Two suggestions to improve the channel: 1) Don't spend so much time on very basic algebra steps. It makes the videos slow and painful to watch. 2) Don't do so many problems with the W function, which imo aren't very interesting compared to problems with elegant, closed-form solutions.
@@kabeeralishaikh1439its not thaught but very useful for competative exams fr,but point should be agreed becoz college students dont need learn just some simple algebra
Thank you everyone for your support. I really appreciate it. Wish you all the best in your life and career. Have a great day! "Kindness costs nothing, but means everything. Don't be afraid to give it away"
Nice video. It would be good to mention that when we get the equation ln y e^(ln y)=½ln 2 then, as ½ln 2>0, it is the main branch W₀ of the Lambert W function that we want to use, and we just get one solution, ln y=W₀(½ln 2), leading to x=e^(2W₀(½ln 2)). If we solve xeˣ=y for -1/e
LambertW Function: If ( y ) e ^ y = a LambertW(a)=LambertW[( y ) e ^ y]= y If y=1 (1)e^1 = e = 2.71 So LambertW(2.71)= 1 If y=2 (2)e^2 = 14.78 So LambertW(14.78)= 2 If x^x = 27, you can easily find x by x^x=3^3 So, x=3 But what about x^x= 20? Here LambertW function will help you find x. x^x=20 ln(x^x) = ln(20) x ln(x) = ln(20) [a can be expressed as e^ln(a) ] [e^ln(x)] ln(x) = ln(20) Now left side is in the form y e^y and you know that LambertW(y e^y)=y So, taking LambertW of both sides LambertW( ln(x) e^ln(x) )= LambertW( ln(20) ) ln(x) = LambertW( ln(20) ) e^ln(x) = e^LambertW( ln(20) ) x = e^LambertW( ln(20) ) You can use LambertW calculator on wolframalpha.com x = 2.8553085 Now verify on your mobile calculator the value of x^x=20 by replacing x with 2.8553085 2.8553085^2.8553085=19.9999999
Thanks for all your videos - I'm enjoying them. One pronunciation suggestion - the 'nat' in 'natural' rhymes with 'hatch' when spoken, so 'natch-urr-al' is how it goes!
Vx ln x = ln 2 2 Vx ln Vx = ln 2 Vx ln Vx = ln2 / 2 This is a transcendental function for which you need a calculator or a table. The reason it has a name (Lambert's W) is because it occurs often in differential equations, which are solutions to real world phenomena. The canonic form f(x) = x ln x has the property that its derivative f'(x) = ln x +1 (f(x))' = chain rule = x' ln x + x (ln x)' = 1 ln x + x 1/x = ln x + 1 This means that the change of the function is logarithmic: as x grows larger, f(x) grows larger but more slowly. This is the inverse effect of an exponential function: as x grows larger, f(x) grows larger and faster.
For a "True" Geniuses. An ironic title, as it fails numeracy. By the way, quotation marks surrounding an adjective (that is, not quoting anything) are used to imply the opposite. Thus, the quotation marks flag the "Geniuses" as not so at all.
@@chadx8269 They mean the english is terrible. It should be "For true geniuses" or "For a true genius". The article "a" is singular and doesn't mix with the plural "geniuses".
Well, this sure was way harder to solve than the first video I saw, the x!=x³-x With a bit of playing around Scalc tells me it would be exactly 1.701332, but I somehow doubt it that this is the exact answer. 😅
Not impressed. In fact, hugely underwhelmed. This is a kitabi problem, and if this is regarded as something Indian students need to learn to pass exams, that is very sad. You get to y ln(y) = 1 in 5 seconds, and after that you have a transcendental equation. You can now make up a function which is the solution to this equation and call it Lambert W or whatever else you want. No insight is gained in doing so. The point of the elementary functions (trigonometric, exp, log, hyperbolics, and their inverses) is that these are defined by solutions to simple integrals, and they capture the behaviour of almost all problems in the physical sciences and engineering.
Was in den Video dargestellt wird, finde ich Blödsinn, allerdings es stimmt rund 1,7 X . Aber was kommt genau raus. Kam sehr schnell drauf das X zwischen 1 und 2 liegen muss. Dann habe ich mit 1,5 gerechnet mit den Taschenrechner , dann mit 1,75 und nach etwa 10 bis 15 mal eingrenzen kam ich auf das genaue Ergebnis X = 1,7011.
@@fernandojackson7207 Schon möglich das die Lösung irrational ist , aber mein Rechner zeigt ein rationales Ergebnis. Nur bestimmt etwa nur 10 Stellen und da weis ich nicht ob dann doch nicht eine 0 kommt.
Lots of tricks... no theory... quiz& problems instead of theoretical insight... such approach is destroying the way Maths is taught in high school all around the world. TH-cam pays tribute to thus trend... There are plenty of videos about equations f(x)=0 where f is a non-algebraic continous real function, whose solution through the compactness property defines the inverse function of f. .
