A Periodic But Not Fixed Point

Geometry Level 5

For every integer k , k, we define a function f k f_k by the formula

f k ( x ) = 100 x − k sin ⁡ x . f_k(x)=100x-k\sin x.

What is the smallest positive integer value of k k such that, for some real α \alpha , we have f k ( f k ( α ) ) = α , f_k\big(f_k(\alpha)\big)=\alpha, but f k ( α ) ≠ α ? f_k(\alpha )\neq \alpha?


Details and Assumptions: The function is evaluated in radians. There is no degree symbol in the problem.


The answer is 102.

View wiki
Calvin Lin by Calvin Lin

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Calvin Lin Staff
Jul 5, 2015

Suppose f k ( α ) = β . f_k(\alpha )=\beta. Then f k ( f k ( α ) ) = α f_k(f_k(\alpha ))=\alpha can be rewritten as the system { 100 α − k sin ⁡ α = β 100 β − k sin ⁡ β = α \begin{cases} 100\alpha -k\sin \alpha =\beta\\ 100\beta-k\sin \beta =\alpha \end{cases}

Subtracting the equations and rearranging, we get 101 α − k sin ⁡ α = 101 β − k sin ⁡ β . 101 \alpha -k\sin \alpha =101\beta-k\sin \beta. If 1 ≤ k ≤ 101 , 1\leq k \leq 101, the function g k ( x ) = 101 x − k sin ⁡ x g_k(x)=101x-k\sin x is increasing (which is easy to check using derivative). So this implies α = β , \alpha =\beta , which contradicts the assumptions.

Now suppose k = 102. k=102. We will show that there exists α ≠ 0 , \alpha \neq 0, such that f k ( α ) = − α . f_k(\alpha )=-\alpha . This is equivalent to 101 α = 102 sin ⁡ α , 101\alpha =102\sin \alpha , which clearly has a non-zero solution. Now note that f k f_k is odd, so f k ( − α ) = α . f_k(-\alpha )=\alpha . Thus, the answer is 102. 102.

3 Helpful 0 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...