I'm tripping on acid right now but I was watching this

You can't, and it doesn't. The argument in that video shows that IF there is an x such that x^x^x^... = 2, then x must be sqrt(2) (or -sqrt(2), which the video ignores). That does not mean that such an x actually exists. (Note that this is the same process going on in your picture.)

In fact, it's not even clear whether x^x^x^... even means anything, much less what it equals. Meaningful interpretations can be given, but it's not cut and dry.

>>7730189
i could tell you, but then i'd have to kill you

>>7730189
For addition:
Any constant, regardless of how small, will not converge to a constant when an infinite number of terms are added.
For multiplication:
The expression at the top can be simplified to
[math]
x= \limit_{k \to \infty} 2^{\frac{1}{k}}
[/math]
which is evidently 1, but only because any number greater than 1 will diverge when multiplied infinite times. The same could be said for any constant.

>>7730222
Any constant, regardless of how small, will diverge* when an infinite number of terms are added
my bad, I'm tired.

>>7730189
Were you a manifold, OP?

>>7730210
>The argument in that video shows that IF there is an x such that x^x^x^... = 2, then x must be sqrt(2) (or -sqrt(2), which the video ignores)
>That does not mean that such an x actually exists.
Okay, but isn't that just the idea behind functions? IF there is an x such that 2x = 3, then x must be 3/2. Why doesn't this mean that 3/2 exists?

>>7730222
>Any constant, regardless of how small, will not converge to a constant when an infinite number of terms are added.
[math]\sum _{n=0}^{\infty }\left(\frac{1}{2}\right)^n[/math]
This is just equal to 2. What do you mean it won't converge to a constant?
I'm also unsure how you simplified the product to x = lim from k to infty 2^1/k.

>>7730222
>Any constant, regardless of how small, will not converge to a constant when an infinite number of terms are added.
But everybody knows that 1+1+1+... = -1/2

>>7730276
exactly. checkmate atheists.

>>7730271
That's not a constant.
That sum's terms converge to 0.
Also,
[math]
\prod_{n=1}^{k} x = x^k \newline
\lim_{k \to \infty} \prod_{n=1}{k} x = \lim_{k \to \infty} x^k = 2 \newline
x = \lim_{k \to \infty} 2^{1/k}
[/math]

>>7730427
Fucking refused to compile
[math]
\prod_{ n = 1 }^{ k } x = x^k
[/math]
[math]
\lim_{k \to \infty} \prod_{n = 1}^{k} x = \lim_{k \to \infty} x^k = 2
[/math]
[math]
x = \lim_{k \to \infty} 2^{1/k}
[/math]

>>7730430
So why doesn't that work for multiplication or addition?

>>7730523
I just explained it in my ORIGINAL POST.
fucking jesus. reading comprehension.

>>7730534
your original post isn't very clear nigger. please clarify.

>>7730626
Because regardless of what constant you put on the right hand side,
[math]
\sum_{n=0}^{\infty} x = c \implies x = 0
[/math]
because for a sum of a constant (which x implicitly is, because x = x for every n), the series only converges if x =0
and similarly
[math]
\prod_{n=0}{\infty} x = c \implies |x| \leq 1
[/math]
because a product only converges for a constant when that constant is less than or equal to 1.

>>7730630
The infinity in the second is a bound, to clarify.

>>7730630
okay. so my question is now, why does this work for exponentiation?

>>7730636
Because even though you have an infinite number of exponents, you have a finite number of terms.

>>7730648
Except for in the case of x^x^x^x.... = 0, which is undefined.

>>7730648
I think you're wrong about that.
Addition
[math]a + n = a + \underbrace{1 + 1 + \cdots + 1}_n[/math]
n copies of 1 added to a.
Multiplication
[math]a \times n = \underbrace{a + a + \cdots + a}_n[/math]
n copies of a combined by addition.
Exponentiation
[math]a^n = \underbrace{a \times a \times \cdots \times a}_n[/math]
n copies of a combined by multiplication.
Tetration
[math]{^{n}a} = \underbrace{a^{a^{\cdot^{\cdot^{a}}}}}_n[/math]
n copies of a combined by exponentiation, right-to-left.

Just as infinite series and infinite products have infinite terms, I'm afraid infinite exponent towers also have infinite terms.

>>7730654
Not necessarily.
If x^x^x^x^x^x.... = 100,
then the product has only 100 terms, because you can have fractional powers.

>>7730654
You would be right if it were
(x^x)^x)^x)^x...., but this is not the same as x^x^x^x^x^x^x.....

