Lecture on Lambert W, 1/5/2021 • Lambert W Function (do... ☀️ Patreon: / blackpenredpen 😃 blackpenredpen... 😃 / blackpenredpen 😃 / blackpenredpen #LambertWfunction #tetration
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@blackpenredpen5 жыл бұрын
Hi all, hopefully this remake makes things more clear! Also try to solve x^2e^x=2 and x+e^x=2
@yonatanzoarets35045 жыл бұрын
Can we solve any exponent equation which contains x in the base and also in the exponent using the w function? If then, can we use this to turn the parametric form of x^y=y^x into catersian form?
@82rah5 жыл бұрын
To solve x^2 * e^x = 2 take sqrt root then apply W to both sides, then x = 2 W(sqrt(2)/2 ). To solve x + e^x = 2 let y =2 -x then y e^y = e^2, apply W to both sides y = W(e^2), x = 2 - W(e^2)
@suraj_mohapatra5 жыл бұрын
hey where you from actually?? Japan?
@yonatanzoarets35045 жыл бұрын
@@suraj_mohapatra Actually I'm from israel
@yonatanbar73115 жыл бұрын
@@yonatanzoarets3504 איזה גבר גם אני מישראל
@sidgar15 жыл бұрын
WWE function requires the coefficient X to smack the e over the head with a chair, while the exponent X jumps off the rope with a cross-body splash. Problem solved.
@blackpenredpen5 жыл бұрын
: )
@PraneshPyaraShrestha4 жыл бұрын
Shane McMahon function jumps from Hell in a Cell
@sxkjknjw23 жыл бұрын
WATCH OUT WATCH OUT! RKO! OUT OF NOWHERE!
@anshulanand023 жыл бұрын
😂😂😂😂😂😂u guys r awesome (btw randy Orton is my fav)
@U014B3 жыл бұрын
Me when trying to do the problems BPRP gives: 🤔🤔🤔 The answer when he reveals it: kzbin.info/www/bejne/mX_KqayCat94d80
@stem61094 жыл бұрын
I like how he always keeps mic in his hand like an apple.
@anshulanand023 жыл бұрын
Pokeball
@justafish55593 жыл бұрын
it fell on his head, just like the apple fell on newton's head.
@Indiandragon2 жыл бұрын
@@justafish5559 🥸
@carlosharmes2378 Жыл бұрын
xchg INT 2Fh & INT 21h (like INT 1Ch & INT 08h)
@93083235 ай бұрын
He's using the Apple of Eden.
@user-vx9xu4qn5q4 жыл бұрын
NO! IN THIS CHANNEL THE VARIABLE FOR W FUNCTION MUST BE FISH!
@cyberbeastry88093 жыл бұрын
xD
@Mnzmanzmz3 жыл бұрын
It is not funny 😒😒😒
@anshulanand023 жыл бұрын
@@Mnzmanzmz lol
@greece87853 жыл бұрын
Our teacher in Greece puts apple as variable
@fredericchopin64453 жыл бұрын
@@Mnzmanzmz what a great way to ruin a joke
@omarifady5 жыл бұрын
X^2*e^x=2 Take the square root Xe^(0.5x)=sqrt(2) Divide by 2 (0.5x)e^(0.5x)=0.5sqrt(2) 0.5x=W(0.5sqrt(2)) X=2W(0.5sqrt(2)) The second one X+e^x=2 Exponent both sides (e^x)*e^(e^x)=e^2 e^x=W(e^2) X=ln(W(e^2))
@blackpenredpen5 жыл бұрын
: )
@tbg-brawlstars Жыл бұрын
Why aren't you taking the -ve value in first one ?
@JacobRy Жыл бұрын
@@tbg-brawlstars you mean from square root?
