solving the tetration equation x^x^x=2 by using Newton's Method

[@OonHan]

But what is the exact value? Time for the *LAMBERT W FUNCTION!!!* ... ?

[@blackpenredpen]

Oon Han I am not sure how to do it tho...

[@OonHan]

@@blackpenredpen thats sad

[@OonHan]

@@fernandogaray1681 nope

[@andi_tafel]

@@OonHan I also thought sqrt(2) first because i thought the equation is (x^x)^x=2 which has the solution sqrt(2)

[@okoyoso]

‽

[@shobhitsinha1754]

The approximation reminds me of an old joke. An engineer and a mathematician are racing towards an attractive girl who is 100m away. In their first step, they travel 50m i.e. half the initial distance. The constraint is that the distance they can travel in successive time-steps should be half of the previous time step. The mathematician gives up, realizing it as an infinite geometric series sum. The engineer continues to move for he is confident that he will soon get close enough for all practical purposes.

[@amaxingmusic9334]

*for all practical purposes* lololol underrated comment

[@neilgerace355]

"Near enough is good enough" -- every engineer

[@l.1244]

But what if the girl is 2m away and the first step is 1, 60m the second one 0,8m > 2m

[@neilgerace355]

@@l.1244 personal experience suggests that neither a mathematician nor an engineer would take such a large first step :)

[@l.1244]

@@neilgerace355 personal experience would refute special relativity. Stay away with your damaging attitude.

[@clubstepdj]

2 is one less than π -Blackpenredpen, 2018

[@papajack2205]

ClubstepDJ that confused me, I did not get the joke...

[@neilgerace355]

To one significant digit, π - 1 = 2, correct

[@icespirit]

clubstepDJ, hello gd player

[@camdamcool6125]

Cant escape gd can ya? :)

[@icespirit]

@@camdamcool6125 lol

[@Bodyknock]

I seem to recall you did a video that √2 ^ √2 ^ √2 ^ √2 ... = 2. If so then it would make sense that √2 ^ √2 ^ √2 is likewise going to be fairly close to 2 so x will be near √2.

[@yge1035]

Doug Rosengard thought the same thing, and because its so close, the best starting point(x1) would'veprobably been 1.41 :)

[@federicovolpe3389]

Doug Rosengard Engineer detected 😂

[@AntimatterBeam8954]

Yes I remember this too

[@spondulix99]

You are absolutely correct. The correct and exact solution is x = sqrt(2). The expression used in the video for the derivative of x^x^x is incorrect. I have left a comment explaining the error in detail. Hopefully, you will be able to locate and read my comment. When the correct expression for the derivative is used, the Newton-Raphson iteration converges to x = sqrt(2). BTW, the correct derivative is f'(x^x^x) = x^x^x [ x (2 ln(x) +1)]. BTW, it is NOT correct that sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2) = 2. When evaluated in the proper order (see my aforementioned comment), sqrt(2)^sqrt(2)^sqrt(2) does equal 2, but sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2) equals 2.665144143... .

[@wolframalpha8634]

Damn! When did I start developing a craze for math...I just cannot miss a day without watching your videos :)

[@neilgerace355]

I think it's fascinating that a simple looking function has such a long expression for the derivative.

[@GrandMoffTarkinsTeaDispenser]

"simple" What

[@BigDBrian]

simple as in 3 terms it doesn't look simple if you know derivation reasonably well

[@light63478]

1:24 most glorious moment in the history of blackredpen

[@santhanam2855]

I must agree

[@jabir5768]

Top 10 rap song

[@semiawesomatic6064]

Guys, he definitely memorized all of those approximations. He was just looking at the camera because he wanted to make eye contact with his audience.

[@yrcmurthy8323]

Nope, i disagree totally. You shouldn't makenfun of him. If you don't know, just don't comment stupid ones.

[@yrcmurthy8323]

Sorry my friend

[@icespirit]

@@yrcmurthy8323 he is just joking

[@angelmendez-rivera351]

semi awesomatic Don’t take @Yrc Murthy seriously. I have had conversations with this child in the past and he is arrogant and disrespectful, incapable of respecting his elders and being civil online. Just forget he even talked to you.

[@angelmendez-rivera351]

semi awesomatic It’s not an authentic apology.

[@fluffymassacre2918]

Aproximations what are we engineers????

[@BigDBrian]

approximations are useful taylor series for instance started off as approximations so did riemann sums, i.e. integrals i.e. fucking everything

[@DrakoonLP]

Gabriel Mesiaa haha that’s a good one xD

[@brandonormsby6297]

Its a method in Differential Equations. Mostly used in mathematic majors

[@peggyfranzen6159]

Gabriel Mesiaa Sometimes.' Watt'? - I think not. Mathematics, science, and, engineering , should be the same What if Issac Newton's theory of gravity is not correct?

[@peggyfranzen6159]

mrBorkD True.

