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?
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" :)
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.
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
"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.
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.
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
@@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+.....
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.
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.
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.
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.
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)
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.
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!
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.
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.
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.
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.
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
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.
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.
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
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 ?
+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.
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!
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 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...
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.
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??
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.
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.
@@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
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.
I'd like to second a comment from 4 years ago: "TheGinginator14 4 years ago (edited) 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 reply: lord_ne 3 years ago I agree with you that it seems like the function should be defined on (0,e^(1/e)] as well" Both of these commentators are correct about what they say. If you graph the function x=y^x, it can be seen that the graph exists before 1/e^e pretty smoothly and y appears to approach to a value close to 0.169... as x approaches to 0.
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!!
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
You put substitution, and made equation - x^2=2 Then I am also doing a substitution x^x^x^x^x^x...=y then x^y=y >>>>>>And what you said about domain range is false as it is not even a function. Then how can you define domain and range?
I think I get it. People are confusing 1.42^1.63 (using rough numbers here) which equals 1.76 with 1.63^1.42 which equals 2. The former allows for infinite number of powers while the other doesn't.
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.
An essential problem here is that you haven't defined what x^x^x^x^... means in the first place. Such operation isn't defined on a standard analysis course so it isn't common knowledge, therefore you should define it in the very beginning, before or immediately after posing the problem from the competition. If you had done that, the question "what's wrong in this reasoning" would be much more meaningful and the discussion about the solution would provide deeper understanding. Without doing that, your explanation is rather brief and shallow.
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" .
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.
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.
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
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.
- "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... ) .
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.
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
I’m reading the comments and I am once again brought to light about how little math I truly know when compared to others. I can do AP Calc AB no prob but what you guys are discussing is a different matter entirely...
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.
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?
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?
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
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.
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).
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
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.
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.
The example I like is using 10 which would yield x = the 10th root of 10. But the 10th root of 10 is actually less than the square root of 2 (raising both to the power of 20 shows 100 < 1024). So with a < b raising "a" to infinite powers of itself is greater than "b" being raised to infinite powers of itself. Clearly impossible.
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) ).
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
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.
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 ?
very nice one --- years ago, I came across the following problem: Suppose S= 1+2+4+8+16+32+.....; therefore 2S=2+4+8+16+32+......; that means 2S=S-1, implying S= - 1. How's that possible? The error here lies in the fact that the series is not convergent, so we can't apply equations by doing math on a non-convergent series, otherwise we will get puzzling (and erroneous) results like this
*_...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..._*
It's very basic and simple : x^x^x^x^x...=4 implies x=sqrt(2) It's an implication, not an equivalence. So after finding a "solution" by a chain of implications, we must check that the reciprocal is true, otherwise you can not conclude you found a solution. If implication gives a solution, and the reciprocal is false, you just show there is no solution (or at least no solution in the domain you are looking at. If your implications only take care of real numbers and the real solution you find doesn't work, you just prooved there is no real solution. But there can be a complex one you forgot about with your implications). Conclusion : Don't try to think in terms of equivalences, you'll fail most of the time. Work with implications, but at the end, never forget to verify the reciprocal !
@@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....
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.
@@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
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.
But there is also another problem: If we use the same trick, we can say that 2 = (x^x)^2 and 2 = x^2 (+ many others) 2 = 2 (x^x)^2 = x^2 x = x^x And this can only be true if x=1 But then 1^1^1^1^... isn’t equal to 2
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?
MindYourDecisions celebrating the channels' 666k subscribers, let's flashback on another video : th-cam.com/video/dPnOI-djLUs/w-d-xo.html
MindYourDecisions Checking for convergence... Does that mean checking for asymptotes?
MindYourDecisions o
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
I wanted to download and read the paper but the membership price is way to expensive. You might have warned us about that.
5x0=0
100.000.000.000x0=0
Wow 5 and 100.000.000.000 are *same*
amanah tigetige actually they’re not
leo bonneville /woosh
5x0 = 2x0. If you “eliminate” the zero you’re actually dividing by zero which is indefinite.
