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An exercice asks me to find the Taylor Polynomial of the function $$f(x)=x^2\ln(1-x^3)$$ on $a=0$.
I understand the Taylor Polynomial as an aproximation of a function in a point in some nth degree which as it gets bigger, the error of the polynomial decreases.
What I would do here is find the first derivatives: $$f'(x)=2x\ln(1-x^3)-2x^4\frac{1}{1-x^3}$$ $$f''(x)=2\left(\ln \left(1-x^3\right)-\frac{3x^3}{1-x^3}\right)-\frac{3\left(4x^3-x^6\right)}{\left(1-x^3\right)^2}$$ And then aplicate the formula of the Taylor Polynomial: $$T(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2 \dots$$ With that i would have the Polynomial of second order, but the exercice doesn't provide any order, so I understand it should be without error. I find that the second derivative is already very tedious, getting the following derivatives would get harder and what is most probable is that i make a mistake at some point.

So this is where i get lost. Maybe I should find some way the nth derivative? How can I do that?
Should i look for a pattern that repeats or see if at some point the derivative is 0?

2 Answers 2

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You have that $log(1-t)=-\sum_1^\infty \frac{t^k}{k}$ so that $t\log(1-t)=-\sum_1^\infty \frac{t^{k+1}}{k}$. Setting $t=x^3$ you get that $x^3\log(1-x^3)=-\sum_1^\infty \frac{x^{3k+3}}{k}$ and now you just need to divide by $x$.

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    Thank you, I didn't thought about it as a power series, now it's quite more simple. Anyway, to use that, I should remember the serie for $\log(1-t)$ Do you think there's another way to do that if I didn't know the particular serie that applies to some other exercice? – 2017-01-23
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    @jogan the easiest (for me at least) to remember this expansion is to note that $(log(1-t))'=-\frac{1}{1-t}=-\sum t^k$ and then integral it back to get the expansion of $log(1-t)$. – 2017-01-23
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Assuming that your $a=0$ is $x=0$, we have $$ x^2\ln(1-x^3)=x^2\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-x^3)^n}{n} $$ and you can cut, where it is needed, to obtain approximation by a Taylor polynomial.