If this gets a post ID of 32768, we advance!

[Orb Fish]

Identify the mob in the red circle:

[PkmnQ]

Identify the mob in the red circle:

They have names?

[Gilbert189]

idk

[Orb Fish]

Identify the mob in the red circle:

They have names?

Technically, they do have names.

[Byron_Inc_TBG]

i got separated

[Orb Fish]

[red herring]

čuentas !!!

[Byron_Inc_TBG]

r

[PkmnQ]

When they pull out scientific notation that's how you know you've either severely won or severely lost

[Gilbert189]

When they pull out scientific notation that's how you know you've either severely won or severely lost

i think that's the behavior of floats

[PkmnQ]

When they pull out scientific notation that's how you know you've either severely won or severely lost

i think that's the behavior of floats

I am familiar with the quirks of floats, like 2**53

[Byron_Inc_TBG]

Weirs

[Gilbert189]

Happy Pi Day!
If you don't already know, I like to celebrate Pi Day by calculating π using methods that other people (that I know) have never tried before. (Well, kinda.)

Since Hyperbolica is going out today, I'm thinking on calculating π using hyperbolic functions. After some researching on Wikipedia, I noticed that the graph for sech(x) looks like a bell shape... And when I integrate that from -∞ to ∞, I noticed that it equals to π!
So that's what I have to make today: integrating sech(x).

I could use arctan, since sech integrates to that, but unfortunately that method already exist. Also, the integrations either uses eˣ or sinh(x) as the input for arctan, so if I use this method, I'd may as well just find arctan(x) and not handling this whole sech thing.

My first, naive attempt is to just integrate it using trapezoidal integration. It didn't go well, with a range of 0 to 2000000, I can only get 2 digits right. And that's after 10 seconds of Python crunching numbers. Here's the code, by the way.

So, I tried integrating it myself, to see if I can get a bit of a performance boost. Again, I know (from Symbolab) that it integrates to arctan, but I want to see if I can somehow dodge that.

Well, I tried.

Code Select Expand

∫ eˣ dx = eˣ
∫ e⁻ˣ dx = -e⁻ˣ
∫ eˣ + e⁻ˣ dx = eˣ - e⁻ˣ

eˣ + e⁻ˣ = 1/0! + x/1! + x²/2! + x³/3! + ...
         + 1/0! - x/1! + x²/2! - x³/3! + ...
         = 1 + x²/2! + x⁴/4! + ... + x²ⁿ/(2n)! + ...

1 / (eˣ + e⁻ˣ) = 1 / (1 + x²/2! + x⁴/4! + ... + x²ⁿ/(2n)! + ...)
               = 1 / (1 + 1/2!*x² + 1/4!*x⁴ + ... + 1/(2n)!*x²ⁿ + ...)

(i give up)
below is nonsense

∫ eˣ + e⁻ˣ dx = ∫ 1 dx + ∫ x²/2! dx + ∫ x⁴/4! dx + ...
              = x + x³/(2!*3) + x⁵/(4!*5) + ... x²ⁿ⁺¹/((2n)!*2n+1) + ...
              = x + x³/3! + x⁵/5! + ... x²ⁿ⁺¹/(2n+1)! + ...

∫ sech(x) = ∫ 2 / (eˣ + e⁻ˣ) dx
          = 2 * ∫ 1 / (eˣ + e⁻ˣ) dx
          = 2 * ln(eˣ + e⁻ˣ)
... but I gave up.
I also tried the infinite series of sech(x) via Euler numbers...

Code Select Expand

(note, i'm still learning how to integrate)

SEE: https://en.wikipedia.org/wiki/Euler_numbers
∫ sech(x) = ∫ ((E₀/0!)*x⁰ + (E₁/1!)*x¹ + (E₂/2!)*x² + (E₃/3!)*x³ + ... + (Eₙ/n!)*xⁿ + ...) dx
          = ∫ ((E₀/0!)*x⁰ + (0/1!)*x¹ + (E₂/2!)*x² + (0/3!)*x³ + ... + (Eₙ/n!)*xⁿ + ...) dx
          = ∫ ((E₀/0!)*x⁰ + (E₂/2!)*x² + ... + (Eₙ/n!)*xⁿ + ...) dx
          = ∫ (E₀/0!)*x⁰ dx + ∫ (E₂/2!)*x² dx + ... + ∫ (E₂ₙ/(2n)!)*x²ⁿ dx + ...

Looking at ∫ (Eₙ/n!)*xⁿ dx:

∫ (Eₙ/n!)*xⁿ dx = Eₙ/n! ∫ xⁿ dx
                = (Eₙ/n!) * (nxⁿ⁻¹)
                = (Eₙnxⁿ⁻¹)/n!

