technically 3 is also in the solution you got. The 2.478... answer is if you use the principal branch of the lambert function i.e. W_0[-(ln3)/3]. If you use the -1 branch you will get the 3 that we expected.
The function x^3*3^(-x) has a maximum 1.01379 at x = 2.73072 .This shows that there are two solutions , x =3 which is obvious , but also a second solution for x < 2.7 , (2.478 ).
@@amornthepmeekangwal9421 For the second solution you have to use the function W[-1,x]. The function W [x] gives the solution x = 2.478 and W[-1 x] gives x = 3 .
@@amornthepmeekangwal9421 If the input is between -1/e and 0, we need to use both the real Lambert W functions, as mentioned in the video. One gives 3, the other gives the other answer
Can you make a video dedicated to non elementary functions? This topic has always fascinated me and I'd love to learn more about it from no one other than you.
If you look at -ln3/3 in the lamber w function you can turn it into -ln3/e^ln3 and then into -ln3*e^-ln3 and since its in the lamber w function you can turn all of it to -ln3 and lastly you get x=(-3*-ln3)/ln3=3 so you can prove 3 is also solution
The general case : a^x = (b x+c)^p can be solved also using the Lambert-function . Setting q = - ln a /(b * p) * a^(-c/ (b * p ) , the three possible solutions are given by x1 = - p/ln a * W(q) - c/b , x2 = - p / ln a * W (-q ) - c/ b , x3 = -p/ ln a * W( -1 , q) - c/b . An example with three solutions : take a = 1.5 ,b= 1.2 ,c = 0.5 ,p = 2 . Then one gets x1= 0.50685 , x2 = 14.0932 , x3 = -1.0854.
Perfect title haha! That's what I think when W function exists. Will you do those special or super hard integrals again? those were super entertaining to watch. Especially with you writing on the -air-
The W function is a way of life. Let f(x) = a for an unreasonably complicated f. Define ominous sounding function = f inverse. The answer is ominous sounding function (a). Problem solved.
I guess, there are more solutions if you use all 3 (complex) cube roots of 1? It really looks like Lambert wanted to solve xe^x, got tired, and invented a function that solves the problem by whatever means necessary. Which.... is not wrong I guess.
For the positive imaginary solution, you use the W_1 branch. For the negative imaginary solution, you use the W_-1 branch. I’d assume that two out of the three representations give the same number.
There's a typo at 6:38: putting in W = -0.9075 gives a positive x = 2.4781, not a negative -2.7231, as mentioned in the video. A minute earlier at 5:38 it was put correctly.
Not true! The exponential functional notation must be functional! As such, there can be only be one equivalent value to any expression of a base number exponentiated. Otherwise statements violate the injective property of all functions. In other words, the equation x^3 = 1 does indeed produce three solutions for x, but the real number 1^(1/3) cannot be equivalent to more than one value, per the basic conventions of functional notation! Common and easy mistake to make, but worth correcting nonetheless!
You don't really explain what you are actually doing to get 2.47805. Though quick math confirms that is correct: 2.48^3 (15.25) = 3^2.48 (15.25). But it still ends up being a mystery how you got there.
This equation has two real solutions: x=3 and x≈2.478 Here is how to derive them properly. x^3 = 3^x | ^(1/3) note: sign is unchanged! x = 3^(x/3) | 3^(x/3) = (e^ln(3))^(x/3) = e^(x*ln(3)/3) x = e^(x*ln(3)/3) x * e^(-x*ln(3)/3) = 1 | * -ln(3)/3 -x*ln(3)/3 * e^(-x*ln(3)/3) = -ln(3)/3 | Lambert W on both sides -x*ln(3)/3 = W(-ln(3)/3) If the entire parameter of Lambert W is greater or equal zero, it is has only one real solution (branch 0). If less than zero and greater or equal -1/e, it has two real solutions (branch 0 and -1). If less than -1/e, no real solutions. Lambert W parameter: -1/e ≈ -0.368 < -ln(3)/3 ≈ -0.366 < 0 => Lambert W has two real solutions! Integer solution: -x*ln(3)/3 = W(-ln(3)/3) | 1/3 = e^ln(1/3) = e^(-ln(3)) -x*ln(3)/3 = W(-ln(3) * e^(-ln(3))) -x*ln(3)/3 = -ln(3) -x/3 = -1 x = 3 Lambert W is also called productlog in WolframAlpha. The first parameter of productlog is optional, and indicates the productlog branch (default is branch 0). The branch is an integer ∈ ℤ. There are infinitely many productlog results in complex numbers for all branches. Real solutions of productlog, however, can only be in branches 0 and -1. Here are both real solutions calculated in WolframAlpha: x = -3 * productlog(0, -ln(3)/3) / ln(3) = 2.47805268028830... x = -3 * productlog(-1, -ln(3)/3) / ln(3) = 3
So x = 3 is also a solution. But my question is how can there be two solutions? Both of the functions are monotonically increasing. And after the exponential function increases the cubic function the rate at which it's increasing is faster so how is it ever possible that the graphs can intersect twice?
