I didn't believe it at first but then I checked with a calculator and it's true. Take 0.5^0.7^0.9 and then take it to the power of higher and higher numbers and it will approach 0.5
@@Parzi_ it must be correct. It seems logically what he has said. 1.1^1.3^1.5^1.7^... equals infinity because it keeps multiplying without stopping. That definetely works with numbers greater than 1. Then we ask ourself what 0.9^infinity actually means. Remember that numbers less than 1 are multiplied, they're getting smaller so 0.9^infinity equals one. Then we know that (every number except 0)^0 equals 1 because it's the same as x^0 = x^(1-1) = x/x = 1 And 0.5^1 is obviously 0.5 so it must be correct.
@@zach4965 Looking at the problem "from the bottom" fundamentally adds brackets where there weren't any before. Since power towers have to be evaluated from the top down, you can't start at the bottom and have the same equation.
I feel the answer would be even more satisfying if there was one more step down, i.e. if the base of it all was 0.3. The answer would be sqrt(0.3), more surprising than just 0.5
As a mathematician, I have seen several examples of infinite power towers. They are well-defined objects whose value is given by the limit of partial towers (when it exists), same as pretty much any infinitely repeated operation.
@@alxjones my only qualm is that unlike the other operations, expotentiation is non-associative and done from the top down. Usually power towers are defined sequentially from an an initial unit at the top. Having infinity as your starting value just strikes me as odd, considering how persnickity math can be when dealing with it -even when it seems perfectly reasonable (intuitively) to have it there. But I'm not a mathmetician, I'm a physicist.
@@insouciantFox I defini8tely get the worry about non-associativity, but there's a difference between the order of partials and the order of operations. The fact that the partials go against the grain here makes the theory more difficult, since we can't go directly from nth partial to (n+1)th partial, but in the end it's just a sequence that either has a limit or doesn't.
@@alxjones Exactly, there's nothing to worry about. Even though I haven't seen power towers before, you can certainly make a sequence of the partial towers, and any sequence either has a limit or doesn't. "Having infinity as your starting value just strikes me as odd". Yes, absolutely. Even though intuitively I'm sure this answer is correct, a more thorough proof would have to define and justify the use of the infinity symbol.
You're thinking about the rule where the base and exponent altogether are being raised to a power: (a³)³ = a³+³ = a⁶ But the equation we have here is one where the exponent itself is being raised to the power. a^(3³) = a²⁷ So the tower of exponents can't break down into a multiplication as it's just the exponent being raised to the proceeding exponent and NOT the base and exponent being raised
Imagine (6/5)^(34/25)^(174/125)^… the nth element in the tower is 7/5-5^-n. Each term is greater than one and increasing, so the tower should grow without bound. Strangely, this is actually finite and less than 2. It is actually very important that the elements of the tower grow fast enough.
Would like to see proof that “any power greater than 1 would go to you infinity”…. I’d have thought it would have to be larger than root2 at least to assert that.
It does have to be over root 2 but the result is the same it diverges from 1.5 and up, and 1.3^divergent = divergent and then 1.1^divergent = divergent and 0.9^ divergent =0
If you actually try in calculator to solve this tower of powers, you see that the answer is 0, because after 0.5^0.7^0.9 you take number less than 1 and when you power it with number higher than 1, you take the less number
theres another way, a^n1^n2...n🛑 = a^(n1*n2...n🛑), and since the value of 'n1*n2...n🛑' = infinite in this case, it would just be 0.5^infinite. This makes it approach 0. this proves that infinite doesn't exist. It doesn't even have an odd or even number because if you have an sum of (-1)^n denoted by A, A = -1,1,-1,1... and actually = 1/2
There is a logic that can be mistakenly understood here. It would be common to say (1.1^1.3^1.5^...) is infinity because it's infinite power of numbers greater than 1. WRONG. Yes, it is infinity but not because what you think. Such logic is based by the fact that (1.3^1.5^1.7^...) is infinity, which is based by the fact that (1.5^1.7^...) is infinity and so on. There is no initial point here. here's an example: (√2^√2^√2^...) This number seems to diverge to infinity but actually is 2. You can try this with your calculator.
Well yeah cutting at the 1.1 because "1.1>1 so it blows up" is wrong. However, the range of convergence for a power tower is [e^(-e), e^(1/e)], so at 1.5 and onwards the power tower starts to diverge.
@@MK-13337 Not exactly wrong as it has monotonically (and linearly) increasing powers, but he just skipped a few steps. No matter what was the step size , as long as it was positive, he'd have inf on all powers > 1. Therefore, had he cut at 1.5, he'd still have 1.3^inf ~ inf, 1.3^inf ~ inf, 1.1^inf ~ inf As the tower doesn't include 1 itself as an exponent, the first power with base < 1 of the tower equals to 0, then the rest is solvable with arithmetic. Had it included 1 itself, the first power with base
@@MaavBR No, his argument was "these powers are bigger than one so they blow up" which is incorrect. Only powers bigger than e^(1/e) (approx 1.44) blow up. I know he ends up correct in the end but getting the correct result with the incorrect reasoning is useless.
Shouldn't we first check whether the series converges following a convergence criterium (and in case which one). Just adding brackets to infinite expressions without checking whether infinite powers still follow associativity, looks a bit sketchy. (Like for infinite sums, where w/o total convergence you get different results depended in which way you sum) Maybe I missed that it's obviously, but I'd be careful in doubt here to just take it as granted. Just a counter example, if I take 0.5^0.7^0.9 then it is a Nr below 1. If I make then this Nr ^ 1.1^1.3^1.5^... it will decrease every time (and by a pretty big potential), so my first guess would have been zero here instead.