@obeyy0urmaster I was a math major and the rest of the video is just algebraic manipulation and identities and whatnot I could basically do as a middle-schooler. But he casually introduces this w function I've never seen before, with no explanation where tf it came from or what it is, or anything. I presume those of us liking this comment relate to this experience I had. We're here to learn something. Everyone watching this video could do HS algebra at one point and don't get much from the rest of the derivation. We *would*, however, like to hear more about this w function that swoops in out of nowhere to save the day here.
@@adogonasidecar1262 well then 19 people and both of you must be underachievers if you have math majors and didn’t research for curiosity any functions
Two suggestions to improve the channel:
1) Don't spend so much time on very basic algebra steps. It makes the videos slow and painful to watch.
2) Don't do so many problems with the W function, which imo aren't very interesting compared to problems with elegant, closed-form solutions.
I agree, especially with the second point, because it is neither taught in schools nor any PU colleges
@@kabeeralishaikh1439its not thaught but very useful for competative exams fr,but point should be agreed becoz college students dont need learn just some simple algebra
Thank you everyone for your support. I really appreciate it. Wish you all the best in your life and career. Have a great day! "Kindness costs nothing, but means everything. Don't be afraid to give it away"
Nice video.
It would be good to mention that when we get the equation
ln y e^(ln y)=½ln 2
then, as ½ln 2>0, it is the main branch W₀ of the Lambert W function that we want to use, and we just get one solution, ln y=W₀(½ln 2), leading to x=e^(2W₀(½ln 2)).
If we solve xeˣ=y for -1/e
Would be great to point to an explanation of the W function there- it's been a minute.
W-function is the anti-function of f(x)=x*exp(x), so that W(x)*exp(W(x)) = x
@@vottka1l I think it is just a local inverse.
x^Sqrt[x]=2 x=e^(2 W(ln(2)/2))=e^(2w(ln(sqrt[2]) It’s in my head.
LambertW Function:
If ( y ) e ^ y = a
LambertW(a)=LambertW[( y ) e ^ y]= y
If y=1
(1)e^1 = e = 2.71
So LambertW(2.71)= 1
If y=2
(2)e^2 = 14.78
So LambertW(14.78)= 2
If x^x = 27, you can easily find x by
x^x=3^3
So, x=3
But what about x^x= 20?
Here LambertW function will help you find x.
x^x=20
ln(x^x) = ln(20)
x ln(x) = ln(20)
[a can be expressed as e^ln(a) ]
[e^ln(x)] ln(x) = ln(20)
Now left side is in the form y e^y and you know that
LambertW(y e^y)=y
So, taking LambertW of both sides
LambertW( ln(x) e^ln(x) )= LambertW( ln(20) )
ln(x) = LambertW( ln(20) )
e^ln(x) = e^LambertW( ln(20) )
x = e^LambertW( ln(20) )
You can use LambertW calculator on wolframalpha.com
x = 2.8553085
Now verify on your mobile calculator the value of
x^x=20 by replacing x with 2.8553085
2.8553085^2.8553085=19.9999999
If you're saying x approx = 1.7, then it's okay to say the answer approx = e - 1 😃
Noo its not possible you maniac
Nice job well done sir. Keep it up sir.
This video could be summarized as 'Use the Lambert W function'. More explanation of this function is all that is needed.
Thanks for all your videos - I'm enjoying them. One pronunciation suggestion - the 'nat' in 'natural' rhymes with 'hatch' when spoken, so 'natch-urr-al' is how it goes!
I like your videos an I am learning from them. Thank you.
Vx ln x = ln 2
2 Vx ln Vx = ln 2
Vx ln Vx = ln2 / 2
This is a transcendental function for which you need a calculator or a table. The reason it has a name (Lambert's W) is because it occurs often in differential equations, which are solutions to real world phenomena.
The canonic form f(x) = x ln x has the property that its derivative f'(x) = ln x +1
(f(x))' = chain rule = x' ln x + x (ln x)' = 1 ln x + x 1/x = ln x + 1
This means that the change of the function is logarithmic: as x grows larger, f(x) grows larger but more slowly. This is the inverse effect of an exponential function: as x grows larger, f(x) grows larger and faster.
Let me guess, W again?
yep!
W function is tricky and confusing.IT amounts to saying that* Anything multiplied by e raised to power that thing is equal to that thing!
For a "True" Geniuses. An ironic title, as it fails numeracy. By the way, quotation marks surrounding an adjective (that is, not quoting anything) are used to imply the opposite. Thus, the quotation marks flag the "Geniuses" as not so at all.
x^(x^0.5) = 2
x^0.5(x^0.5) = 2^0.5
(x^0.5)^(x^0.5) = 2^0.5
ln((x^0.5)^(x^0.5)) = ln(2^0.5)
(x^0.5)ln(x^0.5) = (ln2)/2
ln(x^0.5) = W((ln2)/2)
x^0.5 = e^W((ln2)/2)
x = e^2W((ln2)/2)
(x ➖ 2x+2)
"For a true geniuses"?
