Totally can do Newton method on a slide rule! Since x^x=2 is equiv to x ln(x) = ln2， so iterate x ← x - (x ln(x) - ln(2))/(1+ln(x)) , that is x←(x+ ln(2))/(1+ln(x))
An easier solution in my opinion would be the following: use ln to get x*ln(x)=ln(2) rewrite x as e^ln(x) then apply w(x) to get ln(x)=w(ln(2)) and exponentiate. I think that would be easier. But I can be wrong. Nevertheless, great video and greetings from germany.
Ya that's what I did. We had to get W involved, which motivated the forcing of a ye^y into the expression. It's nice how so many solutions to exponential equations can be neatly expressed in terms of W.
I like your way better. It seems easier to see how each step progresses towards a situation of the form a*e^a. Though I hope I'll just be able to memorize that x^x = a implies x = e^W(ln a).
You know, this video gives me the satisfaction that out there there are people who care about finding the solution to questions like these, for I too often get mocked for asking such things in class.
@@adrianpramadipta8631 Probably cause it doesn't relate to the class and it's distracting them from the actual subject relevant to them: the subject for the test.
There is an amazing method to calculate the w function by newton's method in the calculator! To calculate W(x) , in the calculator enter whatever number and press = So this number will be save into "Ans" And then enter this : Ans^2/(Ans+1)+x/[e^Ans(Ans+1)] And then press = too many times , you will get a very very nice approximation of W(x)!
isn't Newton's method about finding the tangent lines and where those intersect with the line y=0 ? I don't understand how that is what you're doing when typing that into a calculator. The fixed point iteration is about finding the value that won't be changed when plugged into a function, f(x)=x. When you type that into the calculator, you're solving for when your expression becomes equal to ANS, in other words, you're looking for a fixed point. Am I just completely wrong here, or...?
BenDesch to be honest I don’t know much about the fixed point iteration, or even about Newton’s method , but I know that Newton’s method deals with the derivative and the tangent line in a point , so when you set a function equals to zero , the Newton’s method says that you pick whatever x value “ x0” The new x value will be X1=x0-f(x0)/f’(x0) X2=x1-f(x1)/f’(x1) And so on When you apply this many times , you will get closer to the x value that make the function equals to zero This is the idea of newtos’s method! And I recommend you to read it in Wikipedia.
There are already lots of those types of functions floating around even remedial mathematics, it's just that they're taken for granted. sqrt(x) is a function that was defined to solve problems like x^2 - 2 = 0. One would obviously consider sqrt(2) to be a solution in exact form, even though it is still being expressed in terms of a function (the sqrt function) at a specific value: f(2). W(x) is no different from sqrt(x), ln(x), or any trig function, for that matter; they were created to denote solutions to certain types of problems in exact form.
Thats what i was thinking, i feel a bit cheated really. Seems to just write the exact value without using functions you would require an infinite series. I was thinking from the title there would be something weird like sqrt(2/sqrt(e)) or some finite expression like that.
@@PaddedShaman never thought of it like that, I suppose the unfamiliarity of W made me think of it as something less natural than a root or power or log.
Blackpenredpen, your videos are truly brilliant! I hope you can share your enthusiasm and skill in making things clear also as a $-earning job, whether in a school or a University or elsewhere. It’s great how you make the choice of solution strategy clear. The trick of course is: How does one figure out the solution STRATEGY? - What your videos seem to suggest is that math is 25% knowledge (of theorems, properties, etc.), 25% disciplined logic, and 50% creative imagination! - the 50% they never teach you in school...
Just take a log of the equation xlnx = ln2. Now substitution: y = lnx => x=e^y gives equation: ye^y = ln2, where Lambert's function can be applied straight away: y = W(ln2) => x = e^W(ln2). The answer looks different. But multiply and divide by W(ln2): x = W(ln2) e^W(ln2) / W(ln2). Now from the definition of Lambert's function it follows that W(ln2)e^W(ln2) = ln2, so x = ln2/W(ln2).
Wow, exact form! I have an exact form for ∫exp(x²)dx too. What you do is define the function DoubExpInt(x) to be the definite integral of exp(t²)dt from 0 to x, and then the exact solution is ∫exp(x²)dx = DoubExpInt(x) + c!! I find it really irritating when people say they have a "closed form" or whatever of something that they simply defined that way.
Well okay, but then I suppose you will hate log(x) because it can be _defined_ to be the solution to x = e^y, hate sqrt(x) because it's _defined_ as the solution to x = y^2, etc. Defining new operation as the inverses of old operations is a perfectly legitimate way to make more operations. And W(x) is not defined to solve y^y = y, but ye^y = x - so saying it "does not solve it" is not any more correct than saying sqrt(x) cannot be used to "solve" ax^2 + bx + c = 0 because it is "defined" to solve x = y^2. The elementary operations are limited. The trick is finding which operations you can add to extend your solving power that are the most "profitable" - i.e. yes, you can indeed "cheat" on it by defining your problem X as a "new operation", but ideally what you want is a simple operation that solves a great many not-simply-related problems. The Lambert W function can be used for such things as expressing the constant that defines the peak wavelength of radiation emitted by an ideal hot object as a function of temperature, for example - that is, Wien's displacement law, something that is quite different in flavor from solving its original defining equation. And while I suppose one could say the square root is in some sense "simpler" or "more natural", those are notions that themselves are subject to possible challenge and critique. Let me present to you an almost equally logical, if not somewhat slightly more general, operation that could have been used in its stead - the "rectangle root": it is a function R(A, r), where A is the area of a rectangle and r its aspect ratio (the ratio of width to height), and it returns the length of the shorter side: that is, its definition is such that R(A, r) * (r R(A, r)) = A. Why consider this? Consider that computer and mobile device screens are made with certain aspect ratios, or how that camera resolutions are reported in pixels. If I have a 5 megapixel image and I want it in 16:9 aspect ratio, how many pixels on a side? The answer is R(5000000, 16/9), for the shorter side. Does that seem so silly now?? Going to earlier history, we may want to - this would be related to where the idea of square roots started - consider a rectangular piece of farmland, and I want to know how much fencing I will need to enclose it to allow me to grow a given amount of crop or grain. So I would like a function that I can insert the amount of grain (proportionate to the area) and the shape of the rectangle (i.e. r). (Heck, we could even have a different function giving the full perimeter! That seems reasonable, doesn't it?) Of course you see this is "just" sqrt(A/r), but we could just as easily have taken the square root to be defined in terms of it instead, as sqrt(A) = R(A, 1). If you were raised from your school years that R(A, r) was the fundamental function, and, say, the quadratic formula for ax^2 + bx + c = 0 was given to you as x = +/-R(b^2 - 4ac, 4a^2) - b/2a, and your calculator had a little R(A, r) button on it instead of a "sqrt" button, and to find the area of a square you always took R(A, 1), you would have never questioned it. Heck, when presented with sqrt(), this alternatively-socialized parallel-universe "you" might even be retorting that "that's not general enough, it's missing a parameter" or "isn't that just a redundant name for x^(1/2)?" The point is that the things you think are "more valid" when they are not deductions but selections, is more a product of culture, than it is somehow necessarily inherently "better" so much to the point that others should be excluded from consideration at all or denigrated to a "cheat". Indeed, if you drill it at hard enough you can even challenge the whole mathematical edifice itself: why do we even use "numbers"? We could have taken something else as just as basic. In ancient Greece, geometry had a much more fundamental role than it does in modern mathematics. There are, in fact, a number of traditional tribal cultures who do not even have a notion of absolute number, but only of relatively more and less of something. We are so bound up with these cultural norms that it literally not only makes it hard for us to imagine anything different but to immediately dismiss it out of hand as "unnatural" or "clumsy" simply because it doesn't feel "intuitive" or "right", which ignores completely how that that intuition was built up by years of exposure, socialization and education. Indeed, part of critical cultural studies - the stuff that gets blasted like fuck on the Internet as "useless majors" and "ideologue politics" (not to say it can't be used for such since if you criticize culture you can of course use that to advance a political claim, which comes about as "what I want done about it because I see that this other way to do things might actually be 'better'", but there is a _general basic method_ that is present there that can be applied to _every_ human construct that doesn't itself on its own require you to adopt a specific policy position) - is about _doing just this kind of thing_ that I just did above. Saying to things that seem to us as "but that's just _right_ , isn't it?" and seriously asking "Is it?"