You should not need to check convergence. a^x is continuous on R so a^x approaches a^b iff x approaches b. Also your order of operations is wrong. It should be Nr ^ (1.1^(1.3^1.5))
@@arsenypogosov7206 Can you properly note down the series where you take the limes from. I'd expected a_0 = 0.5 and a_n = a_(n-1) ^ (0.5 + n * 0.2) but that doesn't fit your explanation. For me it looks like you start other around, but not with a limited element. But to take a lim, you need a series of finite elements, not like in your explanation where you explain the lim starting with an infinite element (the 1.1 ^ 1.3 ^ ... = \inf). So, a formal notion of a_0 = ..., a_1 = ..., .... and then the specific lim_(n->\inf) (a_i) would really help to understand what we are talking here about exactly. And then I can better check whether the (a_i) converge and in case to which limit.
@@janekschleicher9661 Yeah, sure. Firstly I want to point out that this type of notation is commonly not formally defined, so what I am saying is debatable. But at least it's consistent with what is shown in the video. Now to the point. Let's define sequence a as such: a_0 = 0.5 a_1 = 0.5 ^ 0.7 a_2 = 0.5 ^ (0.7 ^ 0.9) a_3 = 0.5 ^ (0.7 ^ (0.9 ^ 1.1)) a_4 = 0.5 ^ (0.7 ^ (0.9 ^ (1.1 ^ 1.3))) ... a_n = 0.5 ^ (0.7 ^ ... ( ^ (0.5 + 0.2*n))...) Essentially a_n is the first n + 1 terms of the equation: a_n = 0.5^0.7^0.9^...^(0.5+0.2*n). Note the order of ^: a^b^c = a^(b^c) != (a^b)^c. So the parentheses are distributed in this particular way. Also note that a_n != a_{n-1}^{0.5 + 0.2 *n} because of the order of operations. Now to the formal solution. We need this facts: 1) lim a_n = lim a_{n + 1} 2) if f is continuous then: lim f(g) = f(lim g). By f(infinity) we mean the limit of f where it's argument tends to that infinity 3) if a > 0 then f(x) = a^x is continuous. 4) if a > 1 then a^{+infinity} = +infinity 5) if 0
@@arsenypogosov7206 Thanks for the nice explanation! Especially the clear bracketing makes it unambiguous. (With other bracketing, we'd get other results). And I wasn't sure from the video itself.
My guess is that it's 1/2. The portion above 0.9 goes to infinity I think, and so the portion above 0.7 goes to zero, the portion above 0.5 goes to 1, and the entire thing goes to 0.5.
It seems you interpreted the exponents as being equal to infinity, which isn't quite accurate in this scenario. We need to be cautious with numbers less than 1. Here's an important note: (x < 1)^infinity = 0 With this information, we can see that although the exponents approach infinity, 0.9^infinity does not approach infinity; in fact, it approaches 0. This explains the result of the calculations.
@AndyMath your answer is obviously wrong, the correct answer is 0 (try to calculate it by hand to convince yourself). Quick proof: 0.5 ^ 0.7 ^ 0.9 = 0.641... < 1. Then, re-use the first part of your answer: x ^ y with x < 1 and y approaches infinity will approach 0. Nice try though, keep going ☺️
For a slightly more solid proof. Use S the result you are looking for. We have ln(S) = ln(0.5) x (0.7 x 0.9 x 1.1 x 1.3 ...) = ln(0.5) x M We know ln(0.5) ~ -0.69 < 0. So if M approaches infinity, ln(S) will approach minus infinity, so S will approach 0 (use the reverse of ln: exp(X) approaches 0 when X approaches minus infinity) M = 0.7 x 0.9 x 1.1 x 1.3 x ... = 0.63 x (1.1 x 1.3 x ...) > 0.63 x pow(1.1, X) with X approaches infinity. 1.1 being strictly greater than 1, pow(1.1, X) approaches infinity, which implies that M also approaches infinity. Hence, S equals 0 😉
Very nice proof knowledge, even though you made a mistake in the very first step of your argument. Exponentiation rules don't work like that. You worked on the property (x^m)^n = x^mn, which is totally correct. However, the calculation in the problem was of the form x^(m^n) ≠ (x^m)^n, that's how power towers work. Taking into account this correction, the video is right (even though you have to take into account the interval of convergence of power towers, the problem in the video still goes out of it). Nice try though, keep going😊
Cymath, wolframalpha, and GPT-4o all tell me the answer is 0.5 like the video. My phone calculator heads toward zero. I'll assume the phone calc is somehow wrong. 😅
Whats funny is that almost every single one of the people that are on youtube shorts and in this comment section is 11 or under and don’t understand a single shit of what you said there
Anything to the zero power equals 1; yes but the exponent 7 is being raised to the zero power; the base raised to 2nd power then raised to the 7th power is being raised to the zero power. The answer is 1. 7 times 0 equals 0; 2 times 0 equals 0 and 0.5 to the zero power is 1.
No... just no. A power to a power is the same as multiplying the numbers. 0.9 times infinity doesn't equal 0, and would technically to be infinite. Same for everything under it, so the correct answer is infinity.
Why wouldn’t it go to 0? If you simplify the first 2 powers, you get 0.5^0.7^0.9 = a < 1, and then a^inf would be 0? Does this mean that this is a paradox since order of operations can affect it?
