ALL solutions to x^2=2^x

We will find all the solutions to the famous exponential equation x^2=2^x. It is easy to see x=2 and x=4 are the first two solutions, but we will have to use the Lambert W function in order to get the 3rd solution. Lecture on Lambert W function: • Lambert W Function (do...
This is my "equation of the year" in 2019.
To see others, please check out here 👉bit.ly/equationoftheyear
Support the channel: / blackpenredpen

[@blackpenredpen]

Solve this by the super square root kzbin.info/www/bejne/fJC7dGyVjbiqbc0

[@krucyferariusz1813]

what happend if x=0, becouse I can't see 0 in any formula (I mean xe(-∞;0)u(0,∞)

[@Juan-yj2nn]

@@krucyferariusz1813 Shut up

[@davidbroadfoot1864]

@@Juan-yj2nn LOL

[@hybmnzz2658]

@@krucyferariusz1813 dude you can easily see x=0 doesn't solve original equation

[@DatBoi_TheGudBIAS]

0:09 MaLeFiC lAuGh and really? a complex power? brah

[@jzanimates2352]

Best variables for math: X Y Fish

[@remlatzargonix1329]

JZ Animates .....what about star, square, lightning bolt? ........I have seen that before.

[@Pete-Logos]

∞ ∑🙂ⁿ = 1+🙂+🙂²+🙂³+🙂⁴+... ⁿ⁼⁰ I present emoji variable in infinite series.

[@hamsterdam1942]

@@Pete-Logos i see only squares lol

[@That_One_Guy...]

@@Pete-Logos how did you type the sigma symbol along with the index and number ?

[@benjaminrosenberg]

@@That_One_Guy... you can copy and paste the sigma (Σ) from online sources or use a Greek keyboard; for the powers (¹²³⁴⁵⁶⁷⁸⁹⁰) you can get those from copying and pasting as well. For the indices (n=0 at the bottom of the Σ), you can get them to look like they're right below the sigma by using superscripts (aka exponents) on the line beneath, so that they look like they're actually under the sigma. See this example: Σ ⁿ

[@poprockssuck87]

Give a man a fish, and he'll eat for a day. Teach a man WITH fish, and he'll be able to solve for the solutions of x^2=2^x.

[@blackpenredpen]

Aaron L Hahahahahaha!!!

[@RedRad1990]

Give PIERRE DE FERMAT a fish, and he'll FIND A BEAUTIFUL SOLUTION THIS FISH IS NOT ABLE TO CONTAIN

[@dlevi67]

Not only that, but he will be able to continue calculating solutions for all of his lifetime, since there are infinitely many. Not sure what he will eat, but that problem takes care of itself eventually.

[@rorycannon7295]

lol

[@darkseid856]

I am confusion.

[@taflo1981]

"This is not a fish yet, but it's almost a fish." Sounds like evolution at work.🤣

[@blackpenredpen]

Hahahahahaha definitely!!

[@blackpenredpen]

Evolution of equations

[@Icy-ll5ie]

@@blackpenredpen what do you think about "one minus zero-point-nine*repeating"? ( 1-0.(9)=? ) I think matematicians should invent new numbers like 0.(0)1 with the rule that they can not be converted like this: A.b = ab/10

[@Macion-sm2ui]

@@Icy-ll5ie Lol, but 0.(9) is equal to 1. It's not a some kind of magic, neverending irrational number, but it's 1. It's simple proof: 0.(9)*10 = 9.(9) 0.(9)*9=0.(9)*10-0.(9)=9.(9)-0.(9)=9 So, when we divide this by 9, we get 1.

[@hassanakhtar7874]

@@Icy-ll5ie 0.9 repeating is 1 as the guy above me proved. In fact the whole repeating decimal notation is a result of rational numbers existing but our desire to describe it as a decimal (eg: 1/3 =0.333333....). Also in math something that tends or converges to a value is said to be equal to that value. This is a very important thing in analysis so it wont really change or be expanded.