Not gonna read all 500,000 words of that reply, but you seem to gloss over the point where he did no work at all to solve the equation and instead used the function already defined so as to solve this class of equations. He could instead have named the video "introduction to Lambert W function," and that would be more accurate. Indeed, if someone told me they had a closed form representation of the solution to x² = 2, simply saying "it's the square root of two, duh" would not impress me.
The square root is an elementary function and an algebraic function, so of course it is a closed-form solution to the equation, but the entire context of "finding a closed form for the square root of two" implies there is a finite representation in more fundamental terms. There is no creativity in saying that it is what it is.
@SpaghettiToaster : Sure, considering that that is what the standard operation is - defined to solve that equation according to the established body of mathematical practice. But the point is, that there is no reason it could not have been different, and thus it doesn't make sense to say that some operations are "bad" or "invalid" for use even though there is nothing mathematically wrong with them. The trick is though: the Lambert W function is _also_ just as much part of that established body of practice - just that it is not usually given to you in high school. The fact that there are more operations out there, that actually _see use_ , than just the ones you get in high school, or the ones for which there are buttons on your calculator, also should not really be a surprise anyways. At the end of the day, _all_ our definitions are "made up" - the excitement comes about as we play around with them. There might be more of a point if the new operation literally solves one and only one equation, but it doesn't. These definitions, "made up" as they may be, are not made up at random. x^x is not the same expression as xe^x, nor is it the same as the expression in the equation for the Wien constant, nor many other problems for which this new operation has been used to write their solutions.
When you are in hurry 1¹= 1 2²=4 So the number is between these Though 1 is more close to 2 than 4 but this is exponential growth so the answer should be near the arithmetic mean of 1and 2 that is 1.5 I used calculator to verify this and got that the answer is roughly equal to 1.56 Solved in 20 seconds ...
The Lambert function is indeed a bit "remote". If you look it up, however, on Wikipedia, there are many "practical applications" for Lambert. Practical applications should mean something to everyone. BlackPenRedPen concentrates on "pure mathematics"--techniques for solving equations. The Teacher here is not concerned with the engineering/physics applications of equations.
I solved x^x=2 with only ln(...)-> x^x=2, so x^(2x)=4. And 2=x^x, so we substitute; 4=x^((x^x)x)=x^(x^(x+1)). Now take the ln of this; ln(4)=ln(x)(x^(x+1)). Divide by ln(x), and include the ln(x) in the ln(...) on the other side; ln(4-x)=x^(x+1). Now take the ln(...) again-> (x+1)ln(x)=ln(ln(4-x)) Divide by ln(x)->x+1=ln(ln(4-x)-x). Raise e to the power of both sides-> e^(x+1)=ln(4-x)-x. Put in in the division form again-> e^(x+1)=(ln(4)/ln(x))-x Now, ln(2)=ln(x^x)=xln(x). Where can we get this? Including the “-x” in the fraction-> e^(x+1)=(ln(4)-xln(x))/ln(x). Now substitute xln(x) for ln(2), and ln(4) for 2ln(2)-> e^(x+1)=(2ln(2)-ln(2))/ln(x). 2ln(2)-ln(2)=ln(2), and ln(x)=(1/x)(ln(2). Apply these-> e^(x+1)=ln(2)/((1/x)(ln(1)). Cancel the ln(2)’s out-> e^(x+1)=1/(1/x)=x. So, e^(x+1)=x. Take the ln(...) of both sides-> x+1=ln(x). We had earlier x^(x+1)=ln(4)/ln(x), so substitute. ->x^(x+1)=ln(4)/(x+1). Take the ln(...) of both sides; (x+1)ln(x)=ln(ln(4)/(x+1)). We just confirmed 1+x=ln(x), subsitute-> (1+x)^2=ln(ln(4)/(x+1)). Let’s look at (x+1)^2 for a minute.Thats x^2+2x+1=x^2+x+(x+1)= ln(x)+x(x+1)=ln(x)+xln(x)=ln(x)+ln(2) =ln(2x). This was originally ln(ln(4)/(x+1)), so raise e to both sides; 2x=ln(4)/(x+1). Multiply by (x+1) ->2x(x+1)=ln(4). Now we’re talking. 2x^2+2x=ln(4), so 2x^2+2x-ln(4)=0. Now solve the quadratic. I used the quadratic formula, and I got (-1+/-sqrt(1+ln(16))/2. Now Someone just has to find which it is. Wait, I did that using common knowledge; if it’s the negative solution, it’s a negative number to a negative power. The number is then (1/(-m))^(something)= -n. But 2 is positive. Therefore, if x^x=2, x=(-1+Sqrt(1+ln(16))/2.
You made an error by turning (ln 4 / ln x) into ln (4-x). The operation you're probably thinking of works in the inverse fashion: (ln 4 - ln x) would be equal to ln (4/x). Later on this would lead to (x+1) = ln x, and that has no solutions over the real numbers.
Here you go: Expand sqrt (a-sqrt (a-x))=x. You will get x^4-2ax^2-x+a^2-a=0 which can be simplified to (x^-x-1)(x^2+x+1-a). Also note that x^2+x+1-a has two roots, and the positive one equals (sqrt (4a-3)-1)/2.
I mean, technically you're right. That would be equally valid, but much less meaningful. Lambert W is a well known function in mathematics with many applications, not just some arbitrary function
There is a function that is literally equal to the one you described, it is called the super square root function x^x=n , ssrt(x^x)=ssrt(n) , x=ssrt(n) And ssrt(x)= (lnx) / (W(ln(x)))
Merci pour ce contenu génial! Ta méthode est vraiment interessante et comme tu l'as dit, il y a l'autre manière: X^x=2 Xlnx=ln2 Lnx(e)^lnx=ln2 Lnx=w(ln2) X=e^(w(ln2)) J'ai bien aimé l'équivalence!
