Difference between revisions of "1969 IMO Problems/Problem 2"
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<math>f(x)=(b_1\cos{a_1} + b_2\cos{a_2} + \cdots + b_n\cos{a_n})\cos{x} - | <math>f(x)=(b_1\cos{a_1} + b_2\cos{a_2} + \cdots + b_n\cos{a_n})\cos{x} - | ||
− | b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n})\sin{x}</math>. | + | (b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n})\sin{x}</math>. |
Revision as of 14:49, 8 August 2024
Contents
Problem
Let be real constants, a real variable, and Given that prove that for some integer
Solution
Because the period of is , the period of is also . We can get for . Thus, for some integer
Solution 2 (longer)
By the cosine addition formula, This implies that if , Since the period of is , this means that for any natural number . That implies that every value is a zero of .
Remarks (added by pf02, August 2024)
Both solutions given above are incorrect.
The first solution is hopelessly incorrect. It states that (and relies on it) if has period and then for some integer . This is plainly wrong (think of ). There is an obvious "red flag" as far as solutions go, namely the solution did not use that and .
The second solution starts promising, but then it goes on to prove the converse of the given problem, namely that if then for any .
Below, I will give a solution to the problem. I feel a little uneasy about it, it has an "orange flag", namely I make no use of the fact that the coefficients if are the given powers of .
Solution
Do the proof for the slightly more general function
Just like the previous attempt to solve the problem, start by using the formula .
We get .
TO BE CONTINUED. I AM JUST SAVING NOW SO THAT I DON'T LOSE WORK DONE SO FAR.
See Also
1969 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |