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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

$f(x)=o(x^k)$ if:

$underset{xrightarrow 0}{lim} frac{f(x)}{x^k}=0$

Since smaller powers of $x$ matter more when tending to $0$, it essentially means that something is negligible relative to $x^k$ when tending to $0$. Moreover by simple calculus one can see that if $f=o(x^m)$ and $g=o(x^ell)$, then:

$f(x)+g(x)=o(x^{min {m,ell } })$

And for $k<ell$ we have that:

$acdot x^ell = o(x^k)$

Therefore by uniqueness of the Taylor series, you want to write your function of the form:

$cos big( ln(1+x) big)= p_3(x)+o(x^3)$

Where $p_3$ is a polynomial of $x$ with degree at most $3$. If we denote by $T_m^f(x)$ the Taylor polynomial up to order $m$ at $0$. Then by your calculations:

$T_4^{ln(x+1)}(x)=x-frac{x^2}{2}+frac{x^3}{3}-frac{x^4}{4}quad$ and $quad T_6^{cos(x)}(x)=1-frac{x^2}{2}+frac{x^4}{24}-frac{x^6}{720}$

Then the function $T_6^{cos(x)}circ T_4^{ln(x+1)}(x) $ is a polynomial, and if you erase the powers negligible to $x^3$ in the final polynomial, then you obtain the Taylor polynomial at $0$ of order $3$. In particular:

$cos(ln(1+x))= 1 - Big( T_4^{ln(x+1)}(x) Big)/2 + Big( T_4^{ln(x+1)}(x) Big)^4/24+o(x^4)=$

$=1- frac{1}{2} Big( x- frac{1}{2} x^2+ +frac{x^3}{3} +o(x^3) Big)^2 +o(x^4)=$

$=1-frac{1}{2} Big( x^2- x^3 +o(x^3) Big)+ o(x^4)= 1-frac{x^2}{2}+frac{x^3}{2}+o(x^3)$

Where I neglected things that I saw which were going to be of powers strictly greater than $3$.

I think it is helpful to go by this algorithm when unsure.

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, $$f'''(x) = frac{sin log(1+x)}{(1+x)^3} + frac{2cos log(1+x)}{(1+x)^3} + frac{cos log(1+x)}{(1+x)^3} - frac{2 sin log (1+x)}{(1+x)^3}, quad f'''(0) = 3.$$ 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 $$begin{bmatrix}A_n \ B_nend{bmatrix} = M_n begin{bmatrix} 1 \ 0 end{bmatrix},$$ 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, $$M_3 = begin{bmatrix} 3 & 1 \ -1 & 3end{bmatrix}, quad M_4 = begin{bmatrix} -3 & 1 \ -1 & -3end{bmatrix}begin{bmatrix} 3 & 1 \ -1 & 3end{bmatrix} = begin{bmatrix} -10 & 0 \ 0 & -10 end{bmatrix},$$ so that $A_4 = -10$ and the next coefficient is $-10/4! = -5/12$.

Since$$cos x=1-frac{x^2}2+cdots,$$you havebegin{align}cosbigl(log(x+1)bigr)&=1-frac12left(x-frac{x^2}2+frac{x^3}3-cdotsright)^2+cdots\&=1-frac{x^2}2+frac{x^3}2+cdotsend{align}Note that this argument works because the series for $log(x+1)$ has no constant term. A consequence of this is that no matter how more terms you use from both series, you will get no new term whose degree is smaller than $4$.

A heuristic way using complex numbers that's related to one of the other answers: define
$$begin{align}
f(x)&=cosln(1+x) \
g(x)&=sin ln(1+x)text{.}
end{align}$$
Then
$$begin{align}f(0)+mathrm{i}g(0)&=1\
(f+mathrm{i}g)'&=frac{mathrm{i}(f+mathrm{i}g)}{1+x}
end{align}$$
so the left- and right-hand sides of
$$f(x)+mathrm{i}g(x)=(1+x)^{mathrm{i}}$$
coincide as formal power series in $x$. Thus
$$begin{split}cos ln(1+x)&= 1 +Re mathrm{i} x+Re tfrac{mathrm{i}(mathrm{i}-1)}{2} x^2+Re tfrac{mathrm{i}(mathrm{i}-1)(mathrm{i}-2)}{6} x^3+o(x^3)\
&=1-tfrac{x^2}{2}+tfrac{x^3}{2}+o(x^3)end{split}$$