Question

Find the Taylor series of \(f(x) = \sin x\)at \({\rm{a = \pi /6}}\).

Short answer

The Taylor series is \(\frac{{\rm{1}}}{{\rm{2}}}_{{\rm{n = 0}}}^\infty {{\rm{( - 1)}}^{\rm{n}}}\frac{{\rm{1}}}{{{\rm{(2n)!}}}}\;{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^{{\rm{2n}}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{{\rm{(2n + 1)!}}}}\;{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^{{\rm{2n + 1}}}}\).

Step-by-step solution

Step 1:Calculate the derivatives,

\({\rm{f(x) = sinx,}}\;\;\;{\rm{a = \pi /6}}\)

Calculate some derivatives\({{\rm{f}}^{{\rm{(n)}}}}{\rm{(x)}}\) and \({{\rm{f}}^{{\rm{(n)}}}}{\rm{(a)}}\)

The next derivative is \({\rm{sinx}}\).so the same pattern repeats from here. Now assemble the Taylor series.

Use the Taylor series

\[\begin{array}{c}{\rm{f(x) = f(a) + }}\frac{{{{\rm{f}}^{\rm{'}}}{\rm{(a)}}}}{{{\rm{1!}}}}{\rm{(x - a) + }}\frac{{{{\rm{f}}^{\rm{}}}{\rm{(a)}}}}{{{\rm{2!}}}}{{\rm{(x - a)}}^{\rm{2}}}{\rm{ + }}\frac{{{{\rm{f}}^{{\rm{'}}}}{\rm{(a)}}}}{{{\rm{3!}}}}{{\rm{(x - a)}}^{\rm{3}}}{\rm{ + }} \cdots \cdots \\{\rm{sinx = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{x - }}\frac{{\rm{\pi }}}{{\rm{6}}}\\{\rm{ - }}\frac{{\rm{1}}}{{{\rm{2 \times 2!}}}}{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^2}{\rm{ - }}\frac{{\sqrt {\rm{3}} }}{{{\rm{2 \times 3!}}}}\;{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^3}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2 \times 4!}}}}\;{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^4}\;{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{{\rm{2 \times 5!}}}}\;{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^5}\;\;\;{\rm{ + }} \cdots \cdots \end{array}\]

Step 3:In summation form

write in summation form, and look for some patterns. Every two terms above have the same pair of coefficients, but their signs are reversed. merge the two into a single summation term. The factorial part and exponents count along with even numbers for the left term and odd numbers for the right term.

We can represent this as \({\rm{(2n) }}\) and \({\rm{(2n + 1)}}\).

\(\begin{array}{c}{\rm{sinx = }}_{{\rm{n = 0}}}^\infty {{\rm{( - 1)}}^{\rm{n}}}\frac{{\rm{1}}}{{{\rm{2(2n)!}}}}\;{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^{{\rm{2n}}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{{\rm{2(2n + 1)!}}}}\;{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^{{\rm{2n + 1}}}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}_{{\rm{n = 0}}}^\infty {{\rm{( - 1)}}^{\rm{n}}}\frac{{\rm{1}}}{{{\rm{(2n)!}}}}\;{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^{{\rm{2n}}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{{\rm{(2n + 1)!}}}}\;{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^{{\rm{2n + 1}}}}.\end{array}\)

Therefore, The Taylor series is \(\frac{{\rm{1}}}{{\rm{2}}}_{{\rm{n = 0}}}^\infty {{\rm{( - 1)}}^{\rm{n}}}\frac{{\rm{1}}}{{{\rm{(2n)!}}}}\;{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^{{\rm{2n}}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{{\rm{(2n + 1)!}}}}\;{\rm{x - }}{\frac{{\rm{\pi }}}{{\rm{6}}}^{{\rm{2n + 1}}}}\).