Thanks. I had not previously been introduced to the Lambert Function. Interesting that it is referred to as a "function" while for arguments between -1/e and 0 it is not a function. I guess if you limit the domain to be greater than -1/e AND limit the range to be above -1, you could consider it a function.
Functions can have limited domains. In this case, the full function has branches (is multivalued in some places) and a complex domain and range. When dealing with reals, we usually use the principal branch (a "branch cut") and limit the domain. This is similar to Ln(x) vs ln(x) with complex domain. ln(x) is the full multivalued function. Ln(x) is the principal branch. There are many multivalued functions out there, especially when dealing with complex numbers, but even in the reals, there are multivalued functions, like sqrt(x). With sqrt(x), we usually take the principle root, which is what is specifically signified by the radical symbol, and is the positive one for real roots of real numbers. The more general sqrt(x) function always has two values. This is why we write ±√x when we use the square root function. The radical symbol only specifies one of the two values the square root function has.
How is an answer expressed in terms of a function we only compute with newton's method exact? Sure we also use power series to compute exponential functions but still, it's not an elementary function.
It's an exact answer because it's a uniquely specified number under the definition of the W function. It's not much different than saying an answer is exact when it includes pi even though pi is a transcendental number that can only be estimated algorithmically.
It's really arbitrary what functions you choose to include as being useful to call "exact" solutions or better "closed form" solutions. I think the questions that best determine how "acceptable" such functions are is how many equations - that are not specifically contrived to be solved with them - they can solve, even more if such equations are ones that turn up in the course of doing something else that is useful, _and_ compared to the simplicity of expressibility of the functions in different ways like power series, implicit equations, etc. . This "acceptability" criterion is, of course, not something that is easily made mathematically rigorous, but there are some things in maths that you cannot do that with.
In fact, the W function *is* an elementary function because it is the inverse of the elementary function xe^x. Elementary functions are closed on taking inverses.
Goodness this question is difficult. My guess was 3/2, since multiples of square roots cancel the square rooting. And 3 because I wanted it. This results in 1.5 times sqrt(1.5) which is approximately 4/3. 3/2*4/3 = 2, or 1.5 times 1.3 Algebratizing the equation to be y = logbaseX 2, gives a graph where you can just draw a square under it. That also gives nearly exactly 1.5.
This is really inspirating and interesting video. I did not know the LambertW( ) function, which looks very usefull. I have only one remark. It could be better to use only one notation, with f( ) and f^-1() functions, or W( ) and W^-1() functions. But not both symbolics, because it is little bit chaotic. But I agree, it is completly correct, interesting and Cool. Good Job.
What a brilliant derivation, only marred a little by the fact that after all, you still need to look the answer up on a computer. Gee, when I first looked at the problem, I was thinking the same thing, ha ha ha. Now at least I have a new name for the function (grin).
There is a calculator which has the W-function (and a lot more). It is the WP 34 s, a program you can flash to an HP 20 b or 30 b calculator (but difficult because missing cables.) Or use it on apple devices. Or buy it on commerce.hpcalc.org. Emulator for mac, windows and linux on sourceforge available.
I don't know if u guys have payed much attention, but this "proof" means absolutely nothing. It's merely walking around in circles. The end result expresses x in terms of a function that is defined to be the inverse of xe^x, which we then use in order to calculate x, (w(ln2)). But in order for us to compute this result, for which there is no exact solution, we have to resort to numerical methods of calculation, which we could've used to approximate x^x=2 to begin with!
Why do you have to calculate a numerical result here any more than you do for an expression like if you have x^2 = 2 and then take x = sqrt(2)? Usually, we say it's done once we have the "closed form", i.e. when we have reduced it to an expression involving known operations. That is what is done here - we've just added one more operation to our toolset. You just _leave it in the expression form with the Lambert W function_ . The question is, why do you find writing x = sqrt(2) "satisfactory" to solving x^2 = 2, but _not_ x = e^W(ln(2)) "satisfactory" to solving x^x = 2 (or ^2 x = 2)?
@@mike4ty4 I agree that the Lambert W function is nothing but one more operation. But the thing is, if we want to calculate W for any real x, we'll have to find W numerically, Newton-Raphson for instance. And we might as well apply Newton-Raphson to x^x=2 to begin with. Point is, I don't disagrre that what u guys said is correct, I'm just asking what the point of expressing x as W is.
Sir can we differentiate both side i.e X^x=2 .....(1) Differentiating both side wrt X X^x(lnx+1)=0 ......(2) From eq 1 putting X^x=2 in eq 2 We get 2(lnx+1)=0 And from here X=e^-1 Thank you Is this correct, please reply
Much more simple and faster way to find the answer: l = 1; h = 2; while h-l > 0.00000000000001: middle = 0.5*(h+l); if pow(middle, middle) > 2: h = middle; else: l = middle; print(0.5*(h+l));
Is it not true that X = (e ^ W(ln(X))) ^ (e ^ W(ln(X))) for any value of X? Therefore if you get a question in the form: "X ^ X = Y" you can say that X equals e ^ W(ln(Y)) I tried this for a few random numbers in Wolfram Alpha and so far it seems to hold, but I have no idea if it holds for all real numbers (or how you could even prove that).
I think my solution has a more intuitive approach: X^x=2 exp(x*ln(x))=2 x*ln(x)=ln(2) ln(x)*exp(ln(x))=ln(2) ln(x)=W(ln(2)) x=exp(W(ln(2))) Both ways get the same solution obviously tho
A form that can "easily" be written by its power expansion (taylor series). For example, if the solution was, say x = e, then we could express as a power series 1+1/2!+.... as the power series of e^x is well-known. More informally, we should express the solution with "only numbers". If we can express it via elementary functions, like log, exp, sin, cos etc etc, then we can use their well-known power series to write the answer as sums and products with "just numbers". Here, that is not the case. There is a similar "issue" with the integral of a gaussian curve, which is dubbed the error function. In a sense, the way it was solved here was basically going like this: give me some equation to solve: G(x)=0. THE solution can be written as x = f, where we define f to satisfy: G(f)=0. This solves it and is exact (by definiton/construction). As you can imagine/see, we did not really get any further. The only difference here is that the function used is "actually" well-known and useful in other domains (e.g. quantum mechanics) and thus has been well studied.
Melaetas where do you draw the line between well known "elementary" functions like log(x),sin(x),exp(x),etc. and constructed ones like W(x) then?? those other functions were constructed too after all.
Sorry;could you show us the result of x;teaching how to iterate "by hand"until reach that x not varies nearly more? Thanks;im hector Troncoso from Argentina;im 65 years old.
