Fool's problem of the day,

A function $ f $ is defined such that $ f(2) = 60 $ and $ \sum \limits_{i=1} ^n (-1)^i f(i) = nf(n) \forall n 1 \text{ and } n \in \mathbb{N} $. Can you find $ f(257) $ ?

[Solved by @dumbcow]

Good luck!

Question

Fool's problem of the day,

A function $ f $ is defined such that $ f(2) = 60 $ and  $ \sum \limits_{i=1} ^n (-1)^i f(i) = nf(n) \forall n > 1 \text{ and } n \in \mathbb{N} $.  Can you find $ f(257) $ ?

[Solved by @dumbcow]

Good luck!

lgbasallote · Answer

i just came here to give the medal. byeeee

dumbcow · Answer

seems like as n increases f(n) decreases toward 0

perl · Answer

-1 f(1) = 1 f(1)  -> f(1) = 0 
-1f(1) + f(2) = 2 f(2) 
-1 f(1) + 60 = 2 f(2)

apoorvk · Answer

yeha increase and decrease.. but i can t seem to find a trend.. calculating till 57 will take a lot of time... :/ may be we need to go reverse.

dumbcow · Answer

f(11) = f(12) = 10
thats as far as i've gotten

apoorvk · Answer

i found all values till f(9). but we slog more? am sure there's more to it.

perl · Answer

did you get zero for f(1)

apoorvk · Answer

and f(3) = f(4) = 30
no f(1) i got -60 if i am right at all

dumbcow · Answer

no i got f(1) = -60

--> -(-60) +60 = 2*60

apoorvk · Answer

i even have the general equation of every term implicit form.. but what then?

anonymous · Answer

@dumbcow: The original problem ask for f(11) only, but i derived the a closed form for f(n) and revised the constraints ;)

anonymous · Answer

@apoorvk: You would get a recurrence relation then you can use various method to get the closed form in terms of f(2).

Okay no more hints.

dumbcow · Answer

i see, yeah thats the hard part getting the closed form for f(n)

perl · Answer

something seems odd, if you plug in n=1 , or you cant sum from i = 1 to n=1 ?

anonymous · Answer

perl, the function f(n) is defined only for for n>1, my apologies.

dumbcow · Answer

i noticed something else, the sum seems to stay constant at 120 when n is odd
so then it follows
257*f(257) = 120-f(257)

f(257) = 120/258 = .465

is that right?

apoorvk · Answer

$$(\sum_{i=1}^{n-1} (-1)^i f(i)) / (n +1) = f(n)$$ .. when n is odd
$$(\sum_{i=1}^{n-1} (-1)^i f(i)) / (n -1) = f(n)$$ .. when n is even

anonymous · Answer

Congratz @dumbcow that's the right answer.

apoorvk · Answer

yeah sorry when n is 'odd' and so sum of (n-1) terms is 120.

so according to the equation i posted, f(257), which is odd, is 120/(257+1)
=120/258
=0.46512

Dumbcow did it first!!!!!!!!

anonymous · Answer

That's it, just one problem? :/