Fear No Equation

[@BriTheMathGuy]

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[@parthgangwar981]

x=x+1 be like

[@allozovsky]

A typo at 6:38: should be x = 2.4781, not -2.7231

[@BriTheMathGuy]

By the way, x=3 is a solution too :)

[@nzqarc]

🤯

[@ThunderxBoy]

Lol I thought about it the m9ment I've seen this equation

[@Ninja20704]

technically 3 is also in the solution you got. The 2.478... answer is if you use the principal branch of the lambert function i.e. W_0[-(ln3)/3]. If you use the -1 branch you will get the 3 that we expected.

[@fredartson]

x^3= 3^x= k x^k^1/x=3^k^1/3 Hmmm I wonder how 3 could be a solution 😂😂

[@fredartson]

😅 but one should still know his calculus

[@renesperb]

The function x^3*3^(-x) has a maximum 1.01379 at x = 2.73072 .This shows that there are two solutions , x =3 which is obvious , but also a second solution for x < 2.7 , (2.478 ).

[@amornthepmeekangwal9421]

How we know there are 2 solutions when we find 1 solution from Lambert W function and how to find other solutions.

[@renesperb]

@@amornthepmeekangwal9421 For the second solution you have to use the function W[-1,x]. The function W [x] gives the solution x = 2.478 and W[-1 x] gives x = 3 .

[@dageustice]

​@@amornthepmeekangwal9421 If the input is between -1/e and 0, we need to use both the real Lambert W functions, as mentioned in the video. One gives 3, the other gives the other answer

[@namangoyal1297]

Can you make a video dedicated to non elementary functions? This topic has always fascinated me and I'd love to learn more about it from no one other than you.

[@KeithKessler]

You should check out the hypergeometric function and all its special cases.

[@fantasypvp]

I see this as an absolute W

[@zihaoooi787]

nice pun

[@akshayannn9627]

nice

[@Lavasparked]

truly is an |W|

[@-Neko_77-]

​@@Lavasparked😂 beautiful

[@someone._.5333]

Truly a W moment for my mathematical knowledge

[@benetogamerOFC]

Literally

[@yoylecake313]

a W(moment for your mathematical knowledge)

[@RadekBuczkowski-h2y]

This equation has two real solutions: x=3 and x≈2.478 Here is how to derive them properly. x^3 = 3^x | ^(1/3) note: sign is unchanged! x = 3^(x/3) | 3^(x/3) = (e^ln(3))^(x/3) = e^(x*ln(3)/3) x = e^(x*ln(3)/3) x * e^(-x*ln(3)/3) = 1 | * -ln(3)/3 -x*ln(3)/3 * e^(-x*ln(3)/3) = -ln(3)/3 | Lambert W on both sides -x*ln(3)/3 = W(-ln(3)/3) If the entire parameter of Lambert W is greater or equal zero, it is has only one real solution (branch 0). If less than zero and greater or equal -1/e, it has two real solutions (branch 0 and -1). If less than -1/e, no real solutions. Lambert W parameter: -1/e ≈ -0.368 < -ln(3)/3 ≈ -0.366 < 0 => Lambert W has two real solutions! Integer solution: -x*ln(3)/3 = W(-ln(3)/3) | 1/3 = e^ln(1/3) = e^(-ln(3)) -x*ln(3)/3 = W(-ln(3) * e^(-ln(3))) -x*ln(3)/3 = -ln(3) -x/3 = -1 x = 3 Lambert W is also called productlog in WolframAlpha. The first parameter of productlog is optional, and indicates the productlog branch (default is branch 0). The branch is an integer ∈ ℤ. There are infinitely many productlog results in complex numbers for all branches. Real solutions of productlog, however, can only be in branches 0 and -1. Here are both real solutions calculated in WolframAlpha: x = -3 * productlog(0, -ln(3)/3) / ln(3) = 2.47805268028830... x = -3 * productlog(-1, -ln(3)/3) / ln(3) = 3

[@עופרב-ק3י]

If you look at -ln3/3 in the lamber w function you can turn it into -ln3/e^ln3 and then into -ln3*e^-ln3 and since its in the lamber w function you can turn all of it to -ln3 and lastly you get x=(-3*-ln3)/ln3=3 so you can prove 3 is also solution

[@Matthew_Klepadlo]

I probably shouldn’t watch this video until I can solve this myself, since this is good practice for me learning to use the lambert function.