All functions with an exponent can be written in terms of e. Actually, using Fourier transform ( if it exists for f(x) ), everything can be written as an infinite sum of complex coefficients of e.
I don't like this because: Equation is: x^x=2; means: "a number that powered to itself equals 2". Solution given is: ln2/W(ln2); means: "the number that 'e' powered to it becomes 2, divided by the number that 'e' powered to it and then multiplied by it becomes the former (which, by the way, finding it requires solving a similar equation)." So, the solution given is NOT a numerical value, is an expression more complex than the original equation and it depends on knowing a solution for a similar problem. Basically is like saying: "To solve this equation first you need to solve this equation." Which is neither a solution nor helpful. *dramatic fx*
But what is W(x)? We use this unknown function and never even say what it is, never mind how to find it. Maybe this is a well known function to some, but I don’t know it and it seems completely ignored as to what it is algebraically
The more important question is "what do you mean by 'what is'?" It seems this is a common math (and elsewhere!) learning hangup - that there is "one right" expression of something to which all others are inferior, which leads to a lot of misconceptions and discomfort, for example that "pi is infinite" (no it's not, it's just a little more than 3, what's infinite is a specific _representation_ thereof, and not necessarily even the best one). And moreover, learning materials tend to perpetuate this by saying "xyz cannot be expressed in terms of (insert preferred representation X here)" and hammer that in, which make it sound like that that type of representation is invariably the best or "right and true" one, which is not the case. W(x) can be represented in many ways: As the implicit solution of We^W = x, where that x is to range in [-1/e, oo), As the solution of the initial value problem x(1 + W) dW/dx = W, W(e) = 1, As the infinite series W(x) = x - x^2 + (3/2)x^3 - (8/3)x^4 + ... where the general term is (-n)^(n-1)/n! x^n, starting from n = 1, convergent about 0 in a radius of 1/e but then expanded to the full domain by analytic extension, As the result of the given Newton's method converging from a suitable initial guess, As Mező's integral: W(x) = (1/pi) Re[int_{0...pi} ln((e^(e^(it)) - xe^(-it))/(e^(e^(it)) - xe^(it))) dt] and more - infinitely many more, actually. Which is better? Depends on what you want to do. But all of them define the same thing and all of them can be just as validly claimed to _be_ "the function" since they each uniquely specify one function of x and no other, and that is W(x).
mike4ty4 I think a number of people (myself included) find it difficult to visualise/grasp the nature of the function when expressed as the "inverse of xe^x". Anyone of your examples would have helped in the video - perhaps BPRP has a video discussing the function in more detail?
blackpenredpen i was talking about your arc length videos and he asked me where this knowledge came from, and I told him about you and he already knew who you were. My teacher watches your vids
having some trouble with a similar related problem: x^4 = 2^x. I get to the solns as [x: x = +/- e^(ln(2)/4x)] but I cannot get the answers wolframalpha generates following your method with W functions, maybe I am too tired rn...
Can you solve fish^fish=1/2?
No
guy man believe... and u will find a way
would fish just be e^(w(ln(1/2))) ?
daniel actually not (in the real world). Think about the graph of fish^fish
What about x^x = x + 1? The Golden Exponent?
Computer? I used the Log-Log scales on my British Thornton AA010 Comprehensive slide rule.
I got 1.56.
SlideRulePirate : )
@sliderulepirate: but you had to do iterations, too. Am I right?
@Blue Blue. If, by "iterations", you mean resorting to 'tweak and peek' then yes.
Totally can do Newton method on a slide rule!
Since x^x=2 is equiv to x ln(x) = ln2， so iterate x ← x - (x ln(x) - ln(2))/(1+ln(x)) , that is x←(x+ ln(2))/(1+ln(x))
That is just another kind of computer.
An easier solution in my opinion would be the following:
use ln to get x*ln(x)=ln(2)
rewrite x as e^ln(x) then apply w(x) to get ln(x)=w(ln(2))
and exponentiate.
I think that would be easier. But I can be wrong. Nevertheless, great video and greetings from germany.
Ya that's what I did. We had to get W involved, which motivated the forcing of a ye^y into the expression. It's nice how so many solutions to exponential equations can be neatly expressed in terms of W.
I like your way better. It seems easier to see how each step progresses towards a situation of the form a*e^a.
Though I hope I'll just be able to memorize that
x^x = a implies x = e^W(ln a).
@@ZipplyZane if a>(1/e)^(1/e) because it is the absolute minimum of the function(and the only minimum).
We need more LambertW videos. That function is just cool.
You know, this video gives me the satisfaction that out there there are people who care about finding the solution to questions like these, for I too often get mocked for asking such things in class.
Debaditya Bhattacharya why are you mocked for asking these?
@@adrianpramadipta8631 Probably cause it doesn't relate to the class and it's distracting them from the actual subject relevant to them: the subject for the test.
Use a calculator to keep getting close
i always do that for anything i want to calculate
That’s a limit on steroids
There is an amazing method to calculate the w function by newton's method in the calculator!
To calculate W(x) , in the calculator enter whatever number and press =
So this number will be save into "Ans"
And then enter this :
Ans^2/(Ans+1)+x/[e^Ans(Ans+1)]
And then press = too many times , you will get a very very nice approximation of W(x)!
isn't that the fixed point iteration method?
No this is Newton’s method ,but I think this method also works.
isn't Newton's method about finding the tangent lines and where those intersect with the line y=0 ? I don't understand how that is what you're doing when typing that into a calculator. The fixed point iteration is about finding the value that won't be changed when plugged into a function, f(x)=x. When you type that into the calculator, you're solving for when your expression becomes equal to ANS, in other words, you're looking for a fixed point. Am I just completely wrong here, or...?
BenDesch to be honest I don’t know much about the fixed point iteration, or even about Newton’s method , but I know that Newton’s method deals with the derivative and the tangent line in a point , so when you set a function equals to zero , the Newton’s method says that you pick whatever x value “ x0”
The new x value will be
X1=x0-f(x0)/f’(x0)
X2=x1-f(x1)/f’(x1)
And so on
When you apply this many times , you will get closer to the x value that make the function equals to zero
This is the idea of newtos’s method!
And I recommend you to read it in Wikipedia.
Gergő Dénes oh yeah I watched the video , it’s so amazing, I like w function!
Very neat, but really the solution is "exact" in the sense that a function is defined to be exactly the solution of this sort of problem.
"Let _a_ be the number such that this function works. The end. Remember to like, comment, and subscribe!"
There are already lots of those types of functions floating around even remedial mathematics, it's just that they're taken for granted. sqrt(x) is a function that was defined to solve problems like x^2 - 2 = 0. One would obviously consider sqrt(2) to be a solution in exact form, even though it is still being expressed in terms of a function (the sqrt function) at a specific value: f(2). W(x) is no different from sqrt(x), ln(x), or any trig function, for that matter; they were created to denote solutions to certain types of problems in exact form.
