If you can't solve this simple equation without a graph you are a brainlet

If you can't solve this simple equation without a graph you are a brainlet.

Other urls found in this thread:

wolframalpha.com/input/?i=(-2*LambertW[13, -ln(2)/2]/ln(2))^2
wolframalpha.com/input/?i=2^(-2*LambertW[13, -ln(2)/2]/ln(2))
en.wikipedia.org/wiki/Intermediate_value_theorem
twitter.com/NSFWRedditGif

That's some scoliosis right there.

4, 17, 49

0 2 4

0^2 = 0
2^0 = 1

1,2,4

This

1^2 = 1
2^1 = 2

what the fuck does upwards arrow mean?

It means to the power of _

x^2 = x squared.

Sup Forums confirmed for the dumbest board

ok, thanks.
so, 2 and 4 work. cant see any other positive whole number working, is it a fraction/decimal?

Yep

Try rearranging it

you guys are idiots

are you 11 or retarded? x23

the third one is roughly about -0.8, cba posting the exact value with lambert function

-0.766664696

2,4, -0.766

da fuck

good jerb

There are more than three solutions, in fact there are infinitely many solutions.

Let W(z ; n) be then value of LambertW of z on the nth branch:

x^2=2^x

ln(x)/x=ln(2)/2

-ln(x)/x=-ln(2)/2

-ln(x)*e^(-ln(x))=-ln(2)/2

-ln(x)=W(-ln(2)/2 ; n)

1/x=e^W(-ln(2)/2 ; n)

1/x=-ln(2)/(2W(-ln(2)/2 ; n))

x=-2W(-ln(2)/2 ; n)/ln(2)

No there aren't you fucking momo

I kinda just proved it.

Are there any non-real values apart from real three mentioned here?
Or are you being just a huge fucking memer?

x^2=t+h-a*t's(s)o+m/e x=(scoliosis)-r+i(g)-h*t*t/h+e(re)

why the fuck are we doing your school homework for you, you dumb fuck? get the fuck out of here.

11 and retarded.

>Are there any non-real values apart from real three mentioned here?

Yep. Try: -2*LambertW[13, -ln(2)/2]/ln(2)

wolframalpha.com/input/?i=(-2*LambertW[13, -ln(2)/2]/ln(2))^2

wolframalpha.com/input/?i=2^(-2*LambertW[13, -ln(2)/2]/ln(2))

Replace 13 with -1, 0, 2, 3, 4, 5, ... and it will work.

>-1, 0, 2, 3, 4, 5

And 1 works too.

but he called you a momo, so you lose.

K. I am a biologist, so math is not my strongest suite, but Lambert omega appears here
when you take a derivative of -ln(x)*e^(-ln(x))?
Or somehow other way?

We call it the remainder, that's the number that remains.

>but Lambert omega appears here
The omega constant doesn't come up in any of the solutions, as Omega = W(1) where W is taken on the principle branch.

>when you take a derivative of -ln(x)*e^(-ln(x))?
>Or somehow other way?

Explanation of steps:

The Lambert W function satisfies W(x)*e^W(x)=x, all of the identities involving it can be derived from this fact and the properties of ln(x).

x^2=2^x

ln(x)/x=ln(2)/2, take the natural log of each side

-ln(x)/x=-ln(2)/2, take the negative of each side

-ln(x)*e^(-ln(x))=-ln(2)/2, use 1/x=e^(-ln(x)) and -ln(a)=ln(1/a)

-ln(x)=W(-ln(2)/2 ; n), take Lambert W of each side

1/x=e^W(-ln(2)/2 ; n), do e^a to each side

1/x=-ln(2)/(2W(-ln(2)/2 ; n)), use e^W(a)=a/W(a)

x=-2W(-ln(2)/2 ; n)/ln(2), take he reciprocal of each side

infinity

2 and 4

Ok, kinda getting there.
Little question on one step though -
why W(-ln(x)*e^(-ln(x)))=-ln(x)?
how do you get from there to
W(-ln(x))*e^W(-ln(x))? I understand that we get next step from the properties of function, but I am being too daft to understand how do we get W function under ln.Because x=e^ln(x)? I don't really get how to apply that to functions thouhg.
Thanks for the answers btw, appreciate it.

brainlet here, anyone able to tell me how you'd be able to do this? I understand exponential functions and parabolae, but how do you intersect one with the other?

I can't do it analytically, but I can script in R, so I just made a cycle to check if 2^x==x^2 on (-10, 10) with small steps

>why W(-ln(x)*e^(-ln(x)))=-ln(x)

Let W(t)*e^W(t)=t the substitute t=z*e^z to get W(z*e^z)=z.

>brainlet here, anyone able to tell me how you'd be able to do this?

See here .

>I understand exponential functions and parabolae, but how do you intersect one with the other?

I just used identities to do it analytically.

Now to answer this part:

>but I am being too daft to understand how do we get W function under ln


W(a)*e^W(a)=a so we divide each side by W(a) to get e^W(a)=a/W(a). Does that help?

bump

It does.
I think I get it now, but I still wouldn't try that at home without supervision.
I mean, I understand properties and substitutions, but hell if I would arrive at using them when I need to.

how hard is this question mathfags? i dont know much math

looks simple

>but I still wouldn't try that at home without supervision.