@tbg-brawlstars Жыл бұрын
@@JacobRy yes
@thecrazyemu05 Жыл бұрын
@@tbg-brawlstars -sqrt(2)/2 is not in dom(W)
@DonPrizzle Жыл бұрын
I’m a Mechanical Engineering student, and I alway happen to stumble upon your page when I am trying to expand my mathematical knowledge. I absolutely love your stuff, you’re always great at explaining things in an easy to follow manner. Keep it going man, you’re greatly appreciated 🙏
@igorvinicius43053 жыл бұрын
Q1. Solve x^2 e^x = 2 Using square root in both sides, we get: x e^( x/2) = sqrt(2) Then, dividing the equations by 2, (x/2) e^(x/2) = sqrt(2)/2 Hence, applying the Lambert W function, W[ (x/2) e^(x/2) ] = W [ sqrt(2)/2] ==> x/2 = W[ sqrt(2)/2] And, therefore, the solution is x = 2 W[ sqrt(2)/2)]. Q2. Solve for x + e^x = 2. First note that x = ln( e^x) and e^x = ln( e^e^x)! So, we can rewrite the equation as x + e^x = 2 ln(e^x) + ln( e^e^x) = 2 Now, since ln(a) + ln(b) = ln(ab), then we can write ln(e^x * e^e^x) = 2 ==> (e^x) * e^(e^x) = e^2. Let y = ( e^x). Hence we have y* e^y = e ^2 . applying the Lambert W function, W( y* e^y) = W( e^2) y = W(e^2). Substittuing y= e^x, e^x = W(e^2) And therefore the solution is x = ln( W( e^2)).
@gautamgopal35173 жыл бұрын
Wow bro... Thanks a ton! And just to clarify, why can't the second one be done in this way? By rearranging, e^x = 2 - x (2 - x)e^(-x)=1 (2 - x)e^(2 -x) = e² Thus, 2 - x = W(e²) and x = 2 - W(e²) Or are both the same? Please feel free to correct me tho...
@ivan-nm1xn2 жыл бұрын
@@gautamgopal3517 love your solution! It's also correct.
@copperII_6 ай бұрын
Regarding the second one, x + e^x = 2 e^(x+e^x) = e^2 (e^x)e^(e^x) = e^2 e^x = W(e^2) x = ln(W(e^2)) Is this also correct?
@user-mz3tp7cu4q6 ай бұрын
Why do you end up with x when solving the square root of x^2 but not the absolute value of x? |x| ???
@chuckszmanda6603 Жыл бұрын
No problem on Q1 and Q2. Lambert’s W function is also quite useful in solving problems of enzyme kinetics, radiation chemistry, especially ionization kinetics, and solar cell efficiency. Thank you for the nice explanation.
@Arnie101015 жыл бұрын
Thank you, BPRP, that was much clearer! I waited a day and was able to reproduce the solution! I also researched the W function, out of interest, and I think that there won't be a W button on a calculator any time soon! I'll try your other problems in the pinned post but don't hold your breath!
@blackpenredpen5 жыл бұрын
I am glad to hear!! Thank you for your comment Arne.
@epicm9992 жыл бұрын
This makes a lot of sense. -Now we need someone to solve for W-
@reefu2 жыл бұрын
I’d love if you made a video talking about the practical applications of the W lambert function. We use it in electrical engineering to simulate diode circuits, as their load line equation leads to a non linear transcendental equation.
@blackpenredpen2 жыл бұрын
Unfortunately, I do not have experience with that. Do you have an actual setup of the equation?
@reefu2 жыл бұрын
@@blackpenredpen So the load line equation becomes: i_d = -1/R(nV_t ln(i_d / I_s + 1) - V_th), (R, n, V_T, I_s, and V_th are just constants) then we have to solve for i_d. Eventually you can get this of the form x + k = e^x
@reefu2 жыл бұрын
In latex, the code is i_d = - \frac{1}{R} (n V_T \ln (\frac{i_d}{I_s} + 1) - V_{th})
@ShanBojack Жыл бұрын
us mathematicians generally don't care about practical applications most of the time, we're happy with how beautiful the math is lmao XD
@reefu Жыл бұрын
@@ShanBojack Oh yeah I totally understand, I’m doing maths and physics with some electives in electrical engineering, these are cool in their own right, but I was pleasantly surprised to see it used for a practical purpose
@juliengrijalva86065 жыл бұрын
BlackPenRedPen, I am 14, and I read a calculus textbook in seventh grade, and you are my favorite youtuber. If you see this please solve: \int _0^{\infty }\:cos\left(ln\left(x ight) ight)e^{-x}dx+i\int _0^{\infty \:}\:sin\left(ln\left(x ight) ight)e^{-x}dx. that is latex code, it gives the integrals. This expression is equal to i! by the gamma/Pi function, and I have not successfully evaluated it yet. You are awesome.
