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>>101209047
This is about the big O notation, how fast a function grows
I posted it here because it's more used in computer science, about how slow an algorithm becomes
You need to know calc 1 limits

Let me explain it to you
f(x) ∈ O(g(x)) means function f(x) is in O(g(x)) class, "∈" just means "in the set"

A limit supremum it's just the limit or the upper bound in case the limit doesn't exist (like how sin doesn't have a normal limit but the supremum is 1, the lower bound would be a limit infimum -1)

y = 4 is in O(1) class, because:
The limit of f(x)/g(x) is 4/1, which exists and is smaller than infinity

The σ it's just the sigmoid function, only function I thought that would have a limit similar to arctan(x), meaning that it stops growing on a fixed point

lim x->∞ σ(x)/arctan(x) = 1/(pi/2) so 2/pi, which is valid

But after posting I discovered it's he real class is O(1), which also works and it's simpler
lim x->∞ σ(x)/1 = 1/1

>>101207045
What does this have to do with sigmoid?

>>101209219
Only function I could think of that has a similar growth to arctan, less than linear but bigger than constant. Arctan stops growing at pi/2 while the sigmoid stops growing at 1

But as I said in >>101209217, the sigmoid and arctan are considered to have constant growth rate O(1)

>>101207045
Only one I can think of is Union-Find, it's O(α(n))(amortized). α(n) is the inverse of the Ackerman function. I copied this from Wikipedia

theyre going to force me to learn this shit in college aren't they...

>>101210730
I have no clue what that means, but it's cool to know there's other O classes than the most common ones

>>101210796
Probably

>>101207045
>Is there a O(arctan x) algorithm?
I know an algorithm that runs in O(|arctan x|) time. This is probably not an interesting example, but you could easily have a Python program like this:

import math
def f(x):
    y = math.fabs( math.atan(x) )
    while y > 0:
        y -= 1
        print('ahoy')

The program's execution time is O(|arctan x|), because the loop never runs more iterations than that.
There are probably more interesting algorithms out there that run in O(arctan x)

>>101207045
You dont understand how big O notation works. O(arctan x) = O(1)

>>101211161
>>101210730
>>101209254
>>101209217
Stupid and wrong.

>>101212479
>O(arctan x) = O(1)
why?

>>101212636
Because arctan x = O(1) and 1 = O(arctan x)

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>>101212479
Thanks, I understood it after posting >>101209217

>>101212636
Constant growth, limit of arctan x to infinity is pi/2, would not be O(1) only if the limit of arctan x/1 where divergent and without supremum, like log does

limit log(x)/1 is x.ln(10)/0 which definitely diverges, log(x) also has no upper bound so it doesn't have a supremum

Now if you excuse me, I'm going to jerk off and sleep

>>101212729
A correction: log(x)/1 diverges to infinity, I can't use lhopital as it is not an undefined form

>>101207045
Per definition every O(1) is also O(arctan x)
In fact they are the same case, O complexity refers to rate of growth at infinity, logx, x, x^2 all approach infinity at infinity, what changes is how rapidly they do it. O(1) and O(arctanx) approach a constant, so they have the same rate of growth at inf - 0.
Multiplying f by a constant doesn't change which Os it belongs to. Arctanx is 1*arctanx. So for chancking if f is in O(arctan x)
>lim x->inf f(x)/(1*arctan x)
Arctan x at inf is a constant
>lim x->inf f(x)/(pi/2)
>lim x->inf (f(x)*2/pi) / 1
O(arctanx) is O(1)

>>101212491
What is so stupid and wrong? They seem like genuine attempts at answering OP's question. Try being less rude and more constructive