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"Prove" 4 = 2. Can You Spot The Mistake?

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Many people emailed me this apparent paradox showing 4 = 2. Can you figure out where the mistake is?
My blog post for this video
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Award winning paper Knoebel, R. Arthur. "Exponentials reiterated." The American Mathematical Monthly 88.4 (1981): 235-252.
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Wikipedia tetration
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False proof 2 equals 4
jeremykun.com/...
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22 авг 2024

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Комментарии : 1,3 тыс.

@MindYourDecisions 6 лет назад

Links from video description: Award winning paper Knoebel, R. Arthur. "Exponentials reiterated." The American Mathematical Monthly 88.4 (1981): 235-252. www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0 Wikipedia tetration en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights False proof 2 equals 4 jeremykun.com/2012/05/05/false-proof-2-4-as-the-limit-of-an-infinite-power-tower/ Math StackExchange links math.stackexchange.com/questions/87870/are-these-solutions-of-2-xxx-cdot-cdot-cdot-correct/87897#87897 math.stackexchange.com/questions/108288/infinite-tetration-convergence-radius?rq=1?

@farisakmal2722 6 лет назад

MindYourDecisions celebrating the channels' 666k subscribers, let's flashback on another video : ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-dPnOI-djLUs.html

@nvapisces7011 6 лет назад

MindYourDecisions Checking for convergence... Does that mean checking for asymptotes?

@zhanipiet2710 6 лет назад

MindYourDecisions o

@manfredicortonesi8919 6 лет назад

MindYourDecisions sorry if i annoy you but i think that the answer to the 2nd problem (x=4) exists and that it is 1,189207115

@mediaguardian 6 лет назад

I wanted to download and read the paper but the membership price is way to expensive. You might have warned us about that.

@Brooke-rw8rc 5 лет назад

Person: "2 = 4!" Physicist: "That makes sense, they're both basically 3."

@darealpoopster 5 лет назад

And 3 is basically Pi

@yurenchu 5 лет назад

And Pi is basically 3.14 , and 3·14 amounts to 42 . So there you have it: 42

@ultman100 5 лет назад

well sry but 24 is nowhere near 3. Bad joke

@yurenchu 5 лет назад

@ultman100, - "well sry but 24 is nowhere near 3. Bad joke" Oops, it appears that we all overlooked an important _factor(ial)_ in this matter. :-)

@RubyPiec 5 лет назад

@@yurenchu And 42 is nowhere near e, but i'll include it anyways, together its ceilinged down to 44.

@amanahtigetige3375 6 лет назад

5x0=0 100.000.000.000x0=0 Wow 5 and 100.000.000.000 are *same*

@cptn_n3m012 5 лет назад

amanah tigetige actually they’re not

@oplayer9846 5 лет назад

leo bonneville /woosh

@albertocrescini2076 5 лет назад

5x0 = 2x0. If you “eliminate” the zero you’re actually dividing by zero which is indefinite.

@eris4734 5 лет назад

Just gotta divide by 0 on both sides!

@miikey_lol 5 лет назад

Thank you Kanye, very cool!

@GourangaPL 6 лет назад

Ok, so, infinite number of mathematicians get into a bar. First orders a beer. Second orders 1/2 beer, 3rd orders 1/4 beer, 4th orders 1/8 beer, barman gives them two beers and says "guys, you have to know your limits" :)

@farisakmal2722 6 лет назад

GourangaPL lol

@jimqwerasdf 6 лет назад

Ayoub"... First orders A beer... " , which means: 1+½+¼+⅛+...=2

@warwickthekingmaker7281 6 лет назад

actually it is 1+2+3+4+5... divided by itself*2. -1/12/(-1/12*2)=0.5 which just means that 1+2+3+4... is not equal to -1/12

@warwickthekingmaker7281 6 лет назад

It is. Because it was calculated using regular math, despite being a non convergent infinite series. They used the exact same kind of math that I used, so unless the convergent series 1+1/2+1/4... is equal to 0.5 in some way, it is just wrong. It has also been debunked several times.

@nvapisces7011 6 лет назад

Man, a geometric progression where the modulus of the common ratio is less than 1(half) so the series converges. Sum of infinite terms of series=(first term)/(1 - common ratio)=(1)/(1 - 1/2)=2 bottles of beer bought in total

@chinareds54 6 лет назад

"Whenever you have an infinite number of items, you generally need to check for convergence before you simply start doing substitutions and simple arithmetic to get an answer" .... unless your name is Ramanujan.

@mickmccrory8534 5 лет назад

An infinite series is never "Equal"...=... to anything, because you aren't done adding it up yet.. The numbers in the series get smaller & smaller, but they never = zero, so adding up an infinite series of finite numbers = infinity.

@panimbryk 5 лет назад

mick mccrory it’s not a good argument at all because convergent series are equal to a certain number even though “you aren’t adding it up yet”. When you calculate convergent series and you write the sign - equals - it doesn’t mean that you are done adding it up. It’s beacuse it approaches a number, so in this case approaches means equals

@wolfgangster7246 5 лет назад

or reimann

@amit_bisht 5 лет назад

@@mickmccrory8534 An infinite series can be be "Equal" to some finite number. It doesn't require"you" to add it up. Eg. Suppose you have a pie. You can half it, then take one half and half it further and so on. It doesn't matter if you were able to infinitely divide it and then add them or not. The peices were already added up to 1. Similarly 0.9999..... " is equal" to 1 not just tends to. It may also be written as infinite series as 9/10+9/100+9/1000+.....

