Lambert W Function (No real solution???)

[@GillesF31]

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 !!! 🙂

[@edwinwelch1393]

I am very glad I found your channel. They should be teaching this at A level in the UK.

[@intellecta2686]

note: I meant the imaginary part, not the irrational. To practice the application of the Lambert W Function go to: kzbin.info/www/bejne/nqaboIuDe6mBh80 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.

[@elliottmanley5182]

The analytic continuation of W(z). (Since it came up accidentally in this vid! And I can't find any other KZbinrs covering it.

[@moeberry8226]

You are the most gorgeous and elegant woman I have seen, honestly no woman has your beauty.

[@elliottmanley5182]

@@moeberry8226 well thank you

[@intellecta2686]

@@elliottmanley5182 😅

[@intellecta2686]

@@moeberry8226 Thank you for the kind words 🙂

[@johnlinley2702]

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?

[@jamesstrickland833]

someone (I think blackpenredpen) made a couple of viral videos using it and now everyone is hoping to have similar success.

[@kolomun]

This was the best nap I've ever had :)

[@alexanderturoczy7035]

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

[@timless5157]

yes but r < -1/e how did you solve W(-ln(8) / 5) ?

[@alexanderturoczy7035]

@@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

[@vitorsg]

Complex numbers

[@timless5157]

@@vitorsg ok let's see the solution

[@netritos7105]

great video and amazing explanation. I was trying to practice this equation and thank god I found your channel. Thank you so much

[@pythontron8710]

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.

[@RUDYFERNANDOGONZALEZESCOBAR]

Super like desde Guatemala

[@bohharry]

Wow thank you lady you made the explanation very very clear thanks a billion😀

[@arbenkellici3808]

Well done! Keep it up! You are doing an excellent job!

[@ProfeJulianMacias]

I like your class...great

[@iemtruongvan7316]

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.

[@ChavoMysterio]

Which branch of mathematics is the Lambert W function is covered?

[@sberacatalin2250]

Interesant! Frumos! Technic! Next?! 👏👀👍🇦🇩🥇

[@andrewshuttleworth5826]

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.

[@shrishantroy2494]

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.

[@intellecta2686]

Please watch this video from 2:00 min. kzbin.info/www/bejne/pIqaimNnpK1jsJo I hope that's what you are looking for.

[@elliottmanley5182]

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.

[@elliottmanley5182]

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!

[@elliottmanley5182]

@@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.

[@intellecta2686]

All good, it happens. Who works makes mistakes :)

[@Ramkabharosa]

(∀x∈ℝ) [ (8˟ - x⁵) > 0.3636... ], so there is no real-root of the equation (8˟ = x⁵). The minimum value of (8˟ - x⁵) is achieved in the interval (-0.7, -0.5) near -0.59. Just draw the graph of (8˟ - x⁵) with a Desmos Graphing Calculator to see this. .

[@LePhenixGD]

Thank you very much! Your video helped me a lot!

[@edgareduardomatamorosizagu7291]

Beautiful solution.

[@steflabbe]

hi, how to find values a, b in x^a=b^x as the limit between R and C solutions ?

[@michlop452]

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.

[@MartinaHercegovka92]

This helped me a lot, tnx!! Could you explain Fourier series and Fourier transform, please?

[@intellecta2686]

I will for sure in one of the following videos. Thanks for your suggestion.

[@AngeloLaCruz]

🎓(2023)🎓 Greetings From Curaçao (an Island Nation in The Cribbean)

[@kornelije23]

Perfect as always keep up the good work ❤️

[@intellecta2686]

I will, thank you 🙂

[@zaferatasoy3095]

thank you very much...🎉

[@BSnicks]

Which calculator has a built in W(x) function?

[@p4z9m]

Mathematica has productlog.

[@davidbrisbane7206]

Wolfram Alpha

[@lawrencejelsma8118]

​@@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

[@rmw6151]

Thank you for this video - well explained! Perhaps your next topic could be Line Integrals?