[@Oliver_DaNinja]

I mean, you can very simply solve it by realizing x=3 is a solution. 🙃

[@mozd1729]

at 4:00, cube root of 1 is not only 1 but also (-1±i√3)/2

[@happydogeyt1962]

DAYUM

[@zihaoooi787]

For that you need the W_1 branch of the W function.

[@21centuryhippie61]

Not true! The exponential functional notation must be functional! As such, there can be only be one equivalent value to any expression of a base number exponentiated. Otherwise statements violate the injective property of all functions. In other words, the equation x^3 = 1 does indeed produce three solutions for x, but the real number 1^(1/3) cannot be equivalent to more than one value, per the basic conventions of functional notation! Common and easy mistake to make, but worth correcting nonetheless!

[@mozd1729]

@@21centuryhippie61 ah true I forgot about that

[@TechGoat-LMAO]

Thank you so much sir🥹😃

[@ssaamil]

Perfect title haha! That's what I think when W function exists. Will you do those special or super hard integrals again? those were super entertaining to watch. Especially with you writing on the -air-

[@farfa2937]

The W function is a way of life. Let f(x) = a for an unreasonably complicated f. Define ominous sounding function = f inverse. The answer is ominous sounding function (a). Problem solved.

[@symmetricfivefold]

me casually waiting for a product log calculator to be made for students to bring to exams:

[@farfa2937]

I guess, there are more solutions if you use all 3 (complex) cube roots of 1? It really looks like Lambert wanted to solve xe^x, got tired, and invented a function that solves the problem by whatever means necessary. Which.... is not wrong I guess.

[@zihaoooi787]

For the positive imaginary solution, you use the W_1 branch. For the negative imaginary solution, you use the W_-1 branch. I’d assume that two out of the three representations give the same number.

[@SuryaBudimansyah]

The topic that blackpenredpen love

[@solarisone1082]

Yeah, he just posted a video featuring it.

[@renesperb]

The general case : a^x = (b x+c)^p can be solved also using the Lambert-function . Setting q = - ln a /(b * p) * a^(-c/ (b * p ) , the three possible solutions are given by x1 = - p/ln a * W(q) - c/b , x2 = - p / ln a * W (-q ) - c/ b , x3 = -p/ ln a * W( -1 , q) - c/b . An example with three solutions : take a = 1.5 ,b= 1.2 ,c = 0.5 ,p = 2 . Then one gets x1= 0.50685 , x2 = 14.0932 , x3 = -1.0854.

[@zelda1420]

Me at 0:45, already holding a frozen treat from Bahama Buck's: "Way ahead of you."

[@matthewhastings2568]

6:46 says -2.7231 is the solution but that’s a typo: it is indeed 2.47...

[@jackzegas]

So x = 3 is also a solution. But my question is how can there be two solutions? Both of the functions are monotonically increasing. And after the exponential function increases the cubic function the rate at which it's increasing is faster so how is it ever possible that the graphs can intersect twice?

[@advaykumar9726]

I was just thinking about this equation and also about lambert w function yesterday!!!

[@prabhjeevansingh7429]

This Videos Title is misleading. I am now also scared of Lamberts function

[@xyaf.]

You really complicated the whole solution

[@glowstonelovepad9294]

e^x is always greater (or equal at x=e) than x^e.

[@Bangtan_Vibes_7]

My solution steps were like this: X^3 = 3^x →log(x^3) = log(3^x) →3logx = xlog3 →(logx)/x = (log3)/3 So i can absolutely write that x=3.

[@pilotharibo]

x^3 has 3 solutions does this lead to more solutions for the entire thing?

[@ciiil8802]

Me: i have never seen this man in my line

[@XanderSebestyen]

X = 3 3 to the power of 3 is equal to 3 to the power of 3

[@nicolastorres147]

4:07 complex cube roots tho

[@zihaoooi787]

Use different branches of the lambert function.