[@GreenMeansGOF]

For all the people who want an exact form, here you go. However, we must ask: "At what point does defining these productlog type functions become arbitrary?" Let f: R to R, f(x) = x*e^(x*e^x). Define V(x) to be the inverse of f(x). x^(x^x) = 2 ln(x)*x^x = ln(2) ln(x)*e^(ln(x)*e^ln(x)) = ln(2) ln(x) = V(ln(2)) x = e^V(ln(2)) = 1.47668...

[@fantiscious]

V(x) centered at x = 0 is approximately: x⋅[sum from n = 0 to ∞] cₙ⋅xⁿ/n! Where cₙ = 1, -1, 1, 5, -59, 259, 2311, -65479, 594217, 5846471, -313652789, 4707662629, 40989095005, -3994785626677, 88942062286351, 539010699471089, -105872440389821999, 3281478037381787023, 7215557899057278355... (starting from n = 0). Radius of convergence is around 1/4 to 1/2

[@Misteribel]

I'm not a mathematician, but I think your first step is wrong, it should be ln(x^x) x = ln2. Power towers are evaluated right to left. Your end result (1.47668) is still ridiculously close though, but it's not equal to your final e^lambert(ln2), which is 1.5596, and plugged back in the equation is 2.432. So, your final line seems wrong too, it is not an exact solution. It's also not a solution for (x^x)^x, which you seemed to solve instead, so maybe I'm just missing something.

[@TelepathShield]

@@MisteribelI don’t think he is wrong. He did ln[x^(x^x)], and using the product rule for logarithms you can change it to (x^x) * ln(x)

[@gloystar]

Excellent video that demonstrates how useful Newton's method is for approximating solutions very accurately.

[@filip-kochan]

“That is two - so just one less than pi” xd xd

[@blue_blue-1]

I remember my teacher saying that there are a lot of problems which you can‘t e.g. integrate exactly. You have to apply numeric approximations. But thanks for this video. Bprp (almost) integrates everything.

[@mike4ty4]

I agree with where another poster's suggestion below was going. Since tetration is to be considered in some sense a natural extension of the arithmetical operations, then so its inverse operations should be considered as such, too. In particular, for the nth power tower x^^n (or ^n x), we should also have that its inverse, the nth tower root, say twrt(n, x), should be a valid special function alongside of it. Moreover, the square tower root can effectively replace the Lambert W function altogether, since they are exactly equivalent: if x^x = a then x = e^W(ln(a)) i.e. twrt(2, a) = e^W(ln(a)) and thus W(x) = ln(twrt(2, e^x)). So W(x) proves unnecessary/redundant once twrt(n, x) - nth tower root - is accepted, and twrt(n, x) gives you more solving power than old W. The interesting question is if there are any more interesting, non-trivial equations that can be solved with higher-order tower roots like twrt(3, a) than just x^x^x = a and suitably increasing extensions.

[@TechnoRaabe]

nice idea. programming an algorithm that approximates general towerroots must be painful though. 💀

[@kingbeauregard]

Many years ago I came up with a numerical method for finding extrema: take the values of a function at x, x-dx, and x+dx, and figure out where the extremum of the corresponding parabola is. I never found any actual use for it, but it was pretty cool.

[@kingbeauregard]

In case anyone's interested. Suppose you have three data points: (x0-dx, y1), (x0, y2), and (x0+dx, y3) (note that the 0, 1, 2, and 3 are subscripts, not exponents). The three points can be connected with a parabola whose extremum will be at x0 - dx(y3-y1)/(2y3-4y2+2y1). If the denominator is zero, that means the points are on a line and you'll need to try different points.

[@XAE-yc9rr]

I am not sure if someone else in the comments has the solution using Lambert's W-function, but here is mine: Given: x^x^x=2. Raise both sides to the power of 1/x: (x^x^x)^(1/x)=2^(1/x)......x^x=2^(1/x) Apply the natural log to both sides and execute log properties: ln(x^x)=ln(2^(1/x))......x*ln(x)=(1/x)*ln(2) Multiply both sides by x: (x^2)*ln(x)=ln(2) Replace x inside the log by (x^2)^(1/2), then execute its properties: (x^2)*ln((x^2)^(1/2))=ln(2)......(1/2)*(x^2)*ln(x^2)=ln(2) Multiply by 2 and use log property: (x^2)*ln(x^2)=2*ln(2)......(x^2)*ln(x^2)=ln(4) Replace x^2 with e^(ln(x^2)): ln(x^2)*e^(ln(x^2))=ln(4) Use W-function: W(ln(x^2)*e^(ln(x^2)))=W(ln(4))......ln(x^2)=W(ln(4)) Use "e" as a base on both sides and solve for x: e^ln(x^2)=e^W(ln(4))......x^2=e^W(ln(4))...... Thus: x=(+/-)sqrt(e^W(ln(4)))......or (+/-)sqrt(2)

[@i_am_anxious02]

Nope. That’s (x^x)^x. We want x^(x^x). If you raise (x^x)^x to the 1/x, then we get x^x, but with x^(x^x) we get x^(x^(x-1)). It’s unhelpful.