@@arsenypogosov7206 Correct me if I'm wrong, but power towers are just infinite series, and when we evaluate them we are trying to find their limit. When we use this trick, we are sayng that lim(x_{n+1})=lim(0.5^ x_n)=0.5 ^ (lim x_n), which is not the case if the limit we started with does not exists, so we do need to prove the convergance.
@@andnekon if lim x_n exists and is equal to x and f is continuous in x, then lim f(x_n) also exists and is equal to f(x). f(x) = 0.5^x is continuous everywhere.
@@arsenypogosov7206 oh, ok, got it. So we would still have to prove the limit existence for x_n (=1.1^(1.3^(...) in this case, and it's obv inf ), but it's irrelevant for the "trick". Thank you
You are ignoring the properties of exponents that are less that 1 and greater than 0 Also when I put it in the calculator it started rounding to 0.5 when I reached 2.1 in the power tower Also you are incorrectly evaluating exponents. Exponents are always analysed from top to bottom unless stated to do otherwise. Would you evaluate the expression 5/5/5/5 as 1? No because you don’t evaluate it as (5/5)(5/5). Similar principle
that's wrong. exponents are the same as multiplying something by itself a number of times. you can never reach zero by multiplying with numbers other than zero. you can infinitely approach zero but you will never reach it. therefore the correct answer is a number that is infinitely small but greater than zero.
this is simply wrong. firstly you have to define this object: the only method you can define this thing with is by exponentiating the base each time with a new term, so -if i call this object x- i can consider x like the limit for n to infinity of the tower of exponents that grows term by term. Now consider the terms after the 0.9: i'll have a number between 0 and 1 raised to the power of a constantly increasing positive number. so it's provable that this goes to 0, and it goes to 0 really quickly.
@@barndo3141 Formally yes, you'd be more correct in saying that the 0.9 approaches 0, and thus the 0.7 approaches 1, and thus the 0.5 approaches 0.5. Luckily the power tower is infinitely tall, so we can "approach" 0.5 infinitely much.
Any time you see "do infinity things", it really just means "do n things, and take the limit as n goes to infinity". Here, that's the power tower; we _define_ the infinite power tower to be limit as n goes to infinity of the subtowers of height n. All the infinity logic in this solution is just shorthand for limit properties.
You can’t actually have an infinite tower of powers. What you’re actually doing is taking a limit here, and what the video here is saying is that as the “height” of the tower approaches infinity, the value of said tower approaches 0.5
@@SS77S7 ohh 😮 you're right. I ignored those power numbers and thought "if they are going to infinite anyway, why would I even consider them" that kind of thing. Thanks for correcting me 😄
Guess: 0.5 The 1.1 stuff goes to infinity The 0.9 ^ infinity goes to 0 0.7 ^0 is 1 0.5^1 is 0.5 But I feel like you should ideally be more formal about it.
So, because I have a bad habit of rambling, I will have the explanation in three parts, clearly labeled, part 1 explains what pi is and how we got it, part 2 explains why it is infinite (roughly, no mathematical proofs though) and part 3 will be why it doesn't equal infinity 1) pi is the number that you have to multiply the diameter of a circle by to get the circumference. We found it by first taking a circle, measuring the diameter, and then measuring the circumference, and dividing the circumference by the diameter, to get pi 2) pi goes on for infinity, because, roughly, no number exactly fits to multiply the diameter by to equal the circumference, for example, 3 is too low, 4 is too high, 3.1 is too low, 3.2 is too high, and so on, forever 3) pi does go on forever, but doesn't equal infinity, because every added number is after the decimal, so the amount that pi is changing gets smaller every added number. I hope tha helps explain it, and if it doesn't, please let me know!
This isn’t correct. Using the same logic you can start from the bottom and say that (0.5^0.7)^0.9 is less than 1 and the remaining exponent greater than 1. This makes the answer 0. This is also what you see when you evaluate the partial sums of this series.
(.5^.7^.9) is about .646 so .646^infinity goes to zero. For nested exponents you multiply using the power rule, not exponentiate them. At least I think. So it would be .5^infinity ->0
The confusion comes from the order of the exponential, which is understood to be evaluated top down. For top down evaluation, at 1.1, the power above is infinite, but at 0.9 the tower is 0. So for 0.5, there is no 0.5^infinity.
If X = 1.1^1.3^1.5... then X = infinity So 0.5^(0.7^(0.9^X)) = 0.5^(0.7^(0)) = 0.5^1 = 0.5 I may be inclined to say the answer is undefined though as I don't like that infinity in there without any limits.
@@bloopletank2491 But part of the expression diverges to infinity (if expressed as a sequence). And we tend to just say the expression is undefined if it doesn't converge.
@@bloopletank2491 Raising a limit theoretically can work out to be 0 ( or the approached value ) but when using it on a real basis you have to find a convenient value that is close to the approached limit. So yeah, but only here
if you approach the problem in a different way (limits), then yeah, it's legit if not, then we can't make a conclusion, because we don't know the true value of that number, just the way it acts just beware of the function/problem rules and you will be fine
But the expression IS a limit. Almost every time someone writes “…” it means take the limit. For example, 1+1/2+1/4+1/8+… means the limit of the sequence: 1 1+1/2 1+1/2+1/4 1+1/2+1/4+1/8 etc. And in this case 0.5^0.7^0.9^1.1^… means the limit of the following sequence: 0.5 0.5^0.7 0.5^0.7^0.9 0.5^0.7^0.9^1.1 etc.
