Help with Maclaurin series of $\cos(\ln(x+1))$? Hi I've almost completed a maths question I am stuck on I just can't seem to get to the final result. The question is:
Find the Maclaurin series of $g(x) = \cos(\ln(x+1))$ up to order 3.
I have used the formulas which I won't type out as I'm not great with Mathjax yet sorry. But I have obtained:
$$ \ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} $$
$$ \cos x =   1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} $$
I'm sure what I have done above is correct however I can't seem to get to the final solution shown in the answers and would like some help please. Thanks 

Since you also don't need very high orders, the straightforward calculation of derivatives is tractable, though not preferable computationally.  If $f(x) = \cos \log(1+x)$, then $$f'(x) = -\frac{\sin \log (1+x)}{1+x}, \quad f'(0) = 0.$$  Then $$f''(x) = -\frac{\cos \log (1+x)}{(1+x)^2} + \frac{\sin \log(1+x)}{(1+x)^2}, \quad f''(0) = -1.$$  Finally, $$f'''(x) = \frac{\sin \log(1+x)}{(1+x)^3} + \frac{2\cos \log(1+x)}{(1+x)^3} + \frac{\cos \log(1+x)}{(1+x)^3} - \frac{2 \sin \log (1+x)}{(1+x)^3}, \quad f'''(0) = 3.$$  Then $$f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + O(x^4) = 1 - \frac{x^2}{2} + \frac{x^3}{2} + O(x^4).$$
It is worth noting that $$f^{(n)}(x) = \frac{A_n \cos \log (1+x) + B_n \sin \log (1+x)}{(1+x)^n},$$ for suitable constants $A_n$, $B_n$.  A proof by induction is straightforward and yields insight into the recursion relations defining $\{(A_n, B_n)\}_{n \ge 1}$:  $$\begin{align*} f^{(n+1)}(x) &= -A_n \frac{\sin \log (1+x)}{(1+x)^{n+1}} - nA_n \frac{\cos \log (1+x)}{(1+x)^{n+1}} - n B_n \frac{\sin \log(1+x)}{(1+x)^{n+1}} + B_n \frac{\cos \log (1+x)}{(1+x)^{n+1}} \\
&= \frac{(B_n - nA_n) \cos \log(1+x) + (-A_n - nB_n)\sin \log(1+x)}{(1+x)^{n+1}}.
\end{align*}.$$  Therefore, $$\begin{align*} A_{n+1} & = -nA_n + B_n, \\ B_{n+1} &= -A_n - nB_n, \\ A_0 &= 1, \\ B_0 &= 0. \end{align*}$$  In matrix form, this recurrence is equivalent to $$\begin{bmatrix}A_{n+1} \\ B_{n+1}\end{bmatrix} = \begin{bmatrix} -n & 1 \\ -1 & -n \end{bmatrix} \begin{bmatrix} A_n \\ B_n \end{bmatrix},$$ consequently $$\begin{bmatrix}A_n \\ B_n\end{bmatrix} = M_n \begin{bmatrix} 1 \\ 0 \end{bmatrix},$$ where $$M_n = \prod_{k=0}^{n-1} \begin{bmatrix} -k & 1 \\ -1 & -k \end{bmatrix}$$ and $f^{(n)}(0) = A_n$.  This lets us continue our calculation of higher orders with relative ease if we keep track of the matrix product $M_n$, which is always of the form $$M_n = \begin{bmatrix} a & b \\ -b & a \end{bmatrix};$$ for example, $$M_3 = \begin{bmatrix} 3 & 1 \\ -1 & 3\end{bmatrix}, \quad M_4 = \begin{bmatrix} -3 & 1 \\ -1 & -3\end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 3\end{bmatrix} = \begin{bmatrix} -10 & 0 \\ 0 & -10 \end{bmatrix},$$ so that $A_4 = -10$ and the next coefficient is $-10/4! = -5/12$.
