this post was submitted on 08 Mar 2026
27 points (100.0% liked)

askchapo

23237 readers
117 users here now

Ask Hexbear is the place to ask and answer ~~thought-provoking~~ questions.

Rules:

  1. Posts must ask a question.

  2. If the question asked is serious, answer seriously.

  3. Questions where you want to learn more about socialism are allowed, but questions in bad faith are not.

  4. Try !feedback@hexbear.net if you're having questions about regarding moderation, site policy, the site itself, development, volunteering or the mod team.

founded 5 years ago
MODERATORS
 

Am I meant to assume a_i is defined the same way as a_n for each of 1<= i <= n-1 ?

If so, I think I see the proof by induction on n, but the question just says a_i is defined for each 1<=i<=n-1, not that it is defined in that way. Is the question just overly vague or am I missing something obvious?

If only a_n is defined as the greatest integer such that the sum from 0 to n of each a_i/k^i is <=x, then I think there are counterexamples to the hypothesis, right?

Like if x=0.32, k=2, n=2, then a_0=0, and the inequality is satisfied by a_1 = 2 > k-1 = 1 and a_2 = -3 < 0.

you are viewing a single comment's thread
view the rest of the comments
[โ€“] crosswind@hexbear.net 1 points 2 days ago (1 children)

About your question of definitions more broadly, generally when a~n~ is defined, other subscripts a~i~, a~j~, a~m~, a~xyz~, are defined in the same way, even if that isn't specifically stated.
The a is what carries the properties of the sequence, and any variables in the subscript are just different ways of referencing the index. That can be to communicate a separation between concepts, or to set up relationships like i<=n that will be needed later

[โ€“] cosecantphi@hexbear.net 1 points 1 day ago

Thanks, I was thinking the same. I had already wrote a proof that worked if all a_i are defined like that, but wanted to be sure.