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Statement: There exist a polynomial $P$ such that $|P(x) - \cos(x)| \leq 10^{-6}$ for all (real) $x$.

My answer: False. All polynomials of a degree $n \geq 1$ are unbounded as $x$ tends to infinity. A polynomial of degree $n = 0$ is bounded only when it is in the form $y = a$ (horizontal line) but this will not help because $\cos(x)$ varies between $[-1,1]$.

My question: Is my answer reasonable? I am particularly concerned about 'unbound\ed' polynomials as this is the term I just made up.

Please don't give me complete answer, I just want to know the flaws in my argument and get some hints to better solutions.

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Your answer is fine. The term is "unbounded." –  Qiaochu Yuan Oct 27 '10 at 11:46
    
If the polynomial is finite (is this implied when talking about polynomials?) and the inequality has to hold for all $x$ and not just some interval, your answer is fine. –  Dario Oct 27 '10 at 12:10
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@Dario: yes, it is. –  Qiaochu Yuan Oct 27 '10 at 12:40
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Your answer is largely reasonable, but it leaves a gap: how do you know that polynomials are unbounded as x tends to infinity? Can you see an easy way of proving this statement, and better yet, can you make your bounds quantitative? For instance, if I give you a polynomial (e.g., P(X) = 5000 - .0001*X^10) and some value Y (e.g., 10000) and say 'give me an X where the absolute value of this polynomial is greater than Y', how would you go about providing such an X? –  Steven Stadnicki Oct 27 '10 at 18:34
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Also, one small refinement in your argument: you say that the polynomials of degree 0 are bounded 'only when it is in the form y=a', but in fact that describes all polynomials of degree 0 - that section would probably be better phrased as something like 'a polynomial of degree 0 is constant (P(x)=a), and as cos(x) takes on the values 0 and 1 (among others), no value of a can be within 10^-6 of cos(x) everywhere.' –  Steven Stadnicki Oct 27 '10 at 22:05
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3 Answers

You could write $y(x)=a$ instead of $y=a$ to point out you are meaning a constant function, as $y(a)=a$ is not constant. Part (horizontal line) is not necessary.

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up vote 1 down vote accepted

Thanks everyone! The model answer I was given is

False. Roughly speaking, any polynomial can be made as large as you like by taking x to be very large (provided it is of degree greater than zero), whereas $| \cos x| \leq 1$ . There is obviously no polynomial of degree zero (i.e. no constant number) for which the statement holds.

This is from Advanced Problems in Core Mathematics by Siklos. I am not very worried about the details, I am more than happy to be broadly right which I am on this occasion.

Thanks again, everyone!

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It's also false because a continuous function P satisfying the inequality would have infinitely many zeroes.

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