[Legal counting forum game] Counting with four fours

This is really easy to explain, but might be a little harder to do sometimes... The goal is to mathematically make the next number, only using four fours. Of course, you may sue as much mathematical symbols as you want. for ease: Use tags like [Square root: x ]
Things to consider:

i, e, π, φ ... are numbers, but I'd consider e^x as legal, for creativity use

The square root doesn't need a number, the other roots do.

You mayn't use logarithms (x)LOG(x) since you can do this with them:

([Square root: 4]/4)Log( (4)Log ([Square root: 4]) = 1
([Square root: 4]/4)Log( (4)Log ([Square root: [Square root: 4]]) = 2
([Square root: 4]/4)Log( (4)Log ([Square root: [Square root: [Square root: 4]]]) = 3
You get it...

as usual: you may post every secon post = one post between your posts.

For the rest of it: be creative

e^In((4/4)/(4/4)) = 1

(It is not true that the only reason I'm doing this is to explore the edge of the rules, not at all...)

You know you are a terrible human being for posting that, and an even more terrible human being for actually thinking that... *Calls maths police*



Okay... that was not techhnically maths police... but it will do thhe job...

I feel like this deserves to be here

2 = (4*4) / (4+4)

Everyone on EMC: Must get a counting thread going so I can have a popular thread!

3 = (4 + 4 + 4) / 4

∜4x4x4x4

4!/4 + 4/4

Jelle, is what Rhino did allowed? For the a[root: b], does a count as the 4 4's, or can we choose a number there?

I get 7 here?

Either way, let's try 5:

I feel obligated to post here, as I'm studying maths in uni right now, and this is what I was able to produce

8 = lim(x->∞)( 4/x + (4x + 4x)/(x + 4 ) )

Feel like cheating a bit with the x being a real number approaching infinity...

EDIT: Oops, went off Naekals post to go to 8...

Here is my shot at 6:

6 = ( 4 * 4 - 4 ) / [squareroot: 4]

4+4-(4/4)

Or, for those of you who don't like simple answers, you could always do (((e^4)-4)/4)-4 and do some aggressive rounding.

Yeah that was supposed to be a minus sign

4 x 4 / (4 - squareroot(4))

4+4+(4/4)

10 = (4+4/4)*squareroot(4)

11 = 4!/(Squareroot: 4)+4/4

Well... that's something.
Nice job.

Posted from my phone!

Forgot to add this to the other post... This isn't allowed as the 4thh root uses a number in the four. the normal squuare root may be used without a nuumber though, as that is just the normal root. I'd think of all theh othhers as special ones and so need a number

Also: limits are fine... I am terrible at uusing them myself though (I'm way better at pulling things out a vase (Maths joke, sorry))
(Amn't posting 13 duue to alredey havng post the one before the previous number)

13 = (4!/√4)+(4/4)