@Simple What does it represent geometrically?

a hemisphere》

？

I am thinking it represents the surface area of that

(I may be wrong, so do correct me if I am)

and the surface area of a sphere can be found with Google

@SimplyBeautifulArt I don't see how that help me

$r$ is just a constant

I think if you know the surface area of a hemisphere, then it will be equal to the problem

The formula of the area of a sphere is $4\pi r^2$

and a hemisphere is 'blank' of a sphere

@amWhy What to do?

what do you mean a hemisphere is 'blank' of a shpere

What is the surface area of a hemisphere?

$2\pi^2$

@SimplyBeautifulArt I don't think the question should be deleted because of the two upvoted answers; perhaps in time. But I think the question does not need further down-votes!

@Simple $2\pi r^2$

and there you go. I think that's the answer

Since we are integrating respect to y, $r$ and $z$ are constant, I think I can apply the integration formula $\int\frac{1}{a^2-x^2}=\arctan(x)$

Don't think that formula is right

If anything, it should be arcsine I think

That should be $\frac{1}{a^2+x^2}$

Better yet, I think we are basically done here

:-)

If I want to use substitution, what should I let $y$ be?

See the second image

no substitution needed

I saw that, I just want to try substitution :)

oh

You would then need to factor stuff out to make it $\frac c{\sqrt{1-(y/d)^2}}$ for constants $c,d$.

Is this the same process when we use the second formula in the second image?

What I want to do is to let $y=...$, a U-substitution

wait, nevermind, that seem complicated

@Simple Lol, are you good now?

Yeah, I am

└(^o^)┘

Woo!

@SimplyBeautifulArt @Starfall I simply mean that taking the area over some arbitrary path can be done via an integral but it is not necessarily an integral and the order of adding it together shouldn't matter. The integral cares, but the actual act of adding all of the points together shouldn't care. The integral really shouldn't give negative values from different ordering. That seems flawed to me. Addition is commutative so the integral shouldn't care in what order elements are accumulated.

also, if I am merely discussing the idea of the area over a path, then obviously only one of the orderings is actually what I mean (as the other wouldn't give the right area). ;)

@TheGreatDuck I'm not sure I understand. For example, what does area mean for a negative function?

signed area

geez

@SimplyBeautifulArt I merely point out that the integral shouldn't be thought of as the definition of certain area/volume related geometric concepts regarding functions. It can be used to find the values, but it most certainly shouldn't be the definition.

:| Then what was your definition of integral?

hmm?

im saying some geometric concepts shouldn't be defined by the integral even if it can be used to compute them...

when did I ever say anything regarding the definition of the integral?

Well, I was wondering if there was anything in there that explained why direction and integrals are separate.

i just mean that if I give you a parameterized path in the xy plane and a multivariate function z = f(x,y) and ask for the signed area above the path, it isn't ambiguous which direction to integrate along

as only one gives the proper area

I don't think that's necessarily the case

@Ramanujan @AkivaWeinberger Btw, do you know how to code?

I know a little Python

Hm...

No

Would you like to try and make a larger number than S.C.B. Nisman and I in under 256 characters with the restriction of no infinity and no calling in constants (or else you'd just call in some really big number and +1 it)

@Ramanujan Would you like to learn some Ruby?

Yes,shure

@Ramanujan repl.it/F729

@SimplyBeautifulArt I do: The Goodstein number of some large number

Can you figure out what that does? Pretty basic stuff

15

O.O What is that?

Oh, you haven't heard of Goodstein sequences? They're fun!

Can I teach you them?

The basic idea starts with something called "hereditary base-$n$ notation"

@SimplyBeautifulArt what to do?What to type?

So, you know base 2, for example, writes $73$ as $2^6+2^3+1$

@AkivaWeinberger Sure

@Ramanujan You hit "run"

Hereditary base-2 goes a bit further, by rewriting the exponents also:

and it gives you f(3,5)

oh, fun

$73=2^{2^2+2}+2^{2+1}+1$

OK?

Great... so what's the Goodstein?

