/sci/ - Science & Math

>>9638200
x^3=4
x=cube root 4
x=1.5874

(1.5874^1.5874)^3=2^4

9.03!=2^4

Not possible.

x = 1.7216

>>9638220
>assuming x^3=4
PB stronk

>>9638220
>x^3=4
not true, the entire LHS just has to equal 2^4

>>9638271
>>9638248
(x^x)^3=2^4
(x^x)=cuberoot 16
(x^x)=2.5198

I do not know how for to solve this one my friend.

http://www.wolframalpha.com/input/?i=solve+(x%5Ex%5E3)+%3D+2%5E4

can I oldfag now?

>>9638200
x = 1.715 ish

>>9638288
x^x^3 = x^3x
so x^3x must equal 16
find that value

>>9638310
>x^x^3 = x^3x
anon...

>>9638314
But that's correct according to the radical rules. It's just that expressing it that way doesn't really help us to find the answer.

I just briefly looked it up and it seems you need some more advanced mathematics than basic algebra 2 to solve x^x.

>>9638324
x ln 3x = 16

>>9638343
wouldn't it be x^3 ln(x) = ln(16) ?

>>9638359
>Team Minibrain

x^(x^3) does not equal x^(3x). Radical rules imply that (x^x)^3 = x^(3x). The "grammar" matters here.

would It be the third root of 4
x^x^3 = 2^3
take logx of both sides
x^3 = 3logx(2)
sub in x^3 to original equation
x^3logx(2) = 2^3
break it up
x^3 * x^logx(2) = 8
2x^3 = 8
x^3 = 4
x= 4^(1/3)

>>9638376
He's right

Obviously there's no elementary solution to this, if you used pen and paper, using newton's method to find the root of x^x^3 - 2^4 = 0 would work to approximate it, rather tediously.

>>9638372
>>9638416
meant to reply to
>>9638372

>>9638395
where did I mess up

>>9638436
oh I see ok

1.71442

Peep rasse sieg!

>>9638200
1.722 is the answer scrub.

>>9638220
imagine being this retarded

I got 2 this time

x^(x^3)
x^3*log2(x) = 3
x^3 = 3/log2(x)
x^3/log2(x) = 2^3
(x^3.log2(x))^log2 = (2^3)^log2(x)
x^3 = 2^log2(x)

More precisely 1,7217 I could go on foerver...

>>9638200
x = e^(1/3 W(6 i π n + log(4096))) and log(16) + 2 i π n!=0 and n element Z

>>9638533
I don't think that's right

>>9638543
You fucking brainlet don't know how to use a calculator : 1.7217^(1.7217^3) = 16.00372106

>>9638543

I said 1.722 at first then more precisely 1.7217

I've never said 1.714 bullshit

>>9638533
>all that rounding error
This answer is much more accurate
>>9638463

>>9638550
Tganks 4 the answer needed it 4 homework

>>9638562

Ever heard of approximation methods ? dumbass

>>9638558
sorry about that

was your answer rounded up because it overshoots

Would anyone care to explain me how to solve it?

I used a graphing calculator and got 1.624
I have no idea how to solve algebraically

>>9638513
Show how you get there
Dumbass

>>9638568
My answer is given by a method of approximation, I can get all the numbers I want, I choose the precision.

>>9638570
>>9638578
for u

>>9638580
ok

>>9638565
>mad because he can't get an answer accurate past the first decimal
>probably fucked up order of operations somewhere too

lmao @ ur life

>>9638200
x^3 ln(x) = 4ln(2)
e^3ln(x) ln(x) = 4ln(2)
let y = 3ln(x)
ye^y = 12ln(2)
y=W(12ln(2))
x = e^(W(12ln(2))/3) =~ 1.721670170412965998739149039475469284021191706837875947626...

>>9638586
You really understand nothing about math; your calculator and most of the thing you use is approximation based. If I wanted to, I could provide more precison.

>>9638578
nvm I read the question wrong

>>9638589
how did you turn e^3ln(x)ln(x) into 3lnx*e^3lnx

>>9638630

Multiply both sides by 3

>>9638650
where did the lnx come come from

>>9638658
[math]
x^{x^3} = 2^4 \\
x^3 \ln(x) = 4\ln(2) \\
e^{3\ln(x)} \ln(x) = 4\ln(2) \\
{\rm let} ~ y = 3\ln(x) \\
ye^y = 12\ln(2) \\
y=W(12\ln(2)) \\
x = e^{\frac{W(12\ln(2))}{3}} \approx 1.721670170412965998739149039475469284021191706837875947626...

[/math]

http://www.wolframalpha.com/input/?i=nsolve%5Breals%5D+x%5Ex%5E3+%3D+16

Am I smart now?

