Infinite Power Tower Equations Battle!

[@blackpenredpen]

Hi all, thanks for the 500k subs! I will make a post of the winners of the secret giveaway tonight. After that I will be taking a break from YT.

[@itislamin]

Congratulations

[@punjabiguy1296]

I am the second because first are u so can u give me heart

[@integralboi2900]

How long?

[@punjabiguy1296]

How long break??

[@ggeorge02]

I hope you have a nice break!

[@ricsix2.0]

You got me here from the community post

[@peternelson8876]

Can't wait for part 3, I need that domain explanation!

[@ganeshprasad9851]

It's in the description 😂

[@peternelson8876]

@@ganeshprasad9851 oh damn you right, thanks 👍

[@peternelson8876]

kzbin.info/www/bejne/parCnIOvidinp80

[@speedychicken831]

Nerd 😲

[@Kero-zc5tc]

Jjk has ruined me

[@SeeTv.]

Will you explain why the domain is like that in another video?

[@blackpenredpen]

It’s in part 3, see description for the unlisted videos

[@angelmendez-rivera351]

Interestingly, if you have any quantity of the form a^(1/a), where a is positive and real, the corresponding power tower will always converge, because if for all such a not equal to e, a^(1/a) < e^(1/e), since at x = e, x^(1/x) attains its global maximum. So, despite the fact z^^♾ = a only converges if a < e or a = e, the power tower with z = a^(1/a) does converge. It just does not converge to a.

[@thatnhoxiu]

wayt, how is this comment 5 days ago??

[@timka3244]

HOW 5 DAYS AGO????

[@StreetfighterDucati1]

It may not attain that maximum on an open domain, ie the extreme value theory might not hold. X=e is just a critical point

[@dugong369]

If the tower doesn't converge to a, it still has to converge to y such that y^(1/y) = x = a^(1/a), so for a>e, the tower converges to the unique number y (between 1 and e) such that y^(1/y) = a^(1/a). This is the same unique number y such that y^a = a^y. In bprp's example a=3 and y = e^(-productlog(-ln(3)/3)) ~ 2.48 where productlog() is the name used for the Lambert W function on WolframAlpha. Note that 2.48^3 ~ 3^2.48

[@angelmendez-rivera351]

thatnhoxiu The video was unlisted.

[@monke4200]

When are you going to post .. isn't your break over ? 😔

[@spandanhalder9967]

Currently it's 2 am here, and I don't know why I'm watching this at this solemn night, but still I'm enjoying Idk why.

[@gastonsolaril.237]

You're probably one of my fav youtubers of the last 2 years, bro! Your work is awesome. I'm really passionate about maths and you just feed my hype! And in these days of quarantine and isolation, you and Dr. Peyam sometimes even feel like my "math friends"! Hey listen; have you ever thought about a good video series about Stochastic Calculus? I'm quite into it right these days! (financial markets and stuff). Just learned how to derive Black-Scholes' PDE. But I get lost when certain subjects such as Ito's integral or similar come around. It would be excellent if you play some of that sort of things! Long live BPRP!

[@drpeyam]

This is brilliant 😉

[@roberttelarket4934]

I can't believe you've never seen this problem? I first saw it in 1968 as an undergraduate!

[@angelmendez-rivera351]

For anyone who is curious, if you want to know what x^^♾ is equal to for any x in the domain, then notice that y = x^^♾, and x^^♾ = x^(x^^♾) = x^y. Therefore, y = x^y. To solve for y, take the natural logarithm here. We can do this because are assuming x is positive in this initial exercise. Hence ln(y) = y·ln(x), implying ln(y)/y = ln(y)·exp[-ln(y)] = ln(x), hence -ln(x) = -ln(y)·exp[-ln(y)]. If -ln(x) > -1/e or -ln(x) = -1/e, then the above equation implies W[-ln(x)] = -ln(y), where W is the Lambert W function, in this case, the principal branch of the W map. Therefore, y = exp(-W[-ln(x)]) = 1/exp(W[-ln(x)]) = 1/(-ln(x)/W[-ln(x)]) = -W[-ln(x)]/ln(x). The condition that -ln(x) > -1/e or -ln(x) = -1/e implies that ln(x) < 1/e or ln(x) = 1/e, which implies that x = e^(1/e) or x < e^(1/e), which agrees with what was stated during the video.

[@angelmendez-rivera351]

Also, x < e^(1/e) implies ln(x) < 1/e, which implies -ln(x) > -1/e, which implies W[-ln(x)] > -1, which implies -W[-ln(x)] < 1. Therefore, -W[-ln(x)]/ln(x) = y < e, or equal to e if x = e^(1/e), which also agrees with what is stated in the video.

[@timka3244]

4 DAYS ago...

[@angelmendez-rivera351]

Субс Тим The video was unlisted.

[@samueldeandrade8535]

​@timka3244 what a coincidence. 4 years ago you met the worst math content viewer and replied to him with the comment "4 DAYS ago..." Hahahahaha. Life is crazy.

[@dan_was_here9328]

I really like your channel. It is way more entertaining than the other math channels.

[@nikitakipriyanov7260]

Managed to got half of bounds. Namely, x^x^x^...=y means x^y=y, this solves using Lambert W function: y=e^(-W(-ln x)) (I've seen your other videos where you explain what is W and how to solve such equations). But, W(x) domain is x≥-1/e (there are two real branches: W₀ domain is [-1/e; ∞), with values in the range [-1; ∞), W-₁ domain is [-1/e; 0), with values in the range (-∞; -1], that's a multivalued function). This means, there must be -ln(x)≥-1/e, solving that for x gives x≤e^(1/e). Then, if I put that value into equation, I've got y = e^(-W(-ln x)) = e^(-W(-1/e) = e¯¹ = 1/e. Now I need to prove this is lowest possible value of y. Also still haven't figured out how to find out other bounds.

[@nikitakipriyanov7260]

I feel, we need to go from the definition of x^x^x^..., consider series f₀(x)=1, f₁(x)=x^f₀(x)=x, f₂(x)=x^f₁(x)=x^x, and so on, f_n(x)=x^{f_{n-1}(x)}, then define "power tower function" f(x) = lim f_n(x) for n→∞. Now we have to ask ourselves if this ever converges, and if it does, for which values of x? And this shoud give us all bounds.

[@nikitakipriyanov7260]

The solution x=f¯¹(y)=y^(1/y). Let's analyze this formula. It's derivative is df¯¹/dy = y^(1/y)(1-log(y))/y². For df¯¹/dy=0 we get y=e, x=e^(1/e). This is the maximum of the f¯¹(y), that's why when we put 3 in it we have some adequate value, which happened to converge our power series to something less that 3.

[@HoldensBro]

x=y^(1/y). The maximum point of this is x=1.44466786… substituting y=e, you can see that the result is the maximum. You can prove that y=e is the maximum by doing the derivative, which I won’t bother with here. Once y>e, you will see that x=y^(1/y) has two values of x.

[@HoldensBro]

It happens that the cube root of 3 is the same as 2.478^(1/2.478), so the power tower converges to 2.478 instead of 3

[@Aldiyawak]

5:44 *technically speaking* Ah yes, the engineering approximation.

[@TNTacdc]

Congratulations on 500k! Here's to 500k more!

[@alexandrefrancoalcaraz9351]

I'm a math hobbyist, I don't have advanced education in math, but I love them and I'm able to understand most of your videos (they're awesome). I have a question due to my ignorance: When you have a tower exponent of real numbers, it has to be solved up-to-down? I mean, the solution you give to x^x^3 only works if we solve the exponents up-to-down, and then the same answer fits to any x^x^x^...^x^3 form. Is it correct? Thank you very much in advance!

[@balazscsillag6445]

5:40 Engineering students don't know why, because π=e=3

[@chhabisarkar9057]

Lmao

[@lakejizzio7777]

We are using e=2.7 actually.

