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I've seen a proof including this claim:
$$x-\frac{x^3}{6} \le \sin x$$

Now, for my understanding, the series $x-\frac{x^3}{6} +\frac{x^5}{120} -... $ is converging to $\sin x$ in an alternating way. meaning, with each term, the sum of the terms is getting closer to $\sin x$, one time from the bottom and one time from the top.

  • I'd be glad if you could expand about this behavior.
  • Is it true only for $\sin x$ and $\cos x$?

Update:
I want to make my question more clear (but maybe a more general one):

Suppose you're given the first $n$ terms of the expansion of $\cos x$, $\sin x$ (maybe $e^x$ too).
How can you infer if the summation of those terms is bigger or smaller than $f(x)?$
(where $f(x)$ can be $\sin x$, $\cos x$ etc..)

For example, is there a pattern I can rely on? (i.e. alternation)

Thanks.

AnnieOK
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  • it's confusing...do you need the proof of the inequality of alternating behavior of Maclaurin series of trig functions? – Alex Jun 02 '14 at 19:42
  • http://en.wikipedia.org/wiki/Taylor_series

    The first image explains graphically the approach of the taylor expansion of $\sin(x)$ to the function $\sin(x)$. The rest is surely useful and will help you understand what happens for $\cos(x)$

     – Bman72 Jun 02 '14 at 19:44
  • I guess proof would help @Alex. Moreover, I'd like to know which Taylor expansions have this behavior. – AnnieOK Jun 02 '14 at 19:45
  • I take it you are assuming that $x\ge 0$. The "Alternating Series" argument is not right, since the series is technically not an alternating series for all positive $x$, but the inequality does hold. – André Nicolas Jun 02 '14 at 19:50
  • I updated my question, I hope it's more clear now. – AnnieOK Jun 02 '14 at 20:01
  • @AndréNicolas You are using a definition of alternating series that seems to be popular with some calculus textboook authors, but I don't think it is universally accepted. By the definition in the wikipedia article on alternating series, for example, the sine series is certainly alternating for all positive $x$ (but for too large $x$, that does not help with the given inequality, of course). – Harald Hanche-Olsen Jun 02 '14 at 20:12
  • Yes, I am using the definition of standard (North American) calculus texts. In any case for big $x$ we need something like the Mean Value Theorem (for integrals or not), or some remainder estimates based on MVT. – André Nicolas Jun 02 '14 at 20:22

1 Answers1

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You can prove these inequalities by using increasing functions. Consider the inequality,

$$1-\frac{x^2}{2} < \cos x$$ we take the function $f(x)=\cos x -1 +\frac{x^2}{2}$ then $f^{\prime}(x)= -\sin x + x $. Now $0 < f^{\prime}(x)$ because $\sin x < x$ for $0<x$. this means that $f$ is increasing, thus

$$f(0) < f(x)$$ for $0<x$ therefore

$$0 <\cos x -1 +\frac{x^2}{2}$$ Now try the same thing for $$g(x)=\sin x -x +\frac{x^3}{6}$$

You can continue the pattern at the next stage you get

$$\cos x < 1-\frac{x^2}{2}+\frac{x^4}{4!}$$ and

$$\sin x < x-\frac{x^3}{3!}+\frac{x^5}{5!}$$

So the stages alternate between lower and upper bounds.

Rene Schipperus
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