Maclaurin polynomial of $\ln(\cos(x))$ I want to write down $\ln(\cos(x))$  Maclaurin polynomial of degree 6. I'm having trouble understanding what I need to do, let alone explain why it's true rigorously.
The known expansions of $\ln(1+x)$ and $\cos(x)$ gives:
$$\forall x \gt -1,\ \ln(1+x)=\sum_{n=1}^{k} (-1)^{n-1}\frac{x^n}{n} + R_{k}(x)=x-\frac{x^2}{2}+\frac{x^3}{3}+R_{3}(x)$$
$$\cos(x)=\sum_{n=0}^{k} (-1)^{n}\frac{x^{2n}}{(2n)!} + T_{2k}(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+T_{4}(x)$$
Writing $\ln(1+x)$ with $t=x-1$ gives:
$$\forall t \gt 0,\ \ln(t)=\sum_{n=1}^{k} (-1)^{n-1}\frac{(t+1)^n}{n} + R_{k}(t)$$
But now I'm clueless. 


*

*Do I just 'plug' $\cos(x)$ expansion in $\ln(t)$? Can I even do that?

*Isn't it a problem that $\ln(x)$ isn't defined for $x\leq 0$ but $|\cos(x)| \leq 1$?

 A: Since you know the polynomial of $\ln(1+t)$ and you know that $\cos(x)= 1 + (-\frac {x^2}{2!}+\frac{x^4}{4!}+x^4\epsilon(x))$ then you can "plug" it in the polynomial of $\ln(1+t)$. You now have that $\ln(\cos(x)) = P(x) + R(x)$ where P is a polynomial and R is the error such that $\lim _{x\rightarrow 0} \frac{R(x)}{x^k} = 0$ and $k=deg(P)$. 
Every function that has a Taylor polynomial actually has a unique polynomial (for a fixed degree).
If $f(x)=P_1 (x) +x^k \epsilon_1 (x) = P_2 (x)+x^k \epsilon_2 (x)$ are two representations, then taking the limit as x approaches 0 shows that $P_1 (0)=P_2 (0)=a_0$. subtract $a_0$ from both sides, divide by x, and again take the limit as x approaches 0 to see that the coefficients of x in $P_1$ and $P_2$ are the same, and continue this by induction.
This shows that the polynomial that you found is the Taylor polynomial.
In you case, when x is near zero, then $\cos(x)$ is near 1, so $\ln(\cos(x))$ is well defined.
A: Another way to do the problem is to realize that
$$
\frac{d}{dx}\ln(\cos x)=-\tan x
$$
And so the series will just be $\ln(\cos 0)=0$ plus a slightly adjusted series for $-\tan x$.  This also might help you see that the series is valid, since you are doing it out the long way.
A: The MacLaurin series of $f(x)$ is given by
$$f(x)= f(0)+f^{\prime}(0)x + \frac{ f^{\prime \prime }(0)}{2!}x^2 +
\frac{ f^{\prime \prime \prime}(0)}{3!}x^3 + \cdots $$
and so you need to calculate the values of $f(0),f^{\prime}(0),\ldots$ for your function
$f(x)=\log( \cos x).$
We have
$$\begin{align}
f(x) &= \log( \cos x ) \quad \textrm{ and so } f(0)=0, \\
f^{\prime}(x) &= - \tan x \quad \textrm{ and so } f^{\prime}(0)=0, \\
f^{\prime \prime }(x) &= - \sec^2 x \quad \textrm{ and so } f^{\prime \prime }(0)=-1.
\end{align}$$
Just carry on differentiating and evaluating the derivatives at $x=0$ until you reach the term in $x^6 .$
It's worth noting that since $\sec^2 = 1 + \tan^2 x$ we can make the evaluation easier by writing
$$f^{ \prime \prime }(x)=-1-f^{\prime}(x)^2$$
and so, using the chain rule,
$$f^{ \prime \prime \prime }(x)=-2f^{\prime}(x)f^{\prime \prime}(x),$$
and so putting $x=0$ we have $f^{ \prime \prime \prime }(0)=0.$
Differentiating again we get
$$f^{(4)}(x)=-2f^{ \prime \prime }(x)^2-2f^{\prime}(x)f^{\prime \prime \prime}(x)$$
and so $f^{(4)}(0)=-2,$ and so on...
A: I think Joe Johnson's is a good idea:
$$
\frac{d}{dx} \ln\cos x = -\tan x
$$
plus the knowledge of the  Taylor series for the tangent,
$$
\tan x = \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{(2n)!}x^{2n-1} \ , \qquad \text{for} \qquad x \in (-\pi/2 , \pi/2)
$$
(here, $B_s$ are the Bernouilli numbers) gives you everything:
$$
\ln\cos x = C - \int \tan x dx = C - \int \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{(2n)!}x^{2n-1} dx = C - \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{2n(2n)!}x^{2n}
$$
Now, taking $x = 0$ on both sides, you get $C = 0$, so
$$
\ln\cos x = - \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{2n(2n)!}x^{2n} \ , \qquad \text{for} \qquad x \in (-\pi/2 , \pi/2) \ .
$$
A: Try starting from the definition of the MacLaurin series (e.g. as defined here: http://mathworld.wolfram.com/MaclaurinSeries.html)?