Consider the number $4$

That's $2^2$ in hereditary base-2

@Ramanujan See if you can mess around with it to make more interesting things

Change all the 2s to 3s

That gives you $3^3$.

:D

Subtract one

@AkivaWeinberger Wait a minute

That's not going to give relatively large numbers

You get $3^3-1=2\cdot3^2+2\cdot3+2$ in hereditary base-3

@SimplyBeautifulArt Wait.

Lol, ok

Change all the 3s to 4s

You get $2\cdot4^2+2\cdot4+2$

Continue this way, subtracting one, rewriting the number in hereditary base-n, change all the $n$s to $n+1$s

Hm...

@SimplyBeautifulArt Do you think this process will ever terminate?

Let's do a simple example that does terminate relatively quickly.

Oh f###, that sounds horribly insane

Start with $3=2+1$.

It is giving 5^5 =0

Change 2s to 3s, $3+1$, subtract $1$, you get $3$.

@Ramanujan Exponents are 5**5

Change 3s to 4s, $4$, subtract $1$, you get $3$.

Change 4s to 5s, $3$, subtract $1$, you get $2$. :D

Change 5s to 6s, $2$, subtract $1$, you get $1$

Change 7s to 8s, $0$, we're done, it terminated!

@SimplyBeautifulArt So starting with $3$ terminates in seven steps.

how to even program that without calling in some crazy gear?

@SimplyBeautifulArt Now, starting with $4$ also terminates... in $3\cdot2^{402653211}-1$ steps!

It is giving 9** 9 **9 = infinity

Hm...

In fact, no matter what number you start with, it will eventually terminate! Just after an insanely large number of steps.

@Ramanujan You mean 9**9**9?

@AkivaWeinberger yes,how you wrote that?

@SimplyBeautifulArt I didn't think that far ahead

@Ramanujan If it returns infinity, that means you have a number larger than "x" for some programmed "x" that is very large

@AkivaWeinberger I'm thinking that it is like a tree and will grow like crazy in width, but always gets shorter with each step

Not in each step, but essentially yeah

The tree never gets taller

@Ramanujan 9\*\*9\*\*9

@Ramanujan That number is super small

compared to the numbers we're talking about

@SimplyBeautifulArt I'm guessing I'm not allowed to use Google?

9**9**9 :D

@AkivaWeinberger Well, that'd be cheating if you did that...so yeah

D:

Hm.

Oh yeah, and spaces don't count

@SimplyBeautifulArt Oh, also, about Goodstein sequences, not only do they always terminate...

but Peano Arithmetic can't prove that they always terminate!

great...

It's a natural example of a true-but-unprovable statement in PA.

:|

Actually, might look more into that

@Ramanujan

Ackerman?

This function, for example, far exceeds exponentiation

@AkivaWeinberger Hyper operation. Three arguments

Ah

And I mentioned trees earlier because that's how my current number is generated

it slowly loses height, but explodes in other directions

@AkivaWeinberger So, do you think you want to make a larger number than me?

Thinking

:D

Well, I wish you good luck

@SimplyBeautifulArt for f(1,2,3)= 6 void value expression

@Ramanujan When I type it in, I get f(1,2,3)=1

which should be the correct value

How do I write a for loop

@Ramanujan Check indents

And check spelling

@AkivaWeinberger Ask google for something that simple

And I have to go to bed. Good night and good luck!

Goodnight

@AkivaWeinberger What language ??

They were doing Ruby. I changed it to Python, though, since I know some Python

@Ramanujan You misspelled "else" in line 6

@AkivaWeinberger python for loops are kindda different

its uses range

You do for x in range(0,100)

Here's mine:

repl.it/F73O

@SimplyBeautifulArt

Are we still trying to get the biggest number ?