[@Gd_Monsterforce]

X = 3 i don’t even know why this video is 8 minutes unless there are multiple problems

[@StrayChoom]

Coding the newton rhapson method is very easy, but how would you do it by hand?

[@ВикторПоплевко-е2т]

5:35 I'm trying to figure out what am I doing wrong can't we say that -ln(3)=ln(1/3) and -ln(3)/3=1/3*ln(1/3)=e^ln(1/3)*ln(1/3) and W(e^ln(1/3)*ln(1/3))=ln(1/3)=-ln(3) so the entire fraction -3W((-ln(3)/3)/ln(3) simplifies to just 3 which is one of the solutions? correct my mistake please it's probably about the branches of the W function Edit 1: not about branches I'm probably stupid

[@nicolastorres147]

How are the branches of W even ordered (as in the subscript)?

[@francis6888]

-ln3/3 is also equal to ln(1/3)×e^(ln(1/3))

[@gamingtime853]

You can use logarithms also

[@pecavocado4316]

my answer is messed up. x=3^(x/3) 3=x^(x/3) 3=(3^(x/3))^(x/3) 3=3^((x^2)/9) 3^9=3^(x^2) 9=x^2 x=3

[@mohammadjaveed7404]

x equal to 3 is exact solution of equation.

[@anncherian]

Soo cool

[@hrsulabh]

Bro but why there exits an extragenous root Is there a domain for w(xe^x)=x As till second last step 3 satisfies but after introducing lambert function [ whose domain is -1/e to infinity which is satisfies] and using xe^x, 3 is not yet the soln As solving in elegant way always requires each step to be accurate

[@fishHater]

Why did I get deja vu from this?

[@michaelbaum6796]

Very nice equation👍

[@eeeeee8762]

I fear no equation... but this *thing*... it scares me.

[@Speed001]

0:07 hmm, only W i know is the Wrongston. I can’t spell it.

[@h7pretzel]

amazing

[@mayosmayo4738]

Mathematicians are really just people that spend all day making and solving puzzles

[@_Exen_]

there is a much easier way to do: 3^x = x^3 ln (3^x) = ln (x^3) x ln(3) = 3 ln(x) x = 3 ln(x)/ln(3)

[@uwuowo7775]

i agree

[@globalwarrior16]

I did it on desmos I got 2.478

[@VVeiQuek]

How is `3` not the solution in any of the methods? 3^3 = 3^3?

[@supremeclamitas5053]

Yes, 3's a solution to this, it's just not mentioned

[@eprzepiora]

3

[@heniwatisetiono6995]

x=3

[@TheG0ldx]

You should have gone a bit deeper into the details :(

[@JustDeerLol]

Hello!

[@yoshikagekiraaaaa]

yes, that's me:)

[@thevividversatilechannel4807]

x = 9

[@natashok4346]

Why 0^x=x^0 no solution?

[@liamernst9626]

0^x is either indeterminate if x=0 or it is 0, x^0 is either indeterminate if x=0 or it is 1

[@Joneyboloney]

3 days ago🎉

[@voidnull9438]

1000th like

[@redroach401]

too easy

[@playwulf5483]

bro just X * X * X = 3 * 3 * 3 x=3 no thanks easy

[@kakashiamv7837]

X=3,(-3±3√-3)/2

[@maheshprajapati8916]

Plz dubbed in hindi 🙏

[@navsha2]

1:09 the equation is showing f=w if you agree 👇

[@dueie83]

Click ehat video on the screen, wait i dont see it

[@CalculusIsFun1]

x^3 = 3^x 3ln(x) = x(ln(3) (3ln(x))/ln(3) = x 3/ln(3) = x/ln(x) ln(3)/3 = ln(x)/x ln(3) * 3^-1 = ln(x) * x^-1 -ln(3) * e^-ln(3) = -ln(x) * e^-ln(x) apply lambert function. -ln(3) = -ln(x) ln(3) = ln(x) x = 3 Alternative way, alternative solution.

[@TheRook404]

Me: x is 3 !

[@rudransh118]

3 factorial !

[@GodMineptas]

Love your vídeos❤️🇧🇷

[@WK-5775]

Why does the W function not yield to the obvious solution x=3? That does not quite give a reliable impression.