We don't talk about +C. (It is not needed)

∫ sech(x) = (E₀*0x⁰⁻¹)/0! + (E₂*2x²⁻¹)/2! + ... + (E₂ₙnx²ⁿ⁻¹)/(2n)! + ...
          = 0 + E₂x/2 + ... + (E₂ₙnx²ⁿ⁻¹)/(2n)! + ...


...however, after I checked it on Desmos, the series converges very, very slowly. Even with 18 terms, it doesn't fit sech(x) for |x| > ~1.

So yeah, I'm kinda stuck. With no good way of calculating ∫ sech(x), I'm stuck on naively calculating it. Maybe I can take an advantage that sech(x) = sec(ix), but I'm not sure how I'll integrate that.

Also, I've ran out of ideas.


Addendum

So, it turns out that the biggest bottleneck is the fact that I use Decimal to calculate π. If I use regular floats, I'll get 3 more digits with roughly the same duration:
the code

Heck, if I could I can just write the code in C:
the code
Look, I've managed to get none of the digits right!

[Nihonium]

nh!suggest either use BeautifulSoup everywhere or not use it at all

Your suggestion has been recorded.

[Orb Fish]

The TBGs AJAX Chat if I never joined the TBGs:

[Luigis_Pizza]

A wild post appeared...
Trainer sent out pikachu...
Pikachu used Thundershock...
Critical hit...
Its not very effective...
Post used words...
Hit 3 times...
Pikachu used thundershock...
Its not very effective...
Post used dark mode...
Pikachus accuracy fell...
Trainer used Poke ball...
1...
2...
3...
Trainer caught post...
Pikachu gained 327 exp...
Pikachu is now level 6...
Pikachu learned tail whip...


Try and guess which pokemon game this pikachu is from...

[solitaire]

A wild post appeared...
Trainer sent out pikachu...
Pikachu used Thundershock...
Critical hit...
Its not very effective...
Post used words...
Hit 3 times...
Pikachu used thundershock...
Its not very effective...
Post used dark mode...
Pikachus accuracy fell...
Trainer used Poke ball...
1...
2...
3...
Trainer caught post...
Pikachu gained 327 exp...
Pikachu is now level 6...
Pikachu learned tail whip...


Try and guess which pokemon game this pikachu is from...

yellow

[red herring]

-pi day post snip-

Here's how I did it using sech(x) = sec(ix):
First, I calculated the indefinite integral of sec(ix)
If you want to derive Euler's formula first using complex numbers, you can do that first

Substitute t = e^x

Notice that all e^x can be replaced with t

That's not so bad! now, this is a trig sub, so I'm going to draw a triangle to make this a bit more clear

Okay, what is t with respect to θ? tan(θ) is equal to t/1, so t = tan(θ)
What about t² + 1? sec(θ) is equal to √t²+1/1, so t²+1 = sec²(θ). We obtain the following substitutions:

Substitute everything

Cancel out sec²(θ) and integrate

Now we can use inverse trigonometric functions to return this back to the x-world (as bprp calls it)

Finally we can try and find the improper integral

The limit of e to the power of negative infinity is going to 0. the arctan of 0 s 0. The limit of e to the power of infinity is going to infinity.  We have several different options for the arctan's value, but since it's usually defined to be 0 < θ < π, I'm going to use the value π/2

Happy pi day, I hope that explains the integral

[solitaire]

use a magikarp and strange glitch shenanigans to create the result of horrific gene splicing

[PkmnQ]

PF

Code Select Expand

[ - (*>(<)
] - )<*

[solitaire]

1. get 2 things of opposite gender, the 3rd move slot of the pokemon that will hatched from an egg made by the parents will influence what pokemon will be obtained with the glitch
2. get a glitchy mess made by trying to clone a thing
3. get 5 things that are useless
4. get egg
5. deposit egg
6. give glitchy mess to daycare man, then take it back, it will become an even glitchier mess
7. put it at top of party. whatever you do don't move around the pokemon past the 6th slot. moving them will modify the code of the game
8. move pokemon without mail to put a pokemon to the top of the party. you will now have 7 pokemon.
9. put the glitchy mess in the daycare again
10. deposit first and second pokemon in the party
11. move pokemon without mail to move egg to the top of the party
12. deposit remaining pokemon (don't deposit the egg)
13. withdraw anything that was not used for the glitch
14. run around until it hatches
15. congratulations. you now have a thing.

[Gilbert189]

how do I add functions

[PkmnQ]

Code Select Expand

(f + g)(x) = f(x) + g(x)i honestly see no reason to do either:
a. just write out f(x) + g(x)
b. make a new function h(x) which is defined as f(x)+g(x)

[solitaire]

ok so if i breed a male hoppip with a female cleafairy it may know splash as it's 3rd move, allowing me to get mewtwo

[solitaire]

ok so if i breed a male hoppip with a female cleafairy it may know splash as it's 3rd move, allowing me to get mewtwo

because that makes sense