Integrating the Lambert W Function | Interesting Integrals

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In this video, we cover a useful technique when integrating nested functions, and how to integrate the Lambert W function.
I haven't uploaded in a while for a couple of reasons. There've been a lot of administrative-type tasks (applying for University, citizenship, etc.) I'm planning on making an update/200 subscriber milestone video soon; just let you guys know what's going to happen in the future.
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Опубликовано:

20 сен 2024

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Комментарии : 32

@SoppingWetDog 5 лет назад

I can’t wait for the 200 sub special! When will you do a face reveal?? loved the vid btw

@XanderGouws 5 лет назад

one day

@Patrick_BW 5 лет назад

@@XanderGouws Xander Gouws, you got 200 subs now can you please do a face reveal.

@stefanosvasileiadis2732 5 лет назад

Just saw your channel and already loved it! Just have a bonus exercise for the comment section. Prove that any concave function f:IR->(0, ∞) can't exist.

@XanderGouws 5 лет назад

Thank you so much! That's a great problem because it makes intuitive sense, but is kinda hard to put that into mathematical language.

@OtiumAbscondita 5 лет назад

@XanderGouws 5 лет назад

@gabor6259 5 лет назад

blackpenredpen, Dr. Payem and Flammable Maths is like a Holy Trinity Of Math but you're slowly turning it into a "Quaternity".

@XanderGouws 5 лет назад

Thank you so much :)

@MathForLife 5 лет назад

Nice video! I have a small question: at 1:43 you set y = W(x), why? You set before that y = ln(...), so evaluating int y dx = int W(x) dx.

@XanderGouws 5 лет назад

Yeah, at first we set it equal to the function inside the integral: y = ln(...), and then show that if we rearrange it, ln(...) = W(x). To go from "y * e^y = x" to "y = W(x)", we're just taking W of both sides. By the definition of W, W(y * e^y) = y.

@rot6015 5 лет назад

This is great, subbed!

@rot6015 5 лет назад

Came from flammable maths

@XanderGouws 5 лет назад

Thank you so much

@PackSciences 5 лет назад

Why did you change from animated stuff to drawing equations? If I am not mistaken, this is the first time you do that, right? Bonus question: Top boundary: e^(1+e) = k W(k) = W(e*e^e) = e Bottom boundary: e = l W(l) = W(e) = 1 Top boundary: kW(k) - k + exp(W(k)) = e^(1+e)*e - e^(1+e) + e^e = (e^e)*e^2 - (e^e)*e + e^e = (e^e)*(e^2 - e + 1) Bottom boundary: lW(l) - l + exp(W(l)) = e*1 - e + e = e Integration leads to (e^e)*(e^2 - e + 1) - e

@XanderGouws 5 лет назад

In terms of actual math writing, it's a lot quicker to draw it out than to animate the LaTeX. Just to be clear, I don't intend to ever *only* doing one or the other - in future videos, there'll be a mix of animations and blackboard-style stuff. And you got the bonus question right! I think it looks _cuter_ if you write it as "e^(e+2) - e^(e+1) + e^e - e", but your answer is definitely correct :p

@romajimamulo 5 лет назад

Couldn't you use the identity at the end to turn exp(w(x)) into x/(w(x))? Since x=w(x)×exp(w(x))

@XanderGouws 5 лет назад

Yes! It's largely up to preference how you write it, but x/w(x) is definitely correct.

@romajimamulo 5 лет назад

@@XanderGouws ah, the reason I would is that then you can factor out the X

@XanderGouws 5 лет назад

@@romajimamulo yee

@fanyfan7466 5 лет назад

Romaji the only difference I can see is that e^w(x) is defined at x=0 (it is 1), while x/w(x) is not defined at x=0, but the limit as x -> 0 x/w(x) = 1 anyway so it’s probably not a big deal

@ayeshas6580 5 лет назад

Nice video! Btw how'd u learn stuff like this? From school or independent study?

@XanderGouws 5 лет назад

All of the math that's either calculus or higher, I taught thanks to RU-vid and some online notes. I think I might make a video on some useful resources, but idk. This problem, in particular, I actually came up with :)

@ayeshas6580 5 лет назад

@@XanderGouws I'd love to watch a video abt the resources you used because I'm trying to self study but don't really know where to start, and I have a lot of time on my hands rn :)