[@angelmendez-rivera351]

A more careful and rigorous way of handling the equation z^2 = 2^z is by noticing that if z is not an integer, then 2^z is inevitably multivalued. Namely, 2^z := exp[ln(2)·z + 2nπi·z] for any integer n. This implies z^2 = exp([ln(2) + 2nπi]·z). Now, a few cases must be considered. The first case is that |Arg(z)| < π/2, and the second case is that π/2 < |Arg(z)| < π. It is notable that if |Arg(z)| < π/2, then |Arg(z^2)| < π, and if π/2 < |Arg(z)| < π, then also |Arg(z^2)| < π. In the first case, z^2 = exp([ln(2) + 2nπi]·z) implies z = exp([ln(2)/2 + nπi]·z), and z = exp([ln(2)/2 + nπi]·z) implies -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = -[ln(2)/2 + nπi]. Therefore, -[ln(2)/2 + nπi]·z = W(m, -[ln(2)/2 + nπi]), equivalent to z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]). Notice that if n = m = 0, then this simplifies to z = 2. In the second case, z^2 = exp([ln(2) + 2nπi]·z) implies -z = exp([ln(2)/2 + nπi]·z), equivalent to -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = ln(2)/2 + nπi. Therefore, -[ln(2)/2 + nπi]·z = W(m, ln(2)/2 + nπi), hence z = -W(m, ln(2)/2 + nπi)/[ln(2)/2 + nπi]. With this, the remaining cases are Arg(z) = π, Arg(z) = π/2, and Arg(z) = -π/2. In the first of these three, z = -r, with r = |r|, so -r = exp(-[ln(2)/2 + nπi]·r), thus [ln(2)/2 + nπi]·r·exp([ln(2)/2 + nπi]·r) = -[ln(2)/2 + nπi], and [ln(2)/2 + nπi]·r = W(m, -[ln(2)/2 + nπi]). Therefore, z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. If Arg(z) = -π/2, then z = -ri, with r = |r|. Hence -ri = exp(-[ln(2)/2 + nπi]·ri), and [ln(2)/2 + nπi]·ri·exp([ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi]. Therefore, [ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), and z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. Finally, if Arg(z) = π/2, then z = ri, with r = |r|, so ri = exp([ln(2)/2 + nπi]·ri), and -[ln(2)/2 + nπi]·ri·exp(-[ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi], hence -[ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), so z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. In summary, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi], in both cases for arbitrary integers n and m. This gives the complete family of complex solutions with no extraneous solutions. The solution families can be compactified. Notice that exp[W(t)] = t/W(t), so exp[-W(t)] = W(t)/t. Therefore, -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = +exp[-W(m, -[ln(2)/2 + nπi])], and -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = -exp[-W(m, +[ln(2)/2 + nπi])]. As such, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = +exp[-W(m, -[ln(2)/2 + nπi])], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -exp[-W(m, +[ln(2)/2 + nπi])]. Again, this solution is complete and with no extraneous solutions.

[@iloveorganicchem6921]

Wow man

[@maxprofane]

You simply answered all my questions about this solution and much more. Thank you very much.

[@M1m1s]

Can't wait for the day where I can understand half the crap that you wrote- if I remember, I might come back here in a few years.

[@xxocry]

​@@M1m1sNah bro you coming back today

[@kalis4132]

bro took it personally

[@n4p3r0]

8:47 We know that we know that

[@tanishqpandey9564]

Just a composite function f(f(x)) such that f(x)=we know that

[@chrissmith1152]

Eminem in disguise

[@davidchuramayta2067]

J

[@recalprasetyo1723]

Make it by 2x

[@motazfawzi2504]

I didn't see that

[@blackpenredpen]

Lambert W function intro: kzbin.info/www/bejne/qYjKf3aolp5mepo

[@michaelschneider4093]

I can't wait to watch this and learn in a class setting. I recently took my differentiation exam (calculus 1) and I got 96.6% woo

[@subhrajyotidutta4725]

How do we show the following is a good approximation: xe^x~(1-7/9)^(-x)-(1+7/9)^x

[@blackpenredpen]

That’s very well done!! I am happy to hear that!

