Proof that $\left | f(x)-T_{2}(f,0)(x) \right | \leq \frac{2}{3}\left | x \right |^{3}$

Question

For all $x \in \left [ -\frac{\pi}{4},\frac{\pi}{4} \right ]$ proof that: $\left | f(x)-T_{2}(f,0)(x) \right | \leq \frac{2}{3}\left | x \right |^{3}$,

where $f(x) = \ln(\cos x)$

In the previous task (not homework), I have calculated the taylor-polynomial of $2^{nd}$ degree:

$$T_{2}(f,0)(x)=-\frac{1}{2}x^{2}$$

Everything inserted in the term, we get:

$$\left | \ln(\cos x) + \frac{1}{2}x^{2} \right | \leq \frac{2}{3}\left | x \right |^{3}$$

We now need two cases.

Case 1: The expressions between the modulus are all positive:

$$\ln(\cos x)+\frac{1}{2}x^{2}\leq \frac{2}{3}x^{3}$$

$$\ln(\cos x)\leq \frac{2}{3}x^{3}-\frac{1}{2}x^{2}$$

$$\ln(\cos x)\leq x^{2}\cdot(\frac{2}{3}x-\frac{1}{2})$$

$$\cos x\leq e^{x^{2}\cdot(\frac{2}{3}x-\frac{1}{2})}$$

Example input $x=-\frac{\pi}{4}$ $\Rightarrow$

$$0 \leq e^{-\frac{\pi}{4}^{2}\cdot(\frac{2}{3} \cdot -(\frac{\pi}{4})-\frac{1}{2})}$$

Case 2 (negative) is done analogically and its final form is:

$$\ln(\cos x)\geq x^{2}\cdot(\frac{2}{3}x-\frac{1}{2})$$

Doesn't make much sense, actually in the end I get a contradiction / mistake if we compare case 1 with case 2 : /

Answer 1

This looks like a problem about calculating an error bound for a Taylor series.

In general, the error $E_n(x)$ for the $n$th-degree Taylor polynomial $T_n(x)$ of $f(x)$ centered at $x=a$ is $$ |E_n(x)| \le \frac{M}{(n+1)!} |x-a|^{n+1},$$ where $M \ge |f^{(n+1)}(t)|$ on the interval between $0$ and $x$ (note that $x$ can be positive or negative). Actually, since we are told that we want a bound for all $x \in [-\pi/4, \pi/4]$, then we can just take $$M = \max_{-\frac{\pi}{4} \le t \le \frac{\pi}{4}} |f^{(n+1)}(t)|.$$

In our case we have $n = 2$, $a = 0$, and $f(x) = \ln(\cos x)$. Then $f'(x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$, and so $f''(x) = -\sec^2 x$. Then we have $$ T_2(x) = \frac{1}{2!} \cdot f''(0) \cdot (x-0)^2 = -\frac{1}{2}x^2, $$ as you've already found.

To find $M$ we'll need $|f'''(x)|$. $$ |f'''(x)| = |-2\sec x \cdot \sec x \tan x| = 2\sec^2 x |\tan x|$$

So we need to maximize $2\sec^2 t |\tan t|$ on the interval $[-\pi/4, \pi/4]$. First of all, note that $2\sec^2 t |\tan t|$ is an even function. Therefore we only need to consider it on $[0,\pi/4]$. Second, because $\tan t > 0$ for $0 < t < \pi/2$, we can drop the absolute value sign on $|\tan t|$. So we want $M$ such that $$M = \max_{0 \le t \le \pi/4} 2\sec^2 t \tan t.$$

If we plot $2\sec^2 t \tan t$ we see that it's strictly increasing. Therefore the max happens at $t = \pi/4$. And we have $$M = 2 \sec^2(\pi/4) \tan(\pi/4) = 2 \cdot 2 \cdot 1 = 2.$$

Put this all together in the formula for $|E_n(x)|$ given above to get the desired result.

Answer 2

By Taylor's theorem, we have $$ |f(x)-T_2(f,0)(x)|\leq\frac{1}{3!}\left|\sup f^{(3)}(\xi)\right||x|^3, $$ where the supremum is taken over all $\xi$ in the interval $\left[ -\frac{\pi}{4},\frac{\pi}{4} \right ]$. The third derivative of $f$ is $$ f^{(3)}(\xi)=-2\tan \xi \sec^2\xi. $$ This function attains its maximum at $\xi=-\frac{\pi}{4}$ with $f^{(3)}\left(-\frac{\pi}{4}\right)=4,$ so we have $$ |f(x)-T_2(f,0)(x)|\leq\frac{1}{3!}\left|\sup f^{(3)}(\xi)\right||x|^3=\frac{4}{6}|x|^3=\frac{2}{3}|x|^3 $$

Answer 3

$f(x)=\log(\cos x)+\frac{x^2}{2}$ is an analytic function in a neighbourhood of zero and an even function. Moreover, $f^{2m}(0)<0$ for any $m\geq 2$, so $$ f(x) = -c_4 x^4 - c_6 x^6 - c_8 x^8 -\ldots \tag{1} $$ leads to: $$ \left| f(x)\right| \leq x^4 (c_4+c_6+c_8+\ldots) = x^4 \cdot\left(-f(1)\right) \tag{2}$$ for any $|x|<1$, hence we have the stronger inequality:

$$\forall x\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right],\qquad \left|\log(\cos x)+\frac{x^2}{2}\right|\leq \frac{2}{17}x^4.\tag{3} $$

The crucial negativity property is either a consequence of the Weierstrass product for the $\cos$ function or a consequence of the fact that $f(x)=\tan x$ fulfills $f'(x)=1+f(x)^2$.