Just gotta divide by 0 on both sides!
Thank you Kanye, very cool!
Person: "2 = 4!"
Physicist: "That makes sense, they're both basically 3."
And 3 is basically Pi
And Pi is basically 3.14 , and 3·14 amounts to 42 .
So there you have it: 42
well sry but 24 is nowhere near 3. Bad joke
@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.
:-)
@@yurenchu And 42 is nowhere near e, but i'll include it anyways, together its ceilinged down to 44.
the error is the fact that 4 isn't the same as two
good use of backward induction smartypants!
Fitting name...
@ wooosh
that is a legal answer tbh
Teachers explaining subjects
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" :)
GourangaPL lol
Ayoub"... First orders A beer... " , which means: 1+½+¼+⅛+...=2
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
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.
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
"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.
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.
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
or reimann
@@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+.....
To understand infinity, you count your steps while walking on the Escherian stairs.
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.
When we mess with infinity, things get a little awkward
Infinity is not a number therefore the rules change.
Thanos intensifies
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
@@shadrana1 they never said it was a number.
"a little"
Nice problem. Always have to be careful with convergence!!
Problematic(puzzle channel) what does convergence mean?
Noor Khudair I second that I don't understand what it means
it means the expression settles down on a particular value, rather than blowing up to infinity or oscillating between multiple values
nathanisbored ohh thank you 🖒
Yup! Divergence is the opposite word
“1=0!”
Math teacher: “well that makes sense...”
Factorial
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.
@@squidy7771 yeah and to avoid confusion its usually defined like this.
@@Karamuto Exactly, and if you start with 1 it's a natural expansion using the formula (n-1)! * n = n! .
Yes
I don't know about anyone else, but my mistake was hitting the play button at this time of night.
yes, it's 22.44 :(
01:16 ffs
John Lee Pettimore III ю
2:59 AM here ;D
Uhm 4:31 am
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.
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.
cept that it actually is equal..
@@TimJSwan Its not
@@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
And the best part: the curve is actually the inverse of y=x^(1/x) (with the given range and domain).
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)
fr
6:08 "whenever you have an infinite number of items,..." Yeah sure all the time
This video is much better than other recent ones. I learned something new.
I was waiting for the explanation of how you find that convergence range
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.
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!
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.
Nobody:
Engineers:
Pi=3=e
chANgE mY mInDE
27=e^pi>pi^e=27
Engeneers >> you
It's true for small values of Pi or large values of e! /s
Engineer here (chemical one). And no, its not true. Assuming you re changing your mind.
Euller’s constant is 1.72.....
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.
Oof
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.
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?
Well one reason is because then we’d have a paradox! :) another is because the inequalities ((n+1)/n)^n
Nice, you get a correct and a wrong solution for all 1
Theo0x89 the false but seemingly correct solution is called an extrainous solution
Theo0x89 im in fifth grade btw
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.
Yeah, but that wasn't the point here
6:17 Nice text effect!
Morph transition in PowerPoint
Well done! The advice about checking convergency is truly important!
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
I agree with you that it seems like the function should be defined on (0,e^(1/e)] as well.
Wow what a magnificent question. I am a teacher from the interior of Brazil and I am surprised by this question
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.
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.
Adrien Bellaiche ummm.. no. That's true and actually is a foundational idea in relativity
Runstar Homer ?????
Bynokel Lets just "imagine" that that equationg does have a solution. It is i...
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.
"can you spot the mistake? neither can we! but it's buried deep in this paper!"
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
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 ?
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.
+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.
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
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)
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!
Wow, my approach was completely different!
I meant converges to 2 btw
How do we find the domain and range of the function y = x^(x^(x^...))? Just wondering.
Isn't it for all positive real numbers...?
