Calculate the taylor polynom for $a(x)=\ln(\cos x)$

Question

Calculate the taylor polynomial of second degree for $a(x)=\ln(\cos x)$ where $x_{0}=0$

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$$a(x)= \ln(\cos x)$$

$$a'(x)=\frac{1}{\cos x}\cdot(-\sin x)=\frac{-\sin x}{\cos x}=-\tan x$$

$$a''(x)= \frac{-\cos x \cdot \cos x -(-\sin x \cdot (-\sin x))}{\cos^{2}x}=\frac{-cos^{2}x-\sin^{2}x}{\cos^{2}x}=-1-\frac{\sin^{2}x}{\cos^{2}x}=1-\tan^{2}x$$

$$T(_{2,0}a)(x)=\frac{\ln(\cos(0))}{0!}\cdot (x-0)^{0}+\frac{-\tan(0)}{1!}\cdot (x-0)^{1}+\frac{-1-\tan^{2}(0)}{2!}\cdot (x-0)^{2}$$

$$T(_{2,0}a)(x)=\frac{0}{1}\cdot1+\frac{0}{1}\cdot x+\frac{(-1-0)}{2!}\cdot x^{2}$$

$$T(_{2,0}a)(x)=-\frac{1}{2}x^{2}$$

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I hope it's correct because it took me a lot time to learn about taylor and doing this task? There are ways of calculating it faster? This is so long and time consuming...

Answer 1

It is correct. The derivative of tangent comes up often and is worth memorizing.

$$D_\theta \tan \theta = \sec^2 \theta$$

which is equivalent to $\tan^2 \theta + 1$ by the pythagorean theorem.

edit: it is almost correct; you got the right answer in the end, but check your signs when calculating $D_x(-\tan x)$.

Answer 2

we have

$ln(cos(x))=$

$ln(1-(1-cos(x)))=$

$ln(1-2sin^2(\frac{x}{2}))=$

$ln(1-2(\frac{x}{2})^2+x^2\epsilon(x))=$

$\frac{-x^2}{2}+x^2\epsilon(x)$

with $\displaystyle{\lim_{x \to 0}\epsilon(x)=0 }$

and finally, the second order Taylor's polynom is

$T(x,2)=-\frac{x^2}{2}$.