@yijiun7553 5 лет назад

To understand infinity, you count your steps while walking on the Escherian stairs.

@TheDannytaz 4 года назад

I remember when I first showed the initial problem to one of our lecturers, he said: "This safely assumes there is a solution yes?" Which we replied yeah. We then showed him, excitedly, the solution but he replied 'The real question here is why does it work for 2?' I personally shrugged it out, but years later I see what he was trying to tell us.

@yourlocalbeeswarm1942 5 лет назад

the error is the fact that 4 isn't the same as two

@nyanbrox5418 5 лет назад

good use of backward induction smartypants!

5 лет назад

Fitting name...

@yourlocalbeeswarm1942 5 лет назад

@ wooosh

@kishorekumarsathishkumar1562 4 года назад

that is a legal answer tbh

@Adomas_B 4 года назад

Teachers explaining subjects

@andygong6719 4 года назад

“1=0!” Math teacher: “well that makes sense...”

@obamabinladen2206 4 года назад

Factorial

@squidy7771 3 года назад

I mean, it does. Here are two ways to prove it: 1) how many ways can you organize 0 objects? That's right, 1. 2) Because (n-1)!=n!/n, it means that 0!=(1-1)!=1!/1=1. I learned these two proofs oy today, and I thought they were really interesting.

@Karamuto 3 года назад

@@squidy7771 yeah and to avoid confusion its usually defined like this.

@heimdall1973 3 года назад

@@Karamuto Exactly, and if you start with 1 it's a natural expansion using the formula (n-1)! * n = n! .

@Arnikaaa 3 года назад

Yes

@caiogaiotto9873 6 лет назад

When we mess with infinity, things get a little awkward

@shadrana1 5 лет назад

Infinity is not a number therefore the rules change.

@spoont9544 5 лет назад

Thanos intensifies

@asktnegi5723 5 лет назад

Actually man this answer is wrong. Because using calculator 20 times or approx 20 i get answer 1.988 and which is nearly equal to two

@alansmithee419 5 лет назад

@@shadrana1 they never said it was a number.

@hienduongvan8454 5 лет назад

"a little"

@problematicpuzzlechannel6663 6 лет назад

Nice problem. Always have to be careful with convergence!!

@noorkhudair9792 6 лет назад

Problematic(puzzle channel) what does convergence mean?

@MohammedYaseenAhammed235 6 лет назад

Noor Khudair I second that I don't understand what it means

@nathanisbored 6 лет назад

it means the expression settles down on a particular value, rather than blowing up to infinity or oscillating between multiple values

@noorkhudair9792 6 лет назад

nathanisbored ohh thank you 🖒

@problematicpuzzlechannel6663 6 лет назад

Yup! Divergence is the opposite word

@JohnLeePettimoreIII 6 лет назад

I don't know about anyone else, but my mistake was hitting the play button at this time of night.

@cruzer6571 5 лет назад

yes, it's 22.44 :(

@kanekeylewer5704 5 лет назад

01:16 ffs

@d3t3cted23 5 лет назад

John Lee Pettimore III ю

@Gumbly_ 4 года назад

2:59 AM here ;D

@4sap41 4 года назад

Uhm 4:31 am

@felissilvestris4415 5 лет назад

Nobody: Engineers: Pi=3=e chANgE mY mInDE

@ncedwards1234 5 лет назад

27=e^pi>pi^e=27

@JefeVideoCuras 4 года назад

Engeneers >> you

@bobon123 4 года назад

It's true for small values of Pi or large values of e! /s

@omg_that_anjo 4 года назад

Engineer here (chemical one). And no, its not true. Assuming you re changing your mind.

@ferddoesweirdthingsinlife1040 4 года назад

Euller’s constant is 1.72.....

@MrBrain4 6 лет назад

Excellent video and explanation! This reminds me a lot of Mathologer's video, where he explains the problem with other channels' assertions that 1+2+3+4+5+… = -1/12.

@TimJSwan 2 года назад

cept that it actually is equal..

@vaishanthjv2519 2 года назад

@@TimJSwan Its not

@rohangeorge712 Год назад

@@TimJSwan its not. but u can define some other symbol for that "equality" but that is not actually equal to -1/12 it just diverges to infinity

@vladimbond 4 года назад

6:08 "whenever you have an infinite number of items,..." Yeah sure all the time

@billy.7113 6 лет назад

This video is much better than other recent ones. I learned something new.

@user-tk4zh7wk7h 4 года назад

I was waiting for the explanation of how you find that convergence range

@chaklee435 3 года назад

the paper he referenced probably has a clean rigorous answer. But I can convince myself using a computer. For some value for x, raise it to the power of x over and over and over and get the computer to check that the solution converges. Do that for a bunch of values of x, and now you have some results you can plot.

@apuji7555 Год назад

Well, x^y = y. Take reciprocal of both sides (raise to -1st power). x^-y = 1/y. Multiply by -ln(x). -ln(x) * x^-y = -ln(x) / y. Multiply by y on both sides. -y * ln(x) * x^-y = -ln(x). x = e^ln(x) => x^-y = e^(-y ln(x)). So, -y ln(x) * e^(-y ln(x)) = -ln(x). Take Lambert W (Product Log) of both sides: W(xe^x) = x. -y ln(x) = W(-ln(x)). Divide by -ln(x). y = - W(-ln(x)) / ln(x). Find domain and range of that, and there you go!

@maxinator2002 5 лет назад

And the best part: the curve is actually the inverse of y=x^(1/x) (with the given range and domain).