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
No need to sketch a graph...You just need to use differentiation to find max or min value
@Christopher Americanos x^y=y is not that hard to differentiate....just use logarithm
@@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...
actually the mistake in X^x^x^x^x^x^.... = 2 is X^x^x^x^x^.... ≠ 2
simple answer no equation, 100% real
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.
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??
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
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.
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.
Because the interval of convergence is only e^-1 < x e^e^-1
Your method of solving the problem assumes that there is a solution. x^x^x^x...=4 does not have a solution.
how do you prove there is a solution to any problem?
@@wolfgangster7246 it will depend on the problem you're trying to find the solution to.
@@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
Its a 4th root of 4 and that is equal to square root of 2
@@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
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.
This was very interesting. It would be great if you make more videos like that.
Thumps up👍
I'd like to second a comment from 4 years ago:
"TheGinginator14 4 years ago (edited)
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
reply: lord_ne 3 years ago
I agree with you that it seems like the function should be defined on (0,e^(1/e)] as well"
Both of these commentators are correct about what they say. If you graph the function x=y^x, it can be seen that the graph exists before 1/e^e pretty smoothly and y appears to approach to a value close to 0.169... as x approaches to 0.
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!!
jonni2
It is sqrt(2)^(sqrt(2)^sqrt(2))
Not (sqrt(2)^sqrt(2))^sqrt(2)
NoName Mmm yes, like this it works, so the function is y = x^(x^(x^(x^(x^...))))
jakolu
2 roots is less than 2
3 roots are 2
5 roots are 4
NoName Yes
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
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.
I wonder what the range of convergence is over the complex plane. I've never worked with this kind of function
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
x^x^x^x.. Logx = log2
x^x^x^x.. = log2/logx
x^x^x^x.. = 2
log2/logx = 2
Hello. Nice video. How did they found the Df of the function? How do we know is [1/e^e - e^1/e]??
The equation is x^y=y or, x=y^(1/y)
Now just differentiate it and find the maximum and minimum value
I remember think about this very thing when I saw that video! Glad you’re finally addressing it!
Before watching, im assuming this will be 4th root(4) =sqrt(2)
And infinite that
You put substitution, and made equation - x^2=2
Then I am also doing a substitution x^x^x^x^x^x...=y then
x^y=y
>>>>>>And what you said about domain range is false as it is not even a function. Then how can you define domain and range?
Solving those questions naively is like doing physics. Ruling out the “math problem” rigorously is like doing math.
I think I get it. People are confusing 1.42^1.63 (using rough numbers here) which equals 1.76 with 1.63^1.42 which equals 2.
The former allows for infinite number of powers while the other doesn't.
To sum up the resolution of the paradox in one sentence:
The Lambert W function is not a single valued function for all x.
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.
🎉🎊OMG! YOU HAVE 666 666 SUBSCRIBERS! CONGRATULATIONS! 🎊🎉
Mykola Girnyi 666602 as of now
Considering the following sequence: k, sqrt(2)^k, sqrt(2)^(sqrt(2)^k)... It converge when k
Agree
Waitaminute... 4= 2^2, so is it not just two infinite power towers stacked on top one another?
the brackets are different. 4 is (x^(x^(x...)))^(x^(x^(x...)) and 2 is x^(x^(x^(x^(x^(x...)))))
An essential problem here is that you haven't defined what x^x^x^x^... means in the first place. Such operation isn't defined on a standard analysis course so it isn't common knowledge, therefore you should define it in the very beginning, before or immediately after posing the problem from the competition. If you had done that, the question "what's wrong in this reasoning" would be much more meaningful and the discussion about the solution would provide deeper understanding. Without doing that, your explanation is rather brief and shallow.
You could "fix" it by saying the root from 4 = 2
which is 2 = 2
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" .
Great vid! I had no idea about that convergence thing, though. Is it taught at graduate-level math?
how to prove they converge within this range?
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.
Gloweye
Not in this case.
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.
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.
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
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.
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
- "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... ) .
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.
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
You're doing the power from the bottom up. You're supposed to do it from the top down...
