How would you solve this?
ln nigga
>>7942441
ln(x^x)=ln(2) <=> x*ln(x)=ln(2)
Where would you expand it from there?
>>7942438
can't be solved analytically
>>7942438
https://www.wolframalpha.com/input/?i=x^x+%3D+2
I would try Newton's approximation method
>>7942474
UNDERRATE'D
>>7942485
REALY
Honestly? Probably trial and error.
>>7942450
Is there any way to tell if any equation can be solved analytically?
>>7942438
~1.559610
>>7942501
number of unknowns
You can obviously get an exact answer, since wolfram gets one
>>7942438
The exact answer is
1.559610469462369349970388768765
if you don't believe fuck you
>>7942555
I'm pretty sure you can prove the answer can't be rational
>>7942438
Same way you solve 2 * log (2)
You use a calculator.
Idk, Lambert W function or sth
the W function
[math] x^x = 2 [/math]
[math] e^{x\log(x)} = 2 [/math]
[math] x\log(x) = \log(2) [/math]
[math] \log(x) e^{\log(x)} = \log(2) [/math]
[math] \log(x) = W(\log(x)) [/math]
[math] x = e^{W(\log(x))} [/math]
there's probably not a closed form for that
if there is it's probably not pretty
>>7942692
X is on both sides of the equal sign.
>>7942723
x = -ln(0.5)/W(-ln(0.5))
>>7942692
so now W(x) isnt a closed form?
Fuck you
by keeping it as a symbolic expression
#rekd
>>7942815
No it's not.
>>7942501
No. Possible or impossible is vague here. If x^x was a common function in science just like sin(x) then we would say that we can solve this analitically.
>>7942438
lets use numerical analysis! newton's method, runge-kutta, newton-raphson, whatever.
>>7942911
listen , kid. If a solution to a problem cannot be expressed with a finite number of additions and multiplications then it cannot be solved analytically
>>7942555
/thread
How would you solve
[math] x^x^x =2 [math/]
You guys are autistic retards.
It's already been solved here >>7942730
>>7942944
Can you solve sin(x) = c analytically?
>>7942961
what is lim of the solution of
x^x^x^... (n times) = 2
>>7942969
x= arcsin(c)
>>7942970
sqrt(2)
>>7942984
>sqrt(2)
there is no such number
>>7942989
Wildhamburger pls go
>>7942992
>newton-raphson
Mathematics needs a more rigorous foundation anon.
what is lim of the solution of
x^x^x^... (n times) = a
>>7943016
a^(1/a)
>>7943011
No, it doesn't. And anyway Wildhamburger is not rigorous.
I'd take half a log of both sides and a half log by a half exponential cancels out and boom, answer right there
>>7943017
good guess
https://en.wikipedia.org/wiki/Tetration#Other_super-roots
>>7943027
Not a guess. It's easy to see that a^(1/a) tetrated infinitely results in
a^(a/a)^(a/a)^... = a
>>7942448
x=ln(2-x)
>Pic related
>>7943084
"""""""""""""function"""""""""""""""
>>7943084
""""""""""""""Function"""""""""""""
>>7943069
You mean [math] x = \frac{ln(2)}{ln(x)} [/math]
>>7943239
>>7943212
>>7943130
Y=X^0.5
"""""""""Function"""""""""
>>7943243
The other guys had a point, but this one is pretty much a function. Only a literal cry baby autist would get upset over this technicality.
x = 1.559610469 (9 d.p)
>>7943034
that's wrong
btw
a^(1/a) ^ a^(1/a) ^ a^(1/a) ^ ... n times is equal to
a^a^(n/3-1) which does not go to infinity
super-root is is only well-defined if 1/e ≤ ln x ≤ e
https://en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights
[math]x^x = 2[/math]
[math]x \ln x = \ln 2 [/math]
[math]\ln x e^{\ln x} = \ln 2[/math]
[math]\ln x = W(\ln 2)[/math]
[math]x = e^{W(\ln 2)}[/math]
or alternatively
[math]x = \frac{\ln 2}{W(\ln 2)}[/math]
>>7942972
Yea, arcsin(c) because sin(x) just happens to be a very common function. Now define something like anon(c) = x which solves x^x = c. Here you go, it can be solved analytically now.
>>7942438
The solution is transcendental.
x = sqrt2
It's transcendental. You'll need to used newton's method to solve for an answer.
The derivative of [math] x^x [/math] is [math] x^x + x^{x}logx [/math].
-1/12
>>7942438
[math]2^1[/math]
Solved it.