>>9638662
oh That makes more sense. thanks senpai. I think I'm going to try and solve x^x^3 = x^3 with this method

>>9638675
3ln(x) = ln(x)x^3
3 = x^3
x = ∛3

>>9638685
or ln(x) = 0 aka x = 1

>>9638688
I meant to say x^x^3 = 2^3

>>9638688
x could also be 0

>>9638699
0^(0^3) = 1
0^3 = 0

>>9638714
nvm lol

how do you find out the value of W(z)

>>9638695
I got x = e^(w(9ln(2))/3)

>>9638719
https://www.wolframalpha.com/input/?i=lambertW(z)

>>9638722
yup I got 1.622

>>9638699

>0^0^3 = 2^4

>>9638200
1.721670170413
don't be so smug pleb :-P

>>9638395

where does the 2^4 become 2^3

>>9638662
I get to ye^y=12ln(2) but after that, I know you try to isolate y but I just don't know how you go.
At the end I get it you get ln(x)=w(12ln(2))/3 and add e to both sides to get your answer. >x=e^W(12ln(2))3

>>9638662
Fuck this, I'm becoming an English teacher.

>>9638796
its really easy, if you dont know maths it may look like chinese but if I were to explain it to you, you wouldnt find it difficult. The hard part is doing it intuitively quick.

>>9638781
https://en.wikipedia.org/wiki/Lambert_W_function

>>9638662
Nice

>>9638767
I read the question wrong

I've done this wrong but here's my answer:
x^x^3 = x^3x = 2^4
(x^3x)^1/3 = (2^4)^1/3
x^x = 2^4/3
x = logx(2^4/3)
x = 4/3logx(2)
Substituting values of x which get you x
2.3 seems to be closest (shit precision)

>>9638200
Teh function [math] \begin{array}[t]{cccl} f: & \mathbf R_+^* & \longrightarrow & \mathbf R \\ & x & \longmapsto & x^{x^3}\,-\,2^4 \end{array}[/math] is of class [math]\mathscr C^1[/math] over [math]\mathbf R_+^*[/math] and
[eqn]\forall x\,\in\,\mathbf R_+^*,\, f'\left(x\right)\ =\ \left( 3\,x^2\,\ln\,x \,+\, x^3\,\frac1x \right)\, \mathrm e^{x^3\,\ln\,x}\ =\ x^2\,\left( 3\,\ln\,x\,+\,1 \right)\, x^{x^3}.[/eqn]

Additionally,
[eqn]\forall x\,\in\,\mathbf R_+^*,\, f'\left(x\right)\ \leqslant\ 0 \qquad\text{iff}\qquad 3\,\ln\,x\,+\,1\ \leqslant\ 0 \qquad\text{iff}\qquad \ln\,x\ \leqslant\ -\frac13 \qquad\text{iff}\qquad x\ \leqslant\ \mathrm e^{-\frac13}.[/eqn]
This means the minimum of [math]f[/math] is
[eqn]\min\,f\ =\ f\left(\mathrm e^{-\frac13}\right)\ =\ \left( \mathrm e^{-\frac13} \right)^{\left( \mathrm e^{-\frac13} \right)^3}\,-\,2^4\ =\ \left( \mathrm e^{-\frac13} \right)^{\mathrm e^{-1}}\,-\,2^4\ =\ \mathrm e^{-\frac1{3\,\mathrm e}} \,-\, 2^4\ <\ 1\,-\,2^4\ <\ 0.[/eqn]
Then [math]f\left(2\right)\ =\ 2^{2^3}\,-\,2^4\ =\ 2^8\,-\,2^4\ >\ 0[/math]. By the intermediate value theorem, a solution exists. Also note that
[eqn]\forall x\,\in\,\mathbf R_+^*,\, f\left(x\right)\ =\ \mathrm e^{x^3\,\ln\,x} \,-\, 2^4\ \xrightarrow[x\,\to\,0]{}\ 1\,-\,2^4\ <\ 0.[/eqn]
This mean the solution is unique and in the interval [math]\left( \mathrm e^{-\frac13},\, 2\right)[/math]. Using Newton's method as implemented in pic related, an accuracy challenging double precision IEEE754 floating-point numbers can be computed:
[eqn]x\ \approx\ 1.721,670,170,412,966.[/eqn]
Nothing was required to control the convergence better.

>>9639505
>Java

>>>/g/tfo and never return.