[@黃子睿-q5o]

The worker's of Babylon estimate π more accurately than engineering students now! LOL

[@rotabali5338]

We use e equals to 2.8 only

[@naseershaik5818]

Congratulations for got 500k subscribers

[@hydrostrikehd4661]

It's fun to see this video before it's published.

[@hydrostrikehd4661]

Now it has been published lol

[@itislamin]

How do you see these

[@integralboi2900]

ItIsLamin you can find them in playlists, he puts his unlisted videos in playlists before he publishes them.

[@itislamin]

@@integralboi2900 oh

[@nghiaminh7704]

Hey, then what's the problem with tower(x) = 3 ? Is there any clear explaination without using the range of y?

[@egillandersson1780]

Half a million subscribers !!! Nice job !

[@dushyanthabandarapalipana5492]

Thanks !I wish you happy new year!

[@ajinkya2344]

500k hits. Congrats

[@danielmendes2923]

I got a idea. Since now you have 500k subscribes, do 500 integrals in one take to commemorate it. Who more agree with this? 😂

[@cuonghienthaosonbuitrung2841]

that's such a reckless idea. i have never heard it before. but if you want, then split it into many short videos

[@adamkupczyk3826]

Before using x^x^x^x... = 2 you should prove it exists.

[@OCTAGRAM]

Infinite Power Tower was not explicitly defined as limit of finite power towers, so you are free to define it another way, as lim of x from 2 (good value) to 3. This is called continuation

[@blackpenredpen]

OCTAGRAM Oh I mentioned about that in part 3.

[@GreenMeansGOF]

The converging value is x=W_0(a)/a where W_0 is the principle branch of the Lambert W function and a=-ln(cbrt(3)). If I use W_(-1) instead, I get x=3 which does not make sense unless we somehow redefine convergence.

[@mandeltownthekillerfrombab5202]

All numbers go past sqrt(2) is considered as infinite. Therefore two equations are equal.

[@anonymouslies]

A question is this concept releted to the Mandelbrot set? If yes then how? Also at 2:53 you forgot to add the doremon music Really dissapointed!! Great video tough.

[@大大-u3m]

A monk carrying a shark🤙🏻🤙🏻

[@roderickwhitehead]

Perfect followup.

[@dugong369]

If x=a^(1/a) but the tower (x^x^x^x...) doesn't converge to a, it still has to converge to y such that y^(1/y) = x = a^(1/a), so for a>e, the tower converges to the unique number y (between 1 and e) such that y^(1/y) = a^(1/a). This is the same unique number y such that y^a = a^y. In bprp's example a=3 and y = e^(-productlog(-ln(3)/3)) ~ 2.48 where productlog() is the name used for the Lambert W function on WolframAlpha. Note that 2.48^3 ~ 3^2.48

[@DanBurgaud]

Suggestion: Instead of using a whiteboard or blackboard, use a transparent glass. The setup would be you facing the camera and glassboard between you and camera. Obviously, camera will be recording all the writings backward. Then using software, convert/mirror the video. This way, you dont have to keep turning your head to look at camera (to look at the audience); you will always be looking at the audience while writing the equations.

[@nicholasscott3287]

Downside is that you'd have to write baxk-to-front from your perspective to make it look right to the audience

[@DanBurgaud]

@@nicholasscott3287 Software will mirror it. There are some YT streamers doing exactly like this.

[@younesabid5481]

Isn't the infinite power tower a kind of tetration where the "exponent" approaches infinity? And btw, could you please make a video on the inverses of this operation? Namely the super root and the super log. And thanks!

[@angelmendez-rivera351]

Yes, an infinite power tower is "equivalent" to x^^♾. Also, tetration to a fixed exponent only has one inverse map, in this case, the superlogarithm of infinite order, which is expressible in terms of the natural logarithm and the Lambert W function.

[@timka3244]

HOW 5 DAYS AGO

[@cable4751]

@@timka3244 they're probably members

[@peterchan6082]

Hey bprp, 4:38 - 4:58 . . . How do you prove this?