@mike4ty45 жыл бұрын
I will indeed service your solution, as part of ensuring that what happened in my own teen years when I was interested in math and, ironically enough, posing questions about integrals very much like this, does not happen again to someone else on the Internet. To give them the response that I had so much wanted and yet was met in my inquiries with so much grief. First off - I want to get the notation a bit tidied up, as that is a bit hard to read for youtube format. I suppose you mean - so correct if I'm wrong: int_{0...infty} cos(ln(x)) e^(-x) dx + i int_{0...inf} sin(ln(x)) e^(-x) dx. Is that right? If so, then we can proceed to find the integral as follows. First, condense the two integrals by linearity working in reverse: -> int_{0...infty} cos(ln(x)) e^(-x) + i sin(ln(x)) e^(-x) dx Now combine the like terms -> int_{0...infty} [cos(ln(x)) + i sin(ln(x))] e^(-x) dx. Now here's the trick, to relate it to the Gamma function and factorial: _note that cos(a) + i sin(a) = e^(ia)_ , i.e. Euler's formula, and thus this becomes -> int_{0...infty} e^(i ln(x)) e^(-x) dx. But now e^(i ln(x)) = x^i, because anything of the form e^(a ln(x)) is the same as x^a. Thus we have -> int_{0...infty} x^i e^(-x) dx. Now you can figure out what that has to do with the Gamma function and factorials.
@mike4ty45 жыл бұрын
@@zanea7904 aww meehhmmhhrr :) you're welcome.
@juliengrijalva86065 жыл бұрын
@@mike4ty4 I already knew how it relates to the gamma function, I used the gamma function and Euler's identity to get the integrals. I am having trouble evaluating them. I know they dont have elementary antiderivatives, but I am trying to fin the exact answer for the definite integrals.
@mike4ty45 жыл бұрын
@@juliengrijalva8606 So what do you want then by a "solution" to this problem? The exact answer to the definite integals _is_ i! = Gamma(i+1). Are you trying to express this in terms of something else? As I don't think there is any simpler form for i! or Gamma(i+1), just as there is no simpler form for sin(1) (and you certainly won't get fewer symbols than either even if there were.). Usually we just leave those things in that form when writing down equations, and take a numerical approximation to get a mental idea of how big a number that is (e.g. ~0.84 for sin(1) and ~0.50 - 0.15i for i!). Actually, Dr. Peyam has a video on this one, and he mentions about as much though also shows you can represent it in terms of a rather interesting alternative integral. But still no "simpler", exact forms. Not all definite integrals have a representation in terms of "elementary numbers" any more than indefinite ones do in terms of "elementary functions". The only non-trivial value of the Gamma function that looks to have an elementary-number representation is Gamma(1/2) and associated translations by integers Gamma(n+1/2), where Gamma(1/2) = sqrt(pi). Even Gamma(1/3) and Gamma(1/4) do not seem to, though they appear as part of the representation of many other definite integrals, I believe.
@syedmdabid7191 Жыл бұрын
Now, I got it. The Geometrical meaning of W-n( m) where m is an integer. W-n ( m) = ln ( natural logarithm) of the root of the equation x^x= e ^m√√√= This is exact value of W-n (m).
@hanzhang35895 жыл бұрын
If u need to define a new function w, why not define u(x) = inverse function of x^x, and say solution to x^x = 2 is just u(2)?
@yoavcarmel12455 жыл бұрын
That's because the W function has a series expansion so you can calculate it's values
@BeauBreedlove5 жыл бұрын
@@yoavcarmel1245 u(x) does too
@unfetteredparacosmian5 жыл бұрын
@@yoavcarmel1245 but the W function is actually useful in other contexts as well. You could define the W function in terms of the U function as well but the W is more useful on its own. (Also W is defined at x = 0)
@Kes224974 жыл бұрын
I mean, you totally could. That is a valid definition, as long as you are careful about the domain and range of x^x and u(x), and that is the end of it. But the problem now becomes that you have no idea what this function u(x) behaves like. It doesn't really tell you anything useful until you spend a long time studying the properties of u(x). This is the reason you would usually want to reduce the equation in the form of known solutions. If you can get the solution in terms of functions that have already been studied, that's a lot more useful because now you know the behaviour of it and can actually calculate the value by plugging it in, instead of a priori analysing your new function.
@hassanakhtar78744 жыл бұрын
Sure but u(x) sounds pretty arbitrary and meh.
