Taylor series expansion of $f(x) = \sin^3 \left(\ln(1+x) \right)$.

Question

How does one use Taylor series expansion to compute $f^{(3)}(0)$ in which $f(x) = \sin^3 \left(\ln(1+x) \right)$.

Answer 1

In short: by computing the Taylor series expansion of $f$ at $0$ up to order $o(x^3)$, and looking at the coefficient of $x^3$.

Indeed, recall that on a neighborhood of $0$, $$ f(x) = f(0)+f'(0)x+\frac{f^{(2)}(0)}{2!}+\frac{f^{(3)}(0)}{3!}x^3+o(x^3)\tag{1} $$

and to compute the Taylor series, the simplest is to start from known ones and compose them.

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Now, when $x\to 0$ $$\ln(1+x) = x+o(x)$$ (which goes to $0$ when $x\to 0$: we will use that to compose it with $\sin$) and $$\sin x = x+o(x)$$ so that $$ \sin^3 \ln(1+x) = \sin^3\left(x+o(x)\right) = \left(x+o(x)\right)^3 = x^3\left(1+o(1)\right)^3 = x^3+o(x^3) $$ from which $\frac{f^{(3)}(0)}{3!} = 1$.

Here, we only did the Taylor expansion of $\ln(1+x)$ and $\sin x$ to first order, because we observed that the eventual cube on the $\sin$ would ensure this was enough. If you don't see it, you can do a further development (to order $o(x^3)$, say, for all). This will work, but will be longer, and not necessary.

Answer 2

One starts with

$$\begin{align} \log(1+x)=&x-{x^2\over 2}+{x^3\over 3}+o(x^3)\\ \sin{X}=&X-{X^3\over 6}+o(X^3) \end{align}$$

Now make $X=\log(1+x)$ in the development of the sine function to get

$$\begin{align}\sin(\log(1+x))=&x-{x^2\over 2}+{x^3\over 3}-{\left(x-{x^2\over 2}+{x^3\over 3}\right)^3\over 6}+o(x^4)\\ =&x-{x^2\over 2}+{x^3\over 6}+o(x^3)\end{align}$$

Now every Taylor series around $0$ is of the form $f(0)+f'(0)x+f''(0)x^2/2+f'''(0)x^3/6+o(x^3)$ and so $f'''(0)=1$. This is for the sine

Now if we raise it to the cube and do the same type of calculations, noting that we can limit the series of sine and logarithms to the first order, we get

$$\sin^3(\log(1+x))=x^3+o(x)$$

And this means $f'''(0)=6$

Answer 3

Probably easier to use a linearized expression for $\sin^3(\theta)$ :

$$\sin^3(\theta)=\frac{3}{4}\sin(\theta)-\frac{1}{4}\sin(3\theta)$$

Now, for all $x>-1$ :

$$f(x)=\sin^3(\ln(1+x))=\frac{3}{4}\sin(\ln(1+x))-\frac{1}{4}\sin(3\ln(1+x))$$

and the Taylor expansion at $x=0$ is :

\begin{align} f(x)&=&\frac{3}{4}\sin(x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3))-\frac{1}{4}\sin(3x-\frac{3x^2}{2}+x^3+o(x^3))\\ &=&\frac{3}{4}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^3}{6}+o(x^3)\right)-\frac{1}{4}\left(3x-\frac{3x^2}{2}+x^3-\frac{9x^3}{2}+o(x^3)\right)\\ &=&x^3+o(x^3) \end{align}

Finally : $f^{(3)}(0)=3!\,\times\,1$, that is :$$\boxed{f^{(3)}(0)=6}$$