Not necesarly, he skipped trough the part of showing that the sequence a_n=x^x^x...^x n times converges, which is not that trivial, is actually the crux of the problem, finding the limit is the easy part
7:55 again I can't agree this part. When you take the real value of squt(2) you get 1.414.... Then 1.414^1.414^1.414^1.414........... it lead to infinite value, so it's NOT convergence, so e^(1/e) is not the peak value of anything > 1 to it's own infinite power tower. I agree the whole other part of this video for the original equation is the correct process and lead to the right answer 1/4. But it's quite eeric to prove an number > 1 to it's ower infinite power tower lead to convergence. It obviously can be prove to be not possible by just take a check in real value. I think it's a enigma in mathematica.
I agree, there are different ways to solve this problem. If we use yours, we'll still get the same answer (or so I thought...) Let's see: 2^(x^x)^(x^x)^(x^x)...=2^(1/2) (x^x)^(x^x)^(x^x)...=(1/2) (x^x)^(1/2)=(1/2) (x^x)=(1/4) This is where I had to use Wolfram-alpha. :( Chaos Insurgency is correct. This leads to the answer x=-2, which is definitely not x=1/4. So you are also correct. But why? Can anyone else help? Please.
@ Everyone -- It is not the same. For example, (x^x)^(x^x) = x^(x*x^x) = x^[x^(x + 1)]. When x = 2, for example, this equals 2^8. In contrast, x^x^x^x = x^(x^(x^x)). When x = 2 here, this equals 2^16.
@Robert Veith I was answering the original question, but yes, you have to start at the top of a power tower for the correct answer, and grouping can cause issues
I get very frustrated with these problems. I have seen so many problems with powers where the order of operations is not clear. The convention that I'm used to is 2^3^4 means (2^3)^4 and not 2^(3^4). After the problem is solved it becomes clear which convention is being used. Why does there seem to be no standard for this? I post comments every time I see it and never get a clear answer. If I use Matlab and plug x-=1/4 into 2^x^x^x... the answer is 1 and not sqrt(2). This is the same convention I was taught, but obviously many others are taught the other convention. It takes all the fun out of the problems when I have to guess at the convention, or look at the video to figure out the convention before I can take a crack at it.
@@SyberMath Thank you. I will check that. By the way I mean nothing bad against you or others. I just have seen this issue on so many videos. I really want to understand if there is more than one accepted standard, or if I'm missing something obvious that is clear to others.
I am really embarrassed, but how do you calculate your powers of 1/4. When I use (X^n)^m= (X)^(m.n) so for instance (.25)^(1/16) I get a different answer or (.25)^(1/64). Can you please elaborate for us less mathematically talented who are still in high school.
No need to be that way! You're fine 😊 This is not the same as (x^x)^x which is x^(x^2) What we have is x(x^x) For x=1/4 this is (1/4)^{(1/4)^(1/4)} We calculate (1/4)^(1/4) first which 1/sqrt(2) then we raise 1/4 to the power that
I love when seemingly infinite functions have a normal function equivalent
Not necesarly, he skipped trough the part of showing that the sequence a_n=x^x^x...^x n times converges, which is not that trivial, is actually the crux of the problem, finding the limit is the easy part
@@MA-bm9jz Indeed. However, that is the domain of upper lower mathematics. I don't think real analysis is in the scope of this channel.
@@gregsavitt7176 yeah, but its still incomplete, thats the whole point of the exercise
Glad to hear that!
Very interesting. A problem with many conclusions. Nice!!
Prof. William M can you please show how he finds his different powers of (1/4)?
@@thedeathofbirth0763 I meant that this problem suggests some interesting implications, not other solutions. 👍
Glad you liked it!
"We're just going to keep it light." Why I love this channel.
By your explanation, it can be x^x^(1/2) = 1/2, and also x^x^x^(1/2) = 1/2. So confusing.
Correct. They’re all satisfied by the same x value
7:55 again I can't agree this part. When you take the real value of squt(2) you get 1.414....
Then 1.414^1.414^1.414^1.414........... it lead to infinite value, so it's NOT convergence, so e^(1/e) is not the peak value of anything > 1 to it's own infinite power tower.
I agree the whole other part of this video for the original equation is the correct process and lead to the right answer 1/4.
But it's quite eeric to prove an number > 1 to it's ower infinite power tower lead to convergence. It obviously can be prove to be not possible by just take a check in real value.
