Euler was influential in math. He He impacted every currently existing field of math. In fact... This function was studied by Euler. But it was named "Lambert W Function" because calling yet another object after Euler is not going to be useful.
yes, someone should design a a scheme for naming all stuff named after gauss and Euller. Something like this: ET|A|1 Euler's theorem in Algebra 1, EI|CA|1 Euler's identity in Complex Analysis 1 and so on.
Thank you very much for this video! It turns out that there is a lack of videos or resources about this wonderful Lambert W function, so I am very grateful to you for what you've been doing!!
Good simple explanation. I came across x.e^x several years ago in modelling multi-access protocols. The ALOHA protocol, traffic intensity function. I struggled to find the inverse function and concluded it was a "one way function" (trap door function). Wish I had known about the W function. (I am an Engineer not a Mathematician). I eventually manged to plot the inverse function and it demonstrated beautifully the ALOHA characteristic, which limits traffic and folds back on itself, due to collisions in accessing the channel. ALOHA was the first multiaccess protocol until CDMA came along.
I was having difficulty in seeing how to use Newton's method to calculate Lambert function values. I've looked at several youtube Lambert function videos and yours answered my question perfectly. Thank you.
Started watching the video and stopped after 3, maybe 4, minutes BECAUSE I could tell it was gonna be good. Looked up your playlist then got back to it. :) Oh, btw it's funny how you talked about it appearing in all sorts of random places. I'm attempting to generate an analytical solution that predicts every single associative, binary operator of an ordered set of n elements (i.e. identify all semi-group structures...there is no known closed form solution as far as I'm aware; only commutative structures) and my math led me to what looked like a pretty simple equation: a*k + log_n(k) = b, solve for k. Threw it into Wolfram and got a solution in terms of the Lambert W Function, which I hadn't used before, and so yeah, here I am.... :) Mathoma, thanks for the awesome content you've produced. Can't wait to watch the rest of your playlist. I don't have a college education so UA-cam channels like this really give me my mental nutrients haha. Best.
I am a person who has loved things in Mathematics since H.S (in the 1960’s) and had never heard of the Lambert W function until a few weeks ago (I love Mathematics, but am rather slow at learning it Things are not Blinding Glimpses of the Obvious to me). I have seen several UA-cam videos talking about the Lambert W function, but they did not explain it well (they seemed to be confused about it, also). This is an excellent explanation of it. I enjoyed listening to a coherent explanation of its properties. I wonder if you could do a video of the history of this function. Until then, I’ll google it. Thank you for this. I’m subscribing to your channel…..
You could have mentioned that normally when we have an equation, we would solve x, which gives us the formula for calculating the values of the function for any given x, but in the case of x*exp(x), x cannot be solved using any elementary functions, which is why the only approach at calculating it is by approximation using numerical algorithms like the Newton's method.
Great video, thanks a bunch for making it. I had some trouble grasping the recursive definition on my own, but you nailed it. One thing you didn't explicitly mention which I find interesting is that you *have* to solve the W function numerically; its form is not possible to define with elementary functions so you can't solve it symbolically.
What mathematical procedure are you actually doing with W to arrive at x = W(5) approximates 1.326. (@9:11)? Such a simple straightforward question that people can't seem to answer. And worse, people thinking W(5) is the final answer.
I saw it pointed out elsewhere that in using Newton's Method to solve x = W(y); y = xeˣ, given y, that convergence is more rapid if, instead of setting it up as xeˣ - y = 0, you take log first, and make it x + ln(x) - ln(y) = 0 I haven't tried this out to verify it, but it's an intriguing claim. Fred
There are a few missing pieces to this problem. First off, you have to analyze the domain where W is defined. The base function x.e^x is not monotonous, so you cannot define an inverse that trivally. As a matter of fact, W is a function that has two branches, the one you talked about is the main W0 branch. W is by natured multi valued for some values of y (between -1/e and 0) since x.e^x is initially decreasing until -1 and then increasing. There are many other properties of the W function to define, but in my opinion to blatantly omit the space where the function is define is a pretty big mistake.