[@johns.8246]

Cause it's not really a function.

[@RaniDevi-xt4hq]

If you use the -1 branch of Lambert W function it yields the solution.

[@WK-5775]

@@RaniDevi-xt4hq OK, I begin to see. So this branching business has to be explained much better than just mumbling something about the W function being "multivalued". Where is tbe cut? How many branches are there? Is there some "default" branch? Which one was used in the video? Why doesn't the obvious solution of the original equation appear in the default branch? Do the other branches give other values (complex ones, I suppose)? Are there Infinitely many of them? Without such information, that W function remains mysterious magic.

[@RaniDevi-xt4hq]

@@WK-5775 en.m.wikipedia.org/wiki/Lambert_W_function#:~:text=The%20principal%20branch%20W0,%2C%20Hare%2C%20Jeffrey%20and%20Knuth. You can read this page.

[@oreql9843]

@@WK-5775 there are two branches to the lambert w function- 0 and -1, the one in this video is the principle branch or the 0 branch. the solution x = 3 comes in the -1 branch. if -1/e < x < 0, there will be two solutions to the equation.

[@Israel-Niles]

I just noticed that your voice sounds a lot like CGP Grey’s😅😅😅

[@victorribera5796]

I don't understand why are some functions not considered analytical. Logarithms are also defined as the inverse of another function and are not represented using the other function notation but a new one and are considered analytical, while the W(•) is not

[@klausklausen4301]

The lambert W function is analytical. You can see it's taylor series representation at 2:09. You might've mixed up analytical with elementary.

[@geopediashorts]

so, W(e) would just be 1? (bc W(1 x e^1))

[@emelilukas8942]

I exponatiated both sides with 1/3 and then plugged in 3^(1/3*x) x, so i got 3^(1/3*3)^(1/3*3)… wich simplifies to 3^1^1^1…. so i just got 1 with basic algebra

[@grayjphys]

Is there a typo? It seems like you need to divide -2.7231 by -Ln(3) to get the nontrivial solution. Looks like two steps were missed.

[@soulsofspirit9729]

3^3 = 27, 3^3 = 27

[@pianohouse2725]

3:00 cant x=3 ?

[@allozovsky]

There's a typo at 6:38: putting in W = -0.9075 gives a positive x = 2.4781, not a negative -2.7231, as mentioned in the video. A minute earlier at 5:38 it was put correctly.

[@hvgaming8379]

W function

[@soulsofspirit9729]

3.

[@jessiez7006]

x=3.

[@partha0212]

Apply loag on first step and then derivate it to get the value of x

[@outhereplayin]

Valid for any no. x^n = n^x, I used to tease my friends in class while I was in 9th grade.

[@984francis]

27=27

[@shazullahyusufzai5704]

How do we solve the equation to find x=3

[@thecarman3693]

Technically ... just look at it.

[@zihaoooi787]

1. Look at it 2. See that cbrt 1 has 3 solutions. Use the imaginary solutions with the corresponding branch of W to solve.

[@Lomainium]

x=3

[@gregorymagery8637]

This is how i solved it x^3 = 3^x x^(3/x)=3 (3/x)*lnx=ln3 3x^(-1)*lnx=ln3 (-lnx)*e^(-lnx)=(-ln3)/3 -lnx = W[(-ln3)/3] -lnx = W[(-ln3)*3^(-1)] -lnx = W[(-ln3)*e^(-ln3)] -lnx = -ln3 x = 3 I wonder how can same formula give 2 different and correst solutions (2,47805 and 3) x={3W[-(ln3)/3]}/(-ln3) (=2,47805) =3*(-ln3)/(-ln3) = 3

[@pideperdonus2]

The solution for X^Y = Y^X is just X=Y so its just X^X, right?

[@renesperb]

The solution x=3 is immediately obvious . This is also the only real solution ,if one looks at the graphs of x^3 and 3^x.

[@renesperb]

Sorry, I missed the other solution x= 2.47......

[@Merched45]

Something that could have been done was, at the end, write (-ln3)/3 as -ln(3)e^(-ln3), giving 3 as a result