Lambert W Function (No real solution???)
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- Опубліковано 31 гру 2020
- In this Video I am going to show you how to apply the Lambert W Function. I am explaining the basic tool when it comes to application of the Lambert W Function. Watching this Video you will be ably to recognize and solve some basic problems using the Lambert W Function.
For the Lambert W Function Introduction go to: www.youtube.com/watch?v=nY7Y0...
For Application of the Lambert W Function go to: • Application of the Lam...
For Application of the Lambert W Function 2 go to: • Find X (Lambert W Func...
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It's never too late to do the right thing: I discovered in 2024 your video (3 years old) which is a very useful step to understand the right way to solve an equation which requires the use of Lambert W function. I notice the powerful mathematics tool we have to handle: the logarithms and their properties. THANK YOU VERY MUCH !!! 🙂
great video and amazing explanation. I was trying to practice this equation and thank god I found your channel. Thank you so much
note: I meant the imaginary part, not the irrational.
To practice the application of the Lambert W Function go to:
ua-cam.com/video/hu8oXMFDNQk/v-deo.html
I hope you liked this short Lambert W functuon series. Feel free to write which topic would you like to see as next on my channel.
The analytic continuation of W(z). (Since it came up accidentally in this vid! And I can't find any other UA-camrs covering it.
You are the most gorgeous and elegant woman I have seen, honestly no woman has your beauty.
@@moeberry8226 well thank you
@@elliottmanley5182 😅
@@moeberry8226 Thank you for the kind words 🙂
Well done! Keep it up! You are doing an excellent job!
Thank you very much! Your video helped me a lot!
It's nice to see Lambert W Functions getting so much discussed on the Web. Could some one drop me a note as to why this is of such great interest now. Am I missing some new theory in Analysis or Statistics or whatever?
someone (I think blackpenredpen) made a couple of viral videos using it and now everyone is hoping to have similar success.
Wow thank you lady you made the explanation very very clear thanks a billion😀
Perfect as always keep up the good work ❤️
I will, thank you 🙂
Nice explanation ❤❤
Interesant! Frumos! Technic! Next?! 👏👀👍🇦🇩🥇
Beautiful solution.
This was the best nap I've ever had :)
I know it’s kinda nitpicky, but at the end you state the answer contains an irrational part, and while this is technically true, I think you were referring to the imaginary part.
Despite this, very good stuff, easy to understand but still technically complicated.
Thank you for this video - well explained! Perhaps your next topic could be Line Integrals?
Interesting suggestion, I will consider it for sure. Thank you for your reply!
I like your class...great
Would you please consider making another video on how to find the value of Lambert W function?
Amazing explanation in less time. Thanks a lot.
Please watch this video from 2:00 min.
ua-cam.com/video/nY7Y01oH0z8/v-deo.html
I hope that's what you are looking for.
W関数は楽しいですね😄👍
hi, how to find values a, b in x^a=b^x as the limit between R and C solutions ?
This helped me a lot, tnx!!
Could you explain Fourier series and Fourier transform, please?
I will for sure in one of the following videos. Thanks for your suggestion.
Thanks for u, its nice :)
Thank you 🙂
Thanks and too good👍👍👍
Thank you!
Great video, the route I took seemed a little faster:
x^5 = 8^x
x = 8 ^ (x/5) [ raise to 1/5 power ]
x * 8 ^ (-x/5) = 1
xe^(rx) = 1 [ define r as -ln(8 ^ 1/5) = -3/5 ln 2 for ease ]
rxe^(rx) = r
rx = W(r)
So finally x = W(r)/r where r = -3/5 ln 2
yes but r < -1/e how did you solve W(-ln(8) / 5) ?
@@timless5157 Y'know I never actually put this into a calculator. I believe the definition can be extended to be defined on complex numbers however
Complex numbers
@@vitorsg ok let's see the solution
I got a different result e^(-W(-ln(8)/5))
Are the two results equal? I don't know enough about the W to verify myself.
Nicely explained 😊 Thanks 👍
Thank you :)
In your first video you explained that the domain of W(x) is (-1,inf). As -ln(8)/5 is greater than - 1, there must be a real solution. Wolfram Alpha is incorrectly returning the -1 branch of W. If you calculate -ln(8)/5 and use the numerical value, it returns a real answer instead.
I was going to ask if you were going to talk about the analytic continuation of W(z). I wasn't expecting it to turn up here and by your expression, neither were you!
@@intellecta2686 oops! I muddled domain with range. Apologies. -ln(8)/5 < -1/e. I'm not sure what I did to get a real answer but it was wrong.
All good, it happens. Who works makes mistakes :)
Phương trình này vô nghiệm
Đặt f(t) =te^t thì f'(t) = (1+t)e^t+(ln8)/5; khi đó lập bảng biến thiên hàm số sẽ cho ta giá trị cực tiểu cũng là giá trị nhỏ nhất là -1/e. Nên -1/e + (ln8)/5 >0. Vì vậy phương trình f(t)=0 vô nghiệm => phương trình đã cho vô nghiệm.
Which branch of mathematics is the Lambert W function is covered?
Good
🎓(2023)🎓 Greetings From Curaçao (an Island Nation in The Cribbean)
cool!