[@roderickwhitehead]

Thanks!

[@fernandobueno8720]

Okay, you expressed the result in terms of the Lambert function. But we wanted the missing root, 0.77, expressed in rationals and complex roots as well.

[@phyricquinn2457]

Thank you for producing all of this unique content! I really love these videos!

[@daniel-fich]

12:57 Look at the way he looks at his math with passion in his eyes :-D

[@PlutoTheSecond]

BPRP: It's probably irrational... Wolfram: It's transcendental.

[@blackpenredpen]

Hahahah yea I know!! I saw that when I was editing the video and I was like hmmmm should I edit that part out. Haha

[@AndrewBlechinger]

I mean...you're not wrong.

[@jamboree1953]

@@blackpenredpen All real transcendental numbers are irrational. So you are still right.

[@Phantlos]

@@jamboree1953 wrong, e*i its still transcendental but not irrational

[@awelotta]

@@Phantlos if it's not irrational, wouldn't it be representable with division of two intervals and this not transcendental?

[@nibblesdotbas]

7:48 How does one know in advance what indices (in this case 0 and -1) to use for the Lambert W function in order to get real answers?

[@cheesefrogsnail]

I had in mind the exact same question...

[@peapopea]

trial and error my friend

[@vma011]

According to Wikipedia, seems like the indices 0 and -1 are standard when working with real answers with the Lambert W

[@MathswithMuneer]

Respect from one math teacher to another math teacher

[@blackpenredpen]

Thank you!!!

[@MindYourFunds]

The third root from the graph is obviously a negative x...how come it became imaginary... Perhaps the two math teacher will explain it to me

[@050138]

@@MindYourFunds you didn't pay attention to the video.... The third root is a negative real number -0.7666.... the fourth fifth and so on - Solutions are imaginary, and they aren't intuitive

[@user-ss7ud9ye8l]

the fish is a paid actor

[@blackpenredpen]

It is!

[@roidaniel10]

אח יקר

[@thereaction18]

Is he in the Stream Actors Guild?

[@thereaction18]

I guess I'll have to look it up on the Ichthyological Movie Data Base.

[@adirmazhir9159]

אח יקר

[@GreenArsenal]

Who else doesn’t understand anything but still watches

[@j1kuhan21]

Me, me

[@francogonzales4943]

Volcanic i dont speak english but i see it aniway

[@Assassin_Bear]

Bruh I'm still learning about line gradients

[@uzofortune7193]

Meee

[@datcr4zyazn851]

Me

[@dihydrogenmonoxid1337]

I should read a book for english but pen(black+red) math videos are sooo much nicer💯

[@createyourownfuture3840]

That's a nice way to write his name...

[@nou3227]

My teacher explaing math: so x is equal to y Blackpenredpen explaning math: *_F I S H_*

[@JyothiSwaroopM]

I've been trying out this problem since I was in 10th grade of school. Now I'm in the third year of engineering. Thanks to you I got to know about the new function "Lambert W function"

[@blackpenredpen]

Glad to hear!! Thank you for the comment too

[@HasXXXInCrocs]

Prove the Lambert W function! I've never heard of it (haven't taken analysis yet, not sure if thats covered there) and i think its awesome!

[@blackpenredpen]

Matt Heitmann I have an introductory video in the description already.

[@paulfaigl8329]

Cool, very cool. I never knew that Herr Lambert was working on photometry, colours etc. and that his W function has so many other uses! Thank you!

[@spelunkerd]

Great video. I completely forgot about Lambert, this was an excellent case, where I could have used it if I was creative enough to think of the way out.

[@blackpenredpen]

Thank you!!

