұрын

W(x) is not an elementary function, so you can't write it as just a single expression. You can _approximate_ it using a Taylor series, but Taylor series have infinitely many terms.

@egeyaman4074

4 жыл бұрын

@@Nickesponja why should it make sense? You can use newton's method for both of these equations directly, no need to write it as lambert function

May apply the iterative method !

It seems a little arbitrary. What's stopping you from creating a function which is the inverse of whichever expression you're working with, and then saying that's the answer?

@XTheDentist

5 жыл бұрын

Its not arbitrary, it simply means that it is a "non elementary" function, so we dont have a nice representation of its inverse. But, we do know that an inverse will simply have the domain & range swapped & since it doesnt pass the horizontal line test, which it needs to for invertability, we simply cut the range at -1/e. And Newtons method can easily be used on computers to calculate values for W.

@hassanakhtar7874

4 жыл бұрын

Logaritms are also just the inverses of exponentials. But they are clearly useful to have because of their identities and properties. (Eg: ln(ab) = ln(a) + ln(b) )

@mdasgupta4368

4 жыл бұрын

You have 2^h=3 You write h= log 2 (3) Log is an inverse of the exponential function. So why do you use it?

@jatloe

4 жыл бұрын

armin Eh. I guess you CAN say it is arbitrary but technically the Lambert W function is a well known function for unelementary stuff.

@angelmendez-rivera351

4 жыл бұрын

dev02ify It is true for the choice of xe^x.

Radically Awesome dude!!!

Superb #YAY THANKS A LOT

Can you solve x + e^x + e^(2x) = 2 and x + e^x + W(e^x) = 2?

How to calculate Lambert W function with Wolfram alpha? W(x) does not work (it is not recognized).

Can you integrate the Lambert W function please?

For the second equation we can just do it like this, can’t we ? x + e^x = 2 e^(x + e^x) = e^2 e^x e^(e^x) = e^2 (because of the powers rule) W(e^x e^(e^x)) = W(e^2) e^x = W(e^2) x = ln(W(e^2)) And we don’t forget that W(x) is a multiple value fonction so we have infinitely many solutions 😉

Where did you find these two questions? Can u tell me where can I find more problems like this ?

I found a weird answer for the second one: x+e^x=2 e^(x+e^x)=e^2 e^x*e^(e^x)=e^2 with lambert W we get e^x=W(e^2) so x=ln(W(e^2)) Does that mean ln(W(e^2))=2-W(e^2)? I checked on Wolfram and yeah, they're the same: This gives us a new identity: ln(W(e^x))=x-W(e^x) or in other words ln(W(x))=lnx-W(x) Unsurprisingly, from this we can prove this identity easily since ln(W(x))=lnx-W(x) ln(W(x))+W(x)=lnx so left hand side gives us = ln(W(x))+ln(e^W(x)) = ln(W(x)*e^W(x)) = ln x Proved (:

@FearblazeBrawlStars

Ай бұрын

This is so cool dude

Looks amazing! But what's the W(x) function?

@bjornfeuerbacher5514

2 жыл бұрын

As he explained: It's the inverse function of f(x) = x e^x.

This is a nice demonstration of the W function. But frankly, I don't think it's even required. We can simply let y=e^x then we have 2 eqs: x²y=2 x+y=2 Solving the above two, we get an algebraic answer (ie, x and y are roots of a polynomial eq) However we know that e is such a number that e^(algebraic)≠algebraic Contradiction! Hence no solution

@jaimeduncan6167

5 жыл бұрын

Gurkirat Singh what ??

@blackpenredpen

5 жыл бұрын

No no, we were solving two separate equations. 1. solve x^2*e^x=2 then 2. solve x+e^x=2

@GurkiratSingh-ds8dq

5 жыл бұрын

@@blackpenredpen Ooh I thought they were simultaneous 😅

@angelmendez-rivera351

4 жыл бұрын

Gurkirat Singh There is only one variable, there is absolutely no reason to take it as a system of equations. He also listed them as separate equations in the video anyway.