5:35 I'm trying to figure out what am I doing wrong can't we say that -ln(3)=ln(1/3) and -ln(3)/3=1/3*ln(1/3)=e^ln(1/3)*ln(1/3) and W(e^ln(1/3)*ln(1/3))=ln(1/3)=-ln(3) so the entire fraction -3W((-ln(3)/3)/ln(3) simplifies to just 3 which is one of the solutions? correct my mistake please it's probably about the branches of the W function Edit 1: not about branches I'm probably stupid
Bro but why there exits an extragenous root Is there a domain for w(xe^x)=x As till second last step 3 satisfies but after introducing lambert function [ whose domain is -1/e to infinity which is satisfies] and using xe^x, 3 is not yet the soln As solving in elegant way always requires each step to be accurate
I don't understand why are some functions not considered analytical. Logarithms are also defined as the inverse of another function and are not represented using the other function notation but a new one and are considered analytical, while the W(•) is not
@@RaniDevi-xt4hq OK, I begin to see. So this branching business has to be explained much better than just mumbling something about the W function being "multivalued". Where is tbe cut? How many branches are there? Is there some "default" branch? Which one was used in the video? Why doesn't the obvious solution of the original equation appear in the default branch? Do the other branches give other values (complex ones, I suppose)? Are there Infinitely many of them? Without such information, that W function remains mysterious magic.
@@WK-5775 en.m.wikipedia.org/wiki/Lambert_W_function#:~:text=The%20principal%20branch%20W0,%2C%20Hare%2C%20Jeffrey%20and%20Knuth. You can read this page.
@@WK-5775 there are two branches to the lambert w function- 0 and -1, the one in this video is the principle branch or the 0 branch. the solution x = 3 comes in the -1 branch. if -1/e < x < 0, there will be two solutions to the equation.
I exponatiated both sides with 1/3 and then plugged in 3^(1/3*x) x, so i got 3^(1/3*3)^(1/3*3)… wich simplifies to 3^1^1^1…. so i just got 1 with basic algebra
The solution for that is in the solution shown, but its via the W_-1 branch of the Lambert function. The empirical solution is the most obvious way of course, but the point is to "derive" the solution rather than having "a solution" that you don't know how it works.
This is how i solved it x^3 = 3^x x^(3/x)=3 (3/x)*lnx=ln3 3x^(-1)*lnx=ln3 (-lnx)*e^(-lnx)=(-ln3)/3 -lnx = W[(-ln3)/3] -lnx = W[(-ln3)*3^(-1)] -lnx = W[(-ln3)*e^(-ln3)] -lnx = -ln3 x = 3 I wonder how can same formula give 2 different and correst solutions (2,47805 and 3) x={3W[-(ln3)/3]}/(-ln3) (=2,47805) =3*(-ln3)/(-ln3) = 3
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x=x+1 be like
A typo at 6:38: should be x = 2.4781, not -2.7231
By the way, x=3 is a solution too :)
🤯
Lol I thought about it the m9ment I've seen this equation
technically 3 is also in the solution you got. The 2.478... answer is if you use the principal branch of the lambert function i.e. W_0[-(ln3)/3]. If you use the -1 branch you will get the 3 that we expected.
x^3= 3^x= k
x^k^1/x=3^k^1/3
Hmmm I wonder how 3 could be a solution 😂😂
😅 but one should still know his calculus
Truly a W moment for my mathematical knowledge
Literally
a W(moment for your mathematical knowledge)
@@yoylecake313 (moment for your mathematical knowledge)e^(moment for your mathematical knowledge)
The function x^3*3^(-x) has a maximum 1.01379 at x = 2.73072 .This shows that there are two solutions , x =3 which is obvious , but also
a second solution for x < 2.7 , (2.478 ).
How we know there are 2 solutions when we find 1 solution from Lambert W function and how to find other solutions.