[@InvaderMixo]

I love learning, especially math. I love your enthusiasm. It's so uplifting to be in a math moment with you!

[@user-bf5bp8bj4v]

Can you make a video explaining LambertW indexes?

[@hermessantos181]

Yeah, plsss

[@peter-hm9iu]

Yes

[@DavidRodriguezBarrios]

Yes no Also asks for more

[@briandohler8025]

Yeah I missed this last step.

[@psilvakimo]

True. He never describes the LambertW index. I had to use a numerical algorithm to solve it. It converges quite slowly.

[@SpaceSwimmer69]

Thanks a lot! I Just finished school, but never heard of the LamberW equasion. Everything else I understood perfectly. I liked it so much, I found solutions for the x>0 myself.

[@byronvega8298]

From this point on I'll start writing functions that depend on fish rather than x

[@univuniveral9713]

Leave endangered species alone, sir. You have done enough harm to our oceans with plastic bags.

[@atheoristspointofview7059]

8:45 "we know that-we know that"(pro jump cuts)

[@blackpenredpen]

Definitely awesome!

[@24kGoldenRocket]

@@blackpenredpen No. It is not awesome. To make assumptions about what your student "already" knows is a poor teaching technique. Yes I can do the algebra in my head also. But you might soon learn that most people cannot. How fast you learn that depends upon your IQ.

[@z_.v]

@@24kGoldenRocket true but I think the line must be drawn where he stops explaining every detail and assumes the viewers know what he is talking about, because this is a (mostly) calc-based channel. The instance mentioned in which he states that "we know that" is quite basic, though. I doubt anyone watching this wouldn't understand how to turn a negative (x < 0) value into a positive value by multiplying by a negative value.

[@anandsuralkar2947]

@@24kGoldenRocket this channel is not for noobs he wont spend his time explaining u what log and exponentials ate in every video he have to pre assume that u know basic maths if not go and learn some basics first

[@24kGoldenRocket]

o@@anandsuralkar2947 LOL. I was a University Math Instructor.

[@tomvanmoer8202]

"The fish" lmao, cracks me up every time.

[@ggonzalezru]

I love your channel, man! thanks for your videos. Cheers from Chile

[@laughinking5101]

I really enjoyed it! Your enthusiasm is contagious and spurred me on too!

[@lumi2030]

I did expect it to have infinitely many solutions after I remember W(x) was a multi-valued function, but it still surprised me when I found out it was actually true.

[@eric_welch]

I have to say I thoroughly enjoy YT suggesting older BPRP vids so I can watch the epic beard/goatee evolution :)

[@shinyless]

I love those videos. Best maths lessons I've watched in a while.

[@treblaalbert4391]

OMG man, your videos are great. Please keep doing these kind of videos, they are so interesting. Calculus is much more interesting when You explain it. Love you

[@Bayerwaldler]

One can also use the identity 2^x = e^((ln2)*x) and take square roots on both sides of the equation. I think it is a little more simple to work out then. I get the solutions x=-(2/ln(2))*W_0(-ln(2)/2) = 2, x= -(2/ln(2))*W_1(-ln(2)/2) = 4 and -(2/ln(2))*W_0(ln(2)/2) = -0.766664... The terms look a little different but yield the same values as in the video.

[@vincentnguyen2558]

Thank you blackpenredpen for answering my question

[@blackpenredpen]

You’re welcome. In fact many ppl have asked this question in the past. It’s a very popular question

[@DANGJOS]

I feel so much better. I spent so much time thinking about this trying to solve it. Who knew it would be something so crazy?!!

[@erickherrerapena8981]

Buen video, ya dos meses viendo todos tua videos. Buen canal.

[@serein_11]

This question was in my 11th grade math book. They seriously expect 11th grade students to solve this omg.

[@samueldeandrade8535]

Probably you are mistaken. Probably they asked just for positive values.