[@XAE-yc9rr]

@@i_am_anxious02 Yes, I agree with you. I was trying to solve it using the latter grouping, but it became tricky. I am throwing every trick in the book at it as I speak.

[@spondulix99]

Brilliant!! And it provides independent confirmation that the correct way to evaluate x^x^x is to treat the right-most x as the exponent of the left-most x^x. That is, to regard x^x^x as (x^x)^x and NOT as x^(x^x). The latter is the error that was made in the derivation of the derivative of x^x^x presented in the video, which derivative is not correct and that leads to the incorrect result for x in the video. Long ago I drew this error to the attention of blackpenredpen, but never received a response. It is a common error that I have seen, and commented on, in other KZbin videos. Unless blackpenredpen has since updated his derivation of his derivatives of various sizes of "towers" of exponents of x, all of his derivative formulas are incorrect. I have just left a comment here in which I explain why the correct interpretation of x^x^x is (x^x)^x and NOT x^(x^x). I am grateful that you have now shown by an independent means that my interpretation is correct, and that the result in the video is incorrect. I hope you can find my comment here so that you can see my reasoning, which is not based on the Lambert W function. I am going to edit my comment to cite your independent derivation of the result x = sqrt(2). BTW, you are correct that x = - sqrt(2) is also a valid solution. Demonstrating that involves the complex plane, but I have done the exercise and have verified that x = - sqrt(2) also yields x^x^x = (x^x)^x = 2. If you want to see how that goes, let me know and I'll write it up for you. Finally, extending the "tower" to, for example, x^x^x^x is properly evaluated as ((x^x)^x)^x and NOT as x^(x^(x^x)). No matter the number of exponents, the nested parentheses start at the left and NOT on the right. In the usual "tower" representation, the nested parentheses start at the bottom of the tower and NOT at the top.

[@TelepathShield]

@@spondulix99wait so you’re saying x^x^x^x is ((x^x)^x)^x and not x^(x^(x^x))?

[@spondulix99]

@@TelepathShield Yes, because the universally accepted interpretation of A^x, where A is an arbitrary function (including A itself being a tetration) is that x in A^x is the exponent of A. In x^x^x^x, none of the x's themselves is assumed to be a tetration. Therefore, by convention, in the absence of parentheses, the right most x in x^x^x^x must be treated as the exponent of x^x^x. That is, (x^x^x)^x. Any other interpretation contradicts the universal convention that x in A^x is the ordinary exponent of A. The same convention leads further to ((x^x)^x)^x. There is nothing at all improper or incorrect in evaluating x^(x^(x^x)) if that grouping expresses the intended order of operations. But, absent any parentheses to dictate the explicitly intended order of the exponentiations, the only interpretation that comports with convention is to treat the right most x as the exponent of everything to the left of the right most x. To emphasize the point further, consider x^x^x^x^x. Why should this necessarily be treated as x^(x^(x^(x^x)))? Why not x^(x^(x^x^x))? Why select just the right most pair of x's as the upper most exponent? Why not select the right most triplet of x's as the upper most exponent? The point is that, absent parentheses specifying explicitly and unambiguously the intended order of operations, the only sensible recourse is to treat, by universally accepted convention, the right most x as the exponent of all of the x's to the left of the right most x, and likewise with any inner tetrations.

[@thefunpolice]

The subtitles are not very accurate but they are funny: Consider 0:40 where BPRP says "you can use the Newton's method to get a really good approximation" and the autogenerated subtitles read instead "you can use the Newton's method to get a ridiculous emission".

[@blackpenredpen]

😂 😂

[@bobdole69420]

You could just cheat and and say x = super-cubed-root of 2 and leave that as an exact answer!

[@fluffymassacre2918]

Bob Dole ☭ I like your thinking

[@Kjarlo]

4:06 Didn’t know you were an engineer😂

[@awertyuiop8711]

What if you take the limit, as n goes to infinity, of X(n)? Is it even possible to get a limit at all?

[@Democriteo]

What Is the function named at 0:24? Olympic what?

[@nozzel829]

I'm not sure why you would use that method for an approximation, instead of this: x^(x^x)=2 x=2^(1/(x^x)) Therefore we can say x[n+1] = 2^(1/x[n]^x[n]) And then start using x0=1 You should see that this approximation converges slightly faster than the method you used in the video. Also, no complicated derivatives are used.

