PHEW.... finally here.....
f(0)=N g(0)=0
g(n+1)=2{f(n)/2+g(n)/2} (fractional part)
f(n+1) = [f(n)/2+g(n)/2] (greatest integer..)
where n is integer
find f(1)+f(2)+f(3)...... infinity...
N is an integer :)
62haha.. this is what i wanted to hear.. :) :D
See to be true this is not a jee problem..... But it will add a new dimension to your thought process
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I agree with sky and that is why u have deleted that part about it not being that bad bhaiyya
62no it is not that bad :P
but bcos to think of the question took me half an hour... so i thought that i rather remove the phrase :D
1thank goodnes u forgot about writing that negative mark thing about the explanation part [1]
1that should not be the way to see it i know but is the answer correct
62f(0)=N g(0)=0
g(n+1)=2{f(n)/2+g(n)/2} (fractional part)
f(n+1) = [f(n)/2+g(n)/2] (greatest integer..)
where n is integer
N=3
f(0)=3
g(1)=2{3/2}=1
f(1)=[3/2]=1
g(2)=2{1/2+1/2} =0
f(2)=[2/2]=1
f(3)=0
f(4)=0...
hence f(1)+f(2) = 2
does this give some hint?
1ans looks to be N-1
yes N-1
edited
P.S jus read chatbox i hadn't read when i posted this :)
9λproved using induction
F(N) = f1 + f2 ............. for N
then F(N+1)= F(N) +1
this is proved by taking N=4λ + (1or 2 or3or4)
and similarly representing N+1 in same form after 3 steps( f1 ,f2,f3) ull find that theyre sum differs by one but their parameters ie f3,g3 bcom same
so rest of sum is identical and induction finished
also F(1) is 0 so F(N)=N-1
1357hint take
N=a0+2a1+4a2..... + 2kak
Now try to solve
where each ai is either 0 or 1
before that you have to justify that the above can be done!
62celestirne how is that!! :O
I din get how you proved the induction hypothesis!
9manish sir i already gave that hint of rep N as a binary form in my previous post but then i found a simpler not so confusing method so i left of that binary form method.
see take the foll values with corresponding f,g after each step
4λ+1 4λ+2 4λ+3 4λ+4 4λ+5
2λ,1 2λ+1,0 2λ+1,1 2λ+2,0 2λ+2,1
λ,1 λ,1 λ+1,0 λ+1,0 λ+1,1
case λ=2μ
μ,1 μ,1 μ,1 μ,1 μ+1,0
as f + g is same in all cases remaining values are identical
similarly take for λ = 2μ+1 also
now wen u add up till the point were they dont repeat u find succesive F(N) differ by 1
this was the proof i was talkin abt
but i feel u ve a more beautiful proof involving binary nos that im currently not able to think abt due to my ongoing chem exams ;(
62I will give a veyrr very different proof...
There is a knock out tennis match of N player..
What is the number of matches required to chose the winner.
At each level, the player who does not play goes to the next level...
Then the expression is the number of matches in each level....
The number of matches is n-1 because at each game, 1 player is eliminated :)
Dont kill me for this solution... I told that this one is the toughest (may be stupid) question u will ever come across :P)
9great q nish
im sure this q is of inmo level but then its hard to thnk in ur trms cos u already know the sol ;)
ne way i solved it usin long method pls give proof using binary nos
62haha..
well u know what .. this question came to me because i had this question.. in permutation.. and the series that i got was the one i gave...
And i was just like in no position to know what the sum would be..
then i realised that the answer is N-1!!
So i made this question from there! :O
The binary method.. I am not very sure... :(
I will have to think for that... (My gut feeling says that it can be done by that!)
May be one way could be prove for any 2n (which sis straightforward) Then prove for sum of numbers.... (But i am not sure if that will work!)
9even my gut feelin was that but i was unable to proceed further so changed to another method
62yup i have... let me fix it ... before more guys start to think.... too hard.. :(
9k
series = n( 1/2 + 1/4 ...) + (1-1/2) +(1-1/4)+(1-1/8).......
= n + ∞ - 1
;)
r u sure the q is rite or am i rong
9oh sry i din see that as greatest int fn neway will try ,the abv mite be a clue
62let me give an example....
for 5
f(5)=[(5+1)/2]=3
ff(5)=[3/2]=1
fff(5) = 0
sum will be 3+1 = 4