Very steep slope so it veru sensitive more digits.
@@chadx8269 They mean the english is terrible. It should be "For true geniuses" or "For a true genius". The article "a" is singular and doesn't mix with the plural "geniuses".
Well, this sure was way harder to solve than the first video I saw, the x!=x³-x
With a bit of playing around Scalc tells me it would be exactly 1.701332, but I somehow doubt it that this is the exact answer. 😅
а можно вместо" функции ламберта "сразу ввести некую" функцию думперта D" - и ответ: x=D(2). смысл решения(его отсутствие)- тот же.
Not impressed. In fact, hugely underwhelmed. This is a kitabi problem, and if this is regarded as something Indian students need to learn to pass exams, that is very sad.
You get to y ln(y) = 1 in 5 seconds, and after that you have a transcendental equation. You can now make up a function which is the solution to this equation and call it Lambert W or whatever else you want. No insight is gained in doing so. The point of the elementary functions (trigonometric, exp, log, hyperbolics, and their inverses) is that these are defined by solutions to simple integrals, and they capture the behaviour of almost all problems in the physical sciences and engineering.
❤
😀
W doesn't make any sens
check: my calculator gives 1.997... for 1.7^(sqrt 1.7)
Was in den Video dargestellt wird, finde ich Blödsinn, allerdings es stimmt rund 1,7 X . Aber was kommt genau raus. Kam sehr schnell drauf das X zwischen 1 und 2 liegen muss. Dann habe ich mit 1,5 gerechnet mit den Taschenrechner , dann mit 1,75 und nach etwa 10 bis 15 mal eingrenzen kam ich auf das genaue Ergebnis X = 1,7011.
Pretty sure the actual solution is Irrational, so you won't get an exact ( Rational) solution.
@@fernandojackson7207 Schon möglich das die Lösung irrational ist , aber mein Rechner zeigt ein rationales Ergebnis. Nur bestimmt etwa nur 10 Stellen und da weis ich nicht ob dann doch nicht eine 0 kommt.
Without Lambert x = 1.70135
1,7013318
Kind of sloppy writing where you really can’t see if an expression is multiplication or in the exponent, otherwise I like it
Lots of tricks... no theory... quiz& problems instead of theoretical insight... such approach is destroying the way Maths is taught in high school all around the world.
TH-cam pays tribute to thus trend...
There are plenty of videos about equations f(x)=0 where f is a non-algebraic continous real function, whose solution through the compactness property defines the inverse function of f.
.
√x = u => x = u²
u²ᵘ = 2
2ulnu = ln2
ulnu = (1/2)ln2
lnu = W[(1/2)ln2]
u = e^W[(1/2)ln2]
x = e^2W[(1/2)ln2]
Ridiculous. Spends painful amounts of time on stuff one should know at 13 y.o. and not an explanation on the W function which is College stuff.
13 y/o learn about exponents, barely even functions
Wtf are you talking about ?
@obeyy0urmaster I was a math major and the rest of the video is just algebraic manipulation and identities and whatnot I could basically do as a middle-schooler. But he casually introduces this w function I've never seen before, with no explanation where tf it came from or what it is, or anything. I presume those of us liking this comment relate to this experience I had.
We're here to learn something. Everyone watching this video could do HS algebra at one point and don't get much from the rest of the derivation. We *would*, however, like to hear more about this w function that swoops in out of nowhere to save the day here.
@@ClarkPotter Exactly
@@adogonasidecar1262 well then 19 people and both of you must be underachievers if you have math majors and didn’t research for curiosity any functions
Can someone help me with the value of w? please 🥺
Wdym
Straight without substitution...
x^√x = 2
ln(x^√x) = ln(2)
√x ln(x) = ln(2)
[ e^ln(√x )] ln(x) = ln(2)
[ e^ln(√x )] ln[ (√x)^2 ] = ln(2)
[ e^ln(√x )] 2 ln(√x) = ln(2)
[ e^ln(√x )] ln(√x) = ln(2)/2
Now left side is in the form y e^y , y being ln(√x)
W( [ e^ln(√x )] ln(√x) ) = W(ln(2)/2)
ln(√x) = W(ln(2)/2)
e^ln(√x) = e^[W(ln(2)/2)]
√x = e^[W(ln(2)/2)]
x = e^{2[W(ln(2)/2)]}
x = e^[2(0.26571)]
x = 1.7013
I THOUGHT YOU HAD THE FIRST DIGITS OF PI VIEWS
Stop yelling your post in all caps.
Don't listen to naysayers, keep making great videos. Thanks
*Thumbs-down* on this whole video.
Нихуя непонятно что говорит. Это Россия а не Америка. Покажи это в Америке!!!
Tovarisch, you need to go to rutube!
W Function again
1.5596105 is the answer
1.7013319979^√1.7013319
979=2