[@BigDBrian]

You could start with sqrt2 as a guess. You know it's close but a low estimate because sqrt2^((sqrt2)^2) = (sqrt2)^2 = 2 and is greater than sqrt2^sqrt2^sqrt2

[@JJ_TheGreat]

Can’t you isolate the x^x power in that equation since we already know that x^x=2 is W(ln(2))?: So x^x^x = x^(x^x)=2 = x^(W(ln(2)))=2 and then use the Lambert W function AGAIN in order to solve THAT equation? Please let me know whether I am on the right track!

[@antimatter2376]

Ah you can't say that xˣ=W(ln2) because you're saying the entire thing is W(ln2). W(ln2) is just a solution to xˣ=2.

[@spondulix99]

You are not on the right track because the correct interpretation of x^x^x is (x^x)^x and, contrary to the work of blackpenredpen, NOT x^(x^x). I have posted a comment here today (2024-06-21) explaining why (x^x)^x is correct and x^(x^x) is incorrect. I hope you can find it. Using x^x^x = (x^x)^x, I get the solution x = sqrt(2). Furthermore, as blackpenredpen has incorrectly used the assumption that x^x^x = x^(x^x) in his derivation of the derivative of x^x^x wrt x, his formula for the derivative of x^x^x is incorrect. The correct formula is f'(x^x^x) = (x^x^x) [ x (2 ln x +1)]. This used in the Newton-Raphson iteration method converges on x = sqrt(2). @XZE-yc9rr has used, and posted here, the Lambert W function method of solving for x, and has likewise found that x = + sqrt(2) is a solution of x^x^x = 2. This method shows that x = - sqrt(2) is also a solution, which I have verified, though the intermediate computations involve the complex plane. Since the Lambert W function method and my method are independent, yet yield the same result, the Lambert W function method provides independent confirmation that (x^x)^x, as used in my solution, is correct, and that x^(x^x) is not correct.

[@grzegorzkosim547]

Question fo BPRP: how long 1/2in garden hose one can fit on a 30 by 30in table, if the hose is rolled into spiral (single layer)?

[@crazy4jee634]

can x^(x^x) be actually solved using the lambert function?? Or lambert function doesnt apply to this>?? PLease tell me..otherwise I may not be able to sleep at night

[@blackpenredpen]

That was the question I had in the video. I need people to help me as well. But I don’t think so tho.

[@fernandogaray1681]

Why √2 is not a solution?

[@MartinPuskin]

You exponentiate from the top down, unless you have parentheses like so (x^x)^x. The value of x^x^x=x^(x^x) when x =sqrt(2) is 1.7608. Sqrt(2) is a solution to (x^x)^x=2

[@Theraot]

I did try solving it with W, could not do it. If we did, it would require numeric methods to approximate the solution anyway... on that regard, we could invent another function to help us solve this, then the solution would be interms of that invented function and we would use numeric methods to approximate it anyway... so... hmm... not worth it.

[@Bodyknock]

Alfonso J. Ramos Well to be fair you need “numeric methods” to get decimal approximations of things like sin and cos too in most cases. The W lambert function isn’t special in that regard. So in that sense W lambert is just as valid a function as sin or cos.

[@Theraot]

@@Bodyknock I do not disagree, however I do not think that being useful to solve this equation is enough to bring forth a new one. I guess I could anyway... then, why not just define the inverse function of x^x^x? By the way, if I had inverse of x*e^x+ln(x) or inverse of x*e^(x*e^x) I could get it done.

[@bobdole69420]

​@@Theraot The function x^x^x is the same a x^^3 (tetration). The inverse functions of tetration are the super-root and super-logarithm. The inverse function to get x^x^x = 2 is x = super-cubed-root of 2. This can be extended to any power tower e.g if x^x^x^x^x = 7 this is the same as x^^5 = 7 and therefore: x = super-5th-root of 7 or super-log base x of 7 = 5 This could be done for even higher hyper-functions like pentation and so on.

[@quantumcity6679]

Black pen red ✒ man ..... 🤘 Can we find the value of x by Leibniz method? 🤔

[@ev4_gaming]

Whata the inverse of ³X ?

[@i_am_anxious02]

Unhelpful, that’s what it is.

[@ev4_gaming]

@@i_am_anxious02 ofcourse

[@jazzybank]

Super-roots

[@derdotte]

i wonder, is there a general proof that 2^2=2*2 is the only number combination of powers where this works? for example x^x = x*x is simple let x^x = x*x x^x = x^2 bases are the same therefore view powers and that results in x = 2 but how about a^b = a*b ? anyone knows?

[@imadhamaidi]

your pair of numbers should satisfy the following: a^(b-1)=b

[@derdotte]

how exactly would that proof go though? dont quite get why it would need to satify that condition yet.