@davidellis1929 3 года назад

The sequence of finite power towers with x = sqrt(2) starts at sqrt(2) and increases monotonically, converging to 2, not 4. The sequence with x = 1.5 diverges, as does the sequence with x=1.45. As Presh points out in his video, no power tower sequence starting at any value of x converges to a value larger than e.

@hasangarmarudi2178 2 года назад

If you equate x^x^x^...= y and solve for x, you will get that x = y^(1/y) which you can write as the function f(x) = x^(1/x) and find that this function is the flipped version of x^x^x^x....=y, flipped over the x=y line. This mins the maximum of the new function is the maximum range of convergence for the x iteration. And that is as said in the video, equal to e^(1/e)

@rohangeorge712 Год назад

@kouverbingham5997 5 лет назад

"can you spot the mistake? neither can we! but it's buried deep in this paper!"

@DivyanJain 6 лет назад

This is an interesting way to disprove it. I commented on the original video explaining how the exponent diverges, and im glad you released a follow-up.

@catakuri6678 6 лет назад

6:17 Nice text effect!

@adi-sngh 4 года назад

Morph transition in PowerPoint

@vameza1 6 лет назад

Well done! The advice about checking convergency is truly important!

@meureforcodematematicacomp6983 4 года назад

Wow what a magnificent question. I am a teacher from the interior of Brazil and I am surprised by this question

@noahtaul 6 лет назад

The same “paradox” appears with solving x^x^x^...=9/4 and solving x^x^x^...=27/8; they both have the same “solution” since (9/4)^(4/9)=(27/8)^(8/27). In fact, it’s the same “paradox” for any two numbers ((n+1)/n)^n and ((n+1)/n)^(n+1), like (5/4)^4=625/256 and (5/4)^5=3125/1024.

@omrinitecki 6 лет назад

noahtaul How can we prove that there aren't any cases in which both of the numbers generated this way fall within the range of convergence?

@noahtaul 6 лет назад

Well one reason is because then we’d have a paradox! :) another is because the inequalities ((n+1)/n)^n

@Theo0x89 6 лет назад

Nice, you get a correct and a wrong solution for all 1

@thatkidkim 6 лет назад

Theo0x89 the false but seemingly correct solution is called an extrainous solution

@thatkidkim 6 лет назад

Theo0x89 im in fifth grade btw

@nasekiller 5 лет назад

yeah, every "prove" that starts with "let x be a solution of" without justifying the existence of a solution or in this case the well-definedness of the problem is probably wrong.

@tomasbeltran04050 4 года назад

Oof

@dcs_0 6 лет назад

This has been bothering me ever since I watched the previous video Thanks so much! A feel like i was just released from a massive burden! XD

@yijiun7553 5 лет назад

2 = x^2 = x^x^2 = x^x^x^2 = x^x^x^x^2 =... 4 = x^4 = x^x^4 = x^x^x^4 = x^x^x^x^4 =... The above equations are true, proving in all instances, x = 2^0.5 Therefore, y = x^y = x^x^y = x^x^x^y = x^x^x^x^y =... , where x = y^(1/y)

@francisluglio6611 6 лет назад

The real lesson here has nothing to do with the examples you gave. Thank you very much. I might never forget what you're really teaching here about always checking the equation first

@azuarc 6 лет назад

When you have x^4 = 2, it's a fourth degree equation. There are four solutions. Two of them are +-sqrt2, but the other two are imaginary values that have to be solved using DeMoivre's Theorem. And because we arrived at this situation by raising something to an even power, we always have to check for extraneous solutions.

@TheSandkastenverbot 2 года назад

Yeah, but that wasn't the point here

@yurenchu 6 лет назад

Hmm... I haven't been reading too far into this subject, but I think the conclusion depends on the definition of y(x) = x^(x^(x^(x^(x^(x^(x^(...))))))) . If y(x) is defined as the limit of the sequence x, x^x, x^(x^x), x^(x^(x^x)), ... , then the mistake is indeed in the fact that y(x) = 4 has no solution because of a limited range in the convergent domain. However, if y(x) is defined as the multi-branch inverse of x(y) = y^(1/y), then the "mistake" is a logical result of the fact that x(2) = x(4) = sqrt(2) . It would be similar to the case of solving the two equations 5 + sqrt(z) = 3 and 5 + sqrt(z) = 7 . If sqrt(z) is defined as the multi-branch inverse of z = x², then both equations would yield the solution z = 4 . But does that also mean that 3 = 7 ?

@nickpatella1525 6 лет назад

yuri renner What’s neat is if you take the limit of x^b, x^(x^b), x^(x^(x^b)))...... where x is +sqrt(2) and b is 4, you get 4.

@yurenchu 5 лет назад

+Nick Patella, Thanks for your reply. You're correct, with that alternative definition of the sequence and for those values of x and b, all entries in the sequence equal 4, and hence the limit also equals 4. However, one question that arises, is how "robust" this convergence is. For example, if b = 4+e, where e is a very small value around 0, would the sequence still converge? And if yes, would it still converge to 4? Or what if b = 4 and x = sqrt(2)+e ? The answers tell us something about how suitable/usable/meaningful this alternative sequence definition is, compared to the "standard" definition and/or other definitions.

@Jhet 6 лет назад

I love your videos, and I thought you said "I'm Fresh Tall Walker" every video until I saw your name in the bottom corner

@adrienconverset6571 6 лет назад

Its like the sum of all positive integers 's value being -1/12. Just because you find a solution doesn't mean it is a solution.

