Periodicity
For all real numbers x , the function f is periodic, with f ( x + 6 ) = f ( x + 1 0 ) = f ( x ) .
If f ( 2 2 ) = 2 2 , what is the value of f ( 4 4 ) ?
The answer is 22.
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1 solution
Nice observations!
Do you think there are any stronger conclusions that can be made?
In particular, if f ( 2 2 ) = 2 2 , then f ( s ) = 2 2 for all s ∈ S . How general of a set S can you come up with?
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I believe that for real number a , b , by using relation f ( x ) = f ( x + 6 a + 1 0 b ) based on linear progression provided, it is possible to create a general terms for this periodically. For example f ( 4 4 ) = f ( 2 2 + 1 0 ( 1 ) + 6 ( 2 ) )
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That's a good idea. However, are you sure that this is true for any real numbers a , b ? Are there further restrictions on them?
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@Eli Ross – Provided that the question originally defined f ( 2 2 ) , the value of f ( 2 2 ) = f ( 2 2 + 6 a + 1 0 b ) = 2 2 . As g c d ( 2 2 , 6 a , 1 0 b ) = 2 , it can also rewrote into f ( 2 ( 1 1 + 3 a + 5 b ) ) = 2 2 . While we can make 1 1 + 3 a + 5 b into integer, we notice 2 in front of it, which means if we narrow down, only every even number x will satisfy f ( x ) = 2 2
Please kindly advise me in this regard again though, as I am out of practice for quite a long time
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@Kay Xspre – That's correct; great work!
In particular, Bezout's Identity tells us that m ⋅ a + n ⋅ b can be equal any multiple of the Greatest Common Divisor of the integers m and n , and cannot be equal to any non-multiple of g cd ( m , n ) .
Here, m = 6 and n = 1 0 , so g cd ( m , n ) = 2 . Thus, any even number (but no other numbers) can be written in the form 2 2 + 6 a + 1 0 b , so f ( s ) = f ( 2 2 ) = 2 2 for all even s .
We try to replace x with 16, which gives f ( 1 6 ) = f ( 2 2 ) = f ( 2 6 ) = 2 2 , then, from f ( x ) = f ( x + 6 ) , replace x with 26, 32 and 38, which will gives f ( 1 6 ) = f ( 2 2 ) = f ( 2 6 ) = f ( 3 2 ) = f ( 3 8 ) = f ( 4 4 ) = 2 2