@@PaddedShaman Very well said
Thats what i was thinking, i feel a bit cheated really. Seems to just write the exact value without using functions you would require an infinite series. I was thinking from the title there would be something weird like sqrt(2/sqrt(e)) or some finite expression like that.
@@PaddedShaman never thought of it like that, I suppose the unfamiliarity of W made me think of it as something less natural than a root or power or log.
2:36 I love how he uses 2 colors at the same time
Pro tip: the quickest way to solve this is by entering different values of x on a calculator and seeing which one gives x^x = 2.
No 🤣
Yes?
Yes?
@@guepardiez no lmao, you will never get the exact value
@@lorenzosaudito That's true. :) Sorry for the repetition, I missed my previous "yes".
Down-voted for clickbait - those two fish aren't alike! That base-fish looks like a cheesy x-prime to me! /s
Ok!
“Downvoted
/s”
I see you are a fellow reddit brother as well
I learn simultaneously English and maths. Couldn't be more happy
Wait a minute. That pen is green. *GASP!*
What is the world coming too?? ;P
Blackpenredpen, your videos are truly brilliant! I hope you can share your enthusiasm and skill in making things clear also as a $-earning job, whether in a school or a University or elsewhere. It’s great how you make the choice of solution strategy clear. The trick of course is: How does one figure out the solution STRATEGY? - What your videos seem to suggest is that math is 25% knowledge (of theorems, properties, etc.), 25% disciplined logic, and 50% creative imagination! - the 50% they never teach you in school...
Just take a log of the equation xlnx = ln2. Now substitution: y = lnx => x=e^y gives equation: ye^y = ln2, where Lambert's function can be applied straight away:
y = W(ln2) => x = e^W(ln2).
The answer looks different. But multiply and divide by W(ln2): x = W(ln2) e^W(ln2) / W(ln2). Now from the definition of Lambert's function it follows that W(ln2)e^W(ln2) = ln2,
so x = ln2/W(ln2).
Wow, exact form! I have an exact form for ∫exp(x²)dx too. What you do is define the function DoubExpInt(x) to be the definite integral of exp(t²)dt from 0 to x, and then the exact solution is ∫exp(x²)dx = DoubExpInt(x) + c!!
I find it really irritating when people say they have a "closed form" or whatever of something that they simply defined that way.
Well okay, but then I suppose you will hate log(x) because it can be _defined_ to be the solution to x = e^y, hate sqrt(x) because it's _defined_ as the solution to x = y^2, etc. Defining new operation as the inverses of old operations is a perfectly legitimate way to make more operations. And W(x) is not defined to solve y^y = y, but ye^y = x - so saying it "does not solve it" is not any more correct than saying sqrt(x) cannot be used to "solve" ax^2 + bx + c = 0 because it is "defined" to solve x = y^2.
The elementary operations are limited. The trick is finding which operations you can add to extend your solving power that are the most "profitable" - i.e. yes, you can indeed "cheat" on it by defining your problem X as a "new operation", but ideally what you want is a simple operation that solves a great many not-simply-related problems. The Lambert W function can be used for such things as expressing the constant that defines the peak wavelength of radiation emitted by an ideal hot object as a function of temperature, for example - that is, Wien's displacement law, something that is quite different in flavor from solving its original defining equation.
And while I suppose one could say the square root is in some sense "simpler" or "more natural", those are notions that themselves are subject to possible challenge and critique. Let me present to you an almost equally logical, if not somewhat slightly more general, operation that could have been used in its stead - the "rectangle root": it is a function R(A, r), where A is the area of a rectangle and r its aspect ratio (the ratio of width to height), and it returns the length of the shorter side: that is, its definition is such that R(A, r) * (r R(A, r)) = A. Why consider this? Consider that computer and mobile device screens are made with certain aspect ratios, or how that camera resolutions are reported in pixels. If I have a 5 megapixel image and I want it in 16:9 aspect ratio, how many pixels on a side? The answer is R(5000000, 16/9), for the shorter side. Does that seem so silly now?? Going to earlier history, we may want to - this would be related to where the idea of square roots started - consider a rectangular piece of farmland, and I want to know how much fencing I will need to enclose it to allow me to grow a given amount of crop or grain. So I would like a function that I can insert the amount of grain (proportionate to the area) and the shape of the rectangle (i.e. r). (Heck, we could even have a different function giving the full perimeter! That seems reasonable, doesn't it?) Of course you see this is "just" sqrt(A/r), but we could just as easily have taken the square root to be defined in terms of it instead, as sqrt(A) = R(A, 1). If you were raised from your school years that R(A, r) was the fundamental function, and, say, the quadratic formula for ax^2 + bx + c = 0 was given to you as x = +/-R(b^2 - 4ac, 4a^2) - b/2a, and your calculator had a little R(A, r) button on it instead of a "sqrt" button, and to find the area of a square you always took R(A, 1), you would have never questioned it. Heck, when presented with sqrt(), this alternatively-socialized parallel-universe "you" might even be retorting that "that's not general enough, it's missing a parameter" or "isn't that just a redundant name for x^(1/2)?" The point is that the things you think are "more valid" when they are not deductions but selections, is more a product of culture, than it is somehow necessarily inherently "better" so much to the point that others should be excluded from consideration at all or denigrated to a "cheat".
Indeed, if you drill it at hard enough you can even challenge the whole mathematical edifice itself: why do we even use "numbers"? We could have taken something else as just as basic. In ancient Greece, geometry had a much more fundamental role than it does in modern mathematics. There are, in fact, a number of traditional tribal cultures who do not even have a notion of absolute number, but only of relatively more and less of something.
We are so bound up with these cultural norms that it literally not only makes it hard for us to imagine anything different but to immediately dismiss it out of hand as "unnatural" or "clumsy" simply because it doesn't feel "intuitive" or "right", which ignores completely how that that intuition was built up by years of exposure, socialization and education. Indeed, part of critical cultural studies - the stuff that gets blasted like fuck on the Internet as "useless majors" and "ideologue politics" (not to say it can't be used for such since if you criticize culture you can of course use that to advance a political claim, which comes about as "what I want done about it because I see that this other way to do things might actually be 'better'", but there is a _general basic method_ that is present there that can be applied to _every_ human construct that doesn't itself on its own require you to adopt a specific policy position) - is about _doing just this kind of thing_ that I just did above. Saying to things that seem to us as "but that's just _right_ , isn't it?" and seriously asking "Is it?"
Not gonna read all 500,000 words of that reply, but you seem to gloss over the point where he did no work at all to solve the equation and instead used the function already defined so as to solve this class of equations. He could instead have named the video "introduction to Lambert W function," and that would be more accurate. Indeed, if someone told me they had a closed form representation of the solution to x² = 2, simply saying "it's the square root of two, duh" would not impress me.
If you expect anything other than sqrt(2) as a "closed form" solution to x²=2, then you haven't been paying attention in high school.
The square root is an elementary function and an algebraic function, so of course it is a closed-form solution to the equation, but the entire context of "finding a closed form for the square root of two" implies there is a finite representation in more fundamental terms. There is no creativity in saying that it is what it is.