[@josephsmith786]

Alternative solution: x = 2^(x/2), so rearranging gives -ln(sqrt(2)) x exp(- ln(sqrt(2)) x) = -ln(sqrt(2)). Taking W of both sides gives (by rearranging) x = W(-ln(sqrt(2))) / -ln(sqrt(2)). No assumptions on signs needed here.

[@adiaphoros6842]

You already made an implicit assumption on signs. x² = y x = ±sqrt(y) So your first 2 lines should be: x² = 2ˣ x = ±2^(x/2)

[@alberteinstein3612]

How does he manage to always teach me something new every single vid?

[@dontawanpitak]

Incredible!!! Love this, thanks for your videos!

[@tch3n93]

Problem: contain exponents blackpenredpen: it's Lambert-W time

[@ankitdubey9310]

Applies Lambert to fish ( e)^ (fish) Le fish: Allow me to introduce myself

[@thedoublehelix5661]

why is this comment pure gold

[@akasin8202]

No, I won't allow

[@Eternap]

lol

[@cbarnes2160]

Lambert W comes up in calculations of current-voltage relationships in solar cells and diodes. Blew my mind the first time I saw it and fun that it comes up in many places.

[@bobbrown5726]

The fish is So Cool !!! Great illustration!

[@tarsala1995]

In 100th episode, please show your left biceps. I can only imagine how shredded it is after this hard holding-ball workout

[@pgk2022]

This is anxiety-inducing as now I know some of the things I'll have to learn when I'm older.

[@smoorej]

UNDEADWARLORD 17 for me it is anxiety-inducing because it shows me things I have forgotten now that I am older

[@angelmendez-rivera351]

It's not so bad. When it's your turn to learn them, you'll be more prepared than you were when exposed to this video for the first time.

[@neohavixx7164]

Got the hoodie late yesterday afternoon. I love it.

[@cav94rojo]

Nunca imaginé que habriá más soluciones. Exploté❗

[@tajpa100]

dear teacher, could you give a more detailed lesson on the Lambert function?

[@fernandofrio5863]

8:46 Yeah we really knew that

[@akbarrahmatullah6701]

You're very passionate about maths! Good videos.

[@tensor131]

tough topic .. good explanation. I have not met the Lambert W function .. clearly it has some fascinating properties .. to be continued!! Thank you again.

[@TheDanksNewGroove]

A more elegant way to write the solution would be -1/ssqrt(√2), where ssqrt() is the super square root from tetration.

[@blackpenredpen]

I just recorded a video on this today! Thank you so much for the idea!

[@TheDanksNewGroove]

blackpenredpen Wow! Looking forward to it!

[@angelmendez-rivera351]

It is a more *concise* way of writing it, yes, but the ssqrt operator is strictly defined in terms of W(x) in the first place.

[@waltuhputyourdaway6561]

Everyone gangsta until lambert w function

[@Doktoren]

Thank you for making me learn a new useful function I didn't know about, the Lambert function

[@GregBakker]

Awesome video, wonderful to see a clear explanation of this.

[@dublingranam3489]

maybe this could also be for you kzbin.info/www/bejne/aX2ze5h4at-Yetk

[@amansharma2957]

Bro You are Godfather of Mathematics Very Impressed by your knowledge..KEEP POSTING SUCH GREAT 😊

[@blackpenredpen]

Thank you!!

[@joaovitorcampos99]

Lim Cot^2 (x) - 1/x^2 x->0 :)

[@bhriguvats4889]

Thank you very much sir for this explanation when I saw question I thought about graph making graph of both equation and saw where it is common point

[@Viewpoint314]

Thanks for the video. I was just thinking about this problem for a few days and then this video popped up. That always seems to happen.

[@MaxxTosh]

Well now you have to find all the values of x for a given n such that x^n=n^x

[@blackpenredpen]

Hahahahaha. Replace all the 2 by n then we are done!

