Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 11

The cubic Taylor polynomial about $a$ isIf $f$ has a continuous fourth derivative near $a$, Taylor's theorem gives $f(x)$ equal to this polynomial plus $O((x-a{)}^4)$. With continuous derivatives only through third order, the corresponding remainder is $o((x-a{)}^3)$. An equality with an infinite Taylor series would require additional convergence information, rather than just the existence of these four terms.

Factor out $9$ and use the binomial series with exponent $-1/2$:Thus the expansion through cubic order isThe coefficients of $x$ and $x^3$ vanish because the function is an even function. The underlying binomial expansion converges for $|x|<3$.

Near zero the logarithm's argument is positive, so $\ln [(2+x{)}^3]=3\ln (2+x)=3\ln 2+3\ln (1+x/2)$. Use the Maclaurin series of the natural logarithm:The local calculation uses $x>-2$ for the real logarithm; the power series converges for $|x|<2$.

Use $\sin x=x-x^3/6+O(x^5)$ inside the exponential series:The two cubic contributions cancel, leavingThus the coefficient of $x^3$ is zero, even though the function itself is not an even function.