[@davidnobre171]

For b=1 there are infinity solutions. Of you are implying a≠b by writing different letters and that b would have to be different than 1 I don't know would have to put some work but doubt I would get anywhere as that seems beyond my skills

[@davidnobre171]

@@derdotte just divide everthing by a. It's what I most likely did just one step division by a

[@davidnobre171]

A=0 is a universal no matter what b is. A= sqrt 3 or - sqrt 3 and b = 3 is also a solution

[@jeremycole9664]

What about something like: x^(2^((2^(1/2))))=(2^(1/2))^(2^x) √2 is a solution but WolframAlpha also gives 3.13556…as a second solution. How do you find this second solution? Thanks.

[@Reddion73]

POV: Your In Zoom And You Don't Understand What The Teacher Is Saying So You Search It Up.

[@terapode]

Your videos are always great.

[@wjm4elements]

Using lambert w the exact value is e^(W(2 ln 2)/2) x^x^x = 2 x ln (x^x) = ln 2 x^2 ln x = ln 2 e^(ln(x^2)) ln x = ln 2 e^(2 ln x) ln x = ln 2 e^(2 ln x) 2 ln x = 2 ln 2 2 ln x = W(2 ln 2) ln x = W(2 ln 2) / 2 x = e ^ (W(2 ln 2) / 2)

[@weinsterle1999]

You are wrong. I'm sorry ...

[@michaelpurtell4741]

Why is this wrong????

[@weinsterle1999]

@@michaelpurtell4741 One must be very careful when dealing with power towers. The equation x^x^x=2 is usually interpreted as x^(x^x)=2, meaning you generally start from the top and work your way down. If you instead have (x^x)^x=2 then the solution x=sqrt(2) becomes very trivial, so the video is not about that. And yes, there is a big difference between these two interpretations. For example, if I type in (3^3)^3, my calculator says it's 19683 but if I do 3^(3^3), I get 7625597484987.

[@spondulix99]

@@michaelpurtell4741 The Lambert W function result is correct. In fact, it predicts both + sqrt(2) and - sqrt(2) as solutions. I have verified the latter to be true, though the intermediate computations involve the complex plane. The Lambert W function method confirms my independent derivation of x = sqrt(2). As my solution is based upon x^x^x = (x^x)^x and NOT upon x^x^x = x^(x^x), the validity of my interpretation of x^x^x as (x^x)^x is confirmed independently by the Lambert W function method. I hope you can find my comment made here today (2024-06-21) explaining why (x^x)^x is correct and x^(x^x) is incorrect.

[@spondulix99]

@@weinsterle1999 The correct interpretation of x^x^x is as (x^x)^x and NOT, as in the video, x^(x^x). For an explanation of why this is so, I hope you can find the comment that I added today (2024-06-21). The significance of the Lambert W solution is that it confirms by INDEPENDENT means that the correct interpretation is, as I have explained, (x^x)^x and NOT x^(x^x). The assumption of the correctness of (x^x)^x both independently yield the same solution x = sqrt(2). In fact the Lambert solution actually shows that - sqrt(2) is also a solution. I have verified this to be true, though the intermediate computations involve the complex plane.

[@Dreamprism]

Heh. After seeing a couple of your other videos, I actually looked this up for myself several days ago. Interesting coincidence that this video is out now.

[@Dreamprism]

Oh. The version I looked up actually offered another approximation as well, based on the Lambert W solution to x^(x^(x+1)) = 2, if I recall correctly. But I suppose there's nothing wrong with pure Newton's Method.

[@alexanderskladovski]

Can you please tell us how to do non-integer tetration?

[@angelmendez-rivera351]

Alexander Skladovski He would not be able to tell you as that is undefined

[@sagarmajumder7806]

Sir, if we use Lambert W function as productlog,then can we use log property in W(xe^x) to make the in the sum form???🙄🙄

[@sorsbandeam3320]

Can you explain why x^-x^-x is equal to 1 above 4ish? Graphing may be easier to see

[@bryangohmppac6417]

Oh, we have the same secret weapon huh

[@Avighna]

I mean, I feel like Newton's method is overkill for this function. We know that it's always increasing, so we could just use binary search to find an extremely accurate approximate.

[@stjarnornasfred]

Let show the power. X^2=2, than X=root2(2). That's all! Now let's continue and show the tetration. X^^2=2, than X=superroot2(2). That's all! We don't need any other approximations.

[@stevenowensweanxen2406]

Can we get the answer with Lambert w function ?

[@ExpungedFunkyFriday]

Whenever solving a tetration we shall always go till 10?

[@ВладимирПутин-ж9ы6п]

x = 2^^(1/3) (tetration)

[@UltraMC3]

if we root both sides by x-root, then we get x^x=2^1/x And we do it again we get x=2^1/x^2 and then we use a calculator to solve we get x = sqrt(2)

[@i_am_anxious02]

No. you can’t do the xth root. This expression is x^(x^x). Not (x^x)^x.

