<s>[INST] quadratic approximation of $e^{x^3}$ at $x=0$ we know that
$$P(x)=f(a)+f'(a)x+\frac{f''(a)}{2}x^2 $$
therefore
$f(0)=1$
if $x^3=u$, so $$f'(e^u)=e^{x^3}3x^2$$ and $$f''(e^u) = 4(e^u)'3x^2+e^u (3x^2)'=e^{x^3}3x^23x^2+e^{x^3}6x=e^{x^3}(9x^4+6x)$$
$$P(x)=1+e^{x^3}3x^3+\frac{e^{x^3}(9x^4+6x)}{2}x^2 $$
is it correct?
can you please point out my mistake.
cause the system marked my answer as a wrong one.  
and what is the magnitude of the error?
 [/INST] We can just kick it and do: $$e^x = \sum_{n \geq 0}\frac{x^n}{n!} \implies e^{x^3} = \sum_{n \geq 0}\frac{x^{3n}}{n!},$$so that: $$e^{x^3} = 1 + x^3 + \frac{x^6}{2} + \frac{x^9}{6}+\cdots.$$We have no $x$ and $x^2$ terms.
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