>>7943702
nigga what
Speaking of unsolvable problems.
Solve this by hand /sci/(x intercepts)
>>7943712
> solve this transcendental function by hand
Yeahhhhhh no
>>7943734
why is it transcendental? (mfw the only things I know are transcendental are pi and e) is it because of 2^x or 1/x?
>>7943734
what if your life depended on it? I'm sure you could muster the will power
>>7943740
google 'transcendental number' and your answer will be clear. The kinds of equations in that pic are generally transcendental.
>>7943740
I know someone's gonna nitpick my definition but it's pretty much just PEMDAS, and polynomial functions. Any logs or exponentials that contain a variable are transcendental. Anything like integrals, infinite sums, ect, are usually beyond transcendental functiobs, and are elementary functions.
Usually they are very messy if they combined with algebraic functions because they are practically different branches of math.
>>7943743
Will power isn't really a factor, your equation is really close to it's most simplified form as it is. I don't think it's possible to extract an exact answer from that equation that is filled with constants.
>>7942474
BRUH
>>7943711
What's wrong with my answer?
>>7943835
2=/=1, pretty much
The problem is asking what number, put to the power of itself, equals 2.
>>7943839
Shit.
Then what about the square root of 2?
>>7943846
sqrt 2 to the power of 2 is 2
sqrt 2 to the power of sqrt 2 is less than 2
>>7943859
shit
>>7943873
Don't you own a calculator, dumbass?
>>7943890
I don't think before I type.
>>7943904
Im sorry anon, I was snappy :(
>>7943941
I'll be the dumb slut in bed.
>>7943948
I call sub!
>>7943994
>>7944002
Come on baby, the only anomaly I have is a big lump on the front side of my body. And it's solid as lead. Not even radiation can make this bitch go down. If your dick has one hole, I'll make two with my solid block of lead. Or make that one hole bigger. hmhmhmhmhmhmhm all that blood makes me horny
>>7944015
That place is too big, and metallic, and too scary. Please comfort me you sexy beast, comfort me while my throbbing piece of led spreads into your genitals and rectum.
>>7943279
...and when it is well defined it is a^(1/a). So I'm correct.
Also, a^(1/a) ^ a^(1/a) ^ a^(1/a) ^ ... n times is not a^a^(n/3-1) . No idea where you got this but it's wrong.
>>7943234
that's what he fucking said
Superroot of 2
>>7944042
>Also, a^(1/a) ^ a^(1/a) ^ a^(1/a) ^ ... n times is not a^a^(n/3-1)
How do you derive
a^(a/a)^(a/a)^... = a ?
>>7944642
scratch this
i mean, how do you derive
a^(1/a) ^ a^(1/a) ^ a^(1/a) ^ ... = a
>>7942501
Galois theory can probably tell, but the method, if any, would be very convoluted.
>>7943276
Did some snooping around and traced its origins to /int/. Some threads there had been going
>""""""""Great"""""""" Britain
>>7943712
First step is to find its roots at 0.
So set f(x) = 0, then multiply both sides by x, then find the roots. Since the resultant polynomial is sextic, you'll need to use ultraradicals for this one, but it's doable.
>>7944654
>Galois theory can probably tell, but the method, if any, would be very convoluted.
Galois theory doesn't apply to this.
>>7944680
Rather than that, I'd think the method just hasn't been discovered, as in the proof is inaccessible at the moment. Still, I would think that Galois theory holds the key to this.
Guess n' check.
Anyway after two minutes it's around 1.55999
>>7944700
by actually reading the article, I found out that a solution expressed in terms of lambert's function is in fact a closed form expression.
>>7944712
Maybe I'm just not good enough at it. Can LaTeX here handle the expression?
inb4 >function
>>7944716
what expression?
>>7944720
The solution to OP's question, of course.
It must be wrong but why does this not work?
>>7944747
because a=b does not imply d/dx(a)=d/dx(b)
>>7944748
It does.
>>7944749
x = -x (when x = 0) doesn't mean 1 = -1
>>7944749
a=b is a condition for certain values of a parameter, a and b don't always have the same gradient.
>>7944756
You just said when x = 0. That means d(-0)/dx = d(+0)/dx
>>7944775
>>7942992
>>7943021
enumerate x and divide by two.
>>7945167
maybe enumerate x then divide by x2.... or something my intuition tells me.
>>7943243
Couldnt you put an imaginary value on it and turn it into a function? Thinking outside the box here.
>>7945171
No. Enumerate it to the base n. x/(n*x) = 2/(n*x),
x^3*n/2 - 1 = 0
x^3 - 1 = 0
x^3 = 0.