@@bloopletank2491 I guess so? You don't see large towers of powers usually, in the case where you do see them its important to distinguish using brackets - I have never been told or explained that you should presume a tower is evaluated top down. Unlike what you wrote, which is unambiguous because people have intentionally said unless stated otherwise 5+4*2=5+(4*2)
@juliang8676 from my education it was established you always go top down. Like if you have x^(y+1), but ofc written without the parentheses with the expression y+1 being the whole super script, you evaluate y+1 first. If it was x^(y^z+1) written as described, you go and evaluate y^z first, the +1, then the x. If you turn the 1 to a 0, it makes it more obvious.
except the fact that towering powers multiply and you need to start from the base to apply the power rule -sound logic .You all need some highschool matematics
No, having no () means you have to start from the very top, working your way down What you are describing only works for ((n^2)^3)^4 n^2^3^3 for example is not n^18 but n^(2^27) or n^134217728 You need some highschool mathematics
I hate to be that guy, but the correct answer is around 0.603. A power tower of 1.1 and 1.3 actually do converge, and 1.5 is the first term in the tower to diverge. So we have 1.1^1.3^inf = 1.471. Then we evaluate the rest of the chain to get around 0.603
Bro i just plugged 1.1^1.3^inf into a calculator, but instead of using infinity I used like 20 and it gave me 73 million i don't think you know whatchu mean
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Rare footage of multiple boxes, how exciting
Ain’t no way I’m watching math 💀
"We can put a box around it." That was so funny for me.
The way you say "How Exciting" in the most deadpan delivery always cracks me up. Love these videos man
Is this correct? This feels like one of those things my math teacher would say "don't play with infinity like that."
well it's pretty logically what he said so it would be correct in my opinion.
I didn't believe it at first but then I checked with a calculator and it's true.
Take 0.5^0.7^0.9 and then take it to the power of higher and higher numbers and it will approach 0.5
@@Parzi_ it must be correct. It seems logically what he has said. 1.1^1.3^1.5^1.7^... equals infinity because it keeps multiplying without stopping. That definetely works with numbers greater than 1.
Then we ask ourself what 0.9^infinity actually means.
Remember that numbers less than 1 are multiplied, they're getting smaller so 0.9^infinity equals one.
Then we know that (every number except 0)^0 equals 1 because it's the same as x^0 = x^(1-1) = x/x = 1
And 0.5^1 is obviously 0.5 so it must be correct.
But what if you look at it the other way around. 0.5^0.7^0.9 is ~0.64 right? And 0.65^infinity should be 0
@@zach4965 Looking at the problem "from the bottom" fundamentally adds brackets where there weren't any before. Since power towers have to be evaluated from the top down, you can't start at the bottom and have the same equation.
I feel the answer would be even more satisfying if there was one more step down, i.e. if the base of it all was 0.3. The answer would be sqrt(0.3), more surprising than just 0.5
But what if the first limit went to -1/12 😢
That os absolutly not the same problem, this is when doing 1+2+3+4+5... and ∞ ≠ -1/12
@@eliotnungaesser2225also just want to add that 1+2+3... does not equal -1/12
@@toastWs
Well you obviously don't know much if you think that 1 + 2 + 3 + ••• ≠ -1/12
@@MrKoteha the irony makes this absolutely hilarious, please inform me once you actually learn about infinite series
@@toastWs r/woooosh
This is really cool! I think this is the first time I’ve actually solved a problem you’ve shown
The rest of the problems have solutions in long form videos most of the time
Dude same! I was so excited when he put a box around everything above 1.1 because then I knew I had it
I think a mathematician would say this is ill-posed, but sure I'll buy it
If you define the power tower as the limit of the obvious sequence, if it exists, then I think this is fine, albeit skipping some steps
As a mathematician, I have seen several examples of infinite power towers. They are well-defined objects whose value is given by the limit of partial towers (when it exists), same as pretty much any infinitely repeated operation.
@@alxjones my only qualm is that unlike the other operations, expotentiation is non-associative and done from the top down. Usually power towers are defined sequentially from an an initial unit at the top. Having infinity as your starting value just strikes me as odd, considering how persnickity math can be when dealing with it -even when it seems perfectly reasonable (intuitively) to have it there.
But I'm not a mathmetician, I'm a physicist.
@@insouciantFox I defini8tely get the worry about non-associativity, but there's a difference between the order of partials and the order of operations.
The fact that the partials go against the grain here makes the theory more difficult, since we can't go directly from nth partial to (n+1)th partial, but in the end it's just a sequence that either has a limit or doesn't.
@@alxjones Exactly, there's nothing to worry about. Even though I haven't seen power towers before, you can certainly make a sequence of the partial towers, and any sequence either has a limit or doesn't.
"Having infinity as your starting value just strikes me as odd". Yes, absolutely. Even though intuitively I'm sure this answer is correct, a more thorough proof would have to define and justify the use of the infinity symbol.
Keep uploading these random interesting problems bro..
Damn interesting
Nice, keep it up
Oh wow, that's really elegant!
If you just mulitple the powers by rule it should be 0.5^infinity which should be 0. isnt that the case?
You're thinking about the rule where the base and exponent altogether are being raised to a power:
(a³)³ = a³+³ = a⁶
But the equation we have here is one where the exponent itself is being raised to the power.
a^(3³) = a²⁷
So the tower of exponents can't break down into a multiplication as it's just the exponent being raised to the proceeding exponent and NOT the base and exponent being raised
@@agaming5164 (a³)³=a^9
x^(a^b) ≠ (x^a)^b
Its the same as saying that
1+3*2=8
That really is exciting! I'm honsetly amazed
This is why I love maths, its so simple at the end, it really scratches an itch in my brain for satisfaction
Imagine (6/5)^(34/25)^(174/125)^… the nth element in the tower is 7/5-5^-n. Each term is greater than one and increasing, so the tower should grow without bound. Strangely, this is actually finite and less than 2. It is actually very important that the elements of the tower grow fast enough.
