Your chart is misleading regarding the function xe^x. This function has a minimum at x = -1, and the plot suggests otherwise. This can lead to a misunderstanding of why precisely -1 is the left endpoint for which x in xe^x is reversible.
Yes, I agree with Calvin Jackson. Presenters introduce and use Lambert W without really explaining it in a way that is clear. You have done this so well. Congratulations! I am a medical statistician (retired from operational work but still researching) and interested in how W might be relevant in transformations involving clinical variables.
So I watched one video where the Lambert function was used, and now I see them all the time, but it's just this random function everyone uses along with some algebraic trickery to answer a hard exponential equation. Thanks for explaining what it actually is beyond just something you use for answering Math Olympiad problems.
Nice video, really looking forward to that series! I see that function so often on math videos without really explaining what it really is and I'm stoked to learn something new ^^
Hello Ivana, I will be 68 in July. This video has helped me understand what The Lambert W Function is. I have watched other videos about The Lambert W Function and I didn't know how confused I was till I saw your video and the graphs you have drawn of: y = xexpx and The Lambert W Function as the inverse of: y = xexpx I didn't know how to calculate the inverse of a function by replacing the x with a y and the y with an x. I wrote down this differential equation: dv/dt = a((1 -2v/c)/(1 -v/c)) where v is the speed and a is the acceleration and c is the speed of light and when I used Wolfram's Mathematica to solve this first order differential equation the solution is: In[1] := DSolve[{EQN, v[0] == 0}, v[t], {t}] Out[1] = {{v[t] -> 1/2 (c - c ProductLog[E^(1 - (4 a t)/c)])}} The ProductLog is The Lambert W Function. I didn't know anything about The Lambert W Function before. All the best and many thanks, Peter Nolan. Ph.D.(physics). Dublin. Ireland.
Thank you Peter very much. I'm glad I was able to help you, that's why I'm making these videos. When I see that it helped someone the videos got a purpose. Thank you for this wonderful comment.
Lambert is well-known amongst students of physics for his work on the luminance of surfaces that emit light; a so-called Lambertian surface has the same luminance when viewed from any direction. As a former, but now retired, lecturer in physics I was familiar with that aspect of his work but only came across the mathematical function named after him within the last few days. By referring to Wikipedia I found that in both cases it is the same Lambert, i.e Johann Heinrich Lambert. This prompts me to ask: Did he introduce the function that was later named after him because to him it had some particular application in physics or perhaps even in the narrower field of optics?
I think it is a Wonder Woman emblem, that she chose for the occasion since it features the letter W. BlackPenRedPen did a similar thing, completely unintentionally, when he wore a T-shirt with the ∂asics logo, while doing a problem with partial derivatives.
@intellecta2686 Thanks for the clear and good explanation. Just one thing, at 5:36 you say " ... and our e is approximately two-point-seventy-one ...". The individual digits after the decimal point should be stated individually and not collectively. For example, when saying a number like 1.75, it should be said "one-point-seven-five" and not "one-point-seventy-five". So, e is approximately two-point-seven-one and not two-point-seventy-one.
hi, I'm still not clear on how to use this function, can you point me to some real world examples or 'simple' problems to solve so I can get a better understanding of the function?
8^x = x^5 no tiene solucion con numeros reales, pero hay soluciones con numeros complejos. Hay cinco soluciones, y todos las soluciones usan la funccion W de Lambert.
For the application of the Lambert W Function, go to: ua-cam.com/video/hu8oXMFDNQk/v-deo.html
Your chart is misleading regarding the function xe^x. This function has a minimum at x = -1, and the plot suggests otherwise. This can lead to a misunderstanding of why precisely -1 is the left endpoint for which x in xe^x is reversible.
Yes, I agree with Calvin Jackson. Presenters introduce and use Lambert W without really explaining it in a way that is clear. You have done this so well. Congratulations! I am a medical statistician (retired from operational work but still researching) and interested in how W might be relevant in transformations involving clinical variables.
You are the only one who has made this function perfectly clear to me. I want to thank you so much. Thanks for explaining it so clearly!
So I watched one video where the Lambert function was used, and now I see them all the time, but it's just this random function everyone uses along with some algebraic trickery to answer a hard exponential equation. Thanks for explaining what it actually is beyond just something you use for answering Math Olympiad problems.
Thanks for this great playlist! I only recently found your channel which is very good. Please start posting again!
I revirewed many videos on the topic. This is the clearest, Kudos.