@@amornthepmeekangwal9421 For the second solution you have to use the function W[-1,x].
The function W [x] gives the solution x = 2.478 and W[-1 x] gives x = 3 .
@@amornthepmeekangwal9421 If the input is between -1/e and 0, we need to use both the real Lambert W functions, as mentioned in the video. One gives 3, the other gives the other answer
Can you make a video dedicated to non elementary functions? This topic has always fascinated me and I'd love to learn more about it from no one other than you.
You should check out the hypergeometric function and all its special cases.
If you look at -ln3/3 in the lamber w function you can turn it into -ln3/e^ln3 and then into -ln3*e^-ln3 and since its in the lamber w function you can turn all of it to -ln3 and lastly you get x=(-3*-ln3)/ln3=3 so you can prove 3 is also solution
The general case : a^x = (b x+c)^p can be solved also using the Lambert-function . Setting q = - ln a /(b * p) * a^(-c/ (b * p ) , the three possible
solutions are given by x1 = - p/ln a * W(q) - c/b , x2 = - p / ln a * W (-q ) - c/ b , x3 = -p/ ln a * W( -1 , q) - c/b .
An example with three solutions : take a = 1.5 ,b= 1.2 ,c = 0.5 ,p = 2 . Then one gets x1= 0.50685 , x2 = 14.0932 , x3 = -1.0854.
I see this as an absolute W
nice pun
nice
truly is an |W|
@@Lavasparked😂 beautiful
Thank you so much sir🥹😃
Perfect title haha! That's what I think when W function exists.
Will you do those special or super hard integrals again? those were super entertaining to watch. Especially with you writing on the -air-
The W function is a way of life. Let f(x) = a for an unreasonably complicated f. Define ominous sounding function = f inverse. The answer is ominous sounding function (a). Problem solved.
I probably shouldn’t watch this video until I can solve this myself, since this is good practice for me learning to use the lambert function.
I mean, you can very simply solve it by realizing x=3 is a solution. 🙃
I was just thinking about this equation and also about lambert w function yesterday!!!
I guess, there are more solutions if you use all 3 (complex) cube roots of 1? It really looks like Lambert wanted to solve xe^x, got tired, and invented a function that solves the problem by whatever means necessary. Which.... is not wrong I guess.
For the positive imaginary solution, you use the W_1 branch. For the negative imaginary solution, you use the W_-1 branch. I’d assume that two out of the three representations give the same number.
A truly W func fr 0:51
Me at 0:45, already holding a frozen treat from Bahama Buck's: "Way ahead of you."
The topic that blackpenredpen love
Yeah, he just posted a video featuring it.
There's a typo at 6:38: putting in W = -0.9075 gives a positive x = 2.4781, not a negative -2.7231, as mentioned in the video. A minute earlier at 5:38 it was put correctly.
at 4:00, cube root of 1 is not only 1 but also (-1±i√3)/2
DAYUM
For that you need the W_1 branch of the W function.
Not true! The exponential functional notation must be functional! As such, there can be only be one equivalent value to any expression of a base number exponentiated. Otherwise statements violate the injective property of all functions. In other words, the equation x^3 = 1 does indeed produce three solutions for x, but the real number 1^(1/3) cannot be equivalent to more than one value, per the basic conventions of functional notation!
Common and easy mistake to make, but worth correcting nonetheless!
@@21centuryhippie61 ah true I forgot about that
6:46 says -2.7231 is the solution but that’s a typo: it is indeed 2.47...
You don't really explain what you are actually doing to get 2.47805. Though quick math confirms that is correct: 2.48^3 (15.25) = 3^2.48 (15.25). But it still ends up being a mystery how you got there.
This equation has two real solutions: x=3 and x≈2.478
Here is how to derive them properly.
x^3 = 3^x | ^(1/3) note: sign is unchanged!
x = 3^(x/3) | 3^(x/3) = (e^ln(3))^(x/3) = e^(x*ln(3)/3)
x = e^(x*ln(3)/3)
x * e^(-x*ln(3)/3) = 1 | * -ln(3)/3
-x*ln(3)/3 * e^(-x*ln(3)/3) = -ln(3)/3 | Lambert W on both sides
-x*ln(3)/3 = W(-ln(3)/3)
If the entire parameter of Lambert W is greater or equal zero, it is has only one real solution (branch 0).
If less than zero and greater or equal -1/e, it has two real solutions (branch 0 and -1).