@SpaghettiToaster : Sure, considering that that is what the standard operation is - defined to solve that equation according to the established body of mathematical practice. But the point is, that there is no reason it could not have been different, and thus it doesn't make sense to say that some operations are "bad" or "invalid" for use even though there is nothing mathematically wrong with them. The trick is though: the Lambert W function is _also_ just as much part of that established body of practice - just that it is not usually given to you in high school. The fact that there are more operations out there, that actually _see use_ , than just the ones you get in high school, or the ones for which there are buttons on your calculator, also should not really be a surprise anyways.
At the end of the day, _all_ our definitions are "made up" - the excitement comes about as we play around with them.
There might be more of a point if the new operation literally solves one and only one equation, but it doesn't. These definitions, "made up" as they may be, are not made up at random. x^x is not the same expression as xe^x, nor is it the same as the expression in the equation for the Wien constant, nor many other problems for which this new operation has been used to write their solutions.
When you are in hurry
1¹= 1
2²=4
So the number is between these
Though 1 is more close to 2 than 4 but this is exponential growth so the answer should be near the arithmetic mean of 1and 2 that is 1.5
I used calculator to verify this and got that the answer is roughly equal to 1.56
Solved in 20 seconds ...
Well.
This means nothing to me.
Ok
High iq individual struggles
Lol
You can't make heaven
The Lambert function is indeed a bit "remote". If you look it up, however, on Wikipedia, there are many "practical applications" for Lambert. Practical applications should mean something to everyone. BlackPenRedPen concentrates on "pure mathematics"--techniques for solving equations. The Teacher here is not concerned with the engineering/physics applications of equations.
I solved x^x=2 with only ln(...)->
x^x=2, so x^(2x)=4. And 2=x^x, so we substitute; 4=x^((x^x)x)=x^(x^(x+1)).
Now take the ln of this;
ln(4)=ln(x)(x^(x+1)). Divide by ln(x), and include the ln(x) in the ln(...) on the other side; ln(4-x)=x^(x+1). Now take the ln(...) again-> (x+1)ln(x)=ln(ln(4-x))
Divide by ln(x)->x+1=ln(ln(4-x)-x). Raise e to the power of both sides->
e^(x+1)=ln(4-x)-x. Put in in the division form again-> e^(x+1)=(ln(4)/ln(x))-x
Now, ln(2)=ln(x^x)=xln(x). Where can we get this? Including the “-x” in the fraction-> e^(x+1)=(ln(4)-xln(x))/ln(x). Now substitute xln(x) for ln(2), and ln(4) for 2ln(2)->
e^(x+1)=(2ln(2)-ln(2))/ln(x).
2ln(2)-ln(2)=ln(2), and ln(x)=(1/x)(ln(2). Apply these->
e^(x+1)=ln(2)/((1/x)(ln(1)). Cancel the ln(2)’s out-> e^(x+1)=1/(1/x)=x. So, e^(x+1)=x. Take the ln(...) of both sides-> x+1=ln(x). We had earlier x^(x+1)=ln(4)/ln(x), so substitute.
->x^(x+1)=ln(4)/(x+1). Take the ln(...) of both sides;
(x+1)ln(x)=ln(ln(4)/(x+1)). We just confirmed 1+x=ln(x), subsitute->
(1+x)^2=ln(ln(4)/(x+1)). Let’s look at (x+1)^2 for a minute.Thats x^2+2x+1=x^2+x+(x+1)=
ln(x)+x(x+1)=ln(x)+xln(x)=ln(x)+ln(2)
=ln(2x). This was originally
ln(ln(4)/(x+1)), so raise e to both sides; 2x=ln(4)/(x+1). Multiply by (x+1)
->2x(x+1)=ln(4). Now we’re talking. 2x^2+2x=ln(4), so 2x^2+2x-ln(4)=0. Now solve the quadratic. I used the quadratic formula, and I got
(-1+/-sqrt(1+ln(16))/2. Now Someone just has to find which it is. Wait, I did that using common knowledge; if it’s the negative solution, it’s a negative number to a negative power. The number is then (1/(-m))^(something)=
-n. But 2 is positive. Therefore, if x^x=2, x=(-1+Sqrt(1+ln(16))/2.
You made an error by turning (ln 4 / ln x) into ln (4-x). The operation you're probably thinking of works in the inverse fashion: (ln 4 - ln x) would be equal to ln (4/x). Later on this would lead to (x+1) = ln x, and that has no solutions over the real numbers.
Nice calming voice. Honestly a great accent and English.
I just love the way the word Lambert rolls off my tongue and sounds :P
Can you Please proof that formula of infinite nested root with alternate + - sign. In that x²-5=√(5-x) video.
Deepa Pathak someone did the proof already. Check out description in that video.
Can't find.please share the link.
Link: th-cam.com/video/fvhtjK_u5vs/w-d-xo.html
Here you go: Expand sqrt (a-sqrt (a-x))=x. You will get x^4-2ax^2-x+a^2-a=0 which can be simplified to (x^-x-1)(x^2+x+1-a). Also note that x^2+x+1-a has two roots, and the positive one equals (sqrt (4a-3)-1)/2.
@@friendtodino3675 thank u
This is no better than saying G(x) is defined as the inverse of x^x, and the answer is just G(2).
except that W(x) is a proper thing like sin(x), not some randomly chosen function. It is the Lambert W function
+Ryan Tay still equally valid
sqrt(x)
ln(x)
sin(x)
cos(x)
etc...
I mean, technically you're right. That would be equally valid, but much less meaningful. Lambert W is a well known function in mathematics with many applications, not just some arbitrary function
There is a function that is literally equal to the one you described, it is called the super square root function
x^x=n , ssrt(x^x)=ssrt(n) , x=ssrt(n)
And ssrt(x)= (lnx) / (W(ln(x)))
You are my favorite math teacher of all time
If you substitute y :=ln(x), then x ln(x) = ye^y. So 2= x^x = e^(x ln(x) ) turns into ln(2) = ye^y, and so y = W(ln(2)), i.e. x = e^(W(ln(2)).
Merci pour ce contenu génial!
Ta méthode est vraiment interessante et comme tu l'as dit, il y a l'autre manière:
X^x=2
Xlnx=ln2
Lnx(e)^lnx=ln2
Lnx=w(ln2)
X=e^(w(ln2))
J'ai bien aimé l'équivalence!
At 4:42 what was wrong with the first "e" ?
The new "e" looks exactly the same...
Tamir Erez it looked lonely ...
He shifted it over to make room for the W, that's all.
I did it in a Planck Time!!!
x^x=2
²x=2
x=ssrt(2)
where ssrt(x) is super square root of x and ²(ssrt(x)) = ssrt(x)^ssrt(x) = x
Easy!!!
İt is not real...
#YAY #BLACKPEN #REDPEN #HELP #STUDIES #EDUCATION THANK YOU !!
