I believe people don’t like it because it is clear that xe^x can not have well defined inverse because it is not biyective. Now sin(x) is not either, it’s just to properly define the argument and be sure one stays there.
As I see you are a very good mathematician. I was working with a lot of equations back then! One of my favourite exercises like 10-15 years ago was the following. We have a and b where a,b e N! a^b+b^a = 423393 and a^a + b^b = 16780341. What is the value of a and b? Resolve it without just trying out numbers and hope we have luck!
I watched about 10 of your videos asking myself, "Why is this weirdo doing math while holding a PokeBall??" Then I finally saw one and realized it's your microphone with a cover on it.
I don't understand much of any of this, but I really like your enthusiasm and way of teaching, the 10 minutes flew by before I even realised. Very entertaining channel
The Lambert W function is not an analytic function. Thus, one cannot present its formula using basic operations. (The sum and multiplication of analytic functions such as polynomials (The constant function is a special case of a polynomial.), exponential functions and trigonometric functions. (I may have omitted something.)
X=1 and x=2 are easy solutions you can guess and then you can show that g(x)=2^x - 2x > 0 for x>2 So x=2 is the biggest solution Then you can show that g(x)
x=log(i)(2) =ln(2)/ln(i) =ln(2)/(πi/2 + 2πni), n is an integer So the principle value is ln(2)/(πi/2), which is the same as ln(4)/πi That's what I think.
Finally, the branches. The only thing we missed is how you don't need wolfram|alpha to figure out that -W_(0)(-ln2/2)/ln2 = 1. You coulda just gone with -ln2/2 = -ln2 · 2^(-1) = -ln2e^(-ln2)
I am a class 11th students and I just got introduced to calculus few days ago and it's super interesting!!! Am more fascinated by the way this teacher switches to different marker in seconds!!!!!😳👍🏻
Where does the -1 in the "parameter" to the W function come from? What do the other values (not 0, 1) of that parameter represent when they are used? (Are they the complex roots of the original equation?)
Since the Lambert w function is an inverse function and it's not bijective you have to choose the branch. It happens to be that -1,1,0 are the easiest branches to work with. The intervals which the branches are are not consistent and the solutions are countably infinite. I suggest you look at the graph of the function and maybe that will clear up why.
It's not the fact that this has two answers that surprises me. It's that the answer can produce integers, but have no analytic way to reduce it. Is there really no way to take your answer in the box and show those answers are 1 and 2 without approximating the W() function?
This is one beautiful problem that links the obvious 2 solutions of 2x=2^x and the 2 forms of the W function. I wonder if there is a possible generalization here beyond 2.
We can generalize to a ** (x - y) = x ** z. , where y >= 1, a and z > 0. The equation to study is f(x) = ln(x) / (x - y). If a > 1, there always two distinct solutions. If a 1, there is only one solution.
Sir I've been watching your videos and it really helped me develop interest in mathematics...earlier I scored 17/50 marks in previous maths test and now it's been 3 months the last test I got 48/50 and I'm the topper of my class. Thank you Sir......
When I do productlog equations I don't convert the number to base e first. I do it in the original base and convert to base e or whatever afterwards using this change of base formula. W[base b](x)=W(x ln(b))/ln(b) I think its much simpler
I have a question about complex numbers : If I have, m = a + bi & n = c + di , where a, b, c, d are real numbers and (i^2) = -1, then is, n < m or, n > m?
An interesting generalization: a^x=a*x, 1=ax a^(-x)=ax e^(-x ln(a)),-ln(a)/a = -ln(a) x e^(-x ln(a)), so you have W(-ln(a)/a) in general. This means you have no real solution if -ln(a)/a0), one real solution if ln(a)/a=-1/e, two real solutions if -1/e
Poser f(x)=2^x-2x Dériver f f’=0 a une seule solution, et les limites de f en + ou - l’infini sont positives, donc f est décroissante, puis croissante. f=0 a donc 2 solutions au maximum. Ici les solutions sont les valeurs évidentes x=1 ou x=2
the equation 2ˣ= 2x can be solved in a simpler, graphical way: we plot y =2ˣ and y= 2x, after which we look at the intersections of the data with the graph and these points will be solutions to this equation. therefore, x=1; x=2
This is what I missed by not majoring in math in college? Chuck in a W. Chuck in a e. Chuck in a Log or a ln. 1 can be anything, 2 has no meaning. Then out of left field, tan, then sin of theta, the sec. Obvious.