That’s because here all the terms are gonna be smaller than e^(1/e)
But in the video the terms grow without bound so it does go to infinity
@@josenobi3022the constant that determines whether n^n^n^… converges is actually e^(1/e) or approx 1.44
@@dreingames9137 yeah that makes more sense, thanks
This is the type of question that you hope is multiple choice on a test
Would like to see proof that “any power greater than 1 would go to you infinity”…. I’d have thought it would have to be larger than root2 at least to assert that.
It does have to be over root 2 but the result is the same it diverges from 1.5 and up, and 1.3^divergent = divergent and then 1.1^divergent = divergent and 0.9^ divergent =0
So you can evaluate a number to the power of infinity?
Theoretically, you can only take the limit of a^n as n approaches infinity, but practically, yes you can.
lim(n->+inf) a^n is 0 if -1 < a < 1
@@kobalad1118 if -1
@@user-yv2fb4mi1k Negative zero is still zero
@@user-yv2fb4mi1k If -1
i don't know why but i love this problem, and your explanation
If you actually try in calculator to solve this tower of powers, you see that the answer is 0, because after 0.5^0.7^0.9 you take number less than 1 and when you power it with number higher than 1, you take the less number
for a sec there I thought the whole thing would be 0 lol
Damn how it's easy😂Love the math tricks u upload🥰Keep going💜
theres another way, a^n1^n2...n🛑 = a^(n1*n2...n🛑), and since the value of 'n1*n2...n🛑' = infinite in this case, it would just be 0.5^infinite. This makes it approach 0. this proves that infinite doesn't exist. It doesn't even have an odd or even number because if you have an sum of (-1)^n denoted by A, A = -1,1,-1,1... and actually = 1/2
that was incredibly easy for the type of questions you post! Should've made it harder
How exciting 😃
"If you wanna try on your own pause it because I'm gonna solve it."
I'm a guy scrolling away on TH-cam Shorts, you expect me to solve that?
I think we need to show that power of 1.1 has no upper bound to say it is infinity
There is a logic that can be mistakenly understood here. It would be common to say (1.1^1.3^1.5^...) is infinity because it's infinite power of numbers greater than 1. WRONG. Yes, it is infinity but not because what you think. Such logic is based by the fact that (1.3^1.5^1.7^...) is infinity, which is based by the fact that (1.5^1.7^...) is infinity and so on. There is no initial point here.
here's an example: (√2^√2^√2^...) This number seems to diverge to infinity but actually is 2. You can try this with your calculator.
Holy shit thats actually exciting
since 1.1^1.3... goes to infinity . 0.9 goes to 0 . so 0.7 goes to 1 . and so final result is 0.5 . pretty sure thats it
I would have preferred he expressed this in terms of limits because a divergent sequence isn't "equal" to anything. It diverges.
Awesome
How exciting
This doesn’t seem sound? sqrt(2)^sqrt(2)^sqrt(2)^… converges to 2 even though the numbers in the power tower are greater than 1.
Well yeah cutting at the 1.1 because "1.1>1 so it blows up" is wrong. However, the range of convergence for a power tower is [e^(-e), e^(1/e)], so at 1.5 and onwards the power tower starts to diverge.
The end will be something like inf^inf, which will obviously be inf. So it chains down for all numbers greater than 1
@@MK-13337appreciate this explanation. I’d seen something like the before and knew after a watching the video that something was off
@@MK-13337 Not exactly wrong as it has monotonically (and linearly) increasing powers, but he just skipped a few steps. No matter what was the step size , as long as it was positive, he'd have inf on all powers > 1. Therefore, had he cut at 1.5, he'd still have 1.3^inf ~ inf, 1.3^inf ~ inf, 1.1^inf ~ inf
As the tower doesn't include 1 itself as an exponent, the first power with base < 1 of the tower equals to 0, then the rest is solvable with arithmetic. Had it included 1 itself, the first power with base
@@MaavBR No, his argument was "these powers are bigger than one so they blow up" which is incorrect. Only powers bigger than e^(1/e) (approx 1.44) blow up. I know he ends up correct in the end but getting the correct result with the incorrect reasoning is useless.
Shouldn't we first check whether the series converges following a convergence criterium (and in case which one).
Just adding brackets to infinite expressions without checking whether infinite powers still follow associativity, looks a bit sketchy. (Like for infinite sums, where w/o total convergence you get different results depended in which way you sum)
Maybe I missed that it's obviously, but I'd be careful in doubt here to just take it as granted.
Just a counter example, if I take 0.5^0.7^0.9 then it is a Nr below 1. If I make then this Nr ^ 1.1^1.3^1.5^... it will decrease every time (and by a pretty big potential), so my first guess would have been zero here instead.
You should not need to check convergence. a^x is continuous on R so a^x approaches a^b iff x approaches b. Also your order of operations is wrong. It should be Nr ^ (1.1^(1.3^1.5))
@@arsenypogosov7206 Can you properly note down the series where you take the limes from.
I'd expected a_0 = 0.5 and a_n = a_(n-1) ^ (0.5 + n * 0.2) but that doesn't fit your explanation.
For me it looks like you start other around, but not with a limited element. But to take a lim, you need a series of finite elements, not like in your explanation where you explain the lim starting with an infinite element (the 1.1 ^ 1.3 ^ ... = \inf).