If less than -1/e, no real solutions.
Lambert W parameter:
-1/e ≈ -0.368 < -ln(3)/3 ≈ -0.366 < 0 => Lambert W has two real solutions!
Integer solution:
-x*ln(3)/3 = W(-ln(3)/3) | 1/3 = e^ln(1/3) = e^(-ln(3))
-x*ln(3)/3 = W(-ln(3) * e^(-ln(3)))
-x*ln(3)/3 = -ln(3)
-x/3 = -1
x = 3
Lambert W is also called productlog in WolframAlpha.
The first parameter of productlog is optional, and indicates the productlog branch (default is branch 0).
The branch is an integer ∈ ℤ. There are infinitely many productlog results in complex numbers for all branches.
Real solutions of productlog, however, can only be in branches 0 and -1.
Here are both real solutions calculated in WolframAlpha:
x = -3 * productlog(0, -ln(3)/3) / ln(3) = 2.47805268028830...
x = -3 * productlog(-1, -ln(3)/3) / ln(3) = 3
me casually waiting for a product log calculator to be made for students to bring to exams:
x^3 has 3 solutions does this lead to more solutions for the entire thing?
Love your vídeos❤️🇧🇷
So x = 3 is also a solution. But my question is how can there be two solutions? Both of the functions are monotonically increasing. And after the exponential function increases the cubic function the rate at which it's increasing is faster so how is it ever possible that the graphs can intersect twice?
The reason is the multiple branches of the W function I'd presume. It's multivalued, so here we get multiple solutions
Coding the newton rhapson method is very easy, but how would you do it by hand?
5:35 I'm trying to figure out what am I doing wrong
can't we say that -ln(3)=ln(1/3) and -ln(3)/3=1/3*ln(1/3)=e^ln(1/3)*ln(1/3) and W(e^ln(1/3)*ln(1/3))=ln(1/3)=-ln(3) so the entire fraction -3W((-ln(3)/3)/ln(3) simplifies to just 3 which is one of the solutions? correct my mistake please it's probably about the branches of the W function
Edit 1: not about branches I'm probably stupid
Is there a typo? It seems like you need to divide -2.7231 by -Ln(3) to get the nontrivial solution. Looks like two steps were missed.
e^x is always greater (or equal at x=e) than x^e.
You really complicated the whole solution
How are the branches of W even ordered (as in the subscript)?
Bro but why there exits an extragenous root
Is there a domain for w(xe^x)=x
As till second last step 3 satisfies but after introducing lambert function [ whose domain is -1/e to infinity which is satisfies] and using xe^x, 3 is not yet the soln
As solving in elegant way always requires each step to be accurate
3:00 cant x=3 ?
This Videos Title is misleading. I am now also scared of Lamberts function
Apply loag on first step and then derivate it to get the value of x
-ln3/3 is also equal to ln(1/3)×e^(ln(1/3))
I don't understand why are some functions not considered analytical.
Logarithms are also defined as the inverse of another function and are not represented using the other function notation but a new one and are considered analytical, while the W(•) is not
The lambert W function is analytical. You can see it's taylor series representation at 2:09. You might've mixed up analytical with elementary.
How is `3` not the solution in any of the methods? 3^3 = 3^3?
Yes, 3's a solution to this, it's just not mentioned
You can use logarithms also
Very nice equation👍
so, W(e) would just be 1? (bc W(1 x e^1))
0:07 hmm, only W i know is the Wrongston. I can’t spell it.
Why does the W function not yield to the obvious solution x=3? That does not quite give a reliable impression.
Cause it's not really a function.
If you use the -1 branch of Lambert W function it yields the solution.
@@RaniDevi-xt4hq OK, I begin to see. So this branching business has to be explained much better than just mumbling something about the W function being "multivalued". Where is tbe cut? How many branches are there? Is there some "default" branch? Which one was used in the video? Why doesn't the obvious solution of the original equation appear in the default branch? Do the other branches give other values (complex ones, I suppose)? Are there Infinitely many of them?
Without such information, that W function remains mysterious magic.
@@WK-5775 en.m.wikipedia.org/wiki/Lambert_W_function#:~:text=The%20principal%20branch%20W0,%2C%20Hare%2C%20Jeffrey%20and%20Knuth.
You can read this page.
@@WK-5775 there are two branches to the lambert w function- 0 and -1, the one in this video is the principle branch or the 0 branch. the solution x = 3 comes in the -1 branch. if -1/e < x < 0, there will be two solutions to the equation.