[@intellecta2686]

Interesting suggestion, I will consider it for sure. Thank you for your reply!

[@marcohk609]

How calculate the function w of lambert to w(ln(8))/5)? If there'inst method , is bether input The function in The computar program.......

[@SALogics]

Nice explanation ❤❤

[@InfiniteLoveMeditationClub]

Nicely explained 😊 Thanks 👍

[@intellecta2686]

Thank you :)

[@davidbrisbane7206]

Is this the principal value of x? There should be an infinite number of complex values x that satify this equation right?

[@purim_sakamoto]

W関数は楽しいですね😄👍

[@differentboy7779]

Thanks and too good👍👍👍

[@intellecta2686]

Thank you!

[@omarjette3859]

There is no real solution. The x^2 and 8^x curves are within 0.37 at x=-0.6, but do not crross.

[@bezobrazie7607]

Thanks for u, its nice :)

[@intellecta2686]

Thank you 🙂

[@hayekianman]

so why does x not have 5 roots?/solutions if it is x^5?

[@meow2550]

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

[@intellecta2686]

I made a quick video for this, you can expect it in a few hours. Thank you for the idea :)

[@meow2550]

@@intellecta2686 thanks

[@moeberry8226]

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.

[@oahuhawaii2141]

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]

[@georgesbv1]

@@oahuhawaii2141 simplifiable as x= 2*sqrt(W(1/4))

[@DeepakKumar-yt6ol]

How to find w(4)=?

[@bdofficiallitonlife]

Good

[@ferdinandrius6063]

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 .

[@tbg-brawlstars]

Will not the answer just simply can't be written as e^{w[5/ln(8)]} ?

[@gerar2158]

Bella mujer

[@brnico6184]

cool!

[@intellecta2686]

Thank you.

[@olusolaomolayo2687]

There is the need to know how to use calculator to determine or evaluate the value of Lambert W. Function.

[@ynag1223]

i have refused this problem at 6:20 why did you add (/w) ?😒

[@rolandkaschek9722]

How do you know this is the only solution? Maple gives 5 solutions as does Wolfram.

[@billrandle4437]

This application of the W function is instructional. However an immediate solution by inspection is 5

[@lukion27]

Not correct. 5^5 != 8^5, so 5 is not a solution.

[@lapotajunior5072]

the first problem is so easy its like 2

[@syedmdabid7191]

Imaginaire, Sic non non root REALIS, responsi

[@diasao5618]

2^n=2n

[@oahuhawaii2141]

n = 1, 2 .

[@Jack_Callcott_AU]

For the answer you said "irrational part", don't you mean imaginary part. Please explain!

[@intellecta2686]

Yes, of course, imaginary. Lapsus Linguae.

[@Jack_Callcott_AU]

@@intellecta2686 I enjoy your videos on Lambert W function.Thanks!

[@intellecta2686]

I'm very glad to hear that, thank you very much.

[@jorgecastillo1420]

Yes the green board is not good ,or the chalk is not good

[@duartetavares4397]

Impossible to see the calculation on that board

[@gerar2158]

El resultado en un número complejo. Ella lo llama irracional que es otra cosa. “i” es imaginario, por tanto, complejo y no racional...

[@alvarogerardoinsuasticamac5724]

Creo que no tiene solución, corregirlo

[@nelson101]

O giz Nao se Vê Bad visibility of scribe.

[@segayanmx4442]

Hi! Pls change your green board to black board! What you are writing is not clear ! Very difficult to follow you!

[@krazieecko]

The sound quality and tiny chalkboard makes this video close to unwatchable

[@GFlCh]

👍 💯 😀 " ≈ 2.207 - [Irrational part] " I expect you meant " ≈ 2.207 - [Imaginary part] " - ― £ $ € ฿ ± Σ Ω Π Δ µ ← ↑ → ↓ ^ √ ³√ ∞ * ≈ ≠ ≤ ≥ ÷ •

[@vansf3433]

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

[@syedmdabid7191]

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.