Use A Telescope
Let
f
be a function satisfying
f
(
x
+
1
)
−
f
(
x
)
=
1
for all
x
.
What is
f
(
2
0
1
6
)
−
f
(
−
2
0
1
6
)
?
The answer is 4032.
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3 solutions
f ( x + n ) − f ( x ) = n , n ∈ I can be proven by induction.
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This could be alternately proved by adding equations in a similar manner as above also.. :-)
The function f ( x ) can be easily seen as the function y = x . Therefore, the expression turns into 2 0 1 6 − ( − 2 0 1 6 ) = 4 0 3 2 .
Yes, f ( x ) = x is a possible function of f ( x ) , but does it mean that it is the only solution for f ( x ) ? Maybe there is another function of f ( x ) such that f ( 2 0 1 6 ) − f ( − 2 0 1 6 ) = 4 0 3 2 ?
The value of the difference is constant for a fixed difference in the input, x, i.e. 1 in both cases. That means f must be a linear function, and clearly its slope is 1: f(x) = x+b f(2016) = 2016+b f(-2016) = -2016+b
So f(2016)-f(-2016) = 4032
How can we be sure that f ( x ) is not a piecewise function? i.e. f ( x ) = x + ( f r a c ( x ) 2 ) , where f r a c ( x ) denotes the fractional part of x ?
Relevant wiki: Telescoping Series - Sum
f ( 2 0 1 6 ) − f ( 2 0 1 5 ) = 1 f ( 2 0 1 5 ) − f ( 2 0 1 4 ) = 1 f ( 2 0 1 4 ) − f ( 2 0 1 3 ) = 1 ⋯ ⋯ ⋯ ⋯
f ( − 2 0 1 3 ) − f ( − 2 0 1 4 ) = 1 f ( − 2 0 1 4 ) − f ( − 2 0 1 5 ) = 1 f ( − 2 0 1 5 ) − f ( − 2 0 1 6 ) = 1
Adding all these equations we get:-
f ( 2 0 1 6 ) − f ( − 2 0 1 6 ) = 4 0 3 2
Generalisation:- f ( x ) − f ( − x ) = 2 x