>>7730655
x^x^x... means it is endless. that's what the repeating dots denotes. that there is no end.
>>7730661
(x^x)^x)^x)... is the same as the limit of x^xn as n approaches infinity. That's different from x^x^x... what we are talking about.
https://en.wikipedia.org/wiki/Tetration
This wikipedia article may clarify things.

>>7730669
if the tetration converges, there are implicitly a finite number of terms. I don't know where the sticking point is here.
x^x^x^x.... = 2 implies there are two terms,
[math]
\sqrt(2) * \sqrt(2)
[/math]

>>7730669
I'm well aware of what the dots mean, you pretentious prick.
Infinite numbers of exponents does not mean infinite number of terms.

>>7730276
>partial sums do not approach finite limit
>doesn't converge
>only found by analytic continuation

>>7730681
explain to me why you keep distinguishing between number of exponents and number of terms. that is like saying number of factors is different from number of terms in an infinite product or the number of summands is different from the number of terms in an infinite series. what's the difference?

>>7730696
Because if your x is a constant greater than 1 (which it is in any convergent infinite tetration), and any tetration is effectively a product, the product must have a finite number of terms (x*x*x*x*x...) to converge.

>>7730702
I should be using the word factors, to clarify. You can express any convergent tetration x^x^x^x^x.... = n as a product with n factors.

>>7730696
Now I realize why we were having a difference of opinion here. Sorry, my imprecise language caused my meaning to be misunderstood.

>>7730702
I shouldn't have brought in the concept of tetration since that is not what I'm talking about. I was simply referencing how exponentiation is next iteration above multiplication. what we're talking about is exponentiation and why an infinite exponent tower converges while an infinite product does not converge. is that what you're saying?

>>7730708
read
>>7730706
We were using the word "terms" differently.

>>7730430
MathJax can't into newline

>>7730703
I feel dumb as fuck right now. I'm still not understanding this. From my understanding, addition is the most basic math operation. From that we get multiplication. From multiplication we get exponentiation.

An infinite series has an infinite number of terms. An infinite product has an infinite number of terms. An infinite power tower, which is the word I was looking for, has an infinite number of terms as well. The question that I have is simple: why is this dude, in the video, able to manipulate that infinite power tower in such a way to yields a meaningful result, whereas manipulating an infinite product or infinite series, would yield a meaningless result? I hope I made my question as clear as possible to facilitate this discussion. Thanks for taking the time to educate my dumb ass.

>>7730723
>why is this dude, in the video, able to manipulate that infinite power tower in such a way to yields a meaningful result, whereas manipulating an infinite product or infinite series, would yield a meaningless result?
I meant to say this:
>why is this dude, in the video, able to manipulate that infinite power tower in such a way to yields a meaningful result, whereas manipulating an infinite product or infinite series by similar means, would yield a meaningless result?

I'd like add on to this. Obviously, the picture I posted in the OP shows that manipulating the terms of an infinite series or product with the same logic that the guy in the video used, leads to an incorrect answer. The question is, why is this so?

>>7730723
It only does for certain cases. x has to be [eqn] n^{1/n} [/eqn] for it to work.

>>7730726
I'm sorry. I don't know what you mean by that. What does for certain cases?

>>7730725
to answer this second question
think about it this way:
we say a series diverges if
[math]
\forall k>0 \exists N: n>N \implies \sum_{k=0}^{n} a_n > k
[/math]
so, if we define
[math]
a_n = \epsilon
[/math]
then we find
[math]
\forall \epsilon > 0, N = \frac{k}{\epsilon}
[/math]
So any constant greater than 0 implies it will diverge.
For products, we say that a product diverges if
[math]
\forall k > 0 \exists N: n>N \implies \prod_{i=0}^{n}a_n > k
[/math]
so we see that if
[math]
a_n = r >1
[/math]
then
[math]
\prod_{i=0}^{n} r = r^n > k when n > N = \frac{log(k)}{log(r)}
[/math]
The second one is undefined at r <1 because N cannot be negative.

>>7730735
>kwhenn

>>7730749
Yeah. It should read
[math]
r^n > k \quad when \quad n>N = \frac{log(k)}{log(r)}
[/math]

>>7730723
You can manipulate some series like that
for example:
[math]
\sum_{n=0}^{\infty} \frac{1}{2} ^n =s
[math]
Observe:
[math]
S - \frac{1}{2}S = \frac{1}{2}
[/math]
[math]
\frac{1}{2} S = \frac{1}{2}
[/math]
[math]
S = 1
[/math]
The fact that it doesn't work in cases such as your original pic is just proof that there are no x for which that holds.