# Mathematical Food for Thought

Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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Exponents Away. Topic: Algebra/Calculus. Level: AIME/Olympiad. November 30th, 2006

Problem: (Problem-Solving Through Problems – 6.5.12) Suppose is a nonnegative integer and

,

where and are real numbers. Prove that if has more than zeros in , then .

Solution: First, we add the claim that can occur for at most distinct .

We proceed by induction. For , we have which we know is not zero unless , so if it has a root it must be identically . It also takes the value at most once because it is monotonic. Now suppose both hypotheses are true for some nonnegative integer . We want to show that

has at most zeros (except when for all ) and takes the value at most times. If , we apply the inductive hypothesis and it is trivially true, so assume . From , we get

.

Letting and for , we obtain

.

By our inductive hypothesis, we know this occurs at most times, proving that has at most zeros or is identically .

Now we want to show that at most times. Suppose for for . Then, by Rolle’s Theorem, we have

between each consecutive pair of ‘s, giving us pairs. But by what we just proved for having at most zeros, we know at most times, since is in the same family of functions. Hence for at most distinct , completing the induction.

Therefore, we conclude that has at most zeros or is identically .

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Comment: A pretty tough problem, took a couple of looks at smaller cases to determine the method with which to solve it. The key is looking for the inductive step, somehow reducing the case to the case. Adding in the extra condition might not have been necessary but it made it easier to move from case to case. Here’s a “flow chart” of what proves what.

for at most values or is identically –> for at most values –> for at most values or is identically –> for at most values.

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Practice Problem: Show that cannot have more than one zero in , regardless of the value of .