[@UltraMC3]

@@i_am_anxious02 No you are wrong, the expression is x^x^x, no brackets as shown in the video. And also if you try x^x = 2^1/x and try to find x you get sqrt(2)

[@i_am_anxious02]

PEROLER he said in the video; “2^2^2. You do THE 2^2 first, and then you get 4. RAISE TWO TO THE FOURTH POWER ...” two to the fourth, not four to the second. If you were correct then he’d do 4^2 instead, despite it having the same answer this one time. x^x^x is, by default, x^(x^x) if there aren’t any parenthesis. We just do the opposite sometimes because it’s more convenient.

[@UltraMC3]

@@i_am_anxious02 Good job, what if you sqrt(2^2^2) you get 2^2 which is 4. 2^2^2 = 16 and if you square root 16 you get 4 again. Also, a square root can be interpreted as a 1/2 power. which means x-rooting is a power of 1/x so this makes total sense. To further point out, sqrt(2^2^2) = (2^2^2)^1/2 and with you saying the final two powers are squared then we get 2^2^1 which is the same as 2^2.

[@i_am_anxious02]

PEROLER no. Try 3^(3^3), and cube rooting it. Your cherry picking.

[@MegamikazeMoriko]

before i even watch this video x^x^x = 2 x^^2 = 2 and then inverse superpower 2 to both sides

[@fakenightbot1880]

2^^2 = 2^2 = 2x2 = 2+2 = 4

[@sugarfrosted2005]

Have you done the infinite power tower equal to 2?

[@i_am_anxious02]

sugarfrosted yep. That’s an old one he did, but it’s sqrt(2).

[@mhammedtomy545]

Sir can you solved with Lambert w function ? Because i tried but i didn't get anything. Thank u very much your vedio are amazing

[@fstasel]

I'd love to see now how this method is extended to multi-variable calculus :D

[@diversidadecientifica5450]

The best is sure he seeing the computer for the approach

[@SomINSYXIENGMAy]

W(5)=?

[@danielw9542]

By trial and error (not a good method for this but I did it anyway), x is the square root of two.

[@i_am_anxious02]

No. That’s (x^x)^x. This expression is x^(x^x). If (x^x)^x=2, and x=sqrt(2), then (sqrt(2))^(sqrt(2)^sqrt(2))=2. Thus sqrt(2)^sqrt(2)=2 so sqrt(2)^2=2, and thus sqrt(2)=1, because the top powers must be the same.

[@danielw9542]

@@i_am_anxious02 oh ok

[@pholioschenouda5395]

This remind me to an irrational raised to irrational power gives us rational so the answer is exactly root2

[@Pianissimo311]

Cant u substitute 2 in power to get x²=2 nad x=√2

[@oscardelgadofort8807]

What about Taylor series for x^x?

[@angelmendez-rivera351]

Óscar Delgado Fort What do you mean,

[@mike4ty4]

Hmm. I suspect you're thinking of using the Lagrange Inversion Theorem, no?

[@angelmendez-rivera351]

Are you asking about a video about the Taylor series of f(x) = x^x, or are you suggesting that we find the Taylor series of g(x) = x^f(x) and then use the Lagrange inversion theorem to find g^-1(2) ?

[@oscardelgadofort8807]

@@angelmendez-rivera351 I thought that a good idea for a future video could be creating the Taylor series for the function x^x

[@crazy4jee634]

I really need to figure more about the W function.. please make more videos on it and give us a beeter explanation on what exactly it is ..you didnt explain the concepts..please explain them better

[@blackpenredpen]

I didn’t explain in this video because I had some other videos for it.

[@eastonrocket兀]

W(xe^x) = x

[@DaimeG]

So... x arrow arrow x Any ideas about that?

[@FTR0225]

How would calculating 3^^3^3 work? (By ^ I mean tetration)

[@icespirit]

3^^27

[@NoNameAtAll2]

3^^(3^3)=3^^27 (3^^3)^3=(3^27)^3=3^81

[@study2847]

From Morocco thank you

[@doganergul5360]

Please correct me Integral xtanx.cos(tan^2x)/x^2+1dx

[@rishavgupta2117]

Can't we get pi using newtons method

[@peggyfranzen6159]

Rishav Gupta Good question.

[@imadhamaidi]

pi is not an algebraic number (not a root of any polynomial with rational coefficients) so I don't know how you'd use newtons' method to get pi (unless e^(ix)+1 magic)

[@diversidadecientifica5450]

Pi is transcendental, so it is not a root of any equation... And the euler number to. But the Golden Ration is not a Transcendental number, so you would get it

[@imadhamaidi]

​@@diversidadecientifica5450 consider cos(x) + i*sin(x) + 1 = 0 ;)

[@diversidadecientifica5450]

@@imadhamaidi Yeah Bro, But do not go crazy, this is not Real :D :D

[@fakenightbot1880]

I guessed "x ≈ 1.476684336"

[@jzanimates2352]

Isn’t an answer sqrt2? Sqrt2^sqrt2^sqrt2 = sqrt2^2 = 2

[@oscarahlke1585]

@blackpenredpen Now the proof that this number is irrational! :D

[@angelmendez-rivera351]

Oscar Ahlke The proof is rather trivial. We know the inverse of the function f(x) = x^x^x must be a transcendental function by definition, and according to a theorem by Gelfond, whose name I forgot, any algebraic input of a transcendental function which is not 0 or 1 which contains no shifts will result in a transcendental number, which is irrational. And this is the case for the solution of x^x^x - 2 = 0.

