09-10-09 Functional Equation question

f(1) = 1

f(2n) = f(n)

f(2n+1) = f(n)+1

where n is an integer and f is a function defined over integers...

Give a simpler explanation for f...

12 Answers

3
msp ·

f is defined such that the difference between any two consecutive integers is one.

62
Lokesh Verma ·

no f(1)= f(2) =1

3
msp ·

mmm yes sir.

i fell into the trap.

1
Philip Calvert ·

[12] no.

3
msp ·

yups,nishant sir asks us a simpler explanation but its getting complex.

1
Philip Calvert ·

for any k = 2n
f(k) = 1 ... f(k+1) = 2 and f(k-1)= n ..this much is clear

as for the rest [12]

3
msp ·

for positive integers which are of multiple of 2 then f(2k)=1 and f(2k+1)=2 (k is an integer)

for negative integers ????

62
Lokesh Verma ·

only for natural numbers...

Btw.. Hint: think of the binary expansion?

341
Hari Shankar ·

I was looking for an esoteric way of giving that hint :D

62
Lokesh Verma ·

lol.. yeah even i was thiking of some other way.. but could not :D

1
Philip Calvert ·

oops really ..

F(n) is the no. of ones in the binary representation of n

62
Lokesh Verma ·

@Philip.. the proof?

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