Conversation

Notices

1. Gnarley Boot (gnarley_boot@norwoodzero.net)'s status on Friday, 12-Jul-2024 19:12:35 JST Gnarley Boot

   in reply to

   • T man :sex: :puffgiga: :puffpowerroll: (epic music enjoyer)
   • 【Ξｎｉｇｍａｔｉｃｏ】:misskey:
   • (NekoSock) Miya Ironami
   • Der Keymeister :chad_2: 🔐 ⬜
   • Iska
   • Gnarley Boot
   @iska @enigmatico @iro_miya
   
   Yeah hit them up with real math they will know what yer talking about cc @theorytoe @3T
   • † top dog :pedomustdie: likes this.
   • Gnarley Boot (gnarley_boot@norwoodzero.net)'s status on Friday, 12-Jul-2024 19:12:36 JST Gnarley Boot

     in reply to

     • 【Ξｎｉｇｍａｔｉｃｏ】:misskey:
     • (NekoSock) Miya Ironami
     • Iska
     @iska @enigmatico @iro_miya
     
     AFAIK toe and 3T are math brains.
   • Iska (iska@catposter.club)'s status on Friday, 12-Jul-2024 19:12:38 JST Iska

     in reply to

     • 【Ξｎｉｇｍａｔｉｃｏ】:misskey:
     • (NekoSock) Miya Ironami

     @enigmatico@mk.absturztau.be @iro_miya@mk.absturztau.be do you actually think fedi users can comprehend math above elementary school

   • 【Ξｎｉｇｍａｔｉｃｏ】:misskey: (enigmatico@mk.absturztau.be)'s status on Friday, 12-Jul-2024 19:12:39 JST 【Ξｎｉｇｍａｔｉｃｏ】:misskey:

     in reply to

     • (NekoSock) Miya Ironami

     @iro_miya While it's true that you can implement sums using a for loop, it can be optimized using basic summation identities. More specifically
     
     \displaystyle{\sum^k_{n=0}3n = 3\sum^k_{n=0}n = 3\left(\sum^{k-1}_{n=0}n + k\right) = 3\left(\frac{k(k-1)}{2}+k\right)}
     
     Where k \in \mathbb{Z}⁺, k > 0. So you can simply use this generic formula and save yourself a for loop.
     
     The same can be applied to the multiplication as well, although in this case how much computational cost you save will depend on how the machine calculates the factorial.
     
     \displaystyle{\prod^k_{n=1}2n = 2\prod^k_{n=1}n = 2(k!)}

   • (NekoSock) Miya Ironami (iro_miya@mk.absturztau.be)'s status on Friday, 12-Jul-2024 19:12:40 JST (NekoSock) Miya Ironami

     • Freya Holmér

     WHAT
     
     (You can follow Freya Holmér on fedi at @acegikmo@mastodon.social!)