Millennium problem 8

Yeah, sqrt2 is between them. Nobody said sqrt(2) = 1.41...

Are you fucking retarded?

>Asks a question
>Gets the correct answer
>NO LEL IT'S NOT [WRONG ANSWER] NICE TRY THOUGH LEL I GOT YUU hue!

x=2^(1/x)=2^(1/(2^(1/x)))=...

Aww how's high school?

That's how math works when you aren't a gay engineer. x doesn't equal x unless it's defined to and there are plenty of cases in abstract algebra where it doesn't.

undergrad spotted

Show one from a text

A number such that the reciprocal of x is lg(x)
How would you even calculate that number though?

Let X={a,b}X={a,b}. Then the relation {(x,y),(y,x)}{(x,y),(y,x)} is symmetric but not reflexive.

Very simple example.

My bad. Let me rewrite that:

Let X={a,b}. Then the relation {(x,y),(y,x)} is symmetric but not reflexive.

What the fuck are you guys talking about? Are we working outside ZFC now?

But that's not equals. Its not even an equivallence relation.

x^x= 2

x= 2^(1/x)

x/(2^(1/x))=1

(x/(2^(1/x))) * ((2^x)/(2^x)) = 1

((2^x)x)/(2) = 1

(2^x) x = 2

2^x = 2x

thus the only number possible to be x is 2, which is impossible

where the fuck is my error /sci ?

Lol what are you even trying to do here?

You found a single relation that isn't reflexive? You just proved some relations aren't equivalence relations. What do you think you you accomplished?
Why didn't you use X in your relation?

> 2^(1/x)
using transcendental functions

No. We aren't.
That's the point. Congrats.
We're being pedantic about it but that was the point, yes.

Just because you're being retarded doesn't mean you're being pedantic. Nothing you said so far is true.

X=X always

You asked for an example where something doesn't "equal" itself. A non-reflexive relation is one such example. You're the weird one here, I just did what you or someone else told me to here:

the last step you multiplied instead of dividing by x.

>an example where something doesn't "equal" itself
> isn't an equivallence relation or equals
coulda just say > if you were just choosing arbitrary relations

If I'm not mistaken [2^(1/x)]*[2^(x)] = 2^[(1+x)/x]
When multiplying terms with the same base you add the exponents.
(2^1)*(2^1) = 2^2 = 4

You can't solve it algebraically. Trick question I'm afraid. I think someone has a proof of that I think.

How would I go about solving it non-algebraically?

How is it solved non-algebraically? Dumb undergrad here is curious.

I'm not sure. I've only taken up to differential equations and calc 3. I haven't taken anything above that. As far as I've read on Wikipedia it has to do with the Lambert function, but I don't really understand it at all. I don't even know when you would learn that as a math undergrad desu. The fact that nobody in this thread has mentioned that might indicate that it probably isn't taught.

Look up Lambert function if you wanna find out.
2000clicks.com/mathhelp/BasicSimplifyingLambertWFunction.aspx
Here's a good place to start. From my understanding it's kind of like how the natural logarithm lnx is the inverse of the exponential function e^x, W(x) is the inverse of xe^x. Kind of weird.

see

Ok?

>The fact that nobody in this thread has mentioned that

sqrt(2)^2 = 2 so no

I hadn't realized the Lambert function and product log were the same thing.

loll

I don't think that's possible OP

It's obviously between 1 and 2, testing 3/2 got a pretty good approximation, 4/5 is too big so it's between 3/2 and 4/5. Just guessing and found 31/20 to be even better.

What (if there is one) is the general rule for getting a series approximation of n roots?

[math] \frac{94241}{60426} [/math]

Only five 9's? Pathetic. You should be ashamed of yourself.

1.559610465.....

1.559610469459.....

1.559610469462369349970388768765002993284883511843091424719

I really want to know how to do this. Somebody figure it out. Tonight I learned that all irrational numbers have an infinite continued fraction representation. Neat. Still don't know how to get this stupid problem.

See
It has an explanation for how to solve x^x= 16

Have you never seen a parabola that crosses the x axis twice?
X[sub]1, X[sub]2

can u tell me who is this men that i see in many memes recenlty ( i am italian, i never saw this man)

Equality by definition must be an equivalence relation and hence reflexive.

sqrt(2)^sqrt(2)=2

Nice troll

There is no solution.

I've gotten as far as seeing the exponential growth of x**x and decided it must have to do with e. Also I'm not in college so I'll pull out of my ass that I can log(2) this bitch. How am I doing? Something like e**log(2) with...some missing part of the log? I'm stuck unless I either already got it or am way off.

That's 2+(1/root2) so no.

Wait no I'm fucking retarded that's just =2 isn't it? My brain is melting.

So far my aproximation is e^(1-sqrt(1-ln2))

it should be equal to 2 at the end so..

Easy
We look for [math] x \in \mathbb{Z}_3 [/math]
Then x=5 is the solution, as
[math] 5^5 = 5^3*5^2 = 5*5^2 = 5^3 = 5 = 2 [/math]

equal means equivalence relation, genius

OP here, the answer is :

e^(W(ln (2))

Pic related shows how you get there.

This guy was on the right track, the W() function is the same as the Lambent function, you can see how it being the inverse of xe ^x is used in the solution. (In the link he posted there is actually a solution for x^x =16)

I love how simple this question looks and the fact that you need a transcendental function in order to solve it. It was interesting to see how everyone approached it and that the discussion moved to the reflexive property Lol.

Sorry about the image on mobile

But you've forgotten that, since 5 is prime (and hence 5^2 as well), the answer has to be nonprime, in this case either 4 or 6 (for n powers of 5).

The solution is almost certainly irrational which means that OP has already posted the solution in the very formulation of the question. If you can generate some sort of analytically interesting but incalculable infinite sum then I'll buy you a beer but any computer capable of floating point arithmetic can get enough digits in about x^x = 2 seconds.

Probably does seem fun though.

You can do square roots in 'long squares' using a pretty much identical algorithm as long division.

...

wew mate

we're looking for the value of x

You're wrong.

x=1.51 approx.

>Find the exact value of x^x = 2

[math]x^x=2[/math]
[math]x\ln x = \ln 2[/math]
[math]e^{\ln x} \ln x = \ln 2[/math]
[math]\ln x = W(\ln 2)[/math]
[math]x=e^{W(\ln 2)}[/math]
Alternatively,
[math]x=\frac{W(\ln 2)e^{W(\ln 2)}}{W(\ln 2)}=\frac{\ln 2}{W(\ln 2)}[/math]

Thanks for showing your work - from the first reply I was able to look up the W function but was curious what the process looked like as I have very little education in this area. Much more straight-forward than I anticipated.