So, a formal notion of a_0 = ..., a_1 = ..., .... and then the specific lim_(n->\inf) (a_i) would really help to understand what we are talking here about exactly. And then I can better check whether the (a_i) converge and in case to which limit.
@@janekschleicher9661 Yeah, sure. Firstly I want to point out that this type of notation is commonly not formally defined, so what I am saying is debatable. But at least it's consistent with what is shown in the video.
Now to the point. Let's define sequence a as such:
a_0 = 0.5
a_1 = 0.5 ^ 0.7
a_2 = 0.5 ^ (0.7 ^ 0.9)
a_3 = 0.5 ^ (0.7 ^ (0.9 ^ 1.1))
a_4 = 0.5 ^ (0.7 ^ (0.9 ^ (1.1 ^ 1.3)))
...
a_n = 0.5 ^ (0.7 ^ ... ( ^ (0.5 + 0.2*n))...)
Essentially a_n is the first n + 1 terms of the equation: a_n = 0.5^0.7^0.9^...^(0.5+0.2*n). Note the order of ^: a^b^c = a^(b^c) != (a^b)^c. So the parentheses are distributed in this particular way. Also note that a_n != a_{n-1}^{0.5 + 0.2 *n} because of the order of operations.
Now to the formal solution. We need this facts:
1) lim a_n = lim a_{n + 1}
2) if f is continuous then: lim f(g) = f(lim g). By f(infinity) we mean the limit of f where it's argument tends to that infinity
3) if a > 0 then f(x) = a^x is continuous.
4) if a > 1 then a^{+infinity} = +infinity
5) if 0
@@arsenypogosov7206 Thanks for the nice explanation!
Especially the clear bracketing makes it unambiguous.
(With other bracketing, we'd get other results).
And I wasn't sure from the video itself.
Yes thats really exciting
Great
My guess is that it's 1/2. The portion above 0.9 goes to infinity I think, and so the portion above 0.7 goes to zero, the portion above 0.5 goes to 1, and the entire thing goes to 0.5.
I figured that out as soon as he boxed everything bigger than 1!
Dont forget to put box around it
Beautiful
You can use the exponent rule (x^m)^n=x^mn so its basically 0.5^infinity so 0
Power towers work from the top down, so it is x^(m^n) which is not equal to (x^m)^n or x^mn
It seems you interpreted the exponents as being equal to infinity, which isn't quite accurate in this scenario. We need to be cautious with numbers less than 1. Here's an important note:
(x < 1)^infinity = 0
With this information, we can see that although the exponents approach infinity, 0.9^infinity does not approach infinity; in fact, it approaches 0. This explains the result of the calculations.
@AndyMath your answer is obviously wrong, the correct answer is 0 (try to calculate it by hand to convince yourself). Quick proof: 0.5 ^ 0.7 ^ 0.9 = 0.641... < 1. Then, re-use the first part of your answer: x ^ y with x < 1 and y approaches infinity will approach 0.
Nice try though, keep going ☺️
For a slightly more solid proof. Use S the result you are looking for. We have ln(S) = ln(0.5) x (0.7 x 0.9 x 1.1 x 1.3 ...) = ln(0.5) x M
We know ln(0.5) ~ -0.69 < 0. So if M approaches infinity, ln(S) will approach minus infinity, so S will approach 0 (use the reverse of ln: exp(X) approaches 0 when X approaches minus infinity)
M = 0.7 x 0.9 x 1.1 x 1.3 x ... = 0.63 x (1.1 x 1.3 x ...) > 0.63 x pow(1.1, X) with X approaches infinity. 1.1 being strictly greater than 1, pow(1.1, X) approaches infinity, which implies that M also approaches infinity.
Hence, S equals 0 😉
Very nice proof knowledge, even though you made a mistake in the very first step of your argument. Exponentiation rules don't work like that. You worked on the property (x^m)^n = x^mn, which is totally correct. However, the calculation in the problem was of the form x^(m^n) ≠ (x^m)^n, that's how power towers work. Taking into account this correction, the video is right (even though you have to take into account the interval of convergence of power towers, the problem in the video still goes out of it). Nice try though, keep going😊
Tetration is not conmutative.. it needs to be done top to bottom.
Cymath, wolframalpha, and GPT-4o all tell me the answer is 0.5 like the video. My phone calculator heads toward zero. I'll assume the phone calc is somehow wrong. 😅
Whats funny is that almost every single one of the people that are on youtube shorts and in this comment section is 11 or under and don’t understand a single shit of what you said there
Anything to the zero power equals 1; yes but the exponent 7 is being raised to the zero power; the base raised to 2nd power then raised to the 7th power is being raised to the zero power. The answer is 1.
7 times 0 equals 0; 2 times 0 equals 0 and 0.5 to the zero power is 1.
No... just no. A power to a power is the same as multiplying the numbers. 0.9 times infinity doesn't equal 0, and would technically to be infinite. Same for everything under it, so the correct answer is infinity.
Nice one
Why wouldn’t it go to 0? If you simplify the first 2 powers, you get 0.5^0.7^0.9 = a < 1, and then a^inf would be 0? Does this mean that this is a paradox since order of operations can affect it?
Tetration is not commutative, don't mix things up.
I love math
Exponentiation isn't associative, is it clear this problem is well defined?
To do limit tricks you first need to prove that the limit exists, but I guess it's good enough for a short
You don't need to prove convergence to do this type of "tricks". Maybe you think of grouping terms in power series?
