Double functions?

Zeitiax SG

Can anyone help me in solving this question..

f(x) = f(x/2002) + 3x
Find f(2)

And what topic does this kind of question falls under?

Thanks in advance.

eagle SG

Glad your first post was made in Homework Forum Mr. Green

First, if you could not solve such questions or cannot see where it is going, you could actually try to write it out like the following

f(2) = f(2/2002) + 3 (2)
f(2/2002) = f(2/2002^2) + 3 (2/2002)
f(2/2004) = f(2/2002^3) + 3 (2/2002^2)
and so on

You will find that it will end up something similar to

3(2) + 3 (2/2002) + 3 (2/2002^2) + ....

Similar to a GP
first term 6
rth term 1/2002
and Sum to Infinity

Thus,
f(2) = a / (1 - r) = 6.003

Zeitiax SG

That was fast, hope i get a clue about what’s happening after i read your post a few more times.. looks… complicated.. hmm…

eagle SG

write it out. You will understand better

Zeitiax SG

So basically its just continuing the pattern until its is sufficient? ( In this case, to 3 dec. places )

eagle SG

More of continuing till you see the pattern

Zeitiax SG

What does a and r represent?

I'm still quite new to series... so... having some problem. hehe..

EDIT:

Found what they represent,

anyway, I've got the question wrong, no wonder the answer looks weird,

f(x) = f(2002/x) + 3x
Find f(2)

It should be the other way.. and some how... I don't know how to apply the sum to infinity of a G.P., is it because its no longer a G.P.?

eagle SG

I think the question like this more wrong:

f(2) = f(2002/2) + 3(2)
f(2002/2) = f(2) + 3(1001)

f(2) = f(2) + 3 (1001) + 3 (2)
0 = 3 (1003) ?????

Zeitiax SG

lol... Didn't notice that.. maybe i'll check the question with my friend again.

Thanks for your help and time..