@AMAJAR44 6 лет назад

No. You are missing something. It is called mathematics. What you just said is as true as saying that the equation x² = -1 has no solution. Being able to understand this is what separates us from pocket calculators.

@RunstarHomer 6 лет назад

Adrien Bellaiche ummm.. no. That's true and actually is a foundational idea in relativity

@gamerdio2503 6 лет назад

Runstar Homer ?????

@praisethyjeebus 6 лет назад

Bynokel Lets just "imagine" that that equationg does have a solution. It is i...

@chervilious 6 лет назад

Nope it's mathematically proved, thus it have to be valid, a solution doesn't have to be logical, nor does it have to be one. to simply put x+1=x is valid in infinity, if x is infinity even x^x=x is valid in infinity (and if x=1). The Banach-Tarski Paradox and Ted-ed infinity hotel may help you understand. That's the beauty of mathematics, because we can't understand something in our sense, but we can describe it mathematically.

@assimilater-quicktips 4 года назад

I wonder what the range of convergence is over the complex plane. I've never worked with this kind of function

@floyo 6 лет назад

Repeating the function root(2)^x gives the infinite exponent formula. If you start with exactly 4 or 2 the output will always be 4 and 2. So the difference is what you start with. Still 2 is often found the most beautiful solution, because this function diverges to 2 for any starting number (Except for 4). That's what I think. Now, lets watch the rest of the video!

@floyo 6 лет назад

Wow, my approach was completely different!

@floyo 6 лет назад

I meant converges to 2 btw

@user-nx8mc3om9r 4 года назад

There is an alternative method to prove x=√2 x^x^x^...=2 Log on both sides x^x^x^...Log x = log2 2logx=log2 Logx = (1/2)log2 Logx = log√2 x=√2

@rahimeozsoy4244 4 года назад

x^x^x^x.. Logx = log2 x^x^x^x.. = log2/logx x^x^x^x.. = 2 log2/logx = 2

@zi_t 6 лет назад

How do we find the domain and range of the function y = x^(x^(x^...))? Just wondering.

@sam2026 6 лет назад

Isn't it for all positive real numbers...?

@nvapisces7011 6 лет назад

Zi Nean Teoh U sketch the graph and label the assymptotes. From there u determine the domain and range... I'm just guessing, not sure how to sketch the graph

@NA-xh4nb 5 лет назад

No need to sketch a graph...You just need to use differentiation to find max or min value

@NA-xh4nb 5 лет назад

@Christopher Americanos x^y=y is not that hard to differentiate....just use logarithm

@OMGclueless 5 лет назад

@@NA-xh4nb That doesn't work. x = sqrt(2), y = 4 is indeed a valid solution of x^y = y but not of the original equation. Analyzing that function won't tell you anything about the the domain and range of the function y = x^x^x^x...

@GogiRegion 6 лет назад

My guess was something along the lines of no solution, but you can solve it, even if that solution is false, similar to an extraneous solution. It sounds like I was right.

@theginginator1488 6 лет назад

Your x^x^x^... relation is only defined on the domain of (0,e^(1/e)). It is the inverse relation of y=x^(1/x) which has a maximum value at x=e Update: basically correct, but was wrong about the lower bound of the function

@n0ame1u1 4 года назад

I agree with you that it seems like the function should be defined on (0,e^(1/e)] as well.

@aaronleperspicace1704 5 лет назад

Wow, thanks for explaining. Because of the power tower thing, I believed for some time that n is always equal to the nth root of n raised to itself infinite times. I tried it with the cube root of 3 raised to itself infinitely and got the equation ( ∛3)^x = x. Astonishingly, I did get the answer equal to 3! Cube root of 3 equals 1.44 something so it is in the defined range by a very narrow margin but 3 is out of the range. But it seemed to work and it made me lose sleep. But taking a closer look at the graph of this equation reveals that the two lines (y = ( ∛3)^x and y = x) intersect at two separate places quite close to each other. One is of course 3 and the other is around 2.5, the latter of which is in the range.

@eugeneimbangyorteza 3 года назад

In the process, it should have been x^4-4=0. Then, complex solutions are on the table. In fact, this is easier to solve using polar complex form.

@AhmadRaza-le2ow 5 лет назад

actually the mistake in X^x^x^x^x^x^.... = 2 is X^x^x^x^x^.... ≠ 2 simple answer no equation, 100% real

@fgvcosmic6752 6 лет назад

Before watching, im assuming this will be 4th root(4) =sqrt(2) And infinite that

@imadoge5036 2 года назад

While trying to solve this, another answer came up. For better understanding, x to the power of x infinately many times is 2. To get 4, the base simply becomes the exponent and infinately many Xs taking the place of base are added meaning x to the power of infinately many times x to the power of infinately many times x, so while that is equal to 2, if it becomes and exponent for infinately many times x, it becomes 2 to the power of 2 and so on. Same goes for 8, 16....etc. Very interesting problem. My explenation may be a little off, but I hope it gets the point across.

@ekadria-bo4962 6 лет назад

Because the interval of convergence is only e^-1 < x e^e^-1

@bernhard5295 6 лет назад

This was very interesting. It would be great if you make more videos like that. Thumps up👍

@jonni2734 6 лет назад

But if I substitute sqrt 2 to x I obtain something which doesn't converge to any value (if you try with the calculator you obtain (sqrt 2)^(sqrt 2)^(sqrt2)^... which is equal to 2 only after 2 iterations, which is (sqrt 2)^(sqrt 2)^(sqrt 2), otherwise it becomes bigger and bigger). What di you think about this???? Anyway good video!!