[@blackpenredpen]

It's in part 3, you can see the video in the description : )

[@cuonghienthaosonbuitrung2841]

where are you now?

[@pierreabbat6157]

If I set x=3^(1/3) and evaluate a, x^a, x^x^a, ..., if a is in (0,3), then the sequence converges to some number between 2.45 and e. But if a>3, then the sequence goes to infinity. If a is just less than 3, the sequence decreases; 3 is an unstable fixed point. If I set y=e^(1/e) and evaluate a, y^a, y^y^a, ..., then e is a metastable fixed point. For a just less than e, the sequence converges very slowly to e, and for a just greater than e, the sequence diverges very slowly away from e, and eventually shoots off to infinity.

[@yugarthsharma626]

Damn the video's real smooth ;D

[@s4m1rza]

Hi, can you explain again please why the infinite superpower of x can be written as x^2?

[@nahrafe]

Yo x^x^x^x... = 2 So because of x^(x^x^x^x...) = 2 We replace the never ending exponent with 2 as both are the same So thats it, x²=2

[@s4m1rza]

@@nahrafe yh but because its infinite, doesn't that mean u could also write it as x^x^2 = 2, or x^x^x^2 = 2, and so on. Also when u tetrate sqrt2 to infinity, u don't get 2.

[@mokouf3]

Similar technique can be used for nested square roots/fractions equations.

[@MathPhysicsFunwithGus]

Great video!!

[@blackpenredpen]

Thanks

[@Dreamprism]

Awesome follow-up to your other video!

[@error_6o6]

Where do I get that chain chomp plush I need it

[@52.yusrilihsanadinatanegar79]

i forgot that this man was having a break

[@davidgillies620]

Infinite power tower not to be confused with Tower of Power which is a Californian R&B band.

[@黃子睿-q5o]

showing the existence of x^x^...... is very important.

[@BTheBlindRef]

so what is that number that the tower of 3^(1/3) converges to? How do you solve for the value of such an expression?

[@blackpenredpen]

See part 4 in the description

[@MithuBhattacharyaMukhopa-gm4yb]

I still dont understand it where it comes from,but the info is excellent

[@boujdadyouness7084]

the maths is amazing . I love the infinity

[@alonilutowich4505]

x^x^x^x... = 2 (1) Say both sides of the equations are the exponents of x x^(x^x^x^x....) = x^2 (2) plug in (1) into (2) x^2 = x^2 Therefore x can be equal to any number Which means 1^1^1^1.... = 2

[@angelmendez-rivera351]

No, that doesn't mean x can equal any number, and it doesn't mean 1^^♾ = 2. If I have the equation x^3 = 2x^2, and I multiply by 0, I get 0x = 0x, which is true for all x. This does not mean x^3 = 2x^2 is true for all x.

[@alonilutowich4505]

@@angelmendez-rivera351 that's not the same thing as what I did, I can say 2x^2 and x^3 are both exponents of x so that x^(2x^2) = x^(x^3) and sure it adds one more solution of x = 1 but it doesn't mean it's any x. The answer to that is still 2 like the original equation with the remove of 0 as an answer due to the domain

[@MercuriusCh]

hi, bprp! Wanna cool task? Just look at this: tg(sin(x)) or sin(tg(x)) which is bigger solve for x on interval (0; pi/100)? Have fun!)

[@cobyambrose2915]

Mistake at 6:27 - the constraint should apply to y and not x.

[@Codertyu]

How to find mirror image in co-ordinate system ,sir please

[@markstahl5272]

Sir, Rules of tetration say you must work "downward" from the highest exponent evaluating to the base. You are incorrectly working "upward", the wrong direction and will give a totally different value. How can the expression be evaluated if you can never start at the last exponent of an infinite tower power and work "downward"?

[@blackpenredpen]

Hi Mark. Whenever we are dealing with infinity, we should do it in terms of a limit. In this situation, we should define the inf power tower as the limit of a sequence of the functions x, x^x,x^x^x,... For more details, please see part 3 in the description.