@rob8764 жыл бұрын
Thought you might appreciate this: - written in a language that will still be around after all the others have died: -- cannot declare variables in postgresql - we're using 'from ( select 5.0*exp(5.0) as z ) as declarations' instead with recursive lambert_w as ( select z, 1 as n, case when z 1.0e-41 ) select n, w, w*exp(w) from lambert_w And in a language that will die soon (VBA): ' The Lambert W function is the function W(x) such that W(x)*exp(W(x)) = x ' or W(x*exp(x)) = x, since W(W*exp(W)) = W if we take W of both sides of the above equation. Public Function LAMBERTW(x as double) As Double Dim W, eW, WeW, WeW_x, dW As Double ' First guess for Lambert W If x
@ToshDeCamerz4 жыл бұрын
What language?
@banana61083 жыл бұрын
What language?
@mkjaiswal113 жыл бұрын
I am just a 9th grade student and I just know the Complex Numbers, but idk why I really enjoy watching this channel. BTW please check if it's correct for the questions given Answer for Q1 :- x = W(2/x) Answer for Q2 :- x= W(2x - x^2)
@tjc95144 жыл бұрын
Dude this was awesome. I really appreciate you sharing this, great job!
@shashvatshukla9 ай бұрын
Been grinning to myself for days because of this series of videos. Thank you!!!
@RobinHillyard2 жыл бұрын
Thanks, it's good! There's a question related to sorting regarding the fewest comparisons between merge sort (n lg n) and insertion sort (n^2 / 4) for small n. It's easy to determine that the curves cross at n = 16. But they also cross closer to n = 1. I was able to use the Lambert W function (my first real application of it) to find the other solution, approx. 1.24, i.e., W(- ln(2) / 4)
@ricenoodles5831 Жыл бұрын
Same here! I was reading the introduction to algorithms 4th edition and realized I couldn't solve the comparison of insertion sort of 8n^2 < merge sort of 64nlog(n)
@82rah5 жыл бұрын
The LambertW function is very interesting. Can you show how to find its derivative and anti derivative? Also how to solve x ln(x) = c
@blackpenredpen5 жыл бұрын
82rah here's the derivative. kzbin.info/www/bejne/gqqogXVsjN2Wlbc
@82rah5 жыл бұрын
@@blackpenredpen Thank you! I'm subscribed, so how did I miss that?
@82rah5 жыл бұрын
To solve x ln(x) =c note exp( ln(x) ) ln(x) = c apply W to both sides, then ln(x) = W(c), x = exp(W(c) ) or x = c/W(c)
@prabhdensingh87403 жыл бұрын
How do you do the lambert w function on a normal scientific calculator??
@82rah3 жыл бұрын
@@prabhdensingh8740 To do numerical calculations, you need a computer algebra system like Maple or Mathematica, or use WolframAlpha on internet.
@Heavenira4 жыл бұрын
OMG I got the first one! It's 2*W(sqrt(2)/2)!!! Thanks for the amazing problem!
@szshyng55993 жыл бұрын
I was stuck for almost 20 minutes and when i saw your ans i tried to work backwards and then boom got it. Thx a lot 😁
@shokan71785 жыл бұрын
I prefer the #YAY intro
@deadsamurai871328 күн бұрын
After the video was uploaded around 5 years earlier, atlast i found the answer of the problem i had 6 years back when i was a school student.
@aarnaify4 жыл бұрын
Second problem: x + e^x = 2 Take exponential of both sides: e^x * e^e^x = e^2 u*e^u = e^2 Let u = e^x u = W(e^2) Since x = ln u x = ln(W(e^2))
@kimjunsik5403 жыл бұрын
you are so clever
@user-fp6pi6wi5l5 жыл бұрын
Thanks for the video. In the first equation we eliminate x=0 by substitution, and then view two cases x>0 and x0 we get: (x/2)e^(x/2)=(2^(1/2))/2 By applying Lambert W function and rearanging: x=2W((2^(1/2))/2) In the case of x
@h4c_185 жыл бұрын
Just do e(x+e^x)=e^2, you get e^x e^(e^x) = e^2, then e^x=W(e^2) and leaves x=ln(W(e^2)) (x-2)+e*(x-2+2)=0 e^2=(2-x)*e^(2-x) 2-x=W(e^2) x=2-W(e^2). Hope that helped ;)
@gazarkhalid8405 жыл бұрын
This was extremely amazing. It explained a way difficult concept in an exceptionally simple manner. Really loving it. #YAY
@studiousboy6443 жыл бұрын
1. 2w(1/root2) 2.ln(w(e²)) Damn took me around 10 attempts to do these. Nice.