@NoNameAtAll2 6 лет назад

jonni2 It is sqrt(2)^(sqrt(2)^sqrt(2)) Not (sqrt(2)^sqrt(2))^sqrt(2)

@jonni2734 6 лет назад

NoName Mmm yes, like this it works, so the function is y = x^(x^(x^(x^(x^...))))

@NoNameAtAll2 6 лет назад

jakolu 2 roots is less than 2 3 roots are 2 5 roots are 4

@jonni2734 6 лет назад

NoName Yes

@Walkerman379 6 лет назад

There are some good replies already. I'd just like to add that you can do it easily on a scientific calculator doing the following: 1. Type in square root of 2 2. Type sqrt(2)^ANS 3. Keep hitting the enter button and watch it converge

@Zwaks 6 лет назад

How does one check for convergence in such an equation?

@wopich9570 6 лет назад

I remember think about this very thing when I saw that video! Glad you’re finally addressing it!

@gunhasirac 4 года назад

Solving those questions naively is like doing physics. Ruling out the “math problem” rigorously is like doing math.

@jiaminzhu406 5 лет назад

how to prove they converge within this range?

@gswcooper7162 10 месяцев назад

The problem comes that if you graph y=x^y (which you can get from y=x^x^x^x^x... by substituting y for the exponent on the other side), you get a graph that includes this curves, but branches off to the left (it splits and one branch goes to x,y=0,0, the other goes to 0,1) and on the right it CURVES BACK AROUND and approaches the line x=1 asymptotically. The point where x=e^(1/e) is the maximum value at which the infinite tetration converges, but above that all the other points on the line are still valid, but you cannot converge to them; they are unstable equilibria.

@colinrichardday 6 лет назад

Wikipedia has a nice description of hyperexponentiation. It would have been nice if you had a definition of it in the first video.

@thefalconator5971 6 лет назад

Your method of solving the problem assumes that there is a solution. x^x^x^x...=4 does not have a solution.

@wolfgangster7246 5 лет назад

how do you prove there is a solution to any problem?

@alansmithee419 5 лет назад

@@wolfgangster7246 it will depend on the problem you're trying to find the solution to.

@ElZedLoL 4 года назад

@@wolfgangster7246 maybe you assume there is a solution and then can conclude something wrong (like 4=2). So the premise must have been wrong. There is no x such that x^x^... =4

@nekilikizhrvatske3336 4 года назад

Its a 4th root of 4 and that is equal to square root of 2

@noir4659 3 года назад

@@wolfgangster7246 just go on desmos and graph "f(x) = x^x^x^x..." yourself, you'll find f (square root 2) only gives you 2

@mikrokaulis32k68 6 лет назад

Hello. Nice video. How did they found the Df of the function? How do we know is [1/e^e - e^1/e]??

@NA-xh4nb 5 лет назад

The equation is x^y=y or, x=y^(1/y) Now just differentiate it and find the maximum and minimum value

@karoshi2 3 года назад

That's funny because I first learned about the solution with powertower(x) = 8 with a lengthy explanation by some mathematician. Only now I realized the issue with convergence.

@bricepilard5267 6 лет назад

Great video . I had the good intuition, but i wasn't able to find the right demonstration.

@azoznail1223 5 лет назад

The formula x^x^x^x^x ..... It can equal to 2 raised to any positive even power greater than 0 2,4,16,....... , but the only right answer is 2

@yurenchu 5 лет назад

- "The formula x^x^x^x^x ..... It can equal to 2 raised to any positive even power greater than 0 2,4,16,....... , but the only right answer is 2 " Huh, wut? The outcome of x^x^x^x^... (which means x^(x^(x^(x^( ... )))) , by the way) depends on the value of x. For examples: - when x = √2 , then x^x^x^x^... = 2 - when x = 1 , then x^x^x^x^... = 1 - when x = (2/3)√(2/3) , then x^x^x^x^... = 2/3 - when x = 1/4 , then x^x^x^x^... = 1/2 - when x = (³√18)/2 , then x^x^x^x^... = 1.5 However, there exist no values of x for which x^x^x^x^... equals 4, 16, 64, 256, ... etc. In fact, there are no values of x for which x^x^x^x^... is greater than Euler's number _e_ (= 2.7182818... ) .

@shriharshithk8256 6 лет назад

I thought the answer for x^x^x^x^(till infinity)=4 was quadrant or fourth root of four.I mean x=4√4. So x^x^x^x^ till infinite where x=4√4= 4. Am I right?

@yurenchu 6 лет назад

How much is (fourth root of four)² ? Is it not 2?

@htcdezire1770 6 лет назад

pls reply to this comment Mind your decisions....

@yurenchu 6 лет назад

There are _four_ fourth-roots of 4; which one are you guys referring to? If x⁴ = 4 , then x = √2 OR x = -√2 OR x = +i√2 OR x = -i√2 However, the first of these four suggestions does not work, because when x = √2, then x^x^x^x^... (when defined as the limit of the sequence x, x^x, x^(x^x), x^(x^(x^x)), ... etc.) converges to 2, and not to 4. So, are you guys suggesting that the other three roots (or at least one of them) do result in an "infinite power tower" that converges to 4?

@shriharshithk8256 6 лет назад

I mean x=4√4. So x^x^x^x^ till infinite where x=4√4= 4.... Am I right?

@calighast 6 лет назад

No.