[@阿巴阿巴-z5i]

Hello, Mr.Cao, can you do a video on volume of revolution in polar coordinate without using double or triple integral?

[@ILoveMaths07]

So that's his name?

[@쉽게-q4r]

I have a Q. What is infinite series of epsilon?

[@Ryan-gq2ji]

Is the derivative of ln(x!) lnx?

[@sophanarato5373]

Please help this integral (cos(2x) - cos(ax))/(cos(x) - cos(ax)) which (a) is a constant!

[@digitalsnowfall1961]

Hey , if i have a polynomial,should its factors also must be a polynomial ? Please answer

[@ghotifish1838]

I like the chain chomp microphone

[@moregirl4585]

Usual version I see don't say "you try (3^1/3)^... and get 2.4" but "you solve for 4 and get same result as 2"

[@DANGJOS]

In fact, any ath root of a will be smaller than eth root of e except eth root of e itself.

[@Yok_Knnn]

hey how are you? It’s been 2 months since you posted video last time. Are you OK?你还好吗？好久没看到你了！

[@MeeraSingh-gn3ik]

Find the polynomial which when divided by a cubic polynomial gives a biquadratic quotient and a linear remainder

[@perpetualrabbit]

I am wondering: if you have an infinite power tower function f(x)=x^x^x^x..... you can write f(x)=x^f(x). But can you also write f(x)=f(x)^f(x) ? In the first case with x^x^x^x·.... you can never start evaluating the 'highest power' in the tower because the tower is infinite. But the second case is even worse: how do you even start figuring out what (x^x^x^...)^(x^x^x^...) means? The first tower is already infinite, so how can I stack the second on top of it? Do I just get x^x^x^x^..... again? If not, why not? I am confused.

[@abrahamherzl9904]

Plz since 4 years i wondered how integral 1/x=lnx can u make a video about it?

[@blackpenredpen]

Not a proof but it’s kinda nice. twitter.com/blackpenredpen/status/1233863309562011648?s=21

[@abrahamherzl9904]

@@blackpenredpen lol i see... u proved it with a method that only can be true when the whole thing is already proved...so there is no proof after all?

[@rachitjoon3811]

Hi,bprp, can u explain why we can't integrate 1/dx

[@Evan-ne5bu]

Bprp do you think you can do a video about the Bessel's differential equation and it's series solution?

[@markklammerts4772]

Nice. But… why does this reasoning converges to sqrt 2 for tetration and not for normal power from bottom to top (in which 3^3^3 would be 27)?

[@Misteribel]

How can 3^3^3 be 27? If you mean not as normal power tower (without parens it’s top to bottom), but left-to-right evaluation, it’s (3^3)^3 = 27^3 = 19683. The other way around it’s 7625597484987. The sequence doesn’t converge to √2, but an infinite power tower (not tetration) of √2^√2^√2…. actually converges to 2.

[@권영훈-l4f]

Very good!

[@roberttelarket4934]

This is an old problem I first saw as a teen in 1968!

[@emanuellandeholm5657]

Are you on vacation? :) Here's a nice integral for you: Let f(x) = sin(arctan(x)) tan(arcsin(x)). This looks kind of scary, huh? In fact the, integral of f(x) is nonelementary, so I suggest you try the integral of f^2(x) = sin^2(arctan(x)) tan^2(arcsin(x)) instead!

[@lifemantras6099]

Blackpenredpen should be change now into bluewhale

[@rajkamal6062]

Hey I want to ask u one question related to integration.While doing Integration of cos square x or sin square x why don't we use the simple linear Integration formula why we use formula of cos2x in doing their integration.please reply...

[@RoMaths]

Big fan of you from INDIA🇮🇳

[@liranzaidman1610]

Who was the mathematician that proved those crazy thing?

[@usptact]

The trick with color pen wouldn't work with any color. The trick only works for a specific range of colors.