[@slowfreq]

@@angelmendez-rivera351 "We know the inverse of the function f(x) = x^x^x must be a transcendental function by definition" Why?

[@angelmendez-rivera351]

armin Because f is an iterated exponential which is not primitive recursive, and the exponential function and the natural logarithm are both transcendental functions. Hence x^x^x = e^(x^x*Ln(x)) also is transcendental. It cannot be expressed as a finite linear combination of finite polynomials, fractional powers, or finite Laurent series.

[@quark67000]

@Angel Mendez-Rivera : This seems false. Take this equation : x^x^x = 1. At least, one solution is 1, who is rational. But with your "arguments", you say : 1 is transcendental... You never use the fact that the function f(x)=x^x^x is equal to 2. So you cannot conclude that x, for this equation, is transcendental.

[@harshyadav9220]

You chould make garph 🤔🤔

[@DESMMMYYYYYY]

I tried using lambert w function buh it didnt work for me i stopped at xln2z=ln2 well who knows it might be wrong🤧🤧🤧

[@PATRIK67KALLBACK]

I might be very stupid but I get the answer ~2.34. If I'm using sqrt(2)=1.414213... I get the answer exactly 2... Please comment if I'm wrong.

[@ninnus12]

PATRIK67KALLBACK you are doing (x^x)^x =2, which is not the conventional interpretation of x^x^x. When you have an exponent tower, you work top down. x^x^x = x^(x^x) Note 2^2^2 can be misleading as both ways the result is 16, so working 3^3^3 you can see the difference of 27^3 from 3^27. Note the wrong one is 5 digits long and the right one is like 13 or so.

[@PATRIK67KALLBACK]

@@ninnus12 Thank you!

[@bmninfo847]

x*a=(a+2)*a What is answer

[@lucasdepetris5896]

But what is the exact form?!

[@angelmendez-rivera351]

Lucas Depetris There is no exact form. There is no known function in terms of which you can write the solution.

[@lucasdepetris5896]

@@angelmendez-rivera351 but maybe there is some algebraic way to isolate x. Unknown way doesn't mean impossible

[@angelmendez-rivera351]

Lucas Depetris There is no algebraic way of isolating x. The inverse function of x^x^x is transcendental and provably not expressible in terms of elementary functions.

[@GreenMeansGOF]

I think if we define V(x) to be the inverse of x*e^(x*e^x), then the exact answer is e^V(ln(2)). I don't think the exact answer can be written in terms of Lambert W.

[@Math-bz8bw]

Can you solve. !X!=

[@Hjerpower]

Oh hey, I asked this question

[@spondulix99]

The derivation of the derivative of x^x^x used in the Newton-Raphson approximation recursion in this video is incorrect, as I pointed out already some time ago. The correct expression for the the derivative of x^x^x is rather f'(x^x^x) = x^x^x [ x (2 ln (x) + 1)]. The incorrect derivative results from improper interpretation of the exponentiation notation. See also the NOTE added below for independent confirmation of my result here. It is universally recognized and accepted that the notation A^y, where A is a constant or function, means that the constant or function A is to be raised to the power y. In the current case, the function A is x^x and the exponent to which A = x^x is to be taken is the right-most x in x^x^x. That is, the correct interpretation of x^x^x is (x^x)^x. The improper derivative resulted from incorrectly interpreting the meaning of x^x^x as x^(x^x). That is, improperly treating the right-most x^x as an exponent. It is not. Consequently, the logarithm of x^x^x is x ln (x^x) = x^2 ln x, and not (x^x) ln x. The latter leads to the incorrect derivation of the derivative of x^x^x. The correct solution for x in x^x^x = 2 is x = sqrt(2), as may be readily verified by correctly evaluating (sqrt(2)^sqrt(2)^(sqrt(2) = ((sqrt2)^(sqrt(2))^sqrt(2). The only circumstance in which (x^x)^x equals x^(x^x) is when x = 2. Otherwise, the two interpretations are unequal. Going further, the correct interpretation of x^x^x^x is ((x^x)^x)^x and not x^(x^(x^x)). The nested parentheses start at the left and not at the right regardless of the number of exponents. When written in the usual "tower" notation, the parentheses start at the bottom of the tower and not at the top of the tower. NOTE ADDED: @XAE-yc9rr has successfully solved for x using the Lambert W function, and the result is as I have also shown, x = sqrt(2). In fact, @XAE-yc9rr has shown, and I have verified, that - sqrt(2) is also a solution, though the complex plane is involved in verifying that this is so. I strongly recommend viewing his derivation. What is most significant about the Lambert W function solution is that it confirms by INDEPENDENT means that the correct solution follows from evaluating x^x^x as (x^x)^x and not, as blackpenredpen has done, x^(x^x). In general, the nested parentheses begin at the left of the sequence of exponents and NOT at the right. Or, in the usual notion, at the bottom of the "tower" and NOT at the top.