That's not how exponent laws work.
Remember that its Sqrt(2) being stacked, not 1/2.
This is the nicest thing ive ever seen for a while. So many e relations. Thats just beautiful
But , how did you plot the graph 🤔
exactly
I’m reading the comments and I am once again brought to light about how little math I truly know when compared to others. I can do AP Calc AB no prob but what you guys are discussing is a different matter entirely...
I'll give you 4 apples, but suddenly it becomes 2
aww, memories. the first power tower video was one of the first videos I watched when I subscribed to this channel. thanks presh!
Great video! Your explanation was clear and easy to follow, thanks!
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.
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?
How much is (fourth root of four)² ? Is it not 2?
pls reply to this comment Mind your decisions....
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?
I mean x=4√4. So x^x^x^x^ till infinite where x=4√4= 4.... Am I right?
No.
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!
Oooh I was asked this in a maths competition but I just guessed root2 (it was multiple choice)
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
how does a person go and graph something like this? and I mean on a TI type calculator
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.
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).
You can just differentiate x=y^(1/y) and find the maximum and minimum value
Wikipedia has a nice description of hyperexponentiation. It would have been nice if you had a definition of it in the first video.
I didn't get why there are these limits
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
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.
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.
Who's here after 3blue1brown did a video on this?? 😂
The example I like is using 10 which would yield x = the 10th root of 10. But the 10th root of 10 is actually less than the square root of 2 (raising both to the power of 20 shows 100 < 1024). So with a < b raising "a" to infinite powers of itself is greater than "b" being raised to infinite powers of itself. Clearly impossible.
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) ).
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
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.
well because it's complex and cant be explained easily?, my point infinity can't be understand in our common sense
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.
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 ?
03:03 I love it
very nice one --- years ago, I came across the following problem: Suppose S= 1+2+4+8+16+32+.....; therefore 2S=2+4+8+16+32+......; that means 2S=S-1, implying S= - 1. How's that possible? The error here lies in the fact that the series is not convergent, so we can't apply equations by doing math on a non-convergent series, otherwise we will get puzzling (and erroneous) results like this
why all exponents above the base equals to 2???
If x^(x^(x^9=(x... = 2, then you can do x^2=2, since that portion is also has the infinite exponents
*_...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..._*
Did someone say X-ponent?
It's very basic and simple :
x^x^x^x^x...=4 implies x=sqrt(2)
It's an implication, not an equivalence. So after finding a "solution" by a chain of implications, we must check that the reciprocal is true, otherwise you can not conclude you found a solution.
If implication gives a solution, and the reciprocal is false, you just show there is no solution (or at least no solution in the domain you are looking at. If your implications only take care of real numbers and the real solution you find doesn't work, you just prooved there is no real solution. But there can be a complex one you forgot about with your implications).
Conclusion : Don't try to think in terms of equivalences, you'll fail most of the time. Work with implications, but at the end, never forget to verify the reciprocal !
x^x^x^x... =2
I thought the answer would converge to 1 for sure.
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🤦🏻
@@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....
@@tomriddle2257 you are repeating my point😐
@@arpitmalik27 That was sarcastic. My second example has a solution near 1.
@@tomriddle2257 lol it's solution is near 2
It's x=2.2991
OR
x=e^(√ln2)
A class of problem I had not expected..thank you
I understand
Yay
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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.
What is infinite + infinite+ infinite+ infinite+ infinite+ infinite upto infinite times
a bigger infinity☺
Ravi Raj R square No, it’s the same
@@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
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.
@@alice_in_wonderland42 how do you feel about bigger infinity
But there is also another problem:
If we use the same trick, we can say that 2 = (x^x)^2 and 2 = x^2 (+ many others)
2 = 2
(x^x)^2 = x^2
x = x^x
And this can only be true if x=1
But then 1^1^1^1^... isn’t equal to 2
No, you can only say 2 = x^(x^2) because of the order you need to do the operations