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Math Help - Taylor/Maclaurin proof

  1. #1
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    Taylor/Maclaurin proof

    Hi

    I would like to prove the following inequalities,

    1) f(x) = \frac{7-3\cos x-6\sin x}{9-3\cos x-8\sin x}, show that \frac{1}{2}\leq f(x)\leq 1

    2) Given that x > 0, prove that x> \sin x > x - \frac{1}{6}x^3

    It's in part of a book about power series expansion of functions, so I'm pretty sure they're looking for a solution involving those. I've put in the expansions for sin and cos but I don't see the logic in how to prove these statements for all x.

    A nudge in the right direction would be much appreciated, thanks

    Stonehambey
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  2. #2
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     \sin x = x + O(x^{3})
     \sin x = x -\frac{x^{3}}{3!} + O(x^{5})
     \sin x = x -\frac{x^{3}}{3!} + \frac{x^{5}}{5!} + O(x^{7})

    So if x>0, then x overestimates the value of \sin x and  x- \frac{x^{3}}{3!} underestimates the value of \sin x

    But that's not much of a proof.
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  3. #3
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    Yeah I was thinking along the same lines, it was proving that the remaining terms do not affect it which was the bit where I was getting stuck, do I have to prove \lim_{x \to +\infty}\frac{x^n}{n!}=0 anywhere?
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