Well, Yea Forums?

Well, Yea Forums?

Attached: youshouldbeabletosolvethis.jpg (1920x1080, 207K)

Other urls found in this thread:

lmgtfy.com/?q=3^3^3
en.wikipedia.org/wiki/P-adic_number
twitter.com/NSFWRedditVideo

X^2 = X^X^X^...

-1/12

42

I was home-schooled, so I never went to high school. How do you expect me to know high school level math?

If x is greater than or equal to 1 and less or equal to eth root of e then x=sqrt2
For positive x smaller than 1 this symbol is always less than 1. And for other values there is no standard convention

X=1
Since you can't create an infinite number using the first equation and equal a solid number you have to simplify it by adding 1.
Thus x^2=1

Why does being homeschooled precludes knowledge of mathematics

How was the answer actually 2 in the anime? I don't recall high school math teaching me this kind of stuff.

Note that if X

But that's wrong.

Enlighten us

X^2=2 hence X=sqrt2.

When you have a pattern almost always the solution is the mathematical inversor.

I'm not that homeschooled user. Just answering for him. Anyone can be a teacher, not everyone can be an effective one.

sqrt(2) ^ sqrt(2) ^ sqrt(2) ^ sqrt(2) is already greater than 2
sqrt2 ^ sqrt2 ^sqrt2 ^ ... goes to infinity

Don't really have time to show the derivation, but the most approximate answer yields x = 0.5ln(exp(Σ(1/2^n)))

Well yeah I agree

Clearly the ... leads off to a X^X+1
You can't really expect me to believe it goes on forever.

That fancy parentheses is useless. You exponentiate then take the logarithm so they cancel each other out. The sum of inverse powers of 1/2 is 2 or 1 depending on whether n=0 or 1 as starting value. Half those is 1 or 1/2. Neither are solutions

>sqrt(2) ^ sqrt(2) ^ sqrt(2) ^ sqrt(2) is already greater than 2
How sure are you about that?

x^x^x^...=2
x^2=?
Sub x^x^x^... for 2 and you get x^x^x^...=?
But you already know x^x^x^...=2
So x^x^x^...=x^2=2
Which is nonsense. This question is dumb.

He is and he is surely mistaken

You can't be serious.
Please tell me you're trolling.

Attached: doubt.jpg (1351x1078, 82K)

There is no number that satisfies x^x^... = 2, so the basic premise is already incorrect.

if x were 1, then x^x^... = 1
if x were 1.0(repeating)1, then x^x^... = infinity
even the smallest increment we can possibly make already overshoots the target 2.

It literally does not, have you even tried doing the calculation? The sequence goes to 2 asymptotically.
You are probably just multiplying the sqrt2's in the exponent instead.

'x to the bullshit' equals 2
the bullshit goes on forever
so 'x to the bullshit' is equal to the bullshit
so the bullshit equals 2
'x to the bullshit' equals 'x to the 2' equals 2

x^x^x^...
is not the same as
x^(x^(x^...)
the answer is x=sqrt(2) and x^2=2.
you can try it on a calculator if you like

>Exponentials and logarithms don't cancel out
What.

yeah I was doing the thing that pointed out.

>this fucking thread again
Not anime or manga.

Most threads on Yea Forums aren't anime nor manga neither.

lol

Attached: oof.jpg (1080x926, 187K)

>math questions in anime and manga
>what is school in anime and manga
It's allowed.

X^sqrt(x)=X therefore if X= 2 then 2^sqrt (2)^^sqrt (2) ^sqrt (2) ^sqrt (2)= 2

>Answer: X^(x)=sqrt (x) then X^(2)=sqrt (2)

End this thread now.

The gambles on this anime are so retarded. Just make them play poker or something instead of doing high school math in an elaborate maze.

lim1^+

Let's take n = x^x^x^...
Then x^n = 2.
Since x^n also equals n, then n = 2.
So x^n = x^2 = 2.
The answer is 2.

Holy fuck user, he's right
ln(e^x) = x*ln(e) = x

Just google the problem if you don't believe me

the thread ended when the correct answer was posted, and it was not yours.

If x^x^x...=2
(x^x^x...)^2=4
thus x^2 = 4^t (t0), x^2=1
thus x=1 or x=-1
but either of them satisfy the proposition. We have to define the number whose square number is 1, but that exceeds the caliculum of high-school.

