#Got 5/7
34 messages · Page 1 of 1 (latest)
its not c=1
whyd you post this in math help tho
it is the f(r)+f(-r) = 0
its c=2
what could be the answeres
huh
well if you solved it you probably know what are the types of the functions
so you can plug some vals and know the answers
I will give you an example f(x) = x works here
so 0 is one of the solutions for 100% so c>=1
busy right now
this video is nice
#mathematics #olympiad #math
International Mathematical Olympiad (IMO) 2024 Day 2
Solutions and discussion of problem 6
65th International Mathematical Olympiad Bath UK
Problem 6 - Algebra / Functional equation
its a pretty tough func equation
I cannot watch it I am still solving it
yeah i got c=1 by f(r)+f(-r)=0 asw
btw f(r)+f(-r)=0 for all r isnt true
for every aquaesulian
so c=1 would only owrk if every aquaesulian function always satisfied f(r)+f(-r)=0, but there are aquaesulian functions where the sum takes 2 different values
for example
f(q)=2floor(q)-q (q ∈ ℚ)
for this aquae f, {f(r)+f(-r):r ∈ ℚ} ⊇ {0,-2}
aka 2 distinct vals
I also got f(x)=x even tho ik it's wrong
pretty tough functional equation problem
f(S(r))=S(r) for all r
and for all r, f(f(r))=r+S(r)