@@arsenypogosov7206 Correct me if I'm wrong, but power towers are just infinite series, and when we evaluate them we are trying to find their limit. When we use this trick, we are sayng that lim(x_{n+1})=lim(0.5^ x_n)=0.5 ^ (lim x_n), which is not the case if the limit we started with does not exists, so we do need to prove the convergance.
@@andnekon if lim x_n exists and is equal to x and f is continuous in x, then lim f(x_n) also exists and is equal to f(x).
f(x) = 0.5^x is continuous everywhere.
@@arsenypogosov7206 oh, ok, got it. So we would still have to prove the limit existence for x_n (=1.1^(1.3^(...) in this case, and it's obv inf ), but it's irrelevant for the "trick". Thank you
noice!
isnt this solving an ap(arthematic progression )'
Power tower shuckle the minors 💀💀
I don't want to be that guy that ruins the party but i think that's wrong.
0,5^0,7^0,9
?????
Thats exactly what he do
You are ignoring the properties of exponents that are less that 1 and greater than 0
Also when I put it in the calculator it started rounding to 0.5 when I reached 2.1 in the power tower
Also you are incorrectly evaluating exponents. Exponents are always analysed from top to bottom unless stated to do otherwise.
Would you evaluate the expression 5/5/5/5 as 1?
No because you don’t evaluate it as (5/5)(5/5). Similar principle
0.5
It s 0.5 right?
Yeah, cool
that's wrong. exponents are the same as multiplying something by itself a number of times. you can never reach zero by multiplying with numbers other than zero. you can infinitely approach zero but you will never reach it. therefore the correct answer is a number that is infinitely small but greater than zero.
Good
Im going insane over the way he called 0.2 "two tenths" I MEAN WHO DOES THAT
this is simply wrong. firstly you have to define this object: the only method you can define this thing with is by exponentiating the base each time with a new term, so -if i call this object x- i can consider x like the limit for n to infinity of the tower of exponents that grows term by term. Now consider the terms after the 0.9: i'll have a number between 0 and 1 raised to the power of a constantly increasing positive number. so it's provable that this goes to 0, and it goes to 0 really quickly.
That is a piss poor proof.
0.9 to the infinity power is an INDETERMINATE FORM!!
But isnt it impossible to shrink the 0.9 to 0?
This was my thought. No matter how many times you divide a number by 0.9, you'll still get a number above zero.
@@barndo3141 Formally yes, you'd be more correct in saying that the 0.9 approaches 0, and thus the 0.7 approaches 1, and thus the 0.5 approaches 0.5. Luckily the power tower is infinitely tall, so we can "approach" 0.5 infinitely much.
Wouldn't it be 0.5^0.7^(1/∞)? Wich would aproach, but never reach zero
Yeah and then it would be basically the ∞th root of 0.7, approaching 1, so the result would be approaching 0.5
They proved that it is 0
Just like . 9 repeating = 1
Any time you see "do infinity things", it really just means "do n things, and take the limit as n goes to infinity". Here, that's the power tower; we _define_ the infinite power tower to be limit as n goes to infinity of the subtowers of height n. All the infinity logic in this solution is just shorthand for limit properties.
@@HYDRA_0384 I get 0.5 too, but maybe technically its just undefined though.
You can’t actually have an infinite tower of powers. What you’re actually doing is taking a limit here, and what the video here is saying is that as the “height” of the tower approaches infinity, the value of said tower approaches 0.5
good 1
If all those power numbers are going to infinite anyway, wouldn't the answer be 0?
No, why would it be 0? The video explains very well why it's 0.5.
@@SS77S7 ohh 😮 you're right. I ignored those power numbers and thought "if they are going to infinite anyway, why would I even consider them" that kind of thing. Thanks for correcting me 😄
Guess: 0.5
The 1.1 stuff goes to infinity
The 0.9 ^ infinity goes to 0
0.7 ^0 is 1
0.5^1 is 0.5
But I feel like you should ideally be more formal about it.
I think he was formal enough. That's the best approach I can imagine which is also easy to understand.
can someone explain why pi goes on forever but isn't infinite
So, because I have a bad habit of rambling, I will have the explanation in three parts, clearly labeled, part 1 explains what pi is and how we got it, part 2 explains why it is infinite (roughly, no mathematical proofs though) and part 3 will be why it doesn't equal infinity
1) pi is the number that you have to multiply the diameter of a circle by to get the circumference. We found it by first taking a circle, measuring the diameter, and then measuring the circumference, and dividing the circumference by the diameter, to get pi
2) pi goes on for infinity, because, roughly, no number exactly fits to multiply the diameter by to equal the circumference, for example, 3 is too low, 4 is too high, 3.1 is too low, 3.2 is too high, and so on, forever
3) pi does go on forever, but doesn't equal infinity, because every added number is after the decimal, so the amount that pi is changing gets smaller every added number.
I hope tha helps explain it, and if it doesn't, please let me know!
@@Franktanker0 Thanks man, Definitely is helpful,Never been so invested reading math
Did anyone check this? I don’t think it’s the right answer.
Damnn
😂
This isn’t correct.
Using the same logic you can start from the bottom and say that (0.5^0.7)^0.9 is less than 1 and the remaining exponent greater than 1. This makes the answer 0. This is also what you see when you evaluate the partial sums of this series.
Nvm. I misunderstood the series.
0,5
не совсем корректное задание. не понятно что найти.