>neither

You guys are almost as bad as /sci/

Yes

you are mistaking x^(x^(...) for x^x^x just like everyone in this thread. the answer's already been posted.

>(t->0)

If x^x^x...=2
2^x = 2
thus x=1
but it doesn’t satisfy the proposition.
So there is no solution.

How am I wrong, retard. The only way you can elevate a pattern of numbers and get the same number is with its mathematical opposite.Do it with (log), and you'll end up with the same result.

>Since x^n also equals n, then n = 2.
u wot

You're close, but your order of operations is wrong.
Instead of 2^x=2, you should get x^2=2.

Are you idiots doing this on purpose to throw the thread into further chaos?
x^x^x^x^... is by definition equal to x^(x^(x^...)), which is different from ((...^x)^x)^x.

I want you to grab a calculator and punch in sqrt(2).
then repeat (sqrt(2))^answer.

>elevate
>pattern
>mathematical opposite
None of those terms mean anything. You're just spewing random mathbabble.
Just admit that you're bad at creative math, it's not nearly as shameful as what you're doing now.

3^3^3=19683
3^(3^(3)=7.625*10^12

What the fuck is t? Are you trying to take a limit?

Whence the fuck did you pull sqrt(x)?

Wrong.
3^3^3=7.625*10^12
3^(3^3)=7.625*10^12
(3^3)^3=19683
3^(3*3)=19683

the ^ operator doesn't work like you think it does

By mathematical convention, when you see a^b^c, you put parentheses around b and c. That's just how it was defined, because the alternative (putting them around a and b) is less interesting.

That's my line.
For fucks sake, lmgtfy.com/?q=3^3^3

This is wrong. But it's the only way you could get the answer to be 2.

Attached: IMG_20190329_192508.jpg (2304x4608, 1.94M)

Dumbass here watch this

x^x^x = 2
x^2 = ?
= 2^2 = 4
= x^2 = 4
Square both sides so, x = 2

x^x^x... = 2
so all of the exponents are equal to 2
so x^2 is equal to 2
so x is sqrt2

mods get rid of this thread NOW

Which part do you think is wrong?
Also see elementary solution:

>repeat (sqrt(2))^answer
My answer was for the second equation which it doesn't have a pattern. The pattern lets you know the result of the second equation, you just have to clear them with logarithms.

x^x^^^^... = 2 (log) ; x(x^x^^^^...) = log_x(2)
x^x^^^^...=logx(2) : 2 = logx(2)

X^2= 2 ; (sqrt 2)^2 =2

Grab a calculator and compare
3^3^3 and (3^3)^3

Its actually rather easy if you think about x^0.5x^0.5x^0.5x etc

Not reading an unrotated picture. Either post it right or don't post it at all.

Actually this is correct but for an ad hoc reason. It turns out for 1/(e^e)

Undefined, you are not allowed to create such a towers, even if there is an answer which sounds legit.

Did you click the link?
Also, try 1+2*3 on your piece of shit calculator.

Why would it be undefined?
It's not like "the sum of all integers is -1/12", if you only make a tower with a height of 10 with the right number (sqrt (2)) you'd get very close to 2.

So many brainlets in this thread, holy shit.

no matter what excuses you come up with the answer will remain x=sqrt(2), x^2=2

The problem isnt very well put. The truth is that for x in (1/(e^e), e^(1/e)) the function is well defined, in fact it's continuous and increasing

the answer is sqrt2 and you losers know it

The answer is 2.

Attached: IMG_20190329_183757_BURST97.jpg (4032x3024, 561K)

2. You can just substitute the first equation into the second (in place of 2).

correct

yes because 2^2^2... = 2

sqrt2^sqrt2^sqrt2... infinitely converges to 2, therefore x is sqrt 2

What the fuck are you talking about? I'm saying that x^x^x is x^(x^x) and not (x^x)^x and that the answer is x=sqrt(2), you're the one spewing bullshit.

x equals sqrt2 and as it happens x^2=2. Due to the repeating notion of the solution confusion can arise
For example if it said x^x^x^x.....=2.5 then x^2.5=2.5

Read the question.
It's x^2=?, not x=?.