I used to watch your videos in high school and couldnt understand a damn thing, now im in college studying cc and everything is clear now, mostly your calculus videos
Question suggestion: x^2 - y^3 = 1, x and y are all integers, what are x and y? Note that there is only one answer for x and y, and you probably already found out x = 3 and y = 2
X = π In RieMann Geometry, π can be solved ☺️ 🤣 into Whole Numbers... If you believe 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... = 2. You can use this Concept to solve Pi 😜☺️ = 2 in RieMann Geometry Mathematics...
Hello blackpenredpen, how are you? Im sorry but I would like to program the lambert w function, can you help me? Is there a site to visit that could help me. Thanks so mucho for the content by the way, you are so smart! Salute you!
I can’t seem to find more info in that regard. So far I just know n=0 gives the principal branch (like the first answer) and n=-1 gives the other one (if any) on WolframAlpha. Btw, any other n will give complex solutions which I have mentioned in my other videos like 2^x=x^2
We should have different bases for the W function like how we can have different bases for logs. The one issue I could think of is notation because W has multiple real branches (ln has multiple branches but it only has on real branch).
The function f(x) = ln(x) / (x-1) , x > 0 , with f(1) = 1 is strictly decreasing and range all positive numbers. Solutions for a**x = a*x , a > 0 . x = 1 is always a solution, if a > 1 , f(x) = ln(a) is the only one else.
@@stevendeans4211 He specified f(1) = 1 at the discontinuity. He isn’t putting x = 1 in the function. For a > 1, f(a) = ln(a)/(a - 1) , f(2) = ln(2) but I don’t see the point. You can graph this function in Desmos in three parts: 0 < x < 1, x = 1, and x > 1. The discontinuity at x = 1 is removed by specifying f(1) = 1. (It is also true that f(a) = ln(a)/(a - 1) for 0 < a < 1.) At x = 1 the limit of the function from right and left has the form 0/0 so L’Hôpital’s rule applies. (The righthand limit is -1 as is the lefthand limit. If this is confusing, it is the fault of the terminology. A good reference is Olmsted’s Advanced Calculus)
@7:00 OK, so if W0 for the solution gives X = 1, that means that W0(-ln(2)/2) = W(ln(1/sqrt(2))) = ln 2 This is the first time I think I've ever seen you put the result of the W function into something that is not just a Wolfram numerical answer Is there an analytical way to come up with that result?
You missed a minus sign, its actually -ln(2). Now the reason is, technically, you can rewrite -ln(2)/2 as -ln(2)*e^(-ln(2)), now see that this is in the form of xe^x, hence, W(-ln(2)/2)=-ln2. And also notice, if you multiply and divide by 2, we get -2ln(2)/4, which is -ln(4)e^(-ln(4)), hence W(-ln(2)/2)=-ln(4)=-2ln(2) if you restrict the range of W(x) to y
Well its exponential vs linear so u can just plug numbers till it stops working, 0 doesnt work, 1 works, 2 works, 3 doesnt and any number further wont either, hence answer is 1 and 2
1^x = x has only one solution x = 1, which is the trivial solution. The fun solution for 1 is undefined. 2^x = 2x has x = 2, 4^x = 4x has x = 1/2. I think that's all the rational ones.
I can understand that W_0(0.5*ln(0.5)) = ln(0.5), because W_0(a*exp(a)) = a. But how do you figure that W_(-1)(0.5*ln(0.5)) = 2*ln(0.5)=ln(0.25)? Or is it some freaky coincidence?
Got exp(-W(-ln(2)/2) /2 when I did it. Results in the same results when evaluating in wolfram alpha so guess correct. But no clue how to reduce it to bprp solution without just doing his derivation 😞
I actually did find all the solutions. You just have to go to different branches of the lambert w function and you will see them. I explained that in detail in my x^2=2^x video.
When I hear lambert function I instantly change video ,why? Because if I were alone I wouldn't be able to find the answer to lambert function by heart , so If at the end I have to search for a plot why on earth couldn't I plot the original functions and just see where they meet at? It's like we do a lot of effort to do it formally but we end up doing what we were avoiding at the beginning. ....so....
Can someone tell me wether my approach also works: 2^x = 2x | :x x^(-1)*2^x = 2 e^(-lnx)*e^xln2 = 2 | ln(…) -lnx+xln2 = ln2 | :ln2 -lnx:ln2 + x = 1 -log_2(x) + x = 1 | +log_2(x), -1 x-1 = log_2(x) | (…)^2 x^2 -2x +1 = x | -x x^2-3x+1 = 0, and solving this is just a quadratic. Would that be a valid solution? Nvm, it isn‘t but where is the mistake?