My guess is 0.5
The answer would be different if you treat powers as lefr-associative, you treated them as right-associative
usually if parentheses aren't used you treat Power towers as right associative
(.5^.7^.9) is about .646 so .646^infinity goes to zero. For nested exponents you multiply using the power rule, not exponentiate them. At least I think. So it would be .5^infinity ->0
It’s a power tower so u go from the last to the first not from the first to the last
You would only multiply with the power rule if there were parentheses. Otherwise you go top down. That's why 3^3^3 = over 7trillion instead of 19.6k
Your equation needs parentheses
No it doesn't, exponents are always evaluated top to bottom. a^b^c always means a^(b^c)
I think you may have made a slight error,since the numbers are also less than one 0.5^0.7^0.9... will also grow so it will be 0.5^∞=0
Nope if you multiply a number less than one with another number it will shrink
@@Kenvasa332 bruh you can check for it yourself.... exponential term is also less than one
@@solivagant234less than one means it will always shrink. explain how 0.7^0.9^… = infinity
@@solivagant234It’s a power tower so u go from the last to the first not from the first to the last
The confusion comes from the order of the exponential, which is understood to be evaluated top down. For top down evaluation, at 1.1, the power above is infinite, but at 0.9 the tower is 0. So for 0.5, there is no 0.5^infinity.
approaches 1. not is.
If X = 1.1^1.3^1.5... then X = infinity
So 0.5^(0.7^(0.9^X)) = 0.5^(0.7^(0)) = 0.5^1 = 0.5
I may be inclined to say the answer is undefined though as I don't like that infinity in there without any limits.
Do you even need a limit? Isn't there just actually an infinite number of terms.
@@bloopletank2491 But part of the expression diverges to infinity (if expressed as a sequence).
And we tend to just say the expression is undefined if it doesn't converge.
What if ...(((((((0.5)⁰'⁷))))))...
Wrong, 0.9^inf = lim -> 0 , NOT 0.
.09^inf=lim->0=0
I feel like in this case it's safe to say the limit is the actual value for the purposes of solving
@@bloopletank2491 Raising a limit theoretically can work out to be 0 ( or the approached value ) but when using it on a real basis you have to find a convenient value that is close to the approached limit. So yeah, but only here
Bruh
.9^inf will never reach zero. It will aproach it very closely though.
shouldn't 0.9^x go to 0.0...(infinite zeros)1 and therefor rather become an epsilon, infinitely small but never reaching zero?
That 0.0...1 simply is considered nonexistent...besides...how many 0.0...01s are there to bring ?
That's 0
Is it legit though? Coz it doesn’t equal zero, it’s limit is zero but not the actual value…
if you approach the problem in a different way (limits), then yeah, it's legit
if not, then we can't make a conclusion, because we don't know the true value of that number, just the way it acts
just beware of the function/problem rules and you will be fine
But the expression IS a limit.
Almost every time someone writes “…” it means take the limit.
For example, 1+1/2+1/4+1/8+… means the limit of the sequence:
1
1+1/2
1+1/2+1/4
1+1/2+1/4+1/8
etc.
And in this case 0.5^0.7^0.9^1.1^… means the limit of the following sequence:
0.5
0.5^0.7
0.5^0.7^0.9
0.5^0.7^0.9^1.1
etc.
Wth
Depends on the order of operations, going top down givs this answer going bottom up gives a different answer
Order of operations state it goes top down.
If it went bottom up it would just make
a^b^c^d^e=a^bcde
@@bloopletank2491 Well yea, thats the problem right? With no brackets and stuff its technically ambiguous
@@juliang8676 that's like saying 5+4×2 is ambiguous
@@bloopletank2491 I guess so? You don't see large towers of powers usually, in the case where you do see them its important to distinguish using brackets - I have never been told or explained that you should presume a tower is evaluated top down. Unlike what you wrote, which is unambiguous because people have intentionally said unless stated otherwise 5+4*2=5+(4*2)
@juliang8676 from my education it was established you always go top down.
Like if you have x^(y+1), but ofc written without the parentheses with the expression y+1 being the whole super script, you evaluate y+1 first. If it was x^(y^z+1) written as described, you go and evaluate y^z first, the +1, then the x.
If you turn the 1 to a 0, it makes it more obvious.
Kind of proud i did this in my head with no prior knowledge to such questions
except the fact that towering powers multiply and you need to start from the base to apply the power rule -sound logic .You all need some highschool matematics
No, having no () means you have to start from the very top, working your way down
What you are describing only works for ((n^2)^3)^4
n^2^3^3 for example is not n^18 but n^(2^27) or n^134217728
You need some highschool mathematics
well erm actually it is technically 0.5126
that doesnt sound right, if you keep multiplying by a number... it will always be... a number... it can never reach true zero
It's effectively 0. It's infinitely close to to zero. It's zero.
I hate to be that guy, but the correct answer is around 0.603. A power tower of 1.1 and 1.3 actually do converge, and 1.5 is the first term in the tower to diverge. So we have 1.1^1.3^inf = 1.471. Then we evaluate the rest of the chain to get around 0.603
1.1^1.3^inf != 1.471
Bro i just plugged 1.1^1.3^inf into a calculator, but instead of using infinity I used like 20 and it gave me 73 million i don't think you know whatchu mean
What if you evaluate the 0.7 to the 0.9 first and then say that to the power of infinity is 0 so the answer is 1?
You don't "evaluate 0.7 to 0.9 first" not how it works
Nope…
How do u eventually get to 0 by multiplying 0.9 by an infinite, ENDLESS number????
Who's multiplying ? We're doing exponents
but the power rule says that it is 0,5^(0,7*0,9*1,1*...) so 0,5 ^ infinity. So approaching 0.
I don't think that's how limits work
He didn't use a limit tho
but that is how they work
Its 0 since its a multiplication sign and anything times 0=0