Alternatively: x + e^x = 2 e^(x + e^x) = e^2 e^x * e^(e^x) = e^2 W(e^x * e^(e^x)) = W(e^2) e^x = W(e^2) ln(e^x) = ln(W(e^2)) x = ln(W(e^2))

I've instantly asked myself, what if the second question would have 2 as a factor: x + 2e^x = 2. Same calculation => x = 2 - W(2e²) = 2 - 2 = 0. Check: 0 + 2e⁰ = 0 + 2*1 = 2.

I remember, when I was younger, I was trying to solve stuff like 2x-ln(x)=3 and it was said you need calculator.. I thought. What? It looks so innocent... Haha

@liammargetts

4 жыл бұрын

wait, so how do you solve 2x - ln(x) = 3?

@Yuuki3uwu

4 жыл бұрын

How to solve it??

@stewartzayat7526

4 жыл бұрын

ln(x)=2x-3 x=e^(2x-3) 2xe^(-2x)=2e^(-3) -2xe^(-2x)=-2e^(-3) -2x=W(-2e^(-3)) x=-W(-2e^(-3))/2 Which is about 0.0556483219...

@jofx4051

4 жыл бұрын

@@stewartzayat7526 Well, it is a shotgun Lambert (W) function which I don't think there is a calculator that can count it directly

@angelmendez-rivera351

4 жыл бұрын

Jofx What do you mean "count it directly"?

Good video!

@blackpenredpen

5 жыл бұрын

Guy Michaely thanjs

Why is W not counted among the elementary functions?

@angelmendez-rivera351

4 жыл бұрын

Many mathematicians do count it as an elementary function, but I suppose it all depends on your level of education. For a sixth-grade student, the trigonometric functions are far from elementary.

W(number < -1/e) would just be a complex result. Or doesn't W(x) have any values at x

@angelmendez-rivera351

4 жыл бұрын

It would be a complex number.

For the second problem, I used a substitution. e^x=2-x let u = 2-x x=2-u e^(2-u)=u e^2=ue^u u=W(e^2) 2-x=W(e^2) x=2-W(e^2)

For the second one this seems shorter x + e^x = 2 (then apply exp) (e^x)*e^(e^x) = e^2 e^x = W(e^2) x = ln(W(e^2))

Thanks for the math

2:45 why you dont put a - + in the left expression too ?

@JoseFernandes-js7ep

4 жыл бұрын

Putting +- in left side is redundant.

I tried to solve those equations before watching the video. In x+e^x=2, I took the approach to use e^(x+e^x) = e^2, which lead me to x=ln(W(e^2)). I was so disappointed when I saw the solution in the video x=2-W(e^2). But I checked my solution in wolframalpha... and it says ln(W(e^2)) and 2-W(e^2) are alternate forms!!!

how to find w(sqrt(2)/2) without any calculator or wolfram alpha?

So cool!

@blackpenredpen

5 жыл бұрын

Thanks!!!

Isn't ln(W(exp(2))) also a valid answer for x+exp(x)=2 ? Just apply the exponential to the whole equation and set X=exp(x), solve for X and then for x.

@angelmendez-rivera351

4 жыл бұрын

Yes, it is. In fact, this a property that you can prove about the Lambert W function. As z |-> W(z) is the right-inverse of z |-> z·exp(z), it is the case that W(z)·exp[W(z)] = z for all z in C. If |z| > 0, then the above implies log[W(z)·exp[W(z)]] = log(z) = log[W(z)] + W(z) = log(z). This is equivalent to log[W(z)] = log(z) - W(z). If z = exp(2), then log(W[exp(2)]) = 2 - W[exp(2)] is merely an special case of the above identity.

(exp) x =i^2 For first equation x=+√2i or x=-√2i, For second equation x=3.

hey bprp,can u plz tell why the Lambert W function's defined like that ?

@angelmendez-rivera351

4 жыл бұрын

It is just useful to have a well-studied inverse map for that particular function.

For equation one, you can keep the minus sign if u allow complex solutions.

@zephir1812

3 жыл бұрын

In the beginning he said that he wanted to keep everything real in this video 😉

You Can just use substitution e^x =u ln(u)+u=2 u=1.55... and then use the ln(1.55..) and you get 0.44

@blackpenredpen

5 жыл бұрын

And how did you get u=1.55..?

I wonder how you solve for the hyper value. For example, 2 n 4 = 16, n=3. e n i*pi, n=3, 2 n 2 =4, n={0,infinity}

@angelmendez-rivera351

4 жыл бұрын

What are you actually asking? What is a hyper value?