Do you happen to have any idea why @11:30 he says that tetration converging to -W(-lnx)/lnx is "mysterious" ? It doesn't look like he responds to comments so I was hoping someone with knowledge might give some insight as to what that could possibly mean. thanks
Best explanation of W that i have watched in you tube, the only missing part was graficks of x*e to the x and W, domains and ranges, but great, thanks Cent from Turkey
That implies that x = 1/Ω = e^Ω - - so that (x, lnx) = (e^Ω, Ω) = (x, 1/x) = (1/Ω, Ω). In other words, Ωe^Ω = 1, which is the very definition of Ω. Which verifies your assertion. And this, in turn, provides another way to compute y = Ω. Start with a guess for 1/y = 1/Ω, say, x₁= 2. Take ln(x₁) and 1/x₁ , which we want to = Ω, and average those two values to get y₂ . Next, take 1/y₂ and e^y₂ and average those two values to get x₂ . Repeat for x₂ what was done with x₁ , and average those two values to get y₃ . Again, find 1/y₃ and e^y₃ , averaging those two values to get x₃ . Keep doing this until the pair of values being averaged get close enough to each other, and the averaged y is your Ω estimate. The first few estimates of Ω using this process are: .5, .57468..., .56654..., .567193..., .5671392..., .5671436... Fred
@@chromaxd26 No, actually, that wasn't Newton's Method (which uses the derivative of the function whose zero you seek); it's a purely algebraic method. Newton's Method, when used on x² = a to find x = √a, boils down to this sort of algebraic method (applied to x = a/x), but for other functions, it doesn't. Fred
since the 90s i used to chase a series representation of omega. it's a number i stumbled across independently, and grew fond of for that reason. it's lead me on many chases, and i learned so much along the way. Simon Plouffe himself sent me a text file of 1 million digits, thus ending my personal challenge of somehow obtaining a million digits. oh well.
i've chased Omega for quite some time now. it is transcendental because of its intimate relationship with e. It can be computed a variety of ways, one of which is a series representation that involves Stirling Numbers of the 2nd kind. (creepy). 0.567143290409783875360 Higher order Newton's methods exist, and i've found a series that converges to it as well. proof wiki gives you the simple proof of its transcendence.
Not necessarily. The decimal expansion of all irrational numbers can only be obtained by an infinitely iterative process. How would you write down the decimal expansion of Sqrt(2)? Turns out, we tend to use Newton's method for that task as well, and other irrational but algebraic numbers.
Please, don't take me wrong. I loved the video. But for what you showed this specific problem that you have solved did not require a W function. All that is strictly required is the Newton's method. If you write x*e^x = y, then you can write f(x)=x*e^x = y => f(x) - y = 0 => x*e^x - y = 0. Then you can solve searchimg for x arounf a specific y value. For this you have used Newthon's method. This strategy allows to get x numeric value even if you are not able to separate x an 'isolate' it from y in the initial equation. Certainly there are problems where the W function as a concept is essential but that is not the case of the example you have chosen.
Apparently it's either not an elementary function or we don't know an expression for it. If there is a proof of the first (which there probably is) I don't know of it.
Since W(y) is the inverse function of y(x)=x*e^x, you can simply use the inverse function derivative rule: en.wikipedia.org/wiki/Inverse_functions_and_differentiation
Nice video but you could have used Desmos to graph W(x). I am also missing application to solve equations like$e^x=3-2x$, or $x^x=2$. Also, the different branches of W(x) are not mentioned ...
I just keep hoping to understand how W(x) can be multivalued. You mentioned that it's a relation, not a function, so I was hoping you'd go into that. Also, I'd be curious how it is evaluated on the complex numbers. Newton's method, as far as I know, only works on the reals.
Don't know why the helpful guy deleted his post. The fact that y=xe^x has a minimum and then goes back up does kinda help me understand the situation. It doesn't tell me why there are so many complex branches, but at least the real ones make more sense.