[@angelmendez-rivera351]

x^n = n^x -> x^n = e[log(n)·x] -> xω(n)^m = e^[log(n)·x/n], where ω(n) = e^[(2π/n)i], and m = 0, 1, ..., n - 1, meaning there are n cases to consider, one for each value of m. Regardless, xω(n)^m = e^[log(n)·x/n] -> xe^[-log(n)·x/n] = ω(n)^m, since 1/ω(n)^m = ω(n)^(n - 1 - m), which we can reindex to be ω(n)^m, since m is a variable. Then [-log(n)·x/n]e^[-log(n)·x/n] = -log(n)·ω(n)^m/n -> -log(n)·x/n = W[-log(n)·ω(n)^m/n] -> x = -n·W[-log(n)·ω(n)^m]/log(n).

[@ejb7969]

@@angelmendez-rivera351 That is intuitively obvious. :{ ) (Respect!)

[@angelmendez-rivera351]

ejbejbphone Thank you

[@nicolasgoubin]

The mic is back

[@paramrajsingh1539]

I was working on this problem but I didn't knew this vid existed and after few minutes it ramlndomly popped up and seems like my FBI did a good job 😊

[@braznartn5176]

Sir... You are truly VALUABLE to our society.

[@5000jaap]

Could you explain this index of W? Just for curiosity

[@anas8183]

It s a complex number

[@5000jaap]

@@anas8183 but how it works

[@anas8183]

@@5000jaap i d ont know my current level in math is not very good i am just 16 yrs old

[@cuentafake140]

I think it is related to Euler's formula: e^(z) = cosx + isinx Where z is a complex number, if that's the case then there are infinite imaginary solutions.

[@TheSavageTeddy]

all the possible variables you could have chosen and you chose FISH!?!?! respected.

[@blackpenredpen]

Hahahah thanks!

[@InfinityGamingHere]

omg teddy what u doing here??!!!

[@TheSavageTeddy]

InfinityGaming omg infinity what r u doing here !?!?!?!!?!?

[@InfinityGamingHere]

@@TheSavageTeddy idk!!!

[@TheSavageTeddy]

InfinityGaming I’m watching youtube!!!!!

[@deans7538]

I have wondered bout this before. Love MATH. Good to know these different things.

[@lumpi472]

Great demonstration 👌🏻

[@alexming9179]

Hummm for a guy who scored 6 out of 120 in mathematics... good job for recommending KZbin

[@chinmayjoshi3592]

8:47 was absolutely unexpected

[@DanBurgaud]

i've watched this many times... each time, new experience.

[@alanwong197]

Very interesting question thank you so much!

[@seifeldidi41]

Can I approximate the negative solution using maclaurin expansion of 2^x ?

[@malekm.naouach7134]

LAMBERT OF 🐠 EAT 🐠 IS 🐠, SO COOL!

[@blackpenredpen]

Wow. Never thought about it like that.

[@danerman73]

Fish is my favorite variable now. Great video.

[@fernandobueno8720]

Está bien, expresaste el resultado en términos de la función de Lambert. Pero nosotros queríamos la raíz que falta, el 0,77, expresado en racionales y las raíces complejas también.

[@gergodenes6360]

Isn't it just one case of your y^x=x^y vid?

[@hermessantos181]

Well, it is. But does he have a video talking about the y^x=x^y? Isn't x^x=y^y?

[@zohichnazirro8640]

@@hermessantos181 kzbin.info/www/bejne/gmG7immiobitr9E

[@hermessantos181]

@@zohichnazirro8640 thank you

[@gergodenes6360]

@@hermessantos181 He also has this gem: kzbin.info/www/bejne/hnqUf5h9qa-ibdU Although, it's without the Lambert-W function, so the complex cases are out of the question.

[@SM321_]

What is the negative real value?

[@blackpenredpen]

-0.76666

[@SM321_]

@@blackpenredpen wow that was the fastest answer I ever got on KZbin😂 thank you

[@blackpenredpen]

Lol. You are welcome

[@aaronleperspicace1704]

Or approximately -23/30. The actual value is irrational but -23/30 is rational.