Also x^x=1/2, x~0.263+0.5i
thank you for uploading
Liked for the "Yay" at 8:23 (since I'm not smart enough to understand anything else).
You are a nature-born teacher. You make things look easy.
Thanks. I had not previously been introduced to the Lambert Function. Interesting that it is referred to as a "function" while for arguments between -1/e and 0 it is not a function. I guess if you limit the domain to be greater than -1/e AND limit the range to be above -1, you could consider it a function.
Functions can have limited domains. In this case, the full function has branches (is multivalued in some places) and a complex domain and range. When dealing with reals, we usually use the principal branch (a "branch cut") and limit the domain. This is similar to Ln(x) vs ln(x) with complex domain. ln(x) is the full multivalued function. Ln(x) is the principal branch. There are many multivalued functions out there, especially when dealing with complex numbers, but even in the reals, there are multivalued functions, like sqrt(x). With sqrt(x), we usually take the principle root, which is what is specifically signified by the radical symbol, and is the positive one for real roots of real numbers. The more general sqrt(x) function always has two values. This is why we write ±√x when we use the square root function. The radical symbol only specifies one of the two values the square root function has.
How is an answer expressed in terms of a function we only compute with newton's method exact? Sure we also use power series to compute exponential functions but still, it's not an elementary function.
It's an exact answer because it's a uniquely specified number under the definition of the W function. It's not much different than saying an answer is exact when it includes pi even though pi is a transcendental number that can only be estimated algorithmically.
It can be computed with other methods as well.
It's really arbitrary what functions you choose to include as being useful to call "exact" solutions or better "closed form" solutions. I think the questions that best determine how "acceptable" such functions are is how many equations - that are not specifically contrived to be solved with them - they can solve, even more if such equations are ones that turn up in the course of doing something else that is useful, _and_ compared to the simplicity of expressibility of the functions in different ways like power series, implicit equations, etc. . This "acceptability" criterion is, of course, not something that is easily made mathematically rigorous, but there are some things in maths that you cannot do that with.
In fact, the W function *is* an elementary function because it is the inverse of the elementary function xe^x. Elementary functions are closed on taking inverses.
Your pronunciation for w is the same as we from Brazil speak
Wonderful. Stunning math trick (5:05)
If you ever get x^x = y when y is a constant, then x = e^(W(ln(y)))
actually tried it with the e^lnx approach no wonder i got stuck. never even heard of that function before
Goodness this question is difficult. My guess was 3/2, since multiples of square roots cancel the square rooting. And 3 because I wanted it.
This results in 1.5 times sqrt(1.5) which is approximately 4/3.
3/2*4/3 = 2, or 1.5 times 1.3
Algebratizing the equation to be y = logbaseX 2, gives a graph where you can just draw a square under it. That also gives nearly exactly 1.5.
This is really inspirating and interesting video. I did not know the LambertW( ) function, which looks very usefull.
I have only one remark. It could be better to use only one notation, with f( ) and f^-1() functions, or W( ) and W^-1() functions. But not both symbolics, because it is little bit chaotic. But I agree, it is completly correct, interesting and Cool. Good Job.
Finally, you give us a solution that make us be confuse level 255, congratulation.
The answer literally in the question if we use wolframalpha
Bit of a stretch to say this is exact. As for the decimal approximation, you could just use Newton - Raphson.
That was new to me. Thanx!👍☺
Helpful video! Thank you. :D
Plase resolve lim(x^x) when x>0
What a brilliant derivation, only marred a little by the fact that after all, you still need to look the answer up on a computer. Gee, when I first looked at the problem, I was thinking the same thing, ha ha ha. Now at least I have a new name for the function (grin).
I love your thumbnails
In Việt Nam, In high school, I didn't know and study the Lambert W Function
I don't understand anything but i like maths and how you explain them xD
There is a calculator which has the W-function (and a lot more). It is the WP 34 s, a program you can flash to an HP 20 b or 30 b calculator (but difficult because missing cables.) Or use it on apple devices. Or buy it on commerce.hpcalc.org.
Emulator for mac, windows and linux on sourceforge available.
WOW!!! I DID NOT KNOW ABOUT THIS!
Fun fact: You can approximate the solution as x = e^(4/9) and get a result of x^x = 2.00003763... That differs from 2 by less than 4 parts in 100,000.
"We don't want x on both sides of the equation", then takes the xth root on both sides 🙄🙄. Amazing video by the way!
Thank you
Excellent class
Brilliant is a brilliant app
But what is x, if x^^x=10? And how do you calculate that?
Thanks. Good explanation.
Can you do x^(x-1)= 2
And x^a = x + a
This video teaches u how to solve x^x=2 without a calculator, but u do need a calculator 💀
Is this greenpenredpen?
Wait, so what is the inverse of xe^x?
I don't know if u guys have payed much attention, but this "proof" means absolutely nothing. It's merely walking around in circles. The end result expresses x in terms of a function that is defined to be the inverse of xe^x, which we then use in order to calculate x, (w(ln2)). But in order for us to compute this result, for which there is no exact solution, we have to resort to numerical methods of calculation, which we could've used to approximate x^x=2 to begin with!
Well, check this out.
Solve x^2=3
ans: x = +- sqrt(3) exact form
approximation: type sqrt(3) to calculator/computer.
same as e^x=2,
x=ln(2)
Why do you have to calculate a numerical result here any more than you do for an expression like if you have x^2 = 2 and then take x = sqrt(2)? Usually, we say it's done once we have the "closed form", i.e. when we have reduced it to an expression involving known operations. That is what is done here - we've just added one more operation to our toolset. You just _leave it in the expression form with the Lambert W function_ . The question is, why do you find writing
x = sqrt(2)
"satisfactory" to solving x^2 = 2, but _not_
x = e^W(ln(2))
"satisfactory" to solving x^x = 2 (or ^2 x = 2)?
@@mike4ty4 I agree that the Lambert W function is nothing but one more operation. But the thing is, if we want to calculate W for any real x, we'll have to find W numerically, Newton-Raphson for instance.
And we might as well apply Newton-Raphson to x^x=2 to begin with.
Point is, I don't disagrre that what u guys said is correct, I'm just asking what the point of expressing x as W is.
This is so cool!
Sir can we differentiate both side i.e
X^x=2 .....(1)
Differentiating both side wrt X
X^x(lnx+1)=0 ......(2)
From eq 1 putting
X^x=2 in eq 2
We get
2(lnx+1)=0
And from here
X=e^-1
Thank you
Is this correct, please reply
Differentiating will not give x value
Differentaition is used for finding slopes
Much more simple and faster way to find the answer:
l = 1; h = 2;
while h-l > 0.00000000000001:
middle = 0.5*(h+l);
if pow(middle, middle) > 2:
h = middle;
else:
l = middle;
print(0.5*(h+l));
so the answer is like e ^ W(ln2) is e ^ (ln2 ^ (-1) ) ?
Is it not true that X = (e ^ W(ln(X))) ^ (e ^ W(ln(X))) for any value of X?