[@kelecsenyizoltan274]

Thank you!

[@raunakagarwal7485]

It would take lot of time to calculate.....

[@jai_nasu]

isnt that sqrt(2)

[@MartinPuskin]

You exponentiate from the top down, unless you have parentheses like so (x^x)^x. The value of x^x^x=x^(x^x) when x =sqrt(2) is 1.7608.

[@mamtaprajapati2710]

Hi Well it was my question.... Thank you!!!

[@MemesbyWill]

Kakegurui brought me here....an anime...think about that

[@camuchoc5913]

Super root left the chat

[@sohampatil6539]

Couldn't you just use Newton's method and get pi?

[@XAE-yc9rr]

Can someone check out the solution I had earlier in the comments?

[@spondulix99]

I have checked it out and it is indeed correct. In fact, both x = + sqrt(2) and - sqrt(2) are correct solutions. I have left a more detailed comment attached to another of your comments. My own independent (of the Lambert W function method) derivation that x = sqrt(2) appears as a comment here made on 2024-06-21. I hope you can find it.

[@Deegius]

Use excel try square root 2 = answer. 1.476666 whatver does not work, gives 2.3333.

[@Electroneer]

That's because you did (x^x)^x, instead of x^x^x, which is calculated as x^(x^x).

[@lordofcastamere9376]

And now x^x^x^x^....=2

[@medxed2827]

Derek Zuurmond it's x = sqrt(2) bro

[@i_am_anxious02]

Yes. Because if x^x^x^...=2, then we can use this to colander the exponents, and now x^2=2. So x=sqrt(2).

[@LeskovPV]

Super!

[@mihailoopacic3756]

Best friend

[@mihailoopacic3756]

You are a best friend

[@rot6015]

Mans liked and commented on his own comment, give him an award

[@MRender32]

Yes!

[@ranjitnayak3157]

Thanks sir

[@Astrophysicalchemicalmath20000]

Answer is x=³√ 2ₛ

[@meteiyigun8075]

I like X

[@witzedegroote1831]

And x^x^x^x=2 x is 1.446601433

[@witzedegroote1831]

1.4466014325*

[@dreadwolf4927]

interesting

[@funicubing7340]

Newton is cool but remember to subscribe to pewdiepie to save him from t-series!

[@vincentwilliamrodriguez3572]

Lol why did you brought that here

[@siulprime9346]

Yikes

[@navarajpanday68]

Integrate x^x^x Lets see how you do it..

[@bruzanhd]

Wurtt Mapper either you're trolling, or you got baited.

[@wurttmapper2200]

Brandon Teeter I was making a reference to a video in which they solved integrals that were not elementary functions by saying they were equal to themselves

[@angelmendez-rivera351]

x^x^x = e^(e^(x*Ln(x))*Ln(x)) = Sum over all n in the natural numbers of [e^(x*Ln(x))*Ln(x)]^n/n!

[@adnanmohammed5158]

Ok

[@Theraot]

x^x^x

[@bouteilledargile]

how would you get a series out of newton's method?

[@angelmendez-rivera351]

35cut By expanding the recursion into an explicit formula, though granted the formula would be so ugly and complicated it is effectively impossible to use.

I was wandering that too.

[@angelmendez-rivera351]

Zoltán Kürti You are asking if I looked up the argument for claim which does not exist. Good try, but no. I already addressed your argument in the past, and I am not going to bother refuting it again, especially since you never even responded with a valid objection. Time to put this conversation to rest.

[@angelmendez-rivera351]

Zoltán Kürti Yes, and no. You are correct. This does not prove your point. In fact, it demonstrates mine. C and R are not equal. Rather, R is a proper subset of C, it is strictly contained by C. Pretty simple.

[@aidanmcnabb1066]

This is wrong! The correct answer is root 2. The newton approximation doesn't work in this case. If you test it then you get approximately 7/3

[@nathanisbored]

i think you dont know how exponents work. it would be root 2 if it was (x^x)^x, not x^x^x

[@tharushafernando4410]

Yeah, I get 1.76.... if I plug it in a calculator.

[@fluffymassacre2918]

That’s not how exponents work, the exponent is being applied directly to exponent not the base and exponent

[@i_am_anxious02]

x^(x^x)? No. (x^x)^x? Yes.