Comment

wolf_2202@sh.itjust.works ⁨1⁩ ⁨week⁩ ago

That depends on the decay factor of one centaur to the next. If the centaurs shrink by anything more than a factor of two, then no. The creature will converge onto a single length.

source

Sort:hotnew top

Liz@midwest.social ⁨1⁩ ⁨week⁩ ago
What? If it’s geometric it needs to be less than 1, that’s all. 9/10 + 81/100 + 729/1000 + … = 10

C•(1-r)^-1^ = C•x

Where r is the ratio between successive terms.

source
eestileib@sh.itjust.works ⁨1⁩ ⁨week⁩ ago
Should be anything less than a harmonic decrease (that is, the nth centaur is 1/n the size of the original).

The harmonic series is the slowest-diverging series.

source
- kogasa@programming.dev ⁨1⁩ ⁨week⁩ ago
  The assumption is that the size decreases geometrically, which is reasonable for this kind of self similarity. You can’t just say “less than harmonic” though, I mean 1/(2n) is “slower”.
  
  source
  - eestileib@sh.itjust.works ⁨1⁩ ⁨week⁩ ago
    Eh, that’s just 1/2 of the harmonic sum, which diverges.
    
    source
    kogasa@programming.dev ⁨1⁩ ⁨week⁩ ago
    Yes, but it proves that termwise comparison with the harmonic series isn’t sufficient to tell if a series diverges.
    
    source
    -> View More Comments
Hjalamanger@feddit.nu ⁨1⁩ ⁨week⁩ ago
Judging by the image the centaura shrink with about a factor of two so the entire creature should be either infinitely long or just very very long.

source