[@313bennyw3]

Is it possible to do the integration of tan(cosx)dx?

[@313bennyw3]

I mean from 0 to 2pi lol

[@parthkatke6706]

What is your channel's profile photo about?

[@soulsilencer1864]

hey i love your videos. Can you find the maximum and minimum values of f(x,y)= (x^y)/(y^x) by using partial derivatives?

[@ДометдеВоргес]

I don't get why it doesn't work with 3 if e=3=π 🤷‍♂️

[@facitenonvictimarum]

He is a teacher. He does math for fun. He makes KZbin videos to make money and to sell merchandise. He is a typical KZbin money-hungry con man.

[@nilaxibhoot2899]

Heyy when will you upload ur next video?

[@pocojoyo]

LOL, man. Dont you prefer a lapel microphone ? BTW Thanks for your videos

[@blackpenredpen]

You’re welcome. I prefer my chain chomp or Pokéball mic : ))

[@TheKannanmnj]

How to denote independent events on Venn diagram .

[@alejandrodelabarra2838]

Please! Use your incredible knowledge to solve "root locus" problems!!!

[@gdash6925]

can you compute productlog(2)

[@rafaelb.333]

Hey, make a video with the proof that the numbers with form abcd... = a! + b! + c! + d! + ... are finite. Ex: 145 = 1! + 4! + 5! It could be interesting

[@lakejizzio7777]

I understand the math behind Solve 1 but If you use sqrt(2) in infinite power tower it goes to infinity? Why we can pretend it's sqrt(2) in this.

[@はにわ-r7p]

They looks alike but we can’t solve one. How fantastic it is!! (I’m Japanese so I may make some grammatical errors.)

[@pauljackson3491]

You said, "they looks" It should be, "they look". This makes the noun match the verb; "they" is plural but "looks" is singular. What you said is still better than most stuff on Twitter though.

[@saharhaimyaccov4977]

Hi bprp .. can u slove the integral for (x⁵+1)^-1? ..please i cant slove it ..

[@saharhaimyaccov4977]

@blackpenredpen

[@monikagulati3737]

Hi I like your videos very much can u please explain through a video why derivative of lnx is 1/x

[@coolmangame4141]

does anyone remember when he said that he'll do x^x^x^x^... = 2017 next year in 2018 lol

[@blackpenredpen]

I...... remember......

[@jofx4051]

Just realizing that it has no solution since y>e

[@frankdominick2546]

You can write x^x^x^x^... = 2 as a 'limit." A reverse limit. lim (...(2^.5)^((2^.5)^((2^.5)^((2^.5)^(2^.5))))). Start from the right and work your way to the left. If you plug these values into your calculator one at a time, you will see that these values do converge to 2 after a while. So, x does in fact = (2^.5). The way you plug this into your calculator is: (2^.5)^(2^.5),.. then (2^.5)^Answer,.. (2^.5)^Answer,.. (2^.5)^Answer,.. and so on.

[@gz4978]

That's why x^^∞=2 and x^^∞=4 haven't the same result!

[@Nylspider]

Hey BPRP, I know you won't see this but I'm going to ask anyways... How did you hold the camera directly above your paper in some of your older videos?

[@lightyagami6647]

There's a stand for that if u want professionality in ur work I used a huge pile of book for that LMAO😂😂😂😂😂

[@rogerkearns8094]

At London's East End we call it an infinite paah taah.

[@danitigre232]

In the next video, can you solve the following integral? Latex: \int _{-a}^a\sqrt{a^2-x^2}\;dx,\;a>0

[@dshifter7]

What happens when you plug in complex values for x in an Infinite Power Tower?

[@sanseng000]

Gets very complex, cannot be easily visualised

[@Alieaz]

Can you do a video on the Newton-Raphson method for solving something like 3x^4-7x-1=0 etc?

[@kaishang6406]

how are the comments days ago if the video is 7 minutes ago?

[@executorarktanis2323]

Member?

[@rahimeozsoy4244]

Patreon