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Mathematics 20 Online
OpenStudy (anonymous):

I have a question #10

OpenStudy (anonymous):

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 \[f(2548)+f(2549)\]

OpenStudy (mr.math):

What's \(I\)?

OpenStudy (asnaseer):

Integers

OpenStudy (anonymous):

right

OpenStudy (jamesj):

(not standard notation, but ok ;-) )

OpenStudy (anonymous):

it is, isn't it LOL

OpenStudy (anonymous):

\[\mathbb Z\]?

OpenStudy (jamesj):

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

OpenStudy (mr.math):

I hate you James!

OpenStudy (asnaseer):

ah - I see the pattern now James - nice solution

OpenStudy (mr.math):

I saw it, but he was faster!

OpenStudy (asnaseer):

Jesse James was always faster - wasn't he? :-)

OpenStudy (mr.math):

Yeah, that's why I hate him! :D

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

i made it all the way up to f(3) and i have had 3 cups of coffee!

OpenStudy (anonymous):

he's totally right. so the ans is 7/2

OpenStudy (anonymous):

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 ;)

OpenStudy (mr.math):

It's not hard at all, it's very easy actually.

OpenStudy (anonymous):

one "very" is not enough :P

OpenStudy (anonymous):

yeah - but i was so slow seeing it - I must drink some coffee!!

OpenStudy (anonymous):

post it foolformath, the chat is quite slow

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