@@ZipplyZane u would need a 4d graph to see all the real and imaginary points, but it's common for an average function in the real world to be crazy in the imaginary world I suppose that W(x) has a pretty crazy graph in the imaginary world
Yes. By "inventing" the Lambert W function. Because it can't be expressed in finite form in "familiar" functions. But realize that there's a high degree of arbitrariness in forming that list. The exponential function and its inverse, the log function, sine and its inverse the arcsine, and everything that can be made from those with the four arithmetic functions and exponentiation. That encompasses the finite-degree polynomials, all the other trig functions and their inverses, the hyperbolic trig functions and their inverses (which are defined using the exponential and log functions). So who's to say that Bessel functions of the first and second kind shouldn't be included? Elliptic integrals? Or any of a myriad of other "special" functions? And then, why wouldn't we throw the W function into that pot? Each of these has, after all, an effective method of computing its values. And don't forget that, even for the "familiar" functions, special techniques are needed to evaluate, say, sin(1.7), or e^(3.67), or even e itself, for that matter. Fred
@@alxjones Sure, but exp(z) is itself, not formed by a finite combination of the arithmetic operators. Its importance is essentially a consequence of its being the solution to one of the 2 or 3 simplest, non-trivial ODE's: dy/dx = y with initial condition (x,y) = (0,1) Fred
The same way. Newton's method works over C just fine. The starting point determines which root you converge to, with the set of all starting points converging to some root R called the basin of attraction for R.
Hi! Great video, helped explain the the Lambert function to me really well. One thing: at 8:38 , is f'(x) still unchanged? I thought it would have a "-dy/dx" term on the end due to implicit differentiation (but I appreciate that if you're solving for a value of y, I.e. y is a constant, than any dy/dx term will always be 0) would you have to include a "-dy/dx* d(G(y))/dy" term if you wanted to solve for a more general function of y, that is G(y)? Thanks
Alex Pope Thanks for the compliment. With regards to the -dy/dx being there, I guess you could think of it being there when we go to take the derivative of f(x)=xe^x-y as an "intermediate" step. But, at the end of the day, we're finding the derivative of the function output, f, with respect to a small change in x, so d/dx of that y term has to be zero, which you also point out. Even if you did explicitly write out the chain rule for some G(y), that observation that dy/dx=0 implies that -dy/dx*d(G(y))/dy = 0, right? You could write it there, but it would be aesthetically displeasing. Have I understood your question?
Fair! exactly, my question I guess I wanted to ask was: How would you solve something like: xe^x = G(y) , where G is some function of y. (sticking with the previous notation). For example xe^x = y^2 where dy/dx isn't necessarily 0. (although always is when considering that in the case you demonstrated where G(y) was just a constant).
Alex Pope Okay, so we're saying that y is going to have some implicit x-dependence, so we would be solving something like xe^x=G(y(x)). So, my instinct is to say that yeah you would have to have the chain rule written out. Here's how I'm thinking of the problem: you would pick an x-value, then since the y depends on x, you get the y-value, and then the function G maps that y-value to some number and we're looking for the number x such that when you do that two-step process, you get the same thing as doing x*e^x. My hunch is that the way you suggest would work with Newton's method with the -dG/dy*dy/dx term, although there may be convergence problems if the G(y) is a wild function. Now, we're left with the question of what's the best way to numerically solve a complicated equation, which is a field unto itself, but if you take your example with: y=cos(x) (here's the x-dependence for y) G(y)=y^2 then we have to solve: x*e^x=y^2 x*e^x=cos^2(x) dG/dy = 2y dy/dx = -sin(x) which can be done with Newton's method (x ~ 0.4834) using the chain rule, but I wouldn't bet on this technique working for pathological functions. You know, historically the Lambert W came about from solving bizarre algebraic equations, so it's an interesting question.
Fair enough. I was guessing that surely if y has it's own function ( G(y) in this case) and then surely there must be a whole family of solutions of y's with corresponding x's? Hence I thought the answer in general would be y = H(x) where H is some other function of x which depends on what the function G was? Just wondering if that had a name...
Alex Pope Well now it just sounds like this function H would be the inverse relation of G. If you had x*e^x=G(y) and wanted the y value, then y = H(x*e^x).
Series for xe^x is summation of x^(n+1)÷n! Or xe^x = y = x+x^2÷1!+x^3÷2!+x^4÷3!+... So,what is the inverse summation of eq. xe^x=y Or series for W(x). If W(x) is lambert w function. I'm curious to know. Please comment.
Amazing, thank you for explaining. Could you please make a video with less explanation of the first part and the more explanation of the coolest properties?
