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

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    Do you see how to compose two Taylor series? – 2017-01-13
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    $\sin^3(\ln(1+x))\sim\sin^3(x)\sim x^3$ – 2017-01-13

3 Answers 3

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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.

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.

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    I did the exact same process, but the answer as per the answer sheet is 6. I confirmed that the answer is indeed 6 using wolframalpha – 2017-01-13
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    And it is $6$ indeed, and the above does show so. Since $3!=6$. – 2017-01-13
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    Oh, sry. I missed the last step, and missed it again when you wrote the answer. Thanks – 2017-01-13
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    No problem -- glad this helped. – 2017-01-13
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    [I've talked about something similar here.](https://math.stackexchange.com/a/467994/263326) For infinite series, where $\frac{\sin x}{x}-1=-\frac{x^2}{6}+\frac{x^4}{120}-O(x^6)$ if $x\neq 0$ and define $\frac{\sin x}{x}$ as $1$ when $x=0$, notice that $$-\frac{1}{2}\left(-\frac{x^2}{6}+\frac{x^4}{120}-O(x^6)\right)^2\neq $$ $$\neq -\frac{1}{2}\left(-\frac{x^2}{6}\right)^2\left(\frac{-\frac{x^2}{6}+\frac{x^4}{120}-O(x^6)}{-\frac{x^2}{6}}\right)^2$$ To make this correct, you must use Cauchy product, where $$-\frac{x^2}{6}+\frac{x^4}{120}-O(x^6)$$ converges absolutely. – 2017-09-17
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    [See what Franz Mertens has proved.](https://en.wikipedia.org/wiki/Cauchy_product#Convergence_and_Mertens.27_theorem) In this case, you said $(x+o(x))^3=x^3(1+o(1))^3$. Use Cauchy product to make it correct, rigorous. Notice that $x+o(x)$ converges absolutely. – 2017-09-17
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    Err. Sorry, I don't understand your point, what you are trying to say, and how it's relevant here? – 2017-09-17
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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 $f'''(0)=6$

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    You forgot the $6$ in the denominator for the final result. – 2017-01-13
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    I did the calculations initially for the sine and not the sine to the cube – 2017-01-13
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    You mean $$x-{x^2\over 2}+{x^3\over 3}-$$ $$-{\left(x-{x^2\over 2}+{x^3\over 3}\right)^3\over 6}+o(x^3)$$ – 2017-09-17
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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 :$$\boxed{f^{(3)}(0)=6}$$