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.
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wolf_2202@sh.itjust.works 5 months agoThat 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.
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.
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”.
Eh, that’s just 1/2 of the harmonic sum, which diverges.
Yes, but it proves that termwise comparison with the harmonic series isn’t sufficient to tell if a series diverges.
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.
Liz@midwest.social 5 months 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.