As with use of any function, practice with it is required to attain of use and the ability able to derive identities with it. If you're still curious I recommend picking up a book on special functions - there's a lot of crazy sick shit you can do with special functions and a hefty dose of complex analysis.

Thank you, user.
>picking up a book on special functions
I think I would give it a try, probably in the free time. Biologists rarely have to face complex functions at all.

Do you have a maths degree or something?

...

Doing a graduate degree in applied computational mathematics but I'm probably going to reapply to another school for a PhD in number theory.

that sounds pretty smart tbh user, have a rare maisie for your troubles

can you explain imaginary numbers to me pls?

Just 2? People are fucking idiots.

Fuck off satan, you suck at maths.

I'm going to assume you've done algebra before. In algebra there is a theorem called the fundamental theorem of algebra (FTA for short). What FTA says is that "The field of complex numbers is algebraically closed," now perhaps you're wondering how that jargon translates in to algebra, and I'll tell you. The statement of FTA is equivalent to:

Given any polynomial p(x)=a(n)*x^n+a(n-1)*x^(n-1)+...+a(0)*x^0 we call n the degree of the polynomial. FTA says that P(x) will have n (not necessarily distinct) numbers z such that P(z)=0. So for example x^2+3x+1=0 has two solutions, x^3+1=0 has three solutions. This is all pretty esoteric if you're not familiar with algebra like that though, so I'll also give an example of how to do something symbolic with them:

i is just a number such that i^2=-1, and using that property we can do arithmetic with it. So for example if we want to find out how to simplify (3+2i)(5-7i) we can use what we already know about algebra and the property i^2=-1 to work with it.

(3+2i)(5-7i)=15+10i-21i-14i^2, FOIL it

=15+10i-21i+14, use i^2=-1

=29-1i, collect real and imaginary parts.

2^0.5=x^(1/x)
i guess only 2, then

So sqrt(-1) is not a real number, but it is given an imaginary value?

Why could you not define a value as 0/0 in a similar way? Like n = 0/0?

What the fuck is this scoliosis meme I'm seeing everywhere suddenly? It seems forced.

2^2=2^2
2^4=16 and 4^2 =16
(-0.7666)^2=0.587 and 2^(-0.7666) =0.587
And as always
You mom = my cumdumpster

today one user posted a photo of a girl pouting ass and both anons posted at the same time "thats some scoliosis"

That's some scoliosis right there.

That's some scoliosis right there.

I know it was wrong because i've made it a graph.

But...that's not even funny though. God I fucking hate this place.

i could have done a substraction, then derivate it and make bolzano but nah

i is only called imaginary because it doesn't exist among real numbers. There is just no real number exists square of which would be -1.
Real numbers not in layman's terms, but in mathematical terms.
It is not as imaginary, as,say, your gf may be.

at least it is not infested with paid shills as pol is

Bolzano like a city in Italy or what?

meant to type

>29-11i

Carpal tunnel is a bitch.

>So sqrt(-1) is not a real number, but it is given an imaginary value?

Bit of history, they are called imaginary numbers because people didn't like them when they were first being developed. In the time that imaginary numbers first started to appear mathematicians really didn't understand them and considered them nonsensical, so they kinda shunned them and gave them the derogatory label 'imaginary'. The name has nothing to do with how philosophically /real/ the numbers are, so in that sense they are just as real as real numbers.

It was a poor name choice, but it has stuck. The modern way to think about it is that imaginary numbers are on an axis orthogonal to the real numbers and they help us describe rotations. Furthermore, the orthogonality isn't just a convenience, it's actually a natural topology brought about by the definition of i.

>Why could you not define a value as 0/0 in a similar way? Like n = 0/0?

There is a system where you can do this, it's called the trivial ring (sometimes called the zero ring). But it doesn't really do anything useful because it only has one element, 0. The reason we can't have this work with the real numbers is because it would lead to contradictions, watch:

Suppose a=0/0, then multiply each side by 0:

a*0=0, but then a can have any value, so a isn't well defined.

>Why could you not define a value as 0/0 in a similar way? Like n = 0/0?

You have made it but you are too autistic to post it or what?

Solutions are x=x^2/2 and + or - x=2x

Bolzano's theorem used to know when does the function grow

en.wikipedia.org/wiki/Intermediate_value_theorem

i made it in the computer after i made my guess

420, 69 and 666

i never learned how to do math, sorry.

>1
>1^2=2^1
>1=2
what a brainlet

If a continuous function has values of opposite sign inside an interval, then it has a root in that interval (Bolzano's theorem)

How, and most importantly, why do I need it to find the root? It only states it exists, and I know for a fact it exists because I have seen parabolic and exponential functions before. It seems completely unrelated to actually calculating the root.

Asking again, is this problem actually hard or are you memeing?

you calculate >0 and

It's not hard to find a solution, but it is hard to find all the solutions if you have not seen the Lambert W function.