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?
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)
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.
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 🙏
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))
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)).
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...
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.
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).
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!
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.
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.
@@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.
@@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.
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
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)
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.
@@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
@@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
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)
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)
@@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)
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.
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
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 ;)
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)
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))
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
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
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?
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)) //
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
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)
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)).
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))
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
@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))
@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)
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
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,∞].
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..
@@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?
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?
¿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....
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.
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.
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)
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.
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".
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?
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Hi all, hopefully this remake makes things more clear! Also try to solve x^2e^x=2 and x+e^x=2
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?
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)
hey where you from actually?? Japan?
@@suraj_mohapatra
Actually I'm from israel
@@yonatanzoarets3504 איזה גבר גם אני מישראל
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.
: )
Shane McMahon function jumps from Hell in a Cell
WATCH OUT WATCH OUT! RKO! OUT OF NOWHERE!
😂😂😂😂😂😂u guys r awesome (btw randy Orton is my fav)
Me when trying to do the problems BPRP gives: 🤔🤔🤔
The answer when he reveals it: ruclips.net/video/cNgxyL5zEAk/видео.html
I like how he always keeps mic in his hand like an apple.
Pokeball
it fell on his head, just like the apple fell on newton's head.
@@justafish5559 🥸
xchg INT 2Fh & INT 21h (like INT 1Ch & INT 08h)
He's using the Apple of Eden.
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 🙏
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))
: )
Why aren't you taking the -ve value in first one ?
@@tbg-brawlstars you mean from square root?
@@JacobRy yes
@@tbg-brawlstars -sqrt(2)/2 is not in dom(W)
NO! IN THIS CHANNEL THE VARIABLE FOR W FUNCTION MUST BE FISH!
xD
It is not funny 😒😒😒
@@Mnzmanzmz lol
Our teacher in Greece puts apple as variable
@@Mnzmanzmz what a great way to ruin a joke
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)).
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...
@@gautamgopal3517 love your solution! It's also correct.
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?
Why do you end up with x when solving the square root of x^2 but not the absolute value of x? |x| ???
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.
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).
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!
I am glad to hear!! Thank you for your comment Arne.
This makes a lot of sense. -Now we need someone to solve for W-
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.
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.
@@zanea7904 aww meehhmmhhrr :) you're welcome.
@@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.
@@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.
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
What language?
What language?
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.
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)
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.
Unfortunately, I do not have experience with that. Do you have an actual setup of the equation?
@@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
In latex, the code is i_d = - \frac{1}{R} (n V_T \ln (\frac{i_d}{I_s} + 1) - V_{th})
us mathematicians generally don't care about practical applications most of the time, we're happy with how beautiful the math is lmao XD
@@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
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))
you are so clever
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)
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)
OMG I got the first one! It's 2*W(sqrt(2)/2)!!! Thanks for the amazing problem!
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 😁
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)?
That's because the W function has a series expansion so you can calculate it's values
@@yoavcarmel1245 u(x) does too
@@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)
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.
Sure but u(x) sounds pretty arbitrary and meh.
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
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 ;)
1. 2w(1/root2)
2.ln(w(e²))
Damn took me around 10 attempts to do these. Nice.
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)
@@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
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!)
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
82rah here's the derivative. ruclips.net/video/LyEPB6Wxc_U/видео.html
@@blackpenredpen Thank you! I'm subscribed, so how did I miss that?
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)
How do you do the lambert w function on a normal scientific calculator??
@@prabhdensingh8740 To do numerical calculations, you need a computer algebra system like Maple or Mathematica, or use WolframAlpha on internet.
1. 2*W(√2/2)
2. 2-W(e^2)
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))
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
woah
A1: aW(b^(1/a)/a where a=b=2. And A2: 2-W(ee) where ee=e^2
thank you
Been grinning to myself for days because of this series of videos. Thank you!!!
I prefer the #YAY intro
Dude this was awesome. I really appreciate you sharing this, great job!