Therefore if you get a question in the form: "X ^ X = Y" you can say that X equals e ^ W(ln(Y))
I tried this for a few random numbers in Wolfram Alpha and so far it seems to hold, but I have no idea if it holds for all real numbers (or how you could even prove that).
So there was another john napier who spent the rest of his life working out values of the lambert function with mclaurin series without a computer?!
Lol the video starts with the Doraemon theme song. Great vid!
Can u solve fish^(fish^y)=z fish=?
I think my solution has a more intuitive approach:
X^x=2
exp(x*ln(x))=2
x*ln(x)=ln(2)
ln(x)*exp(ln(x))=ln(2)
ln(x)=W(ln(2))
x=exp(W(ln(2)))
Both ways get the same solution obviously tho
For some reason the video length is 9:25 but in the thumbnail it says 10:07
Hey fellow pen fans what do you used to study I don't know where to look for good problem to try and solve so where do you guys go?
When you think about it, it really is simple.
You can also use sagemath.
I miss the piano music in the background tbh
How would you solve x[4]x? (x tetrated to x)
I swear W(x) shouldnt be a function or written like that cos I'm pretty sure it's not a function for - 1/e
It's only used as a function for x > 0.
For -1/e
An exact form, but not an explicit form
what's explicit form?
A form that can "easily" be written by its power expansion (taylor series). For example, if the solution was, say x = e, then we could express as a power series 1+1/2!+.... as the power series of e^x is well-known.
More informally, we should express the solution with "only numbers". If we can express it via elementary functions, like log, exp, sin, cos etc etc, then we can use their well-known power series to write the answer as sums and products with "just numbers". Here, that is not the case. There is a similar "issue" with the integral of a gaussian curve, which is dubbed the error function.
In a sense, the way it was solved here was basically going like this: give me some equation to solve: G(x)=0. THE solution can be written as x = f, where we define f to satisfy: G(f)=0. This solves it and is exact (by definiton/construction). As you can imagine/see, we did not really get any further.
The only difference here is that the function used is "actually" well-known and useful in other domains (e.g. quantum mechanics) and thus has been well studied.
ah thanks i get it
But there's no better way.
Melaetas where do you draw the line between well known "elementary" functions like log(x),sin(x),exp(x),etc. and constructed ones like W(x) then?? those other functions were constructed too after all.
Put "Lambert W Function" next to your title, it would help someone.
I guess we need wolfram alpha then
No u could just use newton-raphson method
Sorry;could you show us the result of x;teaching how to iterate "by hand"until reach that x not varies nearly more?
Thanks;im hector Troncoso from Argentina;im 65 years old.
what is the relationship between x^x and x*exp(x) ?
All functions with an exponent can be written in terms of e. Actually, using Fourier transform ( if it exists for f(x) ), everything can be written as an infinite sum of complex coefficients of e.
can you do a video where you show or prove how to find W (x) like solving for y in x=y*e^y
I don't like this because:
Equation is: x^x=2; means: "a number that powered to itself equals 2".
Solution given is: ln2/W(ln2); means: "the number that 'e' powered to it becomes 2, divided by the number that 'e' powered to it and then multiplied by it becomes the former (which, by the way, finding it requires solving a similar equation)."
So, the solution given is NOT a numerical value, is an expression more complex than the original equation and it depends on knowing a solution for a similar problem.
Basically is like saying: "To solve this equation first you need to solve this equation." Which is neither a solution nor helpful.
*dramatic fx*
Eric Ester
Ok. You don't have to like it.
Was it not Feynman who said "shut up and calculate"?😏
Pls do a video on how to diff x!
Use Gamma function
Dear Steve ,please make a vedio on ramanujan,s identity:1+2+3+4.......=-(1\12).is it possible .please.,..,...,....,..
Sardar Alam I think it only works in relation to a solution of the zeta function.
What about Z^Z = -2? Or X^X^X = 2?
Flamingpaper this one needs insight x^x^x(x^x^...^x)=2
@@skalderman square root of 2
But what is W(x)? We use this unknown function and never even say what it is, never mind how to find it. Maybe this is a well known function to some, but I don’t know it and it seems completely ignored as to what it is algebraically
The more important question is "what do you mean by 'what is'?" It seems this is a common math (and elsewhere!) learning hangup - that there is "one right" expression of something to which all others are inferior, which leads to a lot of misconceptions and discomfort, for example that "pi is infinite" (no it's not, it's just a little more than 3, what's infinite is a specific _representation_ thereof, and not necessarily even the best one). And moreover, learning materials tend to perpetuate this by saying "xyz cannot be expressed in terms of (insert preferred representation X here)" and hammer that in, which make it sound like that that type of representation is invariably the best or "right and true" one, which is not the case.
W(x) can be represented in many ways:
As the implicit solution of We^W = x, where that x is to range in [-1/e, oo),
As the solution of the initial value problem x(1 + W) dW/dx = W, W(e) = 1,
As the infinite series W(x) = x - x^2 + (3/2)x^3 - (8/3)x^4 + ... where the general term is (-n)^(n-1)/n! x^n, starting from n = 1, convergent about 0 in a radius of 1/e but then expanded to the full domain by analytic extension,
As the result of the given Newton's method converging from a suitable initial guess,
As Mező's integral: W(x) = (1/pi) Re[int_{0...pi} ln((e^(e^(it)) - xe^(-it))/(e^(e^(it)) - xe^(it))) dt]
and more - infinitely many more, actually.
Which is better? Depends on what you want to do. But all of them define the same thing and all of them can be just as validly claimed to _be_ "the function" since they each uniquely specify one function of x and no other, and that is W(x).
mike4ty4 I think a number of people (myself included) find it difficult to visualise/grasp the nature of the function when expressed as the "inverse of xe^x". Anyone of your examples would have helped in the video - perhaps BPRP has a video discussing the function in more detail?
hello bro. where can i learn about wolfman. i have it installed but done know others. looking forward to hear from you
Ok i Really like the way you pronounce brilliant.org. Its like brindawog
I actually feel bad for thinking I would understand this.
Using my iPhone and guess and check I got 1.55961046 in about 5 minutes
太爱你了，老师
cruelty_A cruelty_A 謝謝！ 我其實今天剛好還錄了中文版呢。 這在上傳中
DUDE MY TEACHER ALREADY KNEW WHO YOU WERE!
Yaboylemon ?
blackpenredpen i was talking about your arc length videos and he asked me where this knowledge came from, and I told him about you and he already knew who you were. My teacher watches your vids
why am i watching this as a high school kid.....
having some trouble with a similar related problem: x^4 = 2^x. I get to the solns as [x: x = +/- e^(ln(2)/4x)] but I cannot get the answers wolframalpha generates following your method with W functions, maybe I am too tired rn...
That means we need computer to solve this
Hey, Blackpenredpen. Is the sale of the tshirts world wide ? Im from Brazil and really want to buy them! Keep up the good work!
Yes. However I haven't been able to put that gussian integral t shirt up yet since I need to fix something.