@TheRambo0103 жыл бұрын
quite interesting, I found x=2-W(e²) for the second equation, and both are correct results, although I cant work out this identity of ln(w(e²))=2-W(e²)
@infinite1.03 жыл бұрын
@@TheRambo010 ln(W(e^2)) = 2 - W(e^2) ln(W(e^2)) + W(e^2) = 2 ln(W(e^2)) +ln(e^W(e^2)) = 2 ln(W(e^2)*e^W(e^2)) = 2 Remember then that W(x)*e^W(x) = x ln(e^2) = 2
@I_like_pi_5 жыл бұрын
1. 2*W(√2/2) 2. 2-W(e^2)
@silvermica9 ай бұрын
So, that’s the Lambert function! That’s so rad.
@henningnagel19773 жыл бұрын
f(x)=x*e^x is an awesome function! You take the derivatives and get f'(x)=(x+1)*e^x; f''(x)=(x+2)*e^x; f'''(x)=(x+3)*e^x; ...; antiderivate is F(x)=(x-1)*e^x+C
@mmmmmmok5292 Жыл бұрын
woah
@libelldrian1733 жыл бұрын
I can't stop smiling.
@night42725 жыл бұрын
The omega function is so cool, I could say is my favorite one, but, it has a problem, if you want to calculate the omega function of a number, it can have more the one solution, so, if you you Wolfram Alpha, it only gives you one, so, if you want to know the other (in case there is), you have to use the Newtown's method, at least, that's what I see
@mennoltvanalten72604 жыл бұрын
Looking at another video of BPRP, the one where he solves x^2=2^x using this function W, wolframalpha has multivariate functions. Perhaps the Omega function is also a multivariate function in WA?
@manudewi4 жыл бұрын
Wow such a great explanation I just had to try those tasks (Q1 und Q2).
@ambroseaurelian96964 жыл бұрын
We love you man you are the best teacher.
@benjaminparra46722 жыл бұрын
A1: aW(b^(1/a)/a where a=b=2. And A2: 2-W(ee) where ee=e^2
@defect835211 ай бұрын
You are awesome man. It was so easy to understand.
@element11928 ай бұрын
q1: x=W(2/x) =~0.9012 q2: x=W(2x-x^2) =~0.4429 (solving with W for q2 produces an extraneous solution of zero, be careful!)
@Chrisreynolds0724199 Жыл бұрын
Incredible. Excellent teacher, thank you very much.
@abs0lute-zer0615 жыл бұрын
Why are these sooo satisfying to watch?
@spockfan20005 жыл бұрын
WWE: I thought I was gonna see Randy Orton in your video :)
@blackpenredpen5 жыл бұрын
Maybe next time.
@JonathanTot7 ай бұрын
I think it would have been really good for you to add the comment that for e^ln(x) = x, over the real this is valid only for x>0 (the range of the exponential, the domain of logarithm) and similarly, your equations labeled (1) and (2) are valid: (1) for x>=-1 (the domain of the original function) and 2) x>=-1/e (the domain of Lambert-W) the solution of x^x=2 simplifies to ln(2)/W(ln(2))
@Edelce4 жыл бұрын
0:35 gotta love that O.G bra
@crisp-cornflake30162 жыл бұрын
I know it’s an old video but: Q1: Take the sqrt and divide by two, x=2w(sqrt(2)/2) Q2: raise both sides to the e: e^(x+e^x) = e^2; this means that e^(x)*e^(e^x) = e^2; therefore e^x = W(e^2), so x = ln(W(e^2)) which is roughly .44285
@lumnisxate_19210 ай бұрын
Q1. x²e^x = 2 Take sqrt both (x² e^x)^(1/2) = 2^(1/2) Simplify x e^(x/2) = 2^(1/2) Divide both sides by 2 (x/2) e^(x/2) = (√2)/2 W both x/2 = W((√2)/2) * 2 both x = 2(W((√2)/2)) //
@ricenoodles5831 Жыл бұрын
Q1: Starting from x^2*e^x = 2, divide both sides by x^2 e^x = (2)/(x^2) then take ln of both sides x = ln((2)/(x^2)) knowing the log property that ln(a/b) = ln(a)-ln(b), x = ln(2)-ln(x^2) then using the log property that ln(a^b)=bln(a) x = ln(2)-2ln(x^2) divide everything by 2 x/2 = ln(2)/2-ln(x) raise everything by e again e^(x/2)=e^(...) then use the exponent property that a^(b-c) = a^b/a^c to parse the e^(...) e^(x/2)=e^(ln(2)/2)/(x) multiply both sides by x xe^(x/2)=e^(ln(2)/2) divide both sides by 2 (x/2)e^(x/2)=e^(ln(2)/2)/2 at this point we can use the lambert function where w(xe^x)=x x/2=w(e^(ln(2)/2)/2) x=2w(e^(ln(2)/2)/2)