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
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?
Wow such a great explanation I just had to try those tasks (Q1 und Q2).
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)) //
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
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)
Thanks for the explanation! Now Lambert W appears less mysterious 🙂
This was extremely amazing. It explained a way difficult concept in an exceptionally simple manner. Really loving it.
#YAY
I can't stop smiling.
So, that’s the Lambert function! That’s so rad.
Incredible. Excellent teacher, thank you very much.
thank you!
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
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)).
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))
Why are these sooo satisfying to watch?
Answer to question 1 is 2 (w (1/sq root (2)))
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
0:35 gotta love that O.G bra
The solution for the first equation is w(1/√2) and the solution for the second equation is ln(w(e^2))
@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))
@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)
We love you man you are the best teacher.
You are awesome man. It was so easy to understand.
0:35 "The original G"
2:19 "What the F is this though?"
O yeah, it's all adding up.
WWE: I thought I was gonna see Randy Orton in your video :)
Maybe next time.
Equivalently, x= invW(2), the inverse-W function of 2.
I think he should name his son "Lambert W. Redpen". Perfect name for a math enthusiast!
0:20
Ln is the inverse of g-¹
5:35
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??
Is it possible to find the actual value of the lambert w function in terms of x?
I will work out a series expansion of W(x) next week. : )
2:19 was not expecting that lol
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
Why is the domain of xe^x [-1;+infinity], why isn't is just all the real numbers?
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,∞].
@@hema.bhandari Nice explanation 👍
Congratulations teacher verry good
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..
W(f(x))=x
Where f(x)=xe^x
@@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?
Love when you say yes yes yes
i dont word i really thank you
This was great!
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
1. I found x= 2× w((racine2)/2)
2. I found x= e power(w(ln2))
From morroco 🇲🇦🇲🇦
Do you mind doing the integral of
(1-x^2)/(x^4+3x^2+1). Thanks👍
Nice work my friend.
Another form of the answer is:
x = ln(2)/W(ln(2))
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?
Hello from future. Im in Nov 3rd
Today is Nov 3rd> Lol
Today is Nov 3 2020 lol
おー ランバートW関数って始めて聞きました
ネイピア数を使ったテーブルマジックのようだ
ここまで教えて貰えばようやくe便利だ～ってなるね
ぼく高校では「やっぱ底は10がよくね？」ってずっと思ってたからさあ
¿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....
x^x = 1000, x = approximately 4,5555
WWE and Blackpenredpen
Bravo!! :)
Can W(x) be calculated by a normal scientific calculator?
Q1: 2W(sqrt(2)/2)
Q2: ln(W(e^2))
hello
Hi sir. Can we use scientific calculator to solve that W(x) instead of the wolfram website ?
Faster way to solve the equation:
x^x=2
x^^2=2
x=ssqrt(2)
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.
I got Square root of 2 divided by (w(-1))^.5
For x^2e^×=2
6:10
That is the inverse of the lineärithm (x log(x)).
... As you demonstrated at 6:36...
Hi , does the equation W(x) = 0 has a solution ? Thanks for the awesome content
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.
I'm a simple kid. I saw WWE logo on your thumbnail, I watched the video. I pressed like.
Juni Raslin thank you!!
aww meehhmmhhrr :)
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.
Very nice video :D
i saw newest video and came here to know what w(x) is
Wow 😲😲😲
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)
Sir why does domain of Xe^X has a domain starts from -1? Please clarify. Thanks for the video.
Integration of x^6-x^5/x^4-x^3
1/7 x^7 - 1/2 x^2 -1/4 x^4 + C
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.
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.
1) x = 2W(sqrt(2)/2); 2) x = lnW(e^2)
Why is the domain of xe^x (-1, infinity) and not (-infinity, infinity)?
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".
why did we chose the domain of the function as Df = [-1;+00[ even if its actually R ?
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?
Do we need to worry bout which branch of the function we are in? Is the W function multi valued?