[@DinoDiniProductions]

Lambert W! Amazing that 1) The function is useful in all kinds of modelling of actual real physical behaviours, e.g. how long will it take for a bucket of water with a hole in it to drain to a certain depth. 2) The function is not implemented in floating point libraries or FPUs.

[@amish4988]

i never thought there were people on the internet who enjoy math the way i do! man.... i wish you were my friend so we can talk about math forever!

[@waterdragonlucas8263]

i tried x^-1 ln(lxl)=1/2 ln(2 on the desmos graphing calculator and the three answers i got were 2, 4, and -0.76666666666666...

[@inx1819]

Is -0.76(6) correct? because i also got that from my calculator and i wanna know if that's ok

[@XgamersXdimensions]

Can you do a video on what the Lambert W function actually is and how it was derived?

[@blackpenredpen]

I have a video in description

[@Sammy-qt9it]

It's weird that, during the first 2 years of a maths degree, I have not encountered the lambert w function, but on KZbin, it's everywhere

[@Julian-ot8cs]

Can you make a video showing the difference between lambert Ws with different bases!

[@zonex001]

I want to see the w function

[@JJ_TheGreat]

Ah, man... This is very unsatisfying: How do we know what the Lambert W function is - without having to use Wolfram Alpha? And what are these "indexes", what do they mean - can you please do a video about them? Would we need to use Wolfram Alpha to calculate them? Because changing them obviously gets you different answers to the question/problem.

[@Gulyus]

The equation is apparently covered in a video in the description. The indexes are different solutions to it in either real or conplex space.

[@JJ_TheGreat]

@@Gulyus Thank you. However, even after viewing that video, I still have questions. For example, how do we know that when using f(x) to calculate the inverse: "f inverse" or "f -1(x)" that f(x) should equal "xe^x" based on the equation given in the video - which is "x^x = 2"? In addition, the video did not explain anything about "indexes". So I would not have even known that they exist otherwise, without them being mentioned in THIS video.

[@literallyme.realmp4]

"The Lambert W function will give you the fish back" is something I never thought I'd hear

[@saarike]

Simply wau!!!! Thank you for interesting vid.

[@koryukengamer5693]

0:57 minecraft villager "hm"

[@cros108]

"most likely it's irrational" a proof on the rationality of this number would be pretty cool tbh

[@spasibamt]

13:00 and lnsqrt2=O x=-e^-W(O) End -1= H x=He^HW(O)

[@Ssilki_V_Profile]

Is it possible to express the negative solution of this particular equation in the form of a radical? Or is it a transcendental number expressed exclusively through transcendental functions?

[@pedrobizarro2164]

I'm still waiting for the proof of the cubic equation I think you can do it in 15 minutes or less. I did it in 20 minutes (because I'm not good at explaining Lol)

[@fluffymassacre2918]

@@aaryanbhatia4939 you aint

[@pedrobizarro2164]

Aaryan Bhatia What I did was give other values ​​to the variables x. if you want to know the extended formula you just have to change those final values ​​at the roots of x

[@angelmendez-rivera351]

Aaryan Bhatia The answer is you *cannot.* Not in general, anyway. Only for certain values of the coefficients is the result simplifiable. Whenever it is not, this is known as casus irreducibilis. If you have ever wondered why an angle of 1° is not constructible, the answer is because 3° is constructible, and third angles of nontrivial constructible angles are not constructible usually, because the third-angle formula in general is unsolvable, as it requires solving a cubic equation that results in casus irreducibilis. This is why the cubic formula is not taught as opposed to the quadratic formula being taught. It is usually not useful, because casus irreducibilis is more common than not.

[@angelmendez-rivera351]

Aaryan Bhatia Well, no, it is not like the quadratic formula. The quadratic formula is useful and practical. The quadratic formula suffers from no casus irreducibilis. The quadratic discriminant also does not suffer from case deficiency. The cubic formula is, in every sense, analogous to the quadratic formula, but strictly inferior.

