Message from @Covariant Canonical Quantization

Discord ID: 554918116795088896


2019-03-12 06:39:14 UTC  

minecraft chest

2019-03-12 06:39:57 UTC  

Eurobeat is the all powerful genre of music

2019-03-12 06:39:57 UTC  

😂 😂 😂 😂

2019-03-12 06:40:35 UTC  

@Covariant Canonical Quantization you're scaring me stup

2019-03-12 06:40:42 UTC  

tall chest

2019-03-12 06:40:55 UTC  

Chest?

2019-03-12 06:41:08 UTC  

like from minecraft

2019-03-12 06:41:11 UTC  

I think

2019-03-12 06:41:12 UTC  

How long is it

2019-03-12 06:41:16 UTC  

hmmmm

2019-03-12 06:41:30 UTC  

Is it

2019-03-12 06:41:32 UTC  

@Citizen Z unban me in 24/7

2019-03-12 06:41:39 UTC  

L O N G

2019-03-12 06:41:49 UTC  

in 22/7

2019-03-12 06:42:10 UTC  

Oh yes

2019-03-12 06:43:21 UTC  

π = 4 arctan(1)

2019-03-12 06:44:35 UTC  

1/(2k+1) < 10 ^-10 for k 5*10^9 - 1/2 <:whopinged:489114411387191296>

2019-03-12 06:44:52 UTC  

.......

2019-03-12 06:45:39 UTC  

1+1=40238890.34i99230

2019-03-12 06:46:22 UTC  

pi/4= sum^infinity_n=0 (1/(4n+1)-1/(4n+3)=sum^infinity_n=0 2/((4n+1)(4n+3))

2019-03-12 06:46:28 UTC  

<:snapsnap:484956825863585792>

2019-03-12 06:49:07 UTC  

pi/2 - 2 sum^(N/2)_(k=1) (-1)^(k-1)/(2k-1) ~ sum^(infinity)_(m=0) E_2m/(N^(2m+1))

2019-03-12 06:49:12 UTC  

<:sob:507995466659659776>

2019-03-12 06:49:28 UTC  

```pi```

2019-03-12 06:50:00 UTC  

π /(9x+1)/=pie=0.2)(m=7x)--2.0x y(86!)-+()

2019-03-12 06:50:26 UTC  

lemme get deep

2019-03-12 06:50:50 UTC  
2019-03-12 06:51:51 UTC  

s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2}

2019-03-12 06:52:10 UTC  

Far more superior music m8

2019-03-12 06:52:14 UTC  

can yall stop bashing ur head against the keyboard?

2019-03-12 06:52:43 UTC  

ax4+bx3+cx2+dx+f=0

2019-03-12 06:52:50 UTC  

I am not

2019-03-12 06:53:06 UTC  

```pi/4 = ( p≡1 Product (mod 4) p/(p-1))(p≡3 product (mod 4) p/(p+1)) = 3/4 * 5/4 * 7/8 * 11/12 * 13/12 * 17/16 ...```

2019-03-12 06:53:07 UTC  

*says while bashes head on keyboard*

2019-03-12 06:53:14 UTC  

x=−b±b2−4ac√2ax=−b±b2−4ac2a

2019-03-12 06:53:38 UTC  

What are these equations

2019-03-12 06:53:44 UTC  

Fab=2∇[aAb]Fab=2∇[aAb] (mod 4) p/(p-1)
∇[aFbc]=0∇[aFbc]=0
∇aFab=Jb∇aFab=Jb ax4+bx3+cx2+dx+f=0

2019-03-12 06:53:47 UTC  

idk they are retarded

2019-03-12 06:53:58 UTC  

∇⋅B⃗ =0∇⋅B→=0
∇⋅E⃗ =4πρ∇⋅E→=4πρ
∂B⃗ ∂t=−∇×B⃗ ∂B→∂t=−∇×B→
∂E⃗ ∂t=+∇×B⃗ −4πJ⃗ .∂E→∂t=+∇×B→−4πJ→.