Using the same notation as last time we want to show that for , (where A is natrual number larger than 1) maximized at the second crossings. Why?

Definition. Let be the largest integer that .

The above claim is equivalent to claiming that is the global maximum of f.

Theorem. .

Proof. , the result follows.

Theorem. iff .

Proof. Straightforward by direct expansion.

Theorem. maximized at .

Proof. This is equivalent to .

which can be shown using modular arithmetic. (It isn't hard to check: there are only finite case to check.)

How about the case ?

Note that and let . Consider the case which is larger than zero for large enough a_k. Now we can summarize our approach to find our optimal solution:

Check the following value of a: if they are a better solution replace the current one.

1) The (usually the best answer)

2) Find smallest . For each find so that is maximized in the range .

Complexity? Something like . The problem is how often the answer falls on and how should we make the algorithm more efficient? I would leave that part open.

What about irrational case? Let's talk about them next them.

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