@curiousminds2603 4 года назад

Sir when i draw this function graph by using geogebra this show one extreme point or (0,1) and one undefined point But my question is that deleting the one raised to power term after approximately 33th power then function behave differently it changes its functionality according to it even or odd powers pls explain me about this why it is happened

@virtuesimo1535 6 лет назад

Wow... So I've actually explored this problem before on my own and arrived at the same answer as you. Thank you for showing this to the public!

@mykolagirnyi9030 6 лет назад

🎉🎊OMG! YOU HAVE 666 666 SUBSCRIBERS! CONGRATULATIONS! 🎊🎉

@alexismandelias 6 лет назад

Mykola Girnyi 666602 as of now

@ziyad1809 5 лет назад

Oooh I was asked this in a maths competition but I just guessed root2 (it was multiple choice)

@emperior Год назад

Great vid! I had no idea about that convergence thing, though. Is it taught at graduate-level math?

@mohitbramhane7604 5 лет назад

We all are forgetting that x^x^x....∞ =y This implies that x^y≈y ,nearly equal.... And the power is also greater than 1 so small change in power (raised to the power of very large number) is a significant quantity ,hence it cannot be neglected

@lythd 5 лет назад

Um no. It is not nearly equal it is equal. Infinity + 1 is still Infinity not nearly Infinity or around about Infinity it is Infinity. Same thing with subtraction Infinity - 1 is almost Infinity it still is Infinity. However good attempt.

@svenminoptra756 6 лет назад

how does a person go and graph something like this? and I mean on a TI type calculator

@MichaelRothwell1 6 лет назад

Todd Rogers, note that the required function is the inverse of the function y=x^(1/x), so plot this function (window x & y both 0 to e), then its inverse using DrawInv.

@yurenchu 6 лет назад

Michael Rothwell, Note though that it's only the inverse of the function y = x^(1/x) as far as x lies between 1/e and e (which means y lies between e^(-e) and e^(1/e), which is the interval of convergence for z(y) = (y^y^y^y^y^...) when defined as the limit of the sequence y, y^y, y^y^y, y^y^y^y, ... ). For values of x between 0 and 1/e (and hence y between 0 and e^(-e) ), the function y = x^(1/x) does have an inverse but it doesn't coincide with the above definition of z(y) = y^y^y^y^... (which is divergent in that domain/range).

@NA-xh4nb 5 лет назад

You can just differentiate x=y^(1/y) and find the maximum and minimum value

@weckar 5 лет назад

Waitaminute... 4= 2^2, so is it not just two infinite power towers stacked on top one another?

@thimowellner7686 5 лет назад

the brackets are different. 4 is (x^(x^(x...)))^(x^(x^(x...)) and 2 is x^(x^(x^(x^(x^(x...)))))

@KnakuanaRka 5 лет назад

Another way of explaining it is that if you actually calculate the value of increasingly large stacks of powers, they converge to 2; the 4 is essentially a spurious solution created by substituting the original power stack into itself and solving s(2)^y=y.

@rkpetry 6 лет назад

*_...I think the bigger-trouble with this, 'infinity', is that it doesn't start, convergence, in its formulation-there is-no-last-x in ↑x to start the convergence process n'even for x=√2..._* *_...in the opposite direction-if you climb its infinite sequence, to find some start, you're actually integrating back out, so-there-should-be no, start, the way you're expecting it..._*

@thebluepc5 5 лет назад

I'll give you 4 apples, but suddenly it becomes 2

@jatinagrawal9341 5 лет назад

But , how did you plot the graph 🤔

@mosab643 4 года назад

exactly

@thesizzles1337 3 года назад

Billboard: [ 4 = 2 ] Person staring: that does not make se- Literally everyone behind that person: IT DOES

@titfortat4405 5 лет назад

To sum up the resolution of the paradox in one sentence: The Lambert W function is not a single valued function for all x.

@TheDutLinx 6 лет назад

Great video! Your explanation was clear and easy to follow, thanks!

@JohnRandomness105 6 лет назад

I didn't email you, but I was one of those who commented on your earlier video that 4 also satisfied the method with the same square root of two. I think that there is a range (or radius?) of convergence where raising the infinite chain of powers actually converges. Two is inside while four is outside.

@farisakmal2722 6 лет назад

aww, memories. the first power tower video was one of the first videos I watched when I subscribed to this channel. thanks presh!

@chrissekely 6 лет назад

I would guess that x^x^x^x^x^x^...=y is solveable for any y if we allow complex numbers. If so, could someone explain how? If not, then couldn't a new type of number be created so this can be solved for any y? (Similar to how i was suggested to equal sqrt(-1) ).

@chervilious 6 лет назад

you could think like this, the equation of 4=2 is real, but not understandable by our common sense, mathematical is a universal language, we know this because quantum mechanics can be explained mathematically but hard to explain verbally, in "infinite universe" it's beyond our logical sense, which by any number can be anything, and any number is equal, larger, and smaller at the same time. To explain it easily, imagine if you have a hotel that have infinite number of room, the room are fully booked, then a guess came, all you need to do is to move 1 to 2, 2 to 3, and so on, if infinite people came, just let the people in the hotel move to the next even number, and let the new people add odd number, even when infinite^infinite people came, (imagine infinite row, where each row have infinite people) you need to move each people with a prime number. infinite row 1 --goes to room-->2^1,2^2,... infinite row 2 --goes to room-->3^1,3^2,... Since they're infinite number of prime, so this can be true. The infinity has a value, the above answer are correct, but it just not understandable by our sense, but it still work in our world, even a little you can find 'zeta function' which uses -> 1+2+3+... = -1/12 and other example

@SmileyMPV 6 лет назад

Well... Since powers are not always well-defined in the complex numbers, this is not going to work. See (-1)^(1/2)=±i for example. No reason to prefer i over -i. There is no continuous complex function f such that f(z)^2=z for all z, so a^(1/2) is not well-defined. The same goes for all fractional powers, and don't even get me started on irrational powers.

