Solving 8 Equations w/ Lambert W function

These questions are designed by "HelloItsMe". Thank you. : )

A few things to keep in mind:
1. If f(x)=x*e^x, then f^-1(x)=W(x).
2. W(x)e^W(x)=x for x is at least -1/e
3. W(xe^x)=x for x is at least -1
#bprplive

0:00 Getting ready!
1:11 Some main properties of W(x)
5:00 Q1, W(W(x))=1
7:51 Q2, W(x)=ln(2x)
12:00 Q3, W(e^x)=(W(e))^x
15:20 Q4, 1+W((e^2-1)x)=e^2
17:48 Q5, W(9x-7)e^W(9x-7)=-1/e
20:29 Q6, W(e^(e^2+1+x^x))=x^x
26:01 Q7, W(e^(x+1))=x+1
28:41 Q8, W(x+1)=(x+1)^2
37:57 Solve x^x^x=a?

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Thank you,
blackpenredpen

[blackpenredpen]

0:00 Getting ready!
1:11 Some main properties of W(x)
5:00 Q1, W(W(x))=1
7:51 Q2, W(x)=ln(2x)
12:00 Q3, W(e^x)=(W(e))^x
15:20 Q4, 1+W((e^2-1)x)=e^2
17:48 Q5, W(9x-7)e^W(9x-7)=-1/e
20:29 Q6, W(e^(e^2+1+x^x))=x^x
26:01 Q7, W(e^(x+1))=x+1
28:41 Q8, W(x+1)=(x+1)^2
37:57 Solve x^x^x=a?

[SeeTv.]

I love it! I could solve the first 7 questions on my own thanks to your fish 🐠 analogy.

[davidblauyoutube]

It is possible to use W to solve a = x^(x^(x+1)). Solution follows:
1) Take the log of both sides: ln a = x^(x+1) ln x = x^x ln x^x.
2) Apply W: W(ln a) = ln x^x = x ln x.
3) Apply W: W(W(ln a)) = ln x.
4) Finally: x = e^W(W(ln a)). Done!

For example, this solves 4 = x^(x^(x+1)) as x ~ 1.559610469.

[mr.doritos]

I have never though of lambert W function to be so interesting, thank you for teaching me that !, i have developed a new thinking ability !

[curtiswfranks]

These were fun!

This was not the video for it, but it came up twice, so I mention it here: When we encounter a new function, we never have original intuition about the actual values it takes. This is certainly true of things like the logarithm, but there was even a time when you did not know what f(x) = x looks like. Sometimes, we can just plug-and-chug some example inputs. But, usually, we get our best intuition by plotting the function. We can be confident in that intuition and get a more-detailed view of the function's properties by doing things like finding domain, range, intercepts, end-behavior, growth patterns, extrema, saddle points, points of inflection or undulation, etc. Similarly, approximations (Newton's method, Taylor expansion for analytic functions, etc.) can help a student get an actual feel for specific outputs (especially for non-special points). After all of that is completed, people tend to feel that they have a good grasp of the function and no longer question things in the manner like "Isn't W(2) cheating?" (not exactly a paraphrase – more like a stereotypical example). It is a good question to have, but we should be asking it about all functions – and the reason for our not doing so is because we have that improved intuition.

[renardtahar]

vous etes le plus admirable de par la methode, rigueur et maitrise des maths! sincerement vous honnorez le youtube! think you

[blackpenredpen]

HW: Solve ln(W(ln(x)))=1
Good luck on typing up the answer : )

[Yuscha]

I really like this video, it helped me understand the ways to use the W(x) function better than any other sources I could find. I think your way of working through these problems helps teach and demonstrate the usefulness of the W() function tool without getting deep into the weeds like other sources (Wikipedia) do.

To comment on something you said in the middle of the video, I think people consider it "cheating" because it feels like mathematicians just "made up" a function to be the opposite of x*exp(x) instead of "solving it for real." But in reality there are plenty of these types of functions that exist. The "error function" for the integral of exp(-x^2) or the Gamma function for non-real-integer factorials. If you think about it, even the "logarithm" itself is just made up to be the opposite of exponentation. I think we're more comfortable with logs because it feels intuitive, but really it's no different. Thanks for the great videos.

[harleyspeedthrust]

damn, wish I knew about this stream yesterday. I had to derive wien's displacement law from max planck's law, and i arrived at an equation that I thought could not be solved analytically... turns out I needed to use the Lambert W function which I only JUST learned about by googling today

[ianfowler]

Here's a cute identity. Easy to prove but still nice.
W[ ln(a^a) ] = ln(a)

[jaskiraatshah]

Bro congrats this is like the first video of yours which has absolutely easy questions

[bassem.al-ashour]

Nice, first you exit Lumbert and then you enter Lumbert from a different door

[NintendoGamer]

My first time that I participated, this was fun!

[akolangto]

Thank you so much master! From the philippines here

[purim_sakamoto]

うおお　夢のW関数ライブ40分！！
必見すぎる
ちょっとずつ何日かに分けて味わって見ます！！！

[levskomorovsky]

Very interesting! But why should expression (x + 1) e ^2(x + 1) in an exercise 8 be squared? You can solve:
1 = (x + 1) e ^2(x + 1)
2 = 2 (x + 1) e^ 2(x + 1)
W(2) = W [2(x + 1) e^ 2 (x + 1)]
2 (x + 1) = W(2)
x = W(2)/2 -1

[vedantgupta]

They were easy if u know W(xe^x) =x, but the 8th one was a little tough.

[iconic]

For Q8, I got ln(x+1) = W(-(e)^2), then solved for x to get
x = e^(W(-(e)^2)) - 1

[vitalsbat]

I just made a program for calculating lambert w function in my scientific calculator, though numbers too large might not work :( and it takes about 3-5 seconds to evaluate lol

[vansf]

You should have noticed that the expression (1- (x+1)(e^(x+1)^2 = 0
is impossible because (x+1)^2 is always >= 0. Hence, e^(x+1)^2 will always be >=1.
Additionally, no matter whatever value of x might be, (x+1) (e^(x+1)^2 ) will never ever = 1
Checking:
For x = -1, (x+1)e^(x+ 1)^2 = 0---> LHS =1 - 0 = RHS = 0 is nonsense or impossible
For x < -1, say x= -2----> (-2+1)e^(-2+1)^2 = -1(e) = -e. Plugging e in the eq will give:
LHS =1- (-e) = RHS = 0, which is again impossible
For x > -1, say x = 1----> (1+1)(e^(1+1)^2 = 2(e^4). Plugging 2(e^4) in the eq will give:
LHS = 1 - e^4;= RHS = 0, which is again impossible

Therefore, the eq (1-(x+1)(e^(x(1)^2 = 0 is false. In other words, no more solution can be found from that equation.
The only solution is x = -1

You guys will need to learn how to solve probs with logical reasoning to understand such notions of mathemstics, as imaginary or complex numbers, Lambert function, and suchlike, which can be misleading or false, instead of copying them so blindly nonsensically, like a photocopying machine.