Thank you.
Bella mujer
Is this the principal value of x? There should be an infinite number of complex values x that satify this equation right?
Which calculator has a built in W(x) function?
Mathematica has productlog.
Wolfram Alpha
@@davidbrisbane7206... If you are using Wolfram Alpha then just look at 8^x - x^5 and Wolfram Alpha shows all zeros ... Then you need no
so why does x not have 5 roots?/solutions if it is x^5?
I need help x^2+8lnx=0
I'm stuck at 1=e^x^2 * x^8
and I have no idea. I need somehow divide by 4.
please send help
I made a quick video for this, you can expect it in a few hours. Thank you for the idea :)
@@intellecta2686 thanks
meow your almost there, just take the 4th root of both side and then multiply both sides by 1/4 and then your good to go! Take the lambert on both sides and simplify.
I just learned about the W() function, and see you shouldn't have exponentiated the equation. Here's what I did:
x^2 + 8*ln(x) = 0
x^2 + 4*ln(x^2) = 0
x' = -4*ln(x') [x' = x^2]
1/4 = (1/x')*ln(1/x')
x"*ln(x") = 1/4 [x" = 1/x' = x^-2]
ln(x^-2) = W(1/4)
x = e^[-W(1/4)/2]
@@oahuhawaii2141 simplifiable as x= 2*sqrt(W(1/4))
Will not the answer just simply can't be written as e^{w[5/ln(8)]} ?
z = -5 W(n , - log 8 /5 )/ log 8 with n integer . Il y a d' autres solutions obtenues avec les racines cinquièmes complexes de l'unité .
i have refused this problem at 6:20
why did you add (/w) ?😒
How do you know this is the only solution? Maple gives 5 solutions as does Wolfram.
Imaginaire, Sic non non root REALIS, responsi
There is the need to know how to use calculator to determine or evaluate the value of Lambert W. Function.
For the answer you said "irrational part", don't you mean imaginary part. Please explain!
Yes, of course, imaginary. Lapsus Linguae.
@@intellecta2686 I enjoy your videos on Lambert W function.Thanks!
I'm very glad to hear that, thank you very much.
2^n=2n
n = 1, 2 .
This application of the W function is instructional. However an immediate solution by inspection is 5
Not correct. 5^5 != 8^5, so 5 is not a solution.
Yes the green board is not good ,or the chalk is not good
There is no real solution. The x^2 and 8^x curves are within 0.37 at x=-0.6, but do not crross.
Impossible to see the calculation on that board
El resultado en un número complejo. Ella lo llama irracional que es otra cosa. “i” es imaginario, por tanto, complejo y no racional...
Creo que no tiene solución, corregirlo
O giz Nao se Vê
Bad visibility of scribe.
👍 💯 😀
" ≈ 2.207 - [Irrational part] "
I expect you meant " ≈ 2.207 - [Imaginary part] "
- ― £ $ € ฿ ± Σ Ω Π Δ µ ← ↑ → ↓ ^ √ ³√ ∞ * ≈ ≠ ≤ ≥ ÷ •
Hi! Pls change your green board to black board! What you are writing is not clear ! Very difficult to follow you!
X^5=8^x, Aut 5lnx=x ln8, Aut lnx/x=ln8/5, Aut lnx e^_lnx= ln8/5 Aut--lnx e^--lnx= --ln8/5 Aut -- lnx=--Wn(--ln8/5) Vel x=e^--Wn( --ln8/5), Sed Wn negatif demonstrat value imaginaire. Sic non root value REALIS responsi.
The sound quality and tiny chalkboard makes this video close to unwatchable
That's not any correct solution, although you have followed all the steps of the notion of lambda function to end up with such an ambitious answer with the "i" or non-existent value
Looking at this can help you understand the notion of imaginary or complex numbers was derived from the pitfall of human fundamental misunderstanding of number signs in human -invented defective notions of mathemstics
8 x 8 = 64, and -8 x -8 = also 64
---> -8 x -8= 8 x 8 ---> -8 must be = 8, which obviously is nonsense
No human mathematician cannot ever prove
Negative x negative = positive
to apply the nonsensical notion that taking a negative value to an even power will result in a positive value
(-x)* (-x) = (-x)^2 = x^2 is nonsense
x^2 - x +1 = 0 ---> x^2 - x = -1
X^2 is always non-negative, and x^2 -x = always non-negative or >=0
You can manipulate the equation to come up with
(x-1/2)^2 +3/4 = O
But (x-1/2)^2 is always positive, thus
(x-1/2)^ +3/4 = 0 is never ever possible
But you guys keep applying the nonsensical notion of imaginary or complex numbers to arrive such absurd and non-existent solutions with "i", which has been used as a sort of ambiguity to cover up human ignorance of their own invented notionscof mathemstics
Cool video! One good tactic with equations of the from x^A = B^x where A and B are constants is to get the equation to the form:
x^A * B^(-x) = 1
At this point you can always raise both sides to the 1/A power to get:
x * B^(-x/A) = 1
Now change the base B to e:
x * e^(-x/A * ln(B)) = 1
Now just multiple both sides by -ln(B) / A and you are ready to take the product log of both sides and solve for x.
Super like desde Guatemala