Lost me! So is the Lambert function just another name for Newton's iterative method of finding roots? I'm either stupid or this topic is not fully explained, and not being a mathematician I will have to be content with the former.
Both expressions are equivalent, for some reason that isn't immediately evident to me. But Wolfram Alpha can confirm it for you if you input "(-productlog(-ln(x))/ln(x)) / exp(-productlog(-ln(x)))". It will show that that expression simplifies to "1", i.e., the numerator and denominator (respectively, the expression in this video and the expression you gave in your comment) are equivalent. Fwiw, when I went to solve x^y=y, I also arrived at the expression given in the video. I rather like that your expression only mentions x once though, seems more elegant.
It seems that Lambert W function should be put in a complex number scenario rather than limit it to real number only. Besides, you've spend really too much time on Newton's method which does not help understanding the Omega function.
0:37 Grammar gripe here - you reversed the words "a" and "an." It should be "Given an x, could a y be found? Given a y, could x be found?" Yes, grammar is important, even in math.
so let me reiterate, you want to solve an equation for x, and when you do you get omega as a result, how is omega any better than x in this context, it is just replacing one letter by another? :)
I think this function isn’t usefull at all if you need to do a numerical method to find the root anyway. It’s fucking non-sense, it’s like you need to find the inverse of the function that isn’t invertible, so you name another function that would do it’s inverse, but it doesn’t, first you need to compute it’s values.... fuck
Euler was influential in math. He He impacted every currently existing field of math. In fact...
This function was studied by Euler. But it was named "Lambert W Function" because calling yet another object after Euler is not going to be useful.
yes, someone should design a a scheme for naming all stuff named after gauss and Euller. Something like this: ET|A|1 Euler's theorem in Algebra 1, EI|CA|1 Euler's identity in Complex Analysis 1 and so on.
A beautiful Lambert W Function "for dummies" video
Excellent video, highly undervalued (there should be millions of views on this video), I rate 10/10
Thank you very much for this video! It turns out that there is a lack of videos or resources about this wonderful Lambert W function, so I am very grateful to you for what you've been doing!!
I wanted to hear about the graph in the thumbnail.🤨
Maybe it's just the Mandelbrot set with the W function instead of squaring
or it shows the branch of W to which Newtons method converges, given some starting value
Thanks, i won't waste my time on this clickbait.
Good simple explanation. I came across x.e^x several years ago in modelling multi-access protocols. The ALOHA protocol, traffic intensity function. I struggled to find the inverse function and concluded it was a "one way function" (trap door function). Wish I had known about the W function. (I am an Engineer not a Mathematician). I eventually manged to plot the inverse function and it demonstrated beautifully the ALOHA characteristic, which limits traffic and folds back on itself, due to collisions in accessing the channel. ALOHA was the first multiaccess protocol until CDMA came along.
I was having difficulty in seeing how to use Newton's method to calculate Lambert function values. I've looked at several youtube Lambert function videos and yours answered my question perfectly. Thank you.
Excellently presented. Thank you years later my good man.
This video is the best one on UA-cam that gives clear explanation of The Lambert W Function. thank you man.
Started watching the video and stopped after 3, maybe 4, minutes BECAUSE I could tell it was gonna be good. Looked up your playlist then got back to it. :)
Oh, btw it's funny how you talked about it appearing in all sorts of random places. I'm attempting to generate an analytical solution that predicts every single associative, binary operator of an ordered set of n elements (i.e. identify all semi-group structures...there is no known closed form solution as far as I'm aware; only commutative structures) and my math led me to what looked like a pretty simple equation: a*k + log_n(k) = b, solve for k. Threw it into Wolfram and got a solution in terms of the Lambert W Function, which I hadn't used before, and so yeah, here I am.... :) Mathoma, thanks for the awesome content you've produced. Can't wait to watch the rest of your playlist. I don't have a college education so UA-cam channels like this really give me my mental nutrients haha.
Best.
I am a person who has loved things in Mathematics since H.S (in the 1960’s) and had never heard of the Lambert W function until a few weeks ago (I love Mathematics, but am rather slow at learning it Things are not Blinding Glimpses of the Obvious to me). I have seen several UA-cam videos talking about the Lambert W function, but they did not explain it well (they seemed to be confused about it, also). This is an excellent explanation of it. I enjoyed listening to a coherent explanation of its properties. I wonder if you could do a video of the history of this function. Until then, I’ll google it. Thank you for this. I’m subscribing to your channel…..