Yeah

It takes quite a while to give me the number

@Nisman I don't think it will give you the number in a short amount of time

It's a very large number

so a very long computation

If you want to work with huge numbers and calculations, you should think or using something more low level and compiled

Like C or Java

It's annoying that it takes my code so long to compute something as small as f(2,21) (which is 44040192)

Actually, that makes sense

The only real operation there is adding 1

so it's really just counting to 44040192

and it would just count to g(10), which is a huge number

According to Google, this works in Python:

(Just a moment)

i = 9 ** 9 ** 9
b = 2

B = lambda y,x=0:y and y % b * (b*b)**B(x) + B(y/b,x+1)

while i:
    i = B(i)-1
    b *= b
print(b)

That ^

Not something I would have come up with

From the same contest, something using functions I'm more familiar with,

i = 9 ** 9 ** 9
b = 2

def B(n):
	j = 0
	while b**j <= n/b:
		j += 1
	return (b+1)**B(j) + B(n - b**j) if n else 0

while i:
	i = B(i) - 1
	b += 1
print b

(102 chars)

What number does it gives ?

Isnt your old code supposed to decrease x ?

I dunno, something really large @Nisman

Its counting 1 by 1

What number did you say it has to go ?

44040192 ?

f(2,21) is that, yeah

Not g(10), which is what I had it calculate

I dont get how it grows

you call g 10 times

Which calls f

but f only decreases x and passes the same y

First answer=10. Then answer=f(10,10). Then answer=f(f(10,10),f(10,10))

why is the number growing ?

etc

Consider f(3,3)

Yes

First answer=3. Then answer=f(2,3). Then answer=f(2,f(2,3)). Then answer=f(2,f(2,f(2,3))).

@Nisman f(2,3)=24

since f(1,n)=2n (easy to show)

and f(2,3)=f(1,f(1,f(1,3))) by definition

So f(3,3)=f(2,f(2,24))

I think you get an infinite loop

What's f(2,24)? That's f(1,f(1,...,f(1,24)...)), twenty-four times

so that's 24*2^24

so f(3,3)=f(2,24*2^24)

Because, when you say for i in range(5), its 0,1,2,3,4

In the end I think you get $f(3,3)=24\cdot2^{24}\cdot2^{24\cdot 2^{24}}$

@Nisman Yeah, but I never use the variable i

Then you enter a for that starts at 0, and decreases 0 by 1

so x is never 0

@Nisman You're confusing x and i

Oh yeah, my bad

You're not using it

x is from the input. i is just an index that never gets used

I just need the f(x-1,answer) line to happen x times.

In any case, $f(3,n)=n2^{n+n2^{n+n2^{n+n2^n}}}$, I'm pretty sure

or something

Something like that

Probably made a small error there.

Point is, $f(3,3)=44040192\cdot2^{44040192}$.

What I mean is this

imgur.com/xGjXp8e

That is the x,y output of f(3,3)

It shouldnt get stuck like that right ?

@Nisman Notice how there's a 2 24 row

The x column isn't all 0s.

It's just that the vast majority of them are, it seems

I'm curious what the 48th row looks like

2 48

Hm

Sorry

1 48

after that I have 1 96

1 912

1 192*

1 384

1 768

The next 2 would probably be 44040192, I think

@AkivaWeinberger Oh

1 1536

@Nisman They're doubling

And what about the 1 and 2 ?

@Nisman Never mind about the 44040192

That was f(2,21), which is unrelated I think

@Nisman thanks for pointing out

@Nisman I think the next 2 would be at 24*2^24, which is

402653184

@Nisman It probably would take a few minutes to get to that number (if you only have it printing rows that start with 2)

@Nisman Yup. I ran it and it had 2 402653184.

The next one would be $2\quad402653184\cdot2^{402653184}$, I think

which is the final answer

My C program is in y = 83886080

What?

I was running f(3,3)

Yes, but the original program run g(10)

I coded it on C and run it

Ah, OK

In any case, g(10) is f(f(f(f(…,…),f(…,…)),f(f(…,…),f(…,…))),f(f(f(…,…),f(…,…)),f(f(…,…),f(…,…))))

but ten layers deep

and at the bottom layer is a bunch of 10s

And f(10,10) is huge (since it's f(9,f(9,...f(9,10)...)) ), so f(f(10,10),f(10,10)) must be enormous.

So g(10) is so much larger than that.

So I think this definitely does give a large number, yes, albeit slowly.