I think it's a enigma in mathematica.
Yes you are right
Are you following the correct order of operations? Remember x^y^z = x^(y^z), not (x^y)^z.
In fact, (x^y)^z = x^(yz).
@@ZipplyZane No matter which power rule you use, they all lead to infinite in the end.
You are wrong.
sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)^... = 2
@@AllanPoeLover
Look up tetration with infinite heights!
That was very nice...
Thanks for your great video...👍👍👍.
np
How do you make these videos? Please reply
Perfect!!! Thank you ❤❤❤❤
You're welcome 😊🧡
I tried calculating it in my head and got x=4 as an answer because for some reason I thought sqrt(2)=2^(-1/2).
Redid the calculation and got x=1/4
Nice and interesting problem, thanks!
Np. Thank you!
8:10 why compare values with e^(1/e), shouldn't we compare them with e ?
Because it's the image from the function
That's a interesting problem thanks for that
No problem 👍
There is a very interesting vid by 3b1b in infinite power tower.
x=1/4
It caused one more confusion. We can also say square root of x^x equal to 1/2, in this case, x is not 1/2. Who can answer me?
x would equal -2
I agree, there are different ways to solve this problem. If we use yours, we'll still get the same answer (or so I thought...) Let's see:
2^(x^x)^(x^x)^(x^x)...=2^(1/2)
(x^x)^(x^x)^(x^x)...=(1/2)
(x^x)^(1/2)=(1/2)
(x^x)=(1/4)
This is where I had to use Wolfram-alpha. :(
Chaos Insurgency is correct. This leads to the answer x=-2, which is definitely not x=1/4. So you are also correct. But why?
Can anyone else help? Please.
@ Everyone -- It is not the same. For example, (x^x)^(x^x) = x^(x*x^x) = x^[x^(x + 1)]. When x = 2, for example, this equals 2^8. In contrast, x^x^x^x = x^(x^(x^x)). When x = 2 here, this equals 2^16.
@Robert Veith I was answering the original question, but yes, you have to start at the top of a power tower for the correct answer, and grouping can cause issues
I get very frustrated with these problems. I have seen so many problems with powers where the order of operations is not clear. The convention that I'm used to is 2^3^4 means (2^3)^4 and not 2^(3^4). After the problem is solved it becomes clear which convention is being used. Why does there seem to be no standard for this? I post comments every time I see it and never get a clear answer. If I use Matlab and plug x-=1/4 into 2^x^x^x... the answer is 1 and not sqrt(2). This is the same convention I was taught, but obviously many others are taught the other convention. It takes all the fun out of the problems when I have to guess at the convention, or look at the video to figure out the convention before I can take a crack at it.
Sorry if it makes you feel that way.
Check this out: postimg.cc/62Ld1dT1
Source: en.wikipedia.org/wiki/Order_of_operations
@@SyberMath Thank you. I will check that. By the way I mean nothing bad against you or others. I just have seen this issue on so many videos. I really want to understand if there is more than one accepted standard, or if I'm missing something obvious that is clear to others.
x^x^... = y = 0,5
x^y = x^0,5 = 0,5
(x^0,5)² = x = 0,25
To easy but interesting 🙂
I am really embarrassed, but how do you calculate your powers of 1/4. When I use (X^n)^m= (X)^(m.n) so for instance (.25)^(1/16) I get a different answer or (.25)^(1/64). Can you please elaborate for us less mathematically talented who are still in high school.
No need to be that way! You're fine 😊
This is not the same as (x^x)^x which is x^(x^2)
What we have is x(x^x)
For x=1/4 this is (1/4)^{(1/4)^(1/4)}
We calculate (1/4)^(1/4) first which 1/sqrt(2) then we raise 1/4 to the power that
The best answers 👌 for the equation [x=-2],x=[(1/4]
Good video : answer obvious if exists…. great discussion of existence
👍
In your solutions :
(2)^x^x^x... = (2)^1/2
(x)^x^x... = 1/2
(x)^1/2 = 1/2
... ... ... (Can that is?) - Please you show me.
x^x^x...=1/2😂
Again ??
答え x=i
You can NOT define 2^x^x^x.. as a recursive infinity, this is a nonsense. So the only sensible definition of 2^1/4^1/4^1/4.. would give 2^0 = 1 )!