Thanks for presenting the basics of W. So often ODE classes jump into tricks using W without a clear motivation.
You could have mentioned that normally when we have an equation, we would solve x, which gives us the formula for calculating the values of the function for any given x, but in the case of x*exp(x), x cannot be solved using any elementary functions, which is why the only approach at calculating it is by approximation using numerical algorithms like the Newton's method.
Except for a few carefully selected values of x. But this is not different from the trig or log functions; we're just much more familiar with them..
Woah I’m impressed that you’ve talked with Classical Theist!! Math channels tied to philosophy and especially Thomism are few to come by.
Yeah I've known him for about 5-6 years now.
Great video, thanks a bunch for making it. I had some trouble grasping the recursive definition on my own, but you nailed it.
One thing you didn't explicitly mention which I find interesting is that you *have* to solve the W function numerically; its form is not possible to define with elementary functions so you can't solve it symbolically.
What mathematical procedure are you actually doing with W to arrive at x = W(5) approximates 1.326. (@9:11)?
Such a simple straightforward question that people can't seem to answer. And worse, people thinking W(5) is the final answer.
I saw it pointed out elsewhere that in using Newton's Method to solve x = W(y); y = xeˣ, given y, that convergence is more rapid if, instead of setting it up as
xeˣ - y = 0,
you take log first, and make it
x + ln(x) - ln(y) = 0
I haven't tried this out to verify it, but it's an intriguing claim.
Fred
amazing! There is no way I would have read the same from wikipedia and understood it.
There are a few missing pieces to this problem. First off, you have to analyze the domain where W is defined. The base function x.e^x is not monotonous, so you cannot define an inverse that trivally. As a matter of fact, W is a function that has two branches, the one you talked about is the main W0 branch. W is by natured multi valued for some values of y (between -1/e and 0) since x.e^x is initially decreasing until -1 and then increasing.
There are many other properties of the W function to define, but in my opinion to blatantly omit the space where the function is define is a pretty big mistake.
@@jamesnash3274 bruh
Do you happen to have any idea why @11:30 he says that tetration converging to -W(-lnx)/lnx is "mysterious" ? It doesn't look like he responds to comments so I was hoping someone with knowledge might give some insight as to what that could possibly mean. thanks
Best explanation of W that i have watched in you tube, the only missing part was graficks of x*e to the x and W, domains and ranges, but great, thanks Cent from Turkey
lnx and 1/x intersect at (x,omega)
That implies that x = 1/Ω = e^Ω - - so that (x, lnx) = (e^Ω, Ω) = (x, 1/x) = (1/Ω, Ω).
In other words, Ωe^Ω = 1, which is the very definition of Ω. Which verifies your assertion.
And this, in turn, provides another way to compute y = Ω.
Start with a guess for 1/y = 1/Ω, say, x₁= 2. Take ln(x₁) and 1/x₁ , which we want to = Ω, and average those two values to get y₂ .
Next, take 1/y₂ and e^y₂ and average those two values to get x₂ .
Repeat for x₂ what was done with x₁ , and average those two values to get y₃ .
Again, find 1/y₃ and e^y₃ , averaging those two values to get x₃ .
Keep doing this until the pair of values being averaged get close enough to each other, and the averaged y is your Ω estimate.
The first few estimates of Ω using this process are: .5, .57468..., .56654..., .567193..., .5671392..., .5671436...
Fred
@@ffggddss i love your comments, fred
@@rot6015 Thanks!
Fred
@@ffggddss so just use newton's method of roots?
@@chromaxd26 No, actually, that wasn't Newton's Method (which uses the derivative of the function whose zero you seek); it's a purely algebraic method.
Newton's Method, when used on x² = a to find x = √a, boils down to this sort of algebraic method (applied to x = a/x), but for other functions, it doesn't.
Fred
Thank you - this is the best explanation I have found.