@jakobthomsen15959 ай бұрын
Thanks for the explanation! Now Lambert W appears less mysterious 🙂
@RobertHayes-gi6dz11 ай бұрын
thank you
@Titurel Жыл бұрын
I was so confused for the first 15 minutes I thought about this. Then I realized I could just imagine W(x) was LN(x) and xe^x was just e^x. Phewww
@DrQuatsch5 жыл бұрын
x^2 * exp(x) = 2. Square root on both sides --> x * exp(x/2) = sqrt(2). Divide both sides by 2 --> x/2 * exp(x/2) = sqrt(2)/2. W Lambert function on both sides --> W((x/2)*exp(x/2)) = x/2 = W(sqrt(2)/2). Multiply both side by 2 --> x = 2 * W(sqrt(2)/2). x + exp(x) = 2. Substitute x = -t + 2 --> (-t + 2) + exp(-t - 2) = 2, which you can write as exp(-t) * exp(-2) = t by rearranging terms. Multiply both sides with exp(t) to get t * exp(t) = exp(-2). W Lambert function on both sides --> W(t * exp(t)) = t = W(exp(-2)). Substitute t = 2 - x back in --> 2 - x = W(exp(-2)), so x = 2 - W(exp(-2)).
@Harlequin_31414 жыл бұрын
Random typo correction on your second answer a year later :D you substituted -t-2 instead of -t+2 right off the bat so your answer is slightly off. Doing the same thing as you did I was able to show that x = 2 - W(exp(2))
@Pritzelita5 жыл бұрын
Is it possible to find the actual value of the lambert w function in terms of x?
@blackpenredpen5 жыл бұрын
I will work out a series expansion of W(x) next week. : )
@weetabixharry3 жыл бұрын
I think he should name his son "Lambert W. Redpen". Perfect name for a math enthusiast!
@gourabghosh55745 жыл бұрын
Answer to question 1 is 2 (w (1/sq root (2)))
@nitinsanatan2935 жыл бұрын
Sir,What exactly can be the value of W...If we have to use it in another expression like e^productlog(ln3),then how to find it??
@idontknowwhathandle2use3 жыл бұрын
0:35 "The original G" 2:19 "What the F is this though?" O yeah, it's all adding up.
@zestyorangez5 жыл бұрын
This was great!
@alexandreman86013 жыл бұрын
Why is the domain of xe^x [-1;+infinity], why isn't is just all the real numbers?
@hema.bhandari3 жыл бұрын
Because there is minima at -1. If you include numbers lower than -1 the function will become many-one function . i.e two pre images for one image and hence the function is non invertible. So domain is [-1,∞].
@erenyalcn93933 жыл бұрын
@@hema.bhandari Nice explanation 👍
@namrnam54133 жыл бұрын
Love when you say yes yes yes
@jamesrockybullin5250 Жыл бұрын
2:19 was not expecting that lol
@bhavydugar66652 жыл бұрын
I am thinking on putting a petition in the international math committee to change the name of lambert function to Steve function or even cooler chow function
@yonatanzoarets35045 жыл бұрын
The solution for the first equation is w(1/√2) and the solution for the second equation is ln(w(e^2))
@yonatanzoarets35045 жыл бұрын
@Sashank Sriram Well for the first one, you are very right, 1/√2=√2/2 , and I forgot to add the 2 before the w(1/√2) For the second one, I raised both sides to the power of e, so e^(x+e^x)=e^2, but e^(x+e^x)=(e^x)•(e^(e^x))=2 Then I inserted both sides into the w function, so e^x=w(e²) , so x=ln(w(e²))
@yonatanzoarets35045 жыл бұрын
@Sashank Sriram Oh, I have just understood that that the solutions are equal to each other Let w(e²)=t ln(w(e²))=ln(t)=ln(t)+t-t=ln(t)+ln(e^t)-t=ln(t•e^t)-t=ln(w(e²)•e^w(e²))-w(e²)=ln(e²)-w(e²)=2-w(e²)
@henokhagos51442 жыл бұрын
i dont word i really thank you
@citizenalex19345 жыл бұрын
I'm a simple kid. I saw WWE logo on your thumbnail, I watched the video. I pressed like.