Nice video, really looking forward to that series! I see that function so often on math videos without really explaining what it really is and I'm stoked to learn something new ^^
So glad to hear that. Thank you for the comment 🙂
Thanks Sis for this lecture
Clear explanation. Thank you!
Hello Ivana,
I will be 68 in July. This video has helped me understand what The Lambert W Function is. I have watched other videos about The Lambert W Function and I didn't know how confused I was till I saw your video and the graphs you have drawn of:
y = xexpx
and The Lambert W Function as the inverse of:
y = xexpx
I didn't know how to calculate the inverse of a function by replacing the x with a y and the y with an x.
I wrote down this differential equation:
dv/dt = a((1 -2v/c)/(1 -v/c))
where v is the speed and a is the acceleration and c is the speed of light and when I used Wolfram's Mathematica to solve this first order differential equation the solution is:
In[1] := DSolve[{EQN, v[0] == 0}, v[t], {t}]
Out[1] = {{v[t] -> 1/2 (c - c ProductLog[E^(1 - (4 a t)/c)])}}
The ProductLog is The Lambert W Function. I didn't know anything about The Lambert W Function before.
All the best and many thanks,
Peter Nolan. Ph.D.(physics). Dublin. Ireland.
Thank you Peter very much. I'm glad I was able to help you, that's why I'm making these videos. When I see that it helped someone the videos got a purpose. Thank you for this wonderful comment.
@@intellecta2686
Hello Ivana,
You are most welcome.
All the best and many thanks,
Peter. Dublin.
Great video, thanks!
Very helpful video - thanks a lot👍
Very good. Thanks 🙏
Thanks! I finally got it 😙
Thaks for your video!
Nice explanation
beautiful, and smart...
The way you teach is very good
Bom vídeo
I luv it when German women speak English ❤️
Croatian, actually!
@@gary.h.turner
My bad, still, I love it.
@@Tomorrow32gay
That's a mighty Slavic sounding "German" lol.
Thank you for your explanation! Could you explain us Lambert-Tsallies method ?
Very nice explanation 😁
Thank you 🙂🤗
❤️
Thanks for the history lesson on some math. Very cool. ☺️
Thank you for watching :)
Of course. Merry Christmas 🎄 🔔
Merry Christmas to you too 🙂 🤶🎅
I can use with quadratic equations ?
Lambert is well-known amongst students of physics for his work on the luminance of surfaces that emit light; a so-called Lambertian surface has the same luminance when viewed from any direction. As a former, but now retired, lecturer in physics I was familiar with that aspect of his work but only came across the mathematical function named after him within the last few days. By referring to Wikipedia I found that in both cases it is the same Lambert, i.e Johann Heinrich Lambert. This prompts me to ask: Did he introduce the function that was later named after him because to him it had some particular application in physics or perhaps even in the narrower field of optics?
discussing Lambert W function and wearing WW shirt...
Day W mood 🙂
This is a limit of infinite iterated exponential. After analytical extend over complex.
Hi Ivana and thanks for your videos. It's a W Lambert function Tshirt that?😀
I think it is a Wonder Woman emblem, that she chose for the occasion since it features the letter W.
BlackPenRedPen did a similar thing, completely unintentionally, when he wore a T-shirt with the ∂asics logo, while doing a problem with partial derivatives.
@intellecta2686
Thanks for the clear and good explanation.
Just one thing, at 5:36 you say " ... and our e is approximately two-point-seventy-one ...".
The individual digits after the decimal point should be stated individually and not collectively. For example, when saying a number like 1.75, it should be said "one-point-seven-five" and not "one-point-seventy-five".
So, e is approximately two-point-seven-one and not two-point-seventy-one.
hi, I'm still not clear on how to use this function, can you point me to some real world examples or 'simple' problems to solve so I can get a better understanding of the function?
It looks like the function is a reflection rather than an inverse.
Ah Euler won't be upset 😅
y = xe^x attains its minimum value of (-1/e) at x=-1
Pinned u a ques on insta plz answer to that ASAP ...thank u
I'll check as soon as possible.
Bella matemática, sus aportes son muy importante
X^5=8^x no tiene solución, creo que cometió un error con respeto.
8^x = x^5 no tiene solucion con numeros reales, pero hay soluciones con numeros complejos. Hay cinco soluciones, y todos las soluciones usan la funccion W de Lambert.
Maths = Euler
Also known as the UA-cam Function.
I am trying to identify your accent. It seems to be German. Am I right?
She is not German, her accent sounds Slavic
Uhh, you are so beautiful like a non-analytical problem solved by lambert W.