Thank you for this video. You did a great job introducing this function. Keep it up! :)
A beautiful introduction, right pace and structure for me, hard to find such neat ones :)
since the 90s i used to chase a series representation of omega. it's a number i stumbled across independently, and grew fond of for that reason. it's lead me on many chases, and i learned so much along the way. Simon Plouffe himself sent me a text file of 1 million digits, thus ending my personal challenge of somehow obtaining a million digits. oh well.
kharnak crux one way omega is interesting is that it is recursive. It has some interesting complex analogs as well
Why does the base have to be eulers number? Why can’t it be any other number?
if omega can only be obtained with an infinitely iterative process does that make it a transcendental number?
i've chased Omega for quite some time now. it is transcendental because of its intimate relationship with e. It can be computed a variety of ways, one of which is a series representation that involves Stirling Numbers of the 2nd kind. (creepy). 0.567143290409783875360 Higher order Newton's methods exist, and i've found a series that converges to it as well. proof wiki gives you the simple proof of its transcendence.
Not necessarily. The decimal expansion of all irrational numbers can only be obtained by an infinitely iterative process. How would you write down the decimal expansion of Sqrt(2)? Turns out, we tend to use Newton's method for that task as well, and other irrational but algebraic numbers.
What is the graph in the thumbnail?
Something like Mandelbrot set
@Multorum Unum множество Мандельброта является фракталом
Please, don't take me wrong. I loved the video. But for what you showed this specific problem that you have solved did not require a W function. All that is strictly required is the Newton's method. If you write x*e^x = y, then you can write f(x)=x*e^x = y => f(x) - y = 0 => x*e^x - y = 0. Then you can solve searchimg for x arounf a specific y value. For this you have used Newthon's method. This strategy allows to get x numeric value even if you are not able to separate x an 'isolate' it from y in the initial equation. Certainly there are problems where the W function as a concept is essential but that is not the case of the example you have chosen.
Hi, this is a cool video, good explaination.
When you take the derivative of f(x)=x*exp(x)-y shouldn't you also derive the y as it is a function of x?
No, since y is simply a constant, not a function in terms of x.
How can we derive the Lambert W Function?.I could't find any kind of article related the topic.
Apparently it's either not an elementary function or we don't know an expression for it. If there is a proof of the first (which there probably is) I don't know of it.
Since W(y) is the inverse function of y(x)=x*e^x, you can simply use the inverse function derivative rule: en.wikipedia.org/wiki/Inverse_functions_and_differentiation
a really intriguing function...instructive video, thank you
Why is the function x.e^x important ?
It's something about complex math stuff in other branches of math. Useless to the average math guy honestly
Nice video but you could have used Desmos to graph W(x). I am also missing application to solve equations like$e^x=3-2x$, or $x^x=2$. Also, the different branches of W(x) are not mentioned ...
Excellent presentation of the topics. DrRahul Rohtak India.
The thumbnail had me thinking it would be something like the Mandelbrot set, but this is cool too
Very clear. Thank you so much!
Wow that was interesting ! Thanks for the coolest property ;)
You should have discussed the domain of W, since x*e^x is not injective (1 to 1 ).
Can't we the value of W-n(2) direct only iff by Newton Raphson method. What's the meaning and the value of W-n(2)
Muchas gracias desde Argentina;muy claro y didactico.
Hector.
Thanks you very much;very clear and powerfull ;so again THANKS AND LETS GO FORTH.!!!!!
I just keep hoping to understand how W(x) can be multivalued. You mentioned that it's a relation, not a function, so I was hoping you'd go into that.
Also, I'd be curious how it is evaluated on the complex numbers. Newton's method, as far as I know, only works on the reals.
Don't know why the helpful guy deleted his post. The fact that y=xe^x has a minimum and then goes back up does kinda help me understand the situation.
It doesn't tell me why there are so many complex branches, but at least the real ones make more sense.
@@ZipplyZane u would need a 4d graph to see all the real and imaginary points, but it's common for an average function in the real world to be crazy in the imaginary world
I suppose that W(x) has a pretty crazy graph in the imaginary world
Thanks....that was helpful...I have only just stumbled upon the Lambert W function...
Top notch video. I would give reddit gold if I had some.
The newtons method is a numerical method. Is there any way to solve this problem analytically?
Apparently not.
Yes. By "inventing" the Lambert W function.