>>9638324
reading the problem tells us
[[math] x^{x^x} = x^{(x^x)} \neq (x^x)^x [/math]
so it's not correct using your high-school rules

>>9638324
>>9638310
>>9638288

The order of operation is P|F E im M|D A|S

>Parenthesis and Functions (inside out)
>Exponents (done ← RIGHT TO LEFT ←)
>implicate multiplication (optional, use style to determine if applicable. Nobody would write 1/2πi to mean ½πi, they would write πi/2)
>Multiplication and Division (done → Left to Right →)
>Addition and Subtraction (done → Left to Right →)

>>9638831
and I spent like 20min trying to get past that point

>>9639505
>"an accuracy challenging double precision IEEE754 floating-point numbers"
>Not using a system with arbitrarily precision numbers

>>9639773
>using something that's slow

>>9639584
>mfw he's salty that Pajeet outperforms half of /sci/

>>9639778
half of /sci/ are high-schoolers asking what courses to take and bragging about their 105 IQs

>>9638200
2^4=16
-> cuberoot(16)=x^x=2.5198421
-> log_x(2.5198421)=x
-> x ~=1.71435062

>>9639800
>this entire solution
what are you doing? your steps do not follow each other, and your numeric solution isn't what you would have gotten following your steps

>>9638200
Numerical solution with Banach? By making it f(x)=x

~1.72167

Is what I got by guessing.

>>9639777
Have you timed your program?

>>9639777
Yeah that's why I use a JVM lang -- because I care about speed. LOL.

>>9639778
An exact solution was already posted. No need for pajeet code.

Also we C++17 on this board.

[eqn]
\rm
\#include~<iostream>
\\
\#include~<variant>
\\
\#include~<cmath>
\\
using~namespace~std;
\\

\\
int~main()~\{
\\
~~~~auto~Halley~=~[](double~x,~auto~f,~auto~df,~auto~ddf,~variant<size\_t,~double>~control)~constexpr~\{
\\
~~~~~~~~if~(holds\_alternative<size\_t>(control))~\{
\\
~~~~~~~~~~~~size\_t~iterations~=~get<size\_t>(control);
\\
~~~~~~~~~~~~while~(iterations--)~\{
\\
~~~~~~~~~~~~~~~~auto~fx~=~f(x);
\\
~~~~~~~~~~~~~~~~auto~dfx~=~df(x);
\\
~~~~~~~~~~~~~~~~x~-=~2~*~dfx*fx~/~(2~*~dfx*dfx~-~fx~*~ddf(x));
\\
~~~~~~~~~~~~\}
\\
~~~~~~~~\}
\\
~~~~~~~~else~\{
\\
~~~~~~~~~~~~double~threshold~=~get<double>(control);
\\
~~~~~~~~~~~~auto~fx~=~f(x);
\\
~~~~~~~~~~~~while~(fx~>~threshold~||~fx~<~-threshold)~\{
\\
~~~~~~~~~~~~~~~~auto~dfx~=~df(x);
\\
~~~~~~~~~~~~~~~~x~-=~2~*~dfx*fx~/~(2~*~dfx*dfx~-~fx~*~ddf(x));
\\
~~~~~~~~~~~~~~~~fx~=~f(x);
\\
~~~~~~~~~~~~\}
\\
~~~~~~~~\}
\\
~~~~~~~~return~x;
\\
~~~~\};
\\