@chervilious 6 лет назад

well because it's complex and cant be explained easily?, my point infinity can't be understand in our common sense

@Nukestarmaster 6 лет назад

No, complex numbers don't help with divergent equations. Although I do wonder what the domain and range of the function is in the complex plane.

@poppy3879 6 лет назад

You could "fix" it by saying the root from 4 = 2 which is 2 = 2

@yurenchu 6 лет назад

No, that doesn't "fix" it. One result says that when x = √2, then y = x^x^x^x^... = 2 . Another "result" says that when x = ⁴√4 = √2, then y = x^x^x^x^... = 4 . In other words: for the same value of x, one result says that y(x) = 2 and another result says that y(x) = 4. The apparent contradiction between those two results is not simply "fixed" by merely saying "√4 = 2" .

@xn85d2 2 года назад

The very first step appears to be incorrect. Assuming that the exponent = 2 makes no sense at all, except if you assume that the last exponent in your infinite series of exponents does nothing at all. In other words, you've implicitly assumed convergence and got lucky whilst doing the wrong thing.

@romilchaudhary7011 2 года назад

I dont know but thats how we solve these problems in maths I can give you example y=underroot(x+underroot (x +underroot(x + underroot(x......))))-Infinty times then we can just asume that since it has infinite value decuding one underroot x wont be problem so it can be wriiten as y=underroot(x +y) Have you tried these type of questions in school??

@EdwinBiclar 4 года назад

Is this Set Theory subject?

@jasonp500 5 лет назад

Prove that 4=2. 1. First prive that 1=2 2. Becauae 1=2, then 2^1 =2^2. 3. Therefore 2=4, and 4=2.

@PowerhouseCell 4 года назад

Who's here after 3blue1brown did a video on this?? 😂

@Pravin.Shidore 3 года назад

You reminded me of collage days, I remember on very first day of junior college my math teacher proved 1 = - 1.

@amologusmogusmogumogu2535 5 лет назад

ANY CURVE has infinite domain and range if extended to complex plane.The answer to the question is plus minus root 2 i by solving x^4=4 and taking the other values.

@yurenchu 5 лет назад

@Rayeed Hasan, - "The answer to the question is plus minus root 2 i by solving x^4=4 and taking the other values. " Nope, the sequence x, x^x , x^(x^x) , x^(x^(x^x)) , ... etc. doesn't seem to converge towards 4, neither for x = i*√2 nor for x = -i*√2 See here: - www.wolframalpha.com/input/?i=a%5Bn%2B1%5D+%3D+(i*sqrt(2))%5E(a%5Bn%5D),+a%5B1%5D+%3D+1 - www.wolframalpha.com/input/?i=a%5Bn%2B1%5D+%3D+(-i*sqrt(2))%5E(a%5Bn%5D),+a%5B1%5D+%3D+1

@amologusmogusmogumogu2535 5 лет назад

@@yurenchu how does this sequence relate ?

@yurenchu 5 лет назад

@Rayeed Hasan, - "@yuri renner how does this sequence relate ?" The expression y = x^x^x^x^... is defined as the limit of the sequence x , x^x , x^(x^x) , x^(x^(x^x)) , ... etc. (see the 1981 article by R. Arthur Knoebel, which is referenced in the video). In other words, if we'd define the sequence a[n] as a[0] = 1 a[1] = x a[2] = x^x a[3] = x^(x^x) a[4] = x^(x^(x^x)) : a[n+1] = x^(a[n]) then y = x^x^x^x^... corresponds to lim{n-->∞} a[n] (which we could call a[∞] in short.) Your claim is that when x = i*√2 or x = -i*√2 , we would have y = x^x^x^x^... = a[∞] being equal to 4. The wolfram|alpha links in my previous reply show that that doesn't seem to be the case. In other words, your claim appears to be incorrect.

@amologusmogusmogumogu2535 5 лет назад

@@yurenchu OK THEN PLEASE HELP ME , In solving the first problem , 2=x^x^x^...... ,the answer is , √2 right ? So we can say , √2^ √2^ √2^........=2 right ? But in this let me take one √2 from the left to the right. Then it becomes √2^ √2^ √2^........=2^(1/ √2) . Bothe cant be correct at the same time . So all of this is counterintuitive. PLEASE help me understand. If my reply was unclear let me know.Thanks Sir

@Intradiction 6 лет назад

Wait, using exponent laws wouldn't that mean 1/2 * 1/2 * 1/2 ... = 2? I guess with infinite stuff an "answer" is not really intuitive. EDIT: sorry lol I can't belive I thought the exponent was 1/2 and not 2^1/2, thx @Hi OK

@danielauto3767 6 лет назад

You're doing the power from the bottom up. You're supposed to do it from the top down...

@Theo0x89 6 лет назад

That's not how exponent laws work.

@blackmamba1261 6 лет назад

Remember that its Sqrt(2) being stacked, not 1/2.