My solution steps were like this:
X^3 = 3^x
→log(x^3) = log(3^x)
→3logx = xlog3
→(logx)/x = (log3)/3
So i can absolutely write that x=3.
4:07 complex cube roots tho
Use different branches of the lambert function.
X = 3 3 to the power of 3 is equal to 3 to the power of 3
Soo cool
The solution shown in this video a 6:46 ... x=-2,7231 is obviously wrong. A simple test would have brought this up.
Me: i have never seen this man in my line
my answer is messed up.
x=3^(x/3)
3=x^(x/3)
3=(3^(x/3))^(x/3)
3=3^((x^2)/9)
3^9=3^(x^2)
9=x^2
x=3
I exponatiated both sides with 1/3 and then plugged in 3^(1/3*x) x, so i got 3^(1/3*3)^(1/3*3)… wich simplifies to 3^1^1^1…. so i just got 1 with basic algebra
amazing
Why did I get deja vu from this?
I fear no equation... but this *thing*... it scares me.
How do we solve the equation to find x=3
Technically ... just look at it.
1. Look at it
2. See that cbrt 1 has 3 solutions. Use the imaginary solutions with the corresponding branch of W to solve.
x^3 = 3^x
3ln(x) = x(ln(3)
(3ln(x))/ln(3) = x
3/ln(3) = x/ln(x)
ln(3)/3 = ln(x)/x
ln(3) * 3^-1 = ln(x) * x^-1
-ln(3) * e^-ln(3) = -ln(x) * e^-ln(x)
apply lambert function.
-ln(3) = -ln(x)
ln(3) = ln(x)
x = 3
Alternative way, alternative solution.
Mathematicians are really just people that spend all day making and solving puzzles
I just noticed that your voice sounds a lot like CGP Grey’s😅😅😅
Me:
x is 3 !
3 factorial !
x equal to 3 is exact solution of equation.
i agree
I did it on desmos
I got 2.478
Why 0^x=x^0 no solution?
0^x is either indeterminate if x=0 or it is 0, x^0 is either indeterminate if x=0 or it is 1
X^3 = 3^x; the empirical solution is x=3; why this solution is not reflecting anywhere.
The solution for that is in the solution shown, but its via the W_-1 branch of the Lambert function. The empirical solution is the most obvious way of course, but the point is to "derive" the solution rather than having "a solution" that you don't know how it works.
You should have gone a bit deeper into the details :(
W function
yes, that's me:)
Valid for any no. x^n = n^x, I used to tease my friends in class while I was in 9th grade.
Hello!
there is a much easier way to do:
3^x = x^3
ln (3^x) = ln (x^3)
x ln(3) = 3 ln(x)
x = 3 ln(x)/ln(3)
3.
3^3 = 27, 3^3 = 27
X=3
x=3.
1000th like
3 days ago🎉
x=3
1:09 the equation is showing f=w if you agree 👇
X is 3 😭😭😭
3
27=27
The solution for X^Y = Y^X is just X=Y so its just X^X, right?
x = 9
x=3
This is how i solved it
x^3 = 3^x
x^(3/x)=3
(3/x)*lnx=ln3
3x^(-1)*lnx=ln3
(-lnx)*e^(-lnx)=(-ln3)/3
-lnx = W[(-ln3)/3]
-lnx = W[(-ln3)*3^(-1)]
-lnx = W[(-ln3)*e^(-ln3)]
-lnx = -ln3
x = 3
I wonder how can same formula give 2 different and correst solutions (2,47805 and 3)
x={3W[-(ln3)/3]}/(-ln3) (=2,47805)
=3*(-ln3)/(-ln3)
= 3
Мерзкий составитель испортит жизни немалому числу ребят, пока его наконец-то репрессируют.
too easy
bro just
X * X * X = 3 * 3 * 3
x=3
no thanks easy
Something that could have been done was, at the end, write (-ln3)/3 as -ln(3)e^(-ln3), giving 3 as a result
Plz dubbed in hindi 🙏
X=3,(-3±3√-3)/2
The solution x=3 is immediately obvious . This is also the only real solution ,if one looks at the graphs of x^3 and 3^x.
Sorry, I missed the other solution x= 2.47......
“x^3=3^x”
I mean yeah if x=3
“≈2.47805”
*w h a t*
“≈-2.7231”
*w h a t ^ 2*
Click ehat video on the screen, wait i dont see it