Science:Math Exam Resources/Courses/MATH103/April 2011/Question 07 (c)/Solution 2

The first solution assumes that the Taylor series for $\cos (x)$ is not known in advance. The Taylor series for $\sin (x)$ is known since it was given in the statement of a previous question.

If the Taylor series for $\cos (x)$ is known, then we can start the solution from the starting point

$\cos (x)=\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k\frac{x^{2k}}{(2k)!}$.

At this point, we multiply the sum by $x$ to obtain

$x\cos (x)=x\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k\frac{x^{2k}}{(2k)!}=\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k\frac{x^{2k+1}}{(2k)!}$.