Difference between revisions of "1997 AHSME Problems/Problem 27"
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== See also == | == See also == | ||
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Latest revision as of 13:14, 5 July 2013
Problem
Consider those functions that satisfy for all real . Any such function is periodic, and there is a least common positive period for all of them. Find .
Solution
Recall that is the fundamental period of function iff is the smallest positive such that for all .
In this case, we know that . Plugging in in for to get the next equation in the recursion, we also get . Adding those two equations gives after cancelling out common terms.
Again plugging in in for in that last equation (in order to get ), we find that . Now, plugging in for , we get . This proves that , so there is a period of , which gives answer . We now eliminate answers through .
Let for . Plugging in into the initial equation gives , which implies that . Since , the function does not have period .
Continuing, , , and . Once we hit , we have . Since , the function does not have period .
Finally, , , , and . Since , the function does not have period .
To confirm that our original period works, we may see that , , , , , , and . Finally, , which is indeed the same as .
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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All AHSME Problems and Solutions |
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