Because it can't be expressed in finite form in "familiar" functions.
But realize that there's a high degree of arbitrariness in forming that list.
The exponential function and its inverse, the log function, sine and its inverse the arcsine, and everything that can be made from those with the four arithmetic functions and exponentiation.
That encompasses the finite-degree polynomials, all the other trig functions and their inverses, the hyperbolic trig functions and their inverses (which are defined using the exponential and log functions).
So who's to say that Bessel functions of the first and second kind shouldn't be included? Elliptic integrals? Or any of a myriad of other "special" functions?
And then, why wouldn't we throw the W function into that pot?
Each of these has, after all, an effective method of computing its values.
And don't forget that, even for the "familiar" functions, special techniques are needed to evaluate, say, sin(1.7), or e^(3.67), or even e itself, for that matter.
Fred
@@ffggddss To be fair, polynomial, trigonometric, exponential, and logarithmic functions are all in the span of exp(z) with algebraic (+,*,o)
@@alxjones Sure, but exp(z) is itself, not formed by a finite combination of the arithmetic operators.
Its importance is essentially a consequence of its being the solution to one of the 2 or 3 simplest, non-trivial ODE's:
dy/dx = y
with initial condition (x,y) = (0,1)
Fred
How do you calculate complex numbers or the different branches of W(x)
empCarnivore good question. Analytically i think..
The second branch isnt well understood methinks
The same way. Newton's method works over C just fine. The starting point determines which root you converge to, with the set of all starting points converging to some root R called the basin of attraction for R.
amazing function, very unknown. nice video.
So cool!
Hi! Great video, helped explain the the Lambert function to me really well.
One thing: at 8:38 , is f'(x) still unchanged? I thought it would have a "-dy/dx" term on the end due to implicit differentiation (but I appreciate that if you're solving for a value of y, I.e. y is a constant, than any dy/dx term will always be 0) would you have to include a "-dy/dx* d(G(y))/dy" term if you wanted to solve for a more general function of y, that is G(y)?
Thanks
Alex Pope Thanks for the compliment. With regards to the -dy/dx being there, I guess you could think of it being there when we go to take the derivative of f(x)=xe^x-y as an "intermediate" step. But, at the end of the day, we're finding the derivative of the function output, f, with respect to a small change in x, so d/dx of that y term has to be zero, which you also point out. Even if you did explicitly write out the chain rule for some G(y), that observation that dy/dx=0 implies that -dy/dx*d(G(y))/dy = 0, right? You could write it there, but it would be aesthetically displeasing.
Have I understood your question?
Fair! exactly, my question I guess I wanted to ask was:
How would you solve something like:
xe^x = G(y) , where G is some function of y. (sticking with the previous notation).
For example xe^x = y^2 where dy/dx isn't necessarily 0. (although always is when considering that in the case you demonstrated where G(y) was just a constant).
Alex Pope Okay, so we're saying that y is going to have some implicit x-dependence, so we would be solving something like xe^x=G(y(x)). So, my instinct is to say that yeah you would have to have the chain rule written out.
Here's how I'm thinking of the problem: you would pick an x-value, then since the y depends on x, you get the y-value, and then the function G maps that y-value to some number and we're looking for the number x such that when you do that two-step process, you get the same thing as doing x*e^x.
My hunch is that the way you suggest would work with Newton's method with the -dG/dy*dy/dx term, although there may be convergence problems if the G(y) is a wild function. Now, we're left with the question of what's the best way to numerically solve a complicated equation, which is a field unto itself, but if you take your example with:
y=cos(x) (here's the x-dependence for y)
G(y)=y^2
then we have to solve:
x*e^x=y^2
x*e^x=cos^2(x)
dG/dy = 2y
dy/dx = -sin(x)
which can be done with Newton's method (x ~ 0.4834) using the chain rule, but I wouldn't bet on this technique working for pathological functions.
You know, historically the Lambert W came about from solving bizarre algebraic equations, so it's an interesting question.
Fair enough. I was guessing that surely if y has it's own function ( G(y) in this case) and then surely there must be a whole family of solutions of y's with corresponding x's? Hence I thought the answer in general would be y = H(x) where H is some other function of x which depends on what the function G was? Just wondering if that had a name...