\\
~~~~auto~f~=~[](double~x)~\{~return~pow(x,~pow(x,~3))~-~16;~\};
\\
~~~~auto~df~=~[](double~x)~\{~return~pow(x,~pow(x,~3)~+~2)*(3~*~log(x)~+~1);~\};
\\
~~~~auto~ddf~=~[](double~x)~\{
\\
~~~~~~~~auto~x3~=~pow(x,~3);
\\
~~~~~~~~auto~logx~=~log(x);
\\
~~~~~~~~return~pow(x,~x3~+~1)*(x3~+~x3~*~logx*(9~*~logx~+~6)~+~6~*~logx~+~5);~
\\
~~~~\};
\\

\\
~~~~cout.precision(16);
\\
~~~~cout~<<~Halley(2,~f,~df,~ddf,~1.0e-14)<<endl;
\\
~~~~cout~<<~Halley(2,~f,~df,~ddf,~size\_t\{10\})~<<~endl;
\\
~~~~return~0;
\\
\}
\\
[/eqn]

>1.721670170412966
>1.721670170412966

>>9638589
>>9638662
thanks m8

CS shitters are so annoying oh my god.

>>9640012
>we C++17 on this board
Never seen genuine C++ on /sci/; every closet /g/ cuck here pretends the most /sci/ language is Lisp. Oh, and R/MATLAB scripting doesn't count as C++.

>>9640098
Nobody ever posts lisp here. >>>/g/o back to the >>/g/hetto where you belong.

>>9640108
The fact you failed your triple meme arrow must mean you're pretty tired after samefagging 40 times in this thread. Must be tiring to realize even Pajeet makes 100 times your salary too. Man, I'd hate to be a physishit. Anyway, go back to your Lisp fizzbuzz now. I have a compiler to make in C++.

>>9640012
Hmm...

>>9638200
>Simple log problem
>brainlet wall
Right you are.

>>9638200
X=2

>>9638200
>>9638220
>>9638288
>>9638310
>>9638395
>>9638523
at least
>>9638589 did it without explanation

all these brainlets kek
x^x^3=2^4
x^x^3=16
x^3x=16
now retards, ever heard of log?
x*log(x^3)=log(16)
now comes the "complicated" part which all of you are to stupid for because you never worked with logarithms:
>write x as e^log(x):
(e^logx)*(log3x)=log16
>Use the W function
what's that? babbys first combinatorics course?
>the w function is the inverse of y=xe^x
so we use that to further transform the equation:
log(3x)=W(log16)
3x=e^(W(log16))
>x=(e^(W(log(16)))/3
And that is the correct solution if you dont want to be a wolframfag like
>>9638242

>>9640677
How does that sweet (You) taste?

>>9640677
>x^x^3=2^4
>x^x^3=16
>x^3x=16
Kys retard

Good Luck

Fkn plebs just use Newton Rhapson

>>9643273
>Newton Rhapson
>not Halley

Pleb.

>>9641770
>x^x^3=x^3x

lets try it,
1^1^3 = 1
1^3*1 = 1
so far so good
2^2^3 = 256
2^3*2 = 64
==F==

x^x^3 != x^3x, so sorry

>>9643301
meant for >>9640677

>>9640012
Why C++? It's not a very good language and the only thing it seems to offer is speed

>>9643309
>It's not a very good language

It's just as good as any other language. People who hate on C++ just repeat /g/ memes and never have actually coded in it but imagine it to be an unsafe Java++.

>>9643335
Not true at all, have you ever tried an FP language?

>>9643347
>FP language

C++ has lambdas and variadic templates are pretty much like coding with car and cdr

>>9643367
Yeah but C++ is not a pleasant language at all. It's very noisy and inconsistent. Have you tried any of the ML family or derivatives? (OCaml, Haskell)

>>9640677
read your post outloud to yourself and then commit suicide

>>9643384
>OCaml

I am not a dirty surrender monkey.

>>9643401
what about Haskell?

>>9638796
the trick is getting that W function in there
so whenever you see any tetration problem (x^x^...=n), think W, since there's always a way to get that W in there
and there's usually a log in that function somewhere too

W is the inverse of y=xe^x btw

Is x^x^3 equal to x^(x^3) or (x^x)^3?

>>9644089
>so whenever you see any tetration problem (x^x^...=n)

No, it only works for 2nd order and the infinite case. Otherwise you're stuck with super roots

[math]
x^{cx^p} = y \\
\ln(x)x^p=\ln(y)/c \\
e^{W(p\ln(y)/c)/p} \\

x^{x^{x^{...}}} = y \\
x^y = y \\
-1 = -yx^{-y} \\
-\ln(x) = -\ln(x)y e^{-\ln(x)y} \\
-\ln(x)y = W(-\ln(x)) \\
y = W(-\ln(x))/-\ln(x) \\
[/math]

>>9644089
>so whenever you see any tetration problem (x^x^...=n)

No, it only works for 2nd order and the infinite case. Otherwise you're stuck with super roots

[math]
x^{cx^p} = y \\
\ln(x)x^p=\ln(y)/c
\\ x=e^{W(p\ln(y)/c)/p}
\\
\\ x^{x^{x^{...}}} = y
\\ x^y = y
\\ -1 = -yx^{-y}
\\ -\ln(x) = -\ln(x)y e^{-\ln(x)y}
\\ -\ln(x)y = W(-\ln(x)) \\ y = W(-\ln(x))/-\ln(x) \\
[/math]

>>9644089
>so whenever you see any tetration problem (x^x^...=n)

No, it only works for 2nd order and the infinite case. Otherwise you're stuck with super roots

>>9644146
>>9644149
>>9644153
alright i get it sheesh

>>9644131
see >>9639770

>>9644155
/sci/ has a lot of tryhard autists.

>>9643347
Read this again.
>People who hate on C++ just repeat /g/ memes

>tfw no a mathfag ,who jerks off to numbers
>tfw doing comp sci and dont know what ordinal numbers are

>>9638324
>x^x^3 = x^3x
>b-but that's correct according to the r-radical r-rules
...so x^3 = 3x? L0Lno fgt pls