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Let

$$f:(-\frac{\pi}{2},\frac{\pi}{2}) \to \Bbb R$$ $$f(x)=-\ln(\cos (x))$$ Show that $$\left| f(x)-\frac{x^2}2 \right| \le \frac23\left| x^3 \right| \qquad x \in[-\frac{\pi}{4},\frac{\pi}{4}]$$

My attempt:

We have $\left| -\ln(\cos (x))-\frac{x^2}2 \right| \le \frac23\left| x^3 \right| \iff |\ln(\cos (x))+\frac{x^2}2\left| \le \frac23 \right|x^3| $

The first thing that comes to my mind is to show that $\cos(\alpha)\ge \frac{1}{\sqrt2}$ for $\alpha \in[-\frac{\pi}{4},\frac{\pi}{4}]$ (i)

After that, I would have to go on to show that $0 \ge \ln(\beta)$ for $\beta \in[\frac{1}{\sqrt2},1]$ (ii) which I also don't know how to prove.

Once that is done we have $$\left| \frac{x^2}2 \right| \le \frac23\left| x^3 \right|$$ and we are done.

So my question is: how can I go on about (i) and (ii)?

Thanks in advance.

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    The [Wiki]("https://en.wikipedia.org/wiki/Taylor_series#First_example") page has an example of expanding $\log(\cos(x))$ up to $O(8)$, so you can have a look over that. – 2017-02-24
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    Hmm... can you use graphs to prove these? Or maybe to solve (1) show that the minimum value of cos(x) is 1/sqrt(2) and for (2) show that minimum value of ln(x) for the given range is less than 0? – 2017-02-24

2 Answers 2

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Applying Taylor's theorem with remainder to $$ \begin{aligned} f(x) &= -\ln(\cos(x)) & f(0) &= 0 \\ f'(x) &= \tan(x) & f'(0) &= 0\\ f''(x) &= 1 + \tan^2(x) & f''(0) &= 1 \\ f'''(x) &= 2 \tan(x) (1+\tan^2(x)) \end{aligned}$$ gives $$ f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f'''(\xi)}{3!} x^3 = \frac{x^2}{2} + \frac{f'''(\xi)}{6} x^3 $$ for some $\xi$ between $0$ and $x$. The assertion follows because for $|\xi| \le |x| \le \frac{\pi}{4}$ $$ |\tan(\xi)| \le 1 \Longrightarrow |f'''(\xi)| \le 4 $$ and therefore $$ \left\lvert f(x) - \frac{x^2}{2} \right\rvert = \frac{|f'''(\xi)|}{6} |x|^3 \le \frac 23 |x|^3 $$

Remark: Your approach cannot work because $$ \left| \frac{x^2}2 \right| \le \frac23\left| x^3 \right| $$ does not hold for $x$ close to zero.

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HINTS:

For $x\in [0,\pi/4]$, use $\tan(x)\ge x$ to show that $-\log(\cos(x))-\frac12 x^2 \ge 0$.

Let $f(x)=-\log(\cos(x))-\frac12 x^2-\frac23 x^3$.

Use $f(0)=0$ and show that $f'(x)\le 0$.

Exploit symmetry to analyze for $x\in[0,\pi/4]$.

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    Unless I am mistaken, your $f$ is *decreasing* on $[0, \pi/4]$. Or did I misunderstand something? – 2017-02-24
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    @MartinR You are not mistaken. That was a typo and thanks to your catching this, I've edited to correct the erratum. Much appreciative. -Mark – 2017-02-24