>yes because 2^2^2... = 2
Retard.

this is literally 6 / 2 (1+2) over again

That was sarcasm.

you claimed the answer was x^2=sqrt(2) in your first post.

>correct answer given within 3 minutes
>800 replies and 250 images omitted, click reply to view.

No one cares about x.
You don't need to know it to solve this question.

>X^X^X^X...=2
wait what. what x would even satisfy that

also true. you can just substitute the 2 in the second statement.

48÷2(9+3) = ?

That wasn't me.
My first reply to you was , and since then you were saying the correct answer and the wrong interpretation of x^x^x.

How do you prove it though?

>Inb4 hur durr use a calculator

Get a calculator and try finding a number such that x^x^x^x^x is approximately 2. It could be a fun exercise.

I'm not a math teacher to explains it like one but I'm right. Second equation. (X=sqrt2) and the (?) simbol equasl to 2.

1

There's bearely been 100 replies, retard.

Wheres the proofs though

how did I reach the correct answer with a wrong interpretation?

you got lucky

Let X^X^X^X... = P
X^X^X^X... = X^(X^X^X...)
This shows that P = X ^ P by substitution
By raising each side to the 1 / P power
X = P ^ (1 / P)
If P = 2, then X = 2 ^ (1 / 2) = sqrt(2)
X ^ 2 = (2 ^ (1 / 2)) ^ 2 = (2 ^ (1 / 2)) * (2 ^ (1 / 2)) = 2
Anyone commenting on the power tower not converging is a brainlet, infinite power towers converge in the range [e ^ -e, e ^ (1 / e)], not [-1, 1].

yeah, let's go with that

Define f(x, n)=x^f(x, n-1), f(x, 0)=1.
Now try plugging in infinity for n. That way you can get a necessary condition on x, but it even it you find a possible solution it could be wrong, because infinity is weird.
Infinity-1=infinity so you get f(x, infinity)=x^f(x, infinity).
You know that f(x, infinity)=2 so you get 2=x^2.
You can also find f(x, infinity) if you only know x, using the Lambert W function, a.k.a. the product logarithm.

Anyway, to prove that sqrt(2) is indeed the solution to f(x, infinity)=2, you look at the sequence of f(sqrt(2), n). Through some analysis you'll see that x

That's what's baffling me too. I can only guess you you used the correct interpretation in your calculations but for some reason wrote the wrong one in post.
Now please acknowledge your mistake so we can move on.

That's not a valid question.

I watched the episode, the answer is 2.
No idea how she came to that answer though since I'm a NEET and fucked off from studying.

Based

To show how it can go wrong, define f(n)=2*f(n-1), f(0)=1. That's 2^n.
If you were to look for f(infinity)=x, you'd find that x=2*x, which only has the solution x=0. However, the function f(n) does not converge to 0 as n gets large.
HOWEVER however, in a different metric, the 2-adic numbers, it does indeed converge to 0. So the solution we got is not meaningless.
Another one, f(n)=1+2*f(n-1), f(0)=1. That's the sum of 2^k for k from 0 to n.
If you were to look for f(infinity)=x, you'd get x=1+2x, x=-1/2. Again, the function diverges in the canonical metric, and yet again, it converges to -1/2 in the 2-adic metric. (I'm a bit rusty and tryping this on my phone so the second one may contain some errors)

Have you tried reading the thread?

In the second one it should be -1,I don't know what kind of brainfart gave me -1/2, anyway now it should be all correct.

For more info: en.wikipedia.org/wiki/P-adic_number

For all neets who are only able to read in anime pictures

Attached: 1553875291830.jpg (1920x1080, 278K)

sqrt(2) you dummy

it doesnt say solve for x dummy

This. I have no idea how the fuck people overcomplicate things this much. Regardless of whatever the fuck x^x^... is supposed to mean, this should hold as long as the substitution property does.

x^2 = 1,11111...∞...n

Retards

I dont remember ever being taught about the third "x" and the "..." even in college but im gonna try understanding without looking at the thread.

The third "x", gonna dub it "xrd", I assume is the power of the power.
As 2^2 = 4 then you do 4^2 from the xrd.
Then 2^2^2rd = 16

The "..." either means its repeating the xrd sequence by infinite or certain amount of times. Going to assume its the former.