Alex Pope Well now it just sounds like this function H would be the inverse relation of G. If you had x*e^x=G(y) and wanted the y value, then y = H(x*e^x).
Series for xe^x is summation of x^(n+1)÷n! Or xe^x = y = x+x^2÷1!+x^3÷2!+x^4÷3!+...
So,what is the inverse summation of eq. xe^x=y Or series for W(x). If W(x) is lambert w function.
I'm curious to know.
Please comment.
It's a bit disappointing that you didn't talk about where the function is defined
Amazing, thank you for explaining. Could you please make a video with less explanation of the first part and the more explanation of the coolest properties?
Lost me! So is the Lambert function just another name for Newton's iterative method of finding roots? I'm either stupid or this topic is not fully explained, and not being a mathematician I will have to be content with the former.
Basically it's the inverse function of xe^x
Hey I need help in integrating this function
dx/{bx^2-axln(x)-bxc}
a,b,c are some constants.
Please help me out.
Please please please....
How can find lambert W matrix?
Very useful, many thanks
Broooo where did you go:(
in the coolest property number 6, the result it wouldnt be this? e^(-W(-ln(x))) ?? nice vid btw
Both expressions are equivalent, for some reason that isn't immediately evident to me. But Wolfram Alpha can confirm it for you if you input "(-productlog(-ln(x))/ln(x)) / exp(-productlog(-ln(x)))". It will show that that expression simplifies to "1", i.e., the numerator and denominator (respectively, the expression in this video and the expression you gave in your comment) are equivalent.
Fwiw, when I went to solve x^y=y, I also arrived at the expression given in the video. I rather like that your expression only mentions x once though, seems more elegant.
Thanks a lot for the video!
Nice explanation
This function is a fractal?
My calculator shows 0.57102 instead??
A calculator doesn't have the lambert W function lol, how did u get Ω
But what about x^x=5 ?
Bruh, that should be easy to someone who needs to know the lambert W function
It seems that Lambert W function should be put in a complex number scenario rather than limit it to real number only. Besides, you've spend really too much time on Newton's method which does not help understanding the Omega function.
No important result! W-n( 2) = ???, Can't we write direct value but by Newton- Raphson method.
Does anyone know how to solve for x: x * logx = 400
0:37 Grammar gripe here - you reversed the words "a" and "an." It should be "Given an x, could a y be found? Given a y, could x be found?" Yes, grammar is important, even in math.
xe^x = -1 => x=W(-1) ≈ -0.31813+1.33723i
xe^x = 0 => x=0
xe^x = 0.5 => x=W(0.5) ≈ 0.351734
xe^x = 1 => x=W(1) ≈ 0.567143
xe^x = 2 => x=W(2) ≈ 0.852606
xe^x = 3 => x=W(3) ≈ 1.04991
xe^x = 4 => x=W(4) ≈ 1.20217
xe^x = 5 => x=W(5) ≈ 1.32672
click in by mistake. leave with no disappointment. Thank You.
so let me reiterate, you want to solve an equation for x, and when you do you get omega as a result, how is omega any better than x in this context, it is just replacing one letter by another? :)
X is a variable and Ω is a constant. It has a value, wich is w(1), or arround 0.59
fab
Aww. I thought it would be different
Clickbait. I'm here for a set on complex plane :(
clpo
scammed by the thumbnail
Eleven minutes of stating the obvious before less than one minute of interesting stuff
I think this function isn’t usefull at all if you need to do a numerical method to find the root anyway. It’s fucking non-sense, it’s like you need to find the inverse of the function that isn’t invertible, so you name another function that would do it’s inverse, but it doesn’t, first you need to compute it’s values.... fuck
😂You described Newton's method just to show that you know it, while you made no use of it while solving... Are you in grade 2?or what😂😂😂😂
+PhyteR
Grade 3, actually. I used Newton's method here.
Dude, what is the use of Newton's formula, if you have used approximation at the end.
+PhyteR
I used Newton's method (what is Newton's _formula_?) to arrive at that approximation.
😁 ya just like another video. Leave it, i can't discuss it with you.
+PhyteR
I suspect you're confused on what Newton's method is and what it's doing.
your x1 should be .571020439 and not .5672....