Science:Math Exam Resources/Courses/MATH103/April 2015/Question 01 (e) (ii)/Solution 1

We know that the Taylor series of $\sin x$ is given by

$\sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+...+{\frac {(-1)^{n-1}x^{2n-1}}{(2n-1)!}}+...=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{(2n-1)!}}x^{2n-1}.$

If you do not memorize the $\sin x$ series, you can check by eliminating the other options.

We know that $\sin 0=0$, the only series with 0 value at 0 is the second option. Among the given functions $\ln(x+1)$ is also 0 at $x=0$, however, the value of its second derivative at $x=0$ is non-zero, so its Taylor series must have a second degree term, so the only option remains is $\sin x$ .