So if "..." means it's infinite, and X^x^xrd = 2
then the actual equation for it could be
X= 1.414
^x = 2
^xrd = 1
Which results in 1.414 ^2 ^1rd = 2 as since there is a "..." it can only be of one number for it to conclude to 2 and numbers being powered by 1 even in an infinite amount of times will always stay the same. So im going to guess the "..." is meant to throw you off.

So for X^2 = ?, using what I know from what numbers the previous equation used to get 2, the answer would be 2.
1.414 ^2 = 2

If thats not the answer then it has to be some imaginary number or N/A as the infinite is being applied here.

Could be assblasted wrong but was bored.

anyone who can't solve this should be b& from Yea Forums

I did statistics, for me that is the one below is a chi-square sorry

see its high school level math

X=1.258
rough answer, but close enough.

Nope sqrt2=1.412

You should be able to answer this

Attached: Dragon Ball Super - 048 - HOPE!! Again - Awaken in the Present, Trunks [DragonTeam][F1AF644C].mkv_sn (1280x720, 174K)

n = x^x^x^..., so x^n = 2 as per the first line in OPs pic.

X = infinity

x^n = 2.
x^n = n.
Therefore, n = 2.

But the equation is X^X^X

x=sqrt(2); then x^2=2

everybody is forgetting that the answers on the floor doors are always 1 or 2, since that's the number of floors you skip

since 1 doesn't work 2 is the answer, you don't need to solve anything else

brainlets, when will they ever learn

Attached: 1274765994247.jpg (640x480, 45K)

kek, so funny seeing everyone struggling so much with this.
I guess you elitist fags were not that smart after all.
Another problem for you (by VI Arnold):
The hypotenuse of a right-angled triangle (in a standard American examination) is 10 inches, the altitude dropped onto it is 6 inches. Find the area of the triangle.

No such triangle exists in Euclidean geometry. If it did area would be 30 sq inches

>x = 27

Attached: 1507788182980.png (1055x800, 574K)

That's more on the level of Yea Forums

*ahem*
FUCK trig

Is.. is it 24 sq inches?

Learned this in algebra 1 when i was in school. kids these days

This unironically convinced me

19*sqrt2 which is fairly close to 27. Indeed x^2=722 and 27^2=729 so x>27-7/54 so approximately 26.87

The responses to this make me sad for the future of mankind

Attached: 1252436692512.png (363x486, 72K)

19

The real question is if there is some non-Euclidean space wherein an equilateral right triangle exists.

>so funny seeing everyone struggling so much with this
Yeah, yanks aren't very good at maths.

A sphere.

x

Yeah, that was dumbfuck obvious in hindsight.

Except the problem has been solved numerous times in this very thread

(X)^X^X^...=2
X^X^...log(x)=log2
x=[log2]/2

>should
Math is a tool of oppression again the working (and non-working) class. Dont even debate this, it a fact that the science is already settled on this. The fact that this man is allowed to impose such bigoted standards onto his students is everything wrong with civilization. No the fact that he is 2D does not excuse his actions, they still represent the tyrannical nature of mathematics. Besides his 2D self was still created and perpetuated by 3D bigots. I am also appalled by the use of 19s in that picture, a clear reference to Adolf Hittler. "A" being the first letter of the alphabet, and "H" being the 9th. If you or the mods of this site have ANY decency left this post will be deleted and you will be placed in jail.

Based and pastapilled

??? thats stupid

Since you didn't understood anything in this thread, try this

Need the area of the square in the triangle.

Been reading through this thread and can't believe you idiots are actually worse than /sci/.

Not if you're the kind of person that actually watches DBS.

The square is how you represent a right angle.

>there are people who actually believe 0.99999999... = 1
Kill yourselves

Yeah; there is nothing to believe in since it's a fact.

Using purely symbolic logic, x is 2. Outside of that, any number will either diverge to infinity or converge to +-1, and so X does not exist.

>any number will either diverge to infinity or converge to +-1,
The educational system has really fallen

>an irrational number is a rational number
nice """"facts""""

>0.999..... is irrational
I know 'alternative facts' are popular among some groups, but in m/a/th we like to stick to reality.