Question-10

Select all options that give the correct the Taylor series expansion of the function on the left around the point \(0\).

• \(\sin x=x-\cfrac{x^{3}}{3!} +\cfrac{x^{5}}{5!} -\cdots\)
• \(\displaystyle \sin x=1-\frac{x^{2}}{2!} +\frac{x^{4}}{4!} -\cdots\)
• \(\displaystyle \cos x=1-\frac{x^{2}}{2!} +\frac{x^{4}}{4!} -\cdots\)
• \(\displaystyle \cos x=x-\frac{x^{3}}{3!} +\frac{x^{5}}{5!} -\cdots\)

Answer

• \(\sin x=x-\cfrac{x^{3}}{3!} +\cfrac{x^{5}}{5!} -\cdots\)
• \(\displaystyle \sin x=1-\frac{x^{2}}{2!} +\frac{x^{4}}{4!} -\cdots\)
• \(\displaystyle \cos x=1-\frac{x^{2}}{2!} +\frac{x^{4}}{4!} -\cdots\)
• \(\displaystyle \cos x=x-\frac{x^{3}}{3!} +\frac{x^{5}}{5!} -\cdots\)

Solution

The Taylor series expansion of \(\displaystyle f( x)\) around \(\displaystyle x=a\) is given by:

\[ f( x) =\frac{f( a)}{0!} +\frac{f^{\prime }( a)}{1!}( x-a) +\frac{f^{\prime \prime }( a)}{2!}( x-a)^{2} +\cdots \]

For \(\displaystyle \sin x\) and \(\displaystyle \cos x\), this becomes:

\[ \begin{aligned} \sin( x) & =x-\frac{x^{3}}{3!} +\frac{x^{5}}{5!} -\cdots \\ & \\ \cos( x) & =1-\frac{x^{2}}{2!} +\frac{x^{4}}{4!} \end{aligned} \]