@blackpenredpen5 жыл бұрын
Juni Raslin thank you!!
@mike4ty45 жыл бұрын
aww meehhmmhhrr :)
@leif10754 жыл бұрын
Wait at 2:20, the notation seems contradictory, if W(x) equals f^-1(x) aka the inverse of f(x) and f(x) equals xe^x, then why doesnt W(x) equal x...that wiuld be the correct implication of that notation..
@blackpenredpen4 жыл бұрын
W(f(x))=x Where f(x)=xe^x
@leif10754 жыл бұрын
@@blackpenredpen Thanks yea sorry it's confusing notation. Question is the lambert function the only waybtonsolve the x^x^3 porblem or is there some other way that tou know of?
@frogmcribbit87784 жыл бұрын
2:50 Let's say we would want to "solve" xe^(x) = -0,1. If I graph the xe^(x) function with the domain being R, I should have two solutions. The W function is the reciprocal of xe^(x) only for the "right side" of the function (for x>= -1). What do we do for the "left side" of xe^(x)? Would we need to define a "second Lambert W function" that covers the reciprocal of xe^(x) for x
@haithammajid40782 жыл бұрын
Nice work my friend.
@ludovicmatagota48204 жыл бұрын
Congratulations teacher verry good
@shivammalluri64035 жыл бұрын
Do you mind doing the integral of (1-x^2)/(x^4+3x^2+1). Thanks👍
@davidappell31053 жыл бұрын
Equivalently, x= invW(2), the inverse-W function of 2.
@sthinvc4 жыл бұрын
Can W(x) be calculated by a normal scientific calculator?
@ricenoodles5831 Жыл бұрын
Q2: I tried u-subbing with u = e^x and ln(u) = x, then solving the rest of the equation in a similar manner to Q1
@kelecsenyizoltan274 Жыл бұрын
Bravo!! :)
@Zumerjud Жыл бұрын
Very nice video :D
@xcalibur64825 жыл бұрын
Hello from future. Im in Nov 3rd
@gazarkhalid8405 жыл бұрын
Today is Nov 3rd> Lol
@sumantchopde90393 жыл бұрын
Today is Nov 3 2020 lol
@nooruddinbaqual78696 ай бұрын
As per Lambert Function,as I have understood it, anything multipled by e raised to power that thing is equal to that thing. But how can it be so? Does W carry some hidden value?
@ankurrai86775 жыл бұрын
WWE and Blackpenredpen
@Sh4dowbanned2 жыл бұрын
0:20 Ln is the inverse of g-¹ 5:35
@marceloescalantemarrugo63915 жыл бұрын
Another form of the answer is: x = ln(2)/W(ln(2))
@osuNoobCast5 жыл бұрын
i saw newest video and came here to know what w(x) is
@user-se7xy9jv9o2 жыл бұрын
1. I found x= 2× w((racine2)/2) 2. I found x= e power(w(ln2)) From morroco 🇲🇦🇲🇦
@jerichorhodesalambatin5209 Жыл бұрын
Hi sir. Can we use scientific calculator to solve that W(x) instead of the wolfram website ?
@infinite1.03 жыл бұрын
Q1: 2W(sqrt(2)/2) Q2: ln(W(e^2))
@aashsyed12773 жыл бұрын
hello
@pokoknyaakuimut0014 жыл бұрын
Wow 😲😲😲
@akramkssiri264210 ай бұрын
Hi , does the equation W(x) = 0 has a solution ? Thanks for the awesome content
@SimonPegasus10 ай бұрын
You can undo the W from the x and then turn 0 into 0e^0 so it would be x = 0e^0 which is equal to 0.
@alejandrodelabarra28383 жыл бұрын
¿Could you use your expertise to teach us how to solve stability problems through the "root-locus method"?? It uses the Laplace Transfom to see if a circuit oscilates or not....
@shahmatsimplex41444 жыл бұрын
Its been over 40 since I studied advanced math and I can still remember most of it. But the Lambert W Function just draws a blank. I see the theory is centuries old. Was it just too complex and not useful decades ago and only during this micro processor age become a valuable tool ?
