Lambert W Function - Introduction

Solving equations with the Lambert W Function

For more info on the Lambert W function:

Introduction Video:

Playlist:

Many practice problems:

My very primitive Lambert W Calculator (by Newton's Method):

Error corrections:
Around 4:43 I said that we cut the graph of y = xe^x at -1/e. NO! That's wrong. We cut that function at x = -1 because that is the minimum point on that function not -1/e. But when that graph is reflected to our Lambert W Function that function has a domain from -1/e to infinity. The better view of this is in the graph I do later around 10:12 during the problem x^x = 5. I'll also try to clean this up in future videos and possibly do another similar introduction video that explains this part more clearly.

Website:

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#algebra
#lambert
#lambertwfunction

[owlsmath]

Error corrections:
Sorry folks! Around 4:43 I said that we cut the graph of y = xe^x at -1/e. NO! That's wrong. We cut that function at x = -1 because that is the minimum point on that function not -1/e. But when that graph is reflected to our Lambert W Function that function has a domain from -1/e to infinity. The better view of this is in the graph I do later around 10:12 during the problem x^x = 5. I'll also try to clean this up in future videos and possibly do another similar introduction video that explains this part more clearly.

[EdwardSileo]

I have to comment here. This is an excellent video:
(1) It explains how inverses work - f(f-1(x)) and f-1(f(x)).
(2) It explains the symmetry of inverse functions.
(3) It explains the domain issues with the Lambert function (with the correction he mentions below).
(4) He explains the horizontal line test.
(5) He solves x^x=5.

[parinose]

@EdwardSileo
8 months ago
I have to comment here. This is an excellent video:
(1) It explains how inverses work - f(f-1(x)) and f-1(f(x)).
(2) It explains the symmetry of inverse functions.
(3) It explains the domain issues with the Lambert function (with the correction he mentions below)
Thanks.

My comment: Today, I saw a video on LF where someone states clearly that LF (W) is NOT A FUNCTION! -despite the name. You can try to arrange things, but it is not a function. Because a function cannot have two values for a given x.
Many Thanks

[mathsplus]

Thanks a million Owls Math for this very comprehensive lesson on Wolfram Alpha Function. The graph made perfect sense and the example problem also made sense to me. I got it now horizontal line test for inverse fn and vertical lines to test if its a function or not. Really really impressed with this one💯👌

[MikeMagTech]

Watching this as a follow-up to your latest video ; )

[steveschooler]

Alternative : x^3 log(x) = log(2) * 2^14 implies (x^3) * (log[x^3]) = 3 * log(2) * 2^14. Let t = log(x^3) implies t e^t = 3 * log(2) * 2^14. Then wolfram alpha : productlog(3 * log(2) * 2^14) gives log(4096) = log(16^3) implies x = 16.

[alanjareg]

very good
But did you solve the problem or more complicated?
Previously, x was to the power of x, which we simply approximate. The answer is a number between 2 and 3.
But with this solution of yours, the natural logarithm, the inverse function, Lambert's function, and Nepri's number itself entered the equation, which not only does not give an answer, but is also incomprehensible.
Finally, the problem with two x is much more beautiful than twisting like this.
Excuse me.

[jim]

x = e^(W ln(5)). Open wolfram alpha and enter e^Lambert W ln(5). x = 2.129372 . . . . 2.129372^2.129372 = . . . .
2^2 = 4, so we know x in this case is going to be only slightly larger than 2.

[ManojkantSamal]

Applying Lambert w function, I got that
X=2.16 (approximately )

[alanjareg]

Please draw graph the following function for me.
Of course, I won't have any problem if you can generalize the drawing for other real points in the neighborhood of the elements of the domain of the function and create a continuous graph even if this function is not defined at those points.
By the way, I don't know English very well, but I understand your math.

My function:

f(x)=x^x
&
Df={x member IR/ (x=-1÷(2n-1) or x=-2÷(2n-1)) and n is member of IN}

[vansf]

Your solution is x = 2.12937
The discrepancy is:
Error = = 0.000435929% roundedvup = 0.00044%
So, your solution has an error of inaccuracy about .00044%

[obeayobami]

Which calculator can we use to calculate w(ln5) ?

[_tree]

if i solve this problem in my company, i use the nonlinear shooting methods such as newton-rapson, secant etc. it's too complicate to understand the Lambert W Function :(

[jim]

x^x = 5
ln(x)^x = ln(5)
xln(x) = ln(5)
e^(ln(x))ln(x) = ln(5)
ln(x)e^(ln(x)) = ln(5)
Wln(x)e^(ln(x)) = Wln(5)
ln(x) = Wln(5)
e^(ln(x)) = e^(Wln(5))
x =e^(Wln(5))
Wolfram Alpha ≈ 2.12937

[vansf]

x = 2.12937248277

Checking:
x^x = 2.12937248277^2.12937248277 =

Discrepancy test:|(5 = =
rounded up = 0%
Since the discrepancy is extremely small, the solution is x = 2.12937248277 is correct