Can you find a connection between erf(x) and W(x)?

@dogbiscuituk

5 жыл бұрын

I mean, other than erf(x) = W(erf(x).exp(erf(x))), obviously!

@justabunga1

4 жыл бұрын

erf(x) is an error function, which is defined to be the 2/sqrt(pi) integral of e^(-x^2) from 0 to t dt. W(x) is a Lambert W function, which is an inverse function of y=xe^x. If you try to switch and y values, you get x=ye^y. We cannot solve for y in terms of elementary function. Both of these are considered to be non-elementary functions.

Can you solve ANY equation of one variable for that variable with this in your toolbox?

@loatchi_le2099

3 жыл бұрын

sqrt(x) = -1.

Very kewl!

very nice!

what is the value of w??

Hello For the 2nd equation, e^x*e^(e^x)=e^2 Taking the W function e^x=W(e^2) x = ln(W(e^2)) Am I right ?

@lollol-co6ly

3 жыл бұрын

Yes, and in fact, ln(W(e^t)) is equal to t - W(e^t) for all t.

Next video: approx value of w(y) where any y is a constant of its domain.

I still cant find a video where to calculate W(x) I don't know how to evaluate it, just what it does... what is W(1), W(pi), etc? how do i reach the number

@angelmendez-rivera351

4 жыл бұрын

You use the properties of the Lambert W function to find identities that can help find some values. Also, there is a Taylor series for W that you can derive pretty easily and find in Wikipedia.

The solution is very similar to 「limit(sinx/x)」problem.I like this kind of problem

How to do this with just natural log? Because i havent learnt lambert w function

@bjornfeuerbacher5514

2 жыл бұрын

You can't do this with only the natural log, you _have_ to use the Lambert W function.

Let's be honest, he needs more subscribers

@hachemimokrane8013

3 жыл бұрын

Il mérite d'avoir bcp d'abonnés J'ai appris bcp d'astuces de sa part

what is on your hand?

Can you solve x^(3x+1)=2 using Lambert-W function or something else ?

@ebenpresec

Жыл бұрын

Hi did you find a solution? I have encountered a similar problem.

What's that ball you hold?

Plz integrate this: 0 to 2π cos(sinx)•e^cosx

@rasheedmohammed2227

3 жыл бұрын

damn it's been a year have you found a solution. I've been trying to solve this integral and cannot seem to get an answer, even wolframalpha ended up using "evaluating from numerical calculation methods." Thanks

What is W function?

Please make a video on super-logarithms!

Raphson method or graphical method.

Does w have a function?

On second equation I tried going this way e^x = 2-x (x-2)e^x = -1 (x-2) e^(x-2) = -1/e^2 x-2 = W(-1/e^2) x = 2 + W(-1/e^2) = 1.8414... What did I do wrong?

Heyyy can solve by an easy meathod e^(x) =2/x^(squared) =2-x Solving this cubic equation by calgons meathod ig

Suggestion: This type of ecuations have a parallel funny way to be solved by means of numerical methods like fixed point iterations with the form x = f{x}. Probably you only like maths juggling.

Whats the name of the formule un spanish?

Nice, nice, nice! Recommended you to my sliderules.

@blackpenredpen

5 жыл бұрын

Blue Blue Hahahaha how about ur Taylor series?

@blue_blue-1

5 жыл бұрын

blackpenredpen, Taylor series I work out seriell with my circular sliderules. 🌝🤖📴🇩🇪‼️👍

@blue_blue-1

5 жыл бұрын

blackpenredpen, But more serious. I had no problem with Taylor-Series, but I had a debate about the unit rad with somebody. I once gave you notice about that... Sadly we didn‘t find a good consensus. Nick was very ignorant.

how can we solve this (x+24)^x=1/16 the x is shifted here

@angelmendez-rivera351

4 жыл бұрын

You cannot solve this using elementary functions, not even using W(x). You need to numerically approximate the solutions.

But how can I solve W(a..) etc without using any computer programs? Is it possible?

@bjornfeuerbacher5514

2 жыл бұрын

Try Newton's tangent method for an approximation. But even for that, you'll probably want to use a calculator. ;-)