[@mennoltvanalten7260]

@Aaryan Bhatia I'm going to guess the answer is 'try and fail'

[@kewrie1630]

You can avoid the calculations, and just do it with visualisation, 2^x is an exponential function, so its like e^x approximately, we can roughly draw the graph in mind, and if we produce the 2^x furthermore, It will intersect in 2 points ! , You can try it in desmos

[@mirzaghalibkiyaad325]

Your video is very helpful for me thank you brother..

[@tcb3901]

Extremely wholesome content

[@dingueriefinanciere]

This is just the problem x^y = y^x you already solved right

[@blackpenredpen]

Different style. I used a parametric approach in that video and lambert W function in this one.

[@yangli3144]

Are the imaginary solutions really legit? The analysis break cases into x>0 and x

[@angelmendez-rivera351]

A more careful and rigorous way of handling the equation z^2 = 2^z is by noticing that if z is not an integer, then 2^z is inevitably multivalued. Namely, 2^z := exp[ln(2)·z + 2nπi·z] for any integer n. This implies z^2 = exp([ln(2) + 2nπi]·z). Now, a few cases must be considered. The first case is that |Arg(z)| < π/2, and the second case is that π/2 < |Arg(z)| < π. It is notable that if |Arg(z)| < π/2, then |Arg(z^2)| < π, and if π/2 < |Arg(z)| < π, then also |Arg(z^2)| < π. In the first case, z^2 = exp([ln(2) + 2nπi]·z) implies z = exp([ln(2)/2 + nπi]·z), and z = exp([ln(2)/2 + nπi]·z) implies -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = -[ln(2)/2 + nπi]. Therefore, -[ln(2)/2 + nπi]·z = W(m, -[ln(2)/2 + nπi]), equivalent to z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]). Notice that if n = m = 0, then this simplifies to z = 2. In the second case, z^2 = exp([ln(2) + 2nπi]·z) implies -z = exp([ln(2)/2 + nπi]·z), equivalent to -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = ln(2)/2 + nπi. Therefore, -[ln(2)/2 + nπi]·z = W(m, ln(2)/2 + nπi), hence z = -W(m, ln(2)/2 + nπi)/[ln(2)/2 + nπi]. With this, the remaining cases are Arg(z) = π, Arg(z) = π/2, and Arg(z) = -π/2. In the first of these three, z = -r, with r = |r|, so -r = exp(-[ln(2)/2 + nπi]·r), thus [ln(2)/2 + nπi]·r·exp([ln(2)/2 + nπi]·r) = -[ln(2)/2 + nπi], and [ln(2)/2 + nπi]·r = W(m, -[ln(2)/2 + nπi]). Therefore, z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. If Arg(z) = -π/2, then z = -ri, with r = |r|. Hence -ri = exp(-[ln(2)/2 + nπi]·ri), and [ln(2)/2 + nπi]·ri·exp([ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi]. Therefore, [ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), and z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. Finally, if Arg(z) = π/2, then z = ri, with r = |r|, so ri = exp([ln(2)/2 + nπi]·ri), and -[ln(2)/2 + nπi]·ri·exp(-[ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi], hence -[ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), so z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. In summary, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi], in both cases for arbitrary integers n and m. This gives the complete family of complex solutions with no extraneous solutions. The solution families can be compactified. Notice that exp[W(t)] = t/W(t), so exp[-W(t)] = W(t)/t. Therefore, -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = +exp[-W(m, -[ln(2)/2 + nπi])], and -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = -exp[-W(m, +[ln(2)/2 + nπi])]. As such, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = +exp[-W(m, -[ln(2)/2 + nπi])], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -exp[-W(m, +[ln(2)/2 + nπi])]. Again, this solution is complete and with no extraneous solutions.

[@saimonhullu]

Love this energy

[@amologusmogusmogumogu2535]

Where can I learn about the product log function?