How does one simplify such a function?
f ( 2 + x ) = f ( 2 − x ) f ( 7 + x ) = f ( 7 − x )
A function f is defined for all real numbers and, for all real x , satisfies the two equations above.
If f ( 0 ) = 0 , what is the smallest number of solutions that f ( x ) = 0 could have in the interval − 2 0 0 0 ≤ x ≤ 2 0 0 0 ?
The answer is 801.
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3 solutions
Moderator note:
How can we generalize this approach?
IE, what can we say if, for some constants a , b , we have
f ( a + x ) = f ( a − x ) f ( b + x ) = f ( b − x )
Nice solution.I did same.
Since f(x) is symmetric about two lines x=a & x=b, we can say it is symmetric about infinitely many such lines.
My solution is the answer for challenge master note in Sharky's solution:
f ( a + x ) = f ( a − x ) f ( b + x ) = f ( b − x )
Replacing ( a − x ) by t in first we get f ( t ) = f ( 2 a − t ) = f ( b − ( t − 2 a + b ) ) also f ( b − ( t − 2 a + b ) ) = f ( b + ( t − 2 a + b ) ) = f ( t + 2 ( b − a ) ) So f ( t ) = f ( t + 2 ( b − a ) )
Hence f ( x ) is periodic with period T = 2 ( b − a )
Remaining solution have been posted by Sharky.
Moderator note:
Not quite complete. What you have shown is that if f is a function that satisfies the conditions, then 2 ( b − a ) is a period of the function.
What is more interesting, is to ask "What is the smallest value that is a period to all such functions that satisfy the conditions". So, we either need to show
1. There is a function that satisfies the function and has period
2
(
b
−
a
)
, or
2. There is a smaller value (maybe
b
−
a
), which is a period of these functions.
As a further extension, when thinking about 2, also consider the case of a = 1 , b = 2 .
It will be proven that the function f have at least 801 real solution x such that f ( x ) = 0 .
We have f ( 2 − x ) = f ( 2 + x ) and f ( 7 − x ) = f ( 7 + x ) for every x ∈ R . Therefore it is obvious that f ( x ) = f ( x + 1 0 ) = f ( x − 1 0 ) ∀ x ∈ R .
It is easy to show that f ( 0 ) = f ( 4 ) = 0 . From the previous equality we have f ( 0 ) = f ( 1 0 k ) = f ( − 1 0 k ) = 0 ∀ k ∈ N and f ( 4 ) = f ( 4 − 1 0 k ) = f ( 4 + 1 0 k ) = 0 ∀ k ∈ N .
Therefore it is easy to check that there are exactly 801 integers k satisfying the equality f ( k ) = 0 with k ∈ [ − 2 0 0 0 , 2 0 0 0 ] . That is, if the only real number x satisfying f ( x ) = 0 with x ∈ [ − 3 , 2 ] is only x = 0 .
We have
f ( 2 + x ) = f ( 2 − x ) = f ( 7 + ( − 5 − x ) ) = f ( 7 − ( − 5 − x ) ) = f ( 1 2 + x ) .
Thus, f ( 2 + x ) = f ( 1 2 + x ) for all real x , which means that f ( t ) = f ( t + 1 0 ) for all real t .
Also
0 = f ( 0 ) = f ( 2 − 2 ) = f ( 2 + 2 ) = f ( 4 ) .
which shows f ( 4 ) = 0 .
Hence f ( 1 0 k ) = f ( 1 0 k + 4 ) = 0 for all integers k .
In the interval − 2 0 0 0 ≤ x ≤ 2 0 0 0 , there are 801 numbers that are equal to either 1 0 k or 1 0 k + 4 where k is an integer.
Now consider the function f ( x ) such that f ( x ) = 0 when x = 1 0 k or x = 1 0 k + 4 for some integer k , and f ( x ) = 1 for the other values of x . It can be easily checked that this function satisfies f ( 2 + x ) = f ( 2 − x ) and f ( 7 + x ) = f ( 7 − x ) for all real x .
Therefore, the answer is 8 0 1 .