anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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lgbasallote
  • lgbasallote
i just came here to give the medal. byeeee
dumbcow
  • dumbcow
seems like as n increases f(n) decreases toward 0
perl
  • perl
-1 f(1) = 1 f(1) -> f(1) = 0 -1f(1) + f(2) = 2 f(2) -1 f(1) + 60 = 2 f(2)

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More answers

apoorvk
  • apoorvk
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
  • dumbcow
f(11) = f(12) = 10 thats as far as i've gotten
apoorvk
  • apoorvk
i found all values till f(9). but we slog more? am sure there's more to it.
perl
  • perl
did you get zero for f(1)
apoorvk
  • apoorvk
and f(3) = f(4) = 30 no f(1) i got -60 if i am right at all
dumbcow
  • dumbcow
no i got f(1) = -60 --> -(-60) +60 = 2*60
apoorvk
  • apoorvk
i even have the general equation of every term implicit form.. but what then?
anonymous
  • anonymous
@dumbcow: The original problem ask for f(11) only, but i derived the a closed form for f(n) and revised the constraints ;)
anonymous
  • anonymous
@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
  • dumbcow
i see, yeah thats the hard part getting the closed form for f(n)
perl
  • perl
something seems odd, if you plug in n=1 , or you cant sum from i = 1 to n=1 ?
anonymous
  • anonymous
perl, the function f(n) is defined only for for n>1, my apologies.
dumbcow
  • dumbcow
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
  • apoorvk
\[(\sum_{i=1}^{n-1} (-1)^i f(i)) / (n +1) = f(n)\] .. when n is odd \[(\sum_{i=1}^{n-1} (-1)^i f(i)) / (n -1) = f(n)\] .. when n is even
anonymous
  • anonymous
Congratz @dumbcow that's the right answer.
apoorvk
  • apoorvk
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
  • anonymous
That's it, just one problem? :/

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