,
(x)(x)(xn) = 1.
sorry for any mistakes.
>>7942438
The following is the simple method.
Write the function as x^x-2=0.
Find a value of x which yields a negative result of x^x-2, for example, x=1.
Then find a value which yields a positive result of x^x-2, for example, x=2.
You are now aware that one root for x^x-2 is placed between 1 and 2.
Continue the process until you get the an approximation of the accuracy you want.
>>7947254
A fixedpoint computation would be more automatic:
[math] x^x = 2 [/math]
means
[math] \log_2(x) = \frac {1} {x} [/math]
and then
[math] 2^{ \frac {1} {x}} = x [/math]
The solution (given in >>7942555 as 1.559610469...)
is the value of the function [math] f(x) := 2^{ \frac {1} {x}} [/math] iterated infinite times and evaluated for some positive number.
That is, the sequence
[math] 2^{ \frac {1} {x} } [/math]
[math] 2^{ 2^{ - 1 / x } } [/math]
[math] 2^{ 2^{ - 2^{ - 1 / x } } } [/math]
[math] 2^{ 2^{ - 2^{ - 2^{ - 1 / x } } } } [/math]
[math] 2^{ 2^{ - 2^{ - 2^{ - 2^{ - 1 / x } } } } } [/math]
[math] 2^{ 2^{ - 2^{ - 2^{ - 2^{ - 2^{ - 1 / x } } } } } } [/math]
[math] 2^{ 2^{ - 2^{ - 2^{ - 2^{ - 2^{ - 2^{ -1/ x } } } } } } } [/math]
[math] 2^{ 2^{ - 2^{ - 2^{ - 2^{ - 2^{ - 2^{ - 2^{ - 1/ x } } } } } } } } [/math]
[math] 2^{ 2^{ - 2^{ - 2^{ -2^{ - 2^{ - 2^{ - 2^{ - 2^{ - 1 / x } } } } } } } } } [/math]
[math] 2^{ 2^{ - 2^{ - 2^{ -2^{ - 2^{ - 2^{ - 2^{ - 2^{ - 2^{- 1 / x } } } } } } } } } } [/math]
evaluated at e.g. x=1 gives
2
1.41421 = sqrt(2)
1.63253
1.52896
1.57357
1.55347
1.56235
1.5584
1.56015
1.55937
Anyway, in >>7942692 we see that the solution is of the form [math](f^{-1} \circ W \circ f) (2) [/math], suggesting the Taylor expansion might be approachable. I think W has nice coefficients at 0 or 1, so that's one approach for a more closed form, but with all the logs I don't know how good the convergence will be anyway.
I cared for an expansion of [math] x^x [/math] once and came up with
[math] x^x = x \sum_{n=0}^\infty \prod_{k=1}^n (1-x) \left( 1 - \frac { 1 + x } { k } \right) [/math]
I think I ended up there by playing with the following substitution:
Let [math]t = 1 - x[/math], then
[math] x^x = (1-t)^{1-t} = (1-t) \dfrac {1} { (1-t)^t } [/math]
which expands as
[math] = 1 - t + t^2 - \dfrac{1}{2} t^3 + \dfrac{1}{3} t^4 - \dfrac{1}{12} t^5+ O(t^6) [/math]
Setting this to 2 and and using Mathematica, I end up with 1.56043.
>>7942438
log, differentiate, find minimum (this is x). In this case the derivative is 0 anyway.
>>7944647
There are a number of ways to do it. The simplest is probably
x^x^x^... = a
x^x^x^x^... = x^a
a = x^a
x = a^(1/a)
>>7947524
If this would work, the solution would be independent of the constant on the right hand side (2 in this case).
The minimum value of x^x you can remember, it's the percentage 1/e = 36.789% of secretaries (pic related) you shouldn't not hire no matter how skilled they are.
https://en.wikipedia.org/wiki/Secretary_problem
>>7947549
Ah yeah, I see. I should have realised that the minimum being equal to the derivative is a red flag.
>>7947562
Bleh, I fucked that sentence up. Whatever, you know what I mean.
>>7947577
Nah, I actually thought about it for 30 seconds and then let it be.
For future reference, I cooked up this:
[math] \dfrac { { \mathrm d } } { { \mathrm d } x } g(x)^{ f ( x ) } = \left[ f ( x ) \, g'( x ) + f'( x ) \, g ( x ) \cdot \log \left ( g ( x ) \right ) \right] \cdot g(x)^{ f ( x ) - 1 } [/math]