@victoriam6569 6 лет назад

I didn't get why there are these limits

@richbros1 6 лет назад

Victoria M Think about it like this. Any higher x value than in the domain would lead to the value of the number on the right to go to infinity. Try plugging in two for x. it’s 2^2^2... which is a constantly growing number. The range comes from plugging in the domain max and mins

@forgottenforger4469 6 лет назад

Basically put; there are numbers that when raised to the exponent-infinity do not approach infinity but fall inside the set domain and range as their numerical value balances itself out so to speak with every pattern. sqrt(2) works as an answer because sqrt(2)'s nature allows it to be raised unto itself infinitely without creating the following paradox: x^n = y = u, where n equals (x^x)z and u /= y.

@maestro69hz 5 лет назад

I would get owned by this question on a math test

@LudwigvanBeethoven2 6 лет назад

This is the nicest thing ive ever seen for a while. So many e relations. Thats just beautiful

@sebastianzaczek 5 лет назад

Did someone say X-ponent?

@tomriddle2257 5 лет назад

x^x^x^x... =2 I thought the answer would converge to 1 for sure.

@arpitmalik27 3 года назад

It's not that x^x^x^x... is 2. It's the equation whose solution is x And clearly it cannot converge with 2 as 1 is not a solution🤦🏻

@tomriddle2257 3 года назад

@@arpitmalik27 OH it's clear that the solution cannot converge to 1, as 1 is no solution? Then what about x^Inf = 2? What is x? Clearly it can't converge to 1, as 1 is no solution.. I just had that idea for a second....

@arpitmalik27 3 года назад

@@tomriddle2257 you are repeating my point😐

@tomriddle2257 3 года назад

@@arpitmalik27 That was sarcastic. My second example has a solution near 1.

@arpitmalik27 3 года назад

@@tomriddle2257 lol it's solution is near 2 It's x=2.2991 OR x=e^(√ln2)

@marklevin3236 3 года назад

Square of a negative number is positive. For example (-7)^2=49. Let a1,a2, a3..... be a sequence of fractions with odd numerator and denominator that converges to limit of 2. Simplest example would be an=(4n+1)/(2n+1)... Let bn=(-7)^an. Sequence bn consists of negative numbers and converges to a limit of -49.. . How to explain this paradox ?

@Snake_In_The_Box 6 лет назад

Considering the following sequence: k, sqrt(2)^k, sqrt(2)^(sqrt(2)^k)... It converge when k

@pedrammahdavi5863 Год назад

Agree

@evniqueen7581 6 лет назад

I understand

@user-eo5bh2zg2 6 лет назад

Yay

@shuiyanli5000 6 лет назад

ĐĘVŅĮ QůèęÑ I

@Anonymous1.0.8 6 лет назад

03:03 I love it

@graycuzzies5834 4 года назад

a=b multiply by a a^2=ab subtract b^2 a^2-b^2=ab-b^2 factorise (a+b)(a-b)=b(a-b) divide by (a-b) (a+b)=b substitute a for b b+b=b combine like terms 2b=b divide by b 2=1

@all8504 Год назад

If you take the equation y = x^x^x^..., you can change it to y = x^y, removing the infinite power tower but expanding the range. y = x^y actually has two y values for every x value greater than 1 and less than e^(1/e). y = x^y is identical to y=x^x^x^... except that for at x = e^(1/e), it adds to the original curve and goes back to the left and up and asymptotes at x = 1 (you can see what I am saying clearly in desmos). This is the reason that at 2:38, it was seen that (sqrt2, 4) is a solution *because* it is a solution to y = x^y, which has no maximum y value, but is not a solution to y=x^x^x^... which has a finite maximum value, that being e.

@dimitrakisladasi5496 6 лет назад

why all exponents above the base equals to 2???

@inactivewhiteboard940 6 лет назад

If x^(x^(x^9=(x... = 2, then you can do x^2=2, since that portion is also has the infinite exponents

@rv9809 6 лет назад

What is infinite + infinite+ infinite+ infinite+ infinite+ infinite upto infinite times

@alice_in_wonderland42 6 лет назад

a bigger infinity☺

@thesinistermobs1564 6 лет назад

Ravi Raj R square No, it’s the same

@alice_in_wonderland42 6 лет назад

@@thesinistermobs1564 yeah i know i was just joking but real numbers have a bigger infinity than whole numbers or integrrs this thing is not there in general mathematics we study. check the web

@johannesvanderhorst9778 5 лет назад

It is the same when we view "infinity" as a cardinal number, ie. as the size of a set. But it is a bigger infinity when we view "infinity" as the limit of 1/z for z -> 0 when doing analysis.

@rv9809 5 лет назад

@@alice_in_wonderland42 how do you feel about bigger infinity

@gloweye 6 лет назад

for any x > 1, infinity exponent goes to infinity. Therefore, there is no solution. (and 1 remains 1, 0 < 1 < 1 goes to zero. Negative numbers sling around 0 and start oscillating. in the case of -1 < z< 0, it also goes to 0, but x < -1 goes to +/- infinity). Taking something to a power over 1 always increases the value, so any infinite iteration MUST be infinity.

@wanyinleung912 6 лет назад

Gloweye Not in this case.

@katzen3314 6 лет назад

It increases the value with each iteration, yes. But that doesn't mean a sequence can't converge. Try it on a calculator, it converges to 2 as the increments get smaller each iteration. Similar to how the series 1/2, 3/4, 7/8... converges to 1 with infinite iterations.

@PvblivsAelivs 4 года назад

The shortcut method computes a possible value of x which _might_ work. If there were a value (x) for which such an infinite power tower gave the answer 4, it would have to be sqrt(2). That power tower does not give that answer. Therefore there does not exist an x which gives that through power tower.