@General12th3 жыл бұрын
The W-Lambert function is a just tool like anything else, but since solving for f(x)e^f(x) is not as common as solving for e^x, it doesn't show up as often. Interestingly, the Taylor series of e^x is x^n / n! , while the Taylor series of the W function is (-1)^n * x^n / n! . In other words, the W function can be approximated in exactly the same way as e^x, except the terms are alternating. Since pre-computing mathematicians had no problems using the Taylor expansion to calculate e^x, there's no reason they couldn't use the expansion for W(x), either. So it definitely wasn't too complex or useless. I suspect it's a slightly less common function that doesn't get taught before college. That's the only reason why.
@wafimarzouqmohammad80544 жыл бұрын
Why is the domain of xe^x (-1, infinity) and not (-infinity, infinity)?
@Alex_Deam4 жыл бұрын
The domain of xe^x isn't that, but that's the only valid portion of its domain you can use with Lambert W because otherwise you have two y values for some values of x for xe^x, which means its inverse would be undefinable. That's what he means by the "horizontal line test".
@factsheet49304 жыл бұрын
Is the answer you got irrational or is there no way to know? Likewise since I actually got here because of the equation x=e^(x-2), do we know if the solutions are in fact irrational?
Doesn't xe^x seem a bit arbitrary to have it's explicit inverse? I feel like factorial is a more important function that has yet to have an inverse (if you say factorial is not injective then make it have multiple branches like the lambert w function has)
@quahntasy5 жыл бұрын
I was half expecting Randy orton or Jhon cena. Lol
@blackpenredpen5 жыл бұрын
: ) How about Stone Cold Steve Austin?
@PraneshPyaraShrestha4 жыл бұрын
@@blackpenredpen I was expecting a Stone Cold Stunner
@lumbybronzearm3 жыл бұрын
I got: x = W(sqrt(2)) for the first problem, and x=ln(W(exp(2))) for the second problem. Can anyone verify if these are correct or incorrect for me, please? I have never attempted a Productlog problem before.
@walaefitout17173 жыл бұрын
My first time trying too... I got x = 2W(1/sqrt(2)) on the first one and x=ln(W(exp(2))) on the second one.
@m_stifeev3 жыл бұрын
1) x = 2W(sqrt(2)/2); 2) x = lnW(e^2)
@syedmdabid7191 Жыл бұрын
But there no use of solving exponential equation by Lambert Wilson method. It's nothing but a notation. To find the root we 've to apply the Newton- Raphson method.
@whatelseison89703 жыл бұрын
Suppose we lived in an alternate timeline where instead of having W(x) be the inverse function of xe^x they wanted it to be the inverse of x^x. How would we use that function to solve xe^x=2?
@seroujghazarian63432 жыл бұрын
Suppose W(x) WAS the inverse function of x^x xe^x=2 Let u=e^x uln(u)=2 u^u=e^2 u=W(e^2) e^x=W(e^2) x=2/W(e^2) ez
@herardpique73027 ай бұрын
x^x = 1000, x = approximately 4,5555
@uzz49433 жыл бұрын
why w(x) is equal to productlog(x)? what is the formula to calculate w(ln2)? if we are calculating it by online calculator, we can also calculate the answer of "x^x=2" by online calculator. there would not require any w(x) function. so we need a formula to calculate w(ln2) by general calculation not by online calculator.
@diptodas50459 ай бұрын
Sir why does domain of Xe^X has a domain starts from -1? Please clarify. Thanks for the video.
@omopsingh39925 жыл бұрын
How this new function. Benefits us ? I mean why is this definition so good that we adopted this in mathematics .like I know the uses of ln but what about W?
@joluju23754 жыл бұрын
I wondered too. Wikipedia gives some hints. However, among all the functions we could invent for solving a given class of problems, it's handy to choose only one, so, why not W(x) even if others would do the job too. Ok, I didn't really answer your question :)
@shahman15 жыл бұрын
mind blown ♥️🔥
@bowielam78662 жыл бұрын
But 1.5596 ^ 2 = 2.4317 which is not too close to 2? And even worse if we try 1.5592 ^ 2 = 2.4311. What I mean is that why the output is 1.5596 when there is other numbers that is closer to 2?
@stem61094 жыл бұрын
My calculator solved! X=1.5596
@khaoulaelkeibi7013 жыл бұрын
plz mathematiens tel me the method how can i solve the equation like arctan(x)=2xarctan'(x)