Evaluating $\ln(\cos x))$ using Taylor expansion

Question

Evaluate $\ln(\cos x)$ at $x_0=0$ and with the order of $n=4$.

Noticing that $\ln(\cos x) = \ln(1+ \cos x - 1)$ we can use $\ln(1+x)$ Taylor series.

Now, I've read I should use:
$$\ln(1+x) = x - \frac{x^2}{2} + R_2(x)$$ $$\cos x -1 = -\frac{x^2}{2} + \frac{x^4}{4!} + R_4(x)$$

Questions:

1. If the demand was $n=4$, why did the author expand $\ln(1+x)$ for only two terms?
2. Following the author's way, how to plug $\cos x$ evaluation?

$$\ln(1+\cos x -1) = -\frac{x^2}{2} + \frac{x^4}{4!} + R_4(x) - \frac{\left(-\frac{x^2}{2} + \frac{x^4}{4!} + R_4(x) \right)^2}{2} + R_2(-\frac{x^2}{2} + \frac{x^4}{4!} + R_4(x))$$

Is this what I should do? Because it's getting kinda messy..

Answer 1

So far so good---the key observations remaining are:

1. In the contribution from the $u^2$ term of $\ln (1 + u)$, we can see that the only way to get a term of order $\leq 4$ in $x$ is $-\tfrac{1}{2}\left(-\frac{x^2}{2}\right)^2 = -\frac{x^4}{8}$.
2. The term $R_2\left(-\frac{x^2}{2} + \cdots\right)$ is already order $6$ in $x$, and so does not contribute to the Taylor polynomial of order $4$.

Applying these facts to simplify the second and third terms gives that the polynomial you seek is $\left(-\frac{x^2}{2} + \frac{x^4}{4!}\right) - \left(\frac{x^4}{8}\right) = -\frac{x^2}{2} - \frac{x^4}{12}$.