I have a question #10
define \[f:I \rightarrow R\] satisfied with\[f(x)=\frac{1+f(x-1)}{1-f(x-1)}\] for all \[x \in I\] where \[f(1)=3\] find \[f(2548)+f(2549)\]
What's \(I\)?
Integers
right
(not standard notation, but ok ;-) )
it is, isn't it LOL
\[\mathbb Z\]?
correct me if I haven't had enough coffee, but don't we have f(1) = 3 f(2) = -2 f(3) = -1/3 f(4) = 1/2 f(5) = 3 So now f(2548) = f(4n) = f(4) = 1/2 and f(4n+1) = f(1) = 3
I hate you James!
ah - I see the pattern now James - nice solution
I saw it, but he was faster!
Jesse James was always faster - wasn't he? :-)
Yeah, that's why I hate him! :D
lol
i made it all the way up to f(3) and i have had 3 cups of coffee!
he's totally right. so the ans is 7/2
I dunno why you thought that this problem is a hard one,this type of trivial functional equations needs nothing more than some elementary induction ;)
It's not hard at all, it's very easy actually.
one "very" is not enough :P
yeah - but i was so slow seeing it - I must drink some coffee!!
post it foolformath, the chat is quite slow
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