Question

How do you find the nth Taylor polynomials for \[f(x) = \sin x\] centred about \[a = 0\] ?

Answer 1

Hint: Here in this question we have to find the nth polynomials of the trigonometric function. By using the formula of Taylor's series \[f(x) = f(a) + f'(a)(x - a) + \dfrac{{f''(a)}}{{2!}}{(x - a)^2} + \dfrac{{f'''(a)}}{{3!}}{(x - a)^3} + ...\] Substituting the value of a to the formula hence we can determine the solution for the given question.

Complete step by step solution:
We have to determine the nth Taylor’s polynomial by using the Taylor’s series expansion.Here the function is \[f(x) = \sin x\]. The formula for the Taylor’s series expansion is given by \[f(x) = f(a) + f'(a)(x - a) + \dfrac{{f''(a)}}{{2!}}{(x - a)^2} + \dfrac{{f'''(a)}}{{3!}}{(x - a)^3} + ...\] the value of a is 0. The general form of Taylor’s series expansion is written as \[f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(a)}}{{n!}}} {(x - a)^n}\]

Now consider the function \[f(x) = \sin x\], the value of f at a=0 is \[f(0) = 0\].Let we find the derivatives of the function and determine the value at a = 0. The first derivative of the function is \[f'(x) = \cos x\]. The value of the first derivative at a = 0 is \[f'(0) = \cos (0) = 1\].

The second derivative of the function is \[f''(x) = - \sin x\], the value of second derivative at a = 0 is \[f''(0) = - \sin (0) = 0\]
The third derivative of the function is \[f'''(x) = - \cos x\], the value of third derivative at a = 0 is \[f'''(0) = - \cos (0) = - 1\]
The fourth derivative of the function is \[f''''(x) = \sin x\], the value of third derivative at a = 0 is \[f''''(0) = \sin (0) = 0\]
The fifth derivative of the function is \[{f^v}(x) = \cos x\], the value of third derivative at a = 0 is \[{f^v}(0) = \cos (0) = 1\]
Therefore the function is written as,
\[\sin x = 0 + 1(x - 0) + \dfrac{0}{{2!}}{(x - 0)^2} + \dfrac{{ - 1}}{{3!}}{(x - 0)^3} + \dfrac{0}{{4!}}{(x - 0)^4} + \dfrac{1}{{5!}}{(x - 0)^5}\]
On simplifying we get
\[\sin x = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} \pm ...\]
In generally it is written as
\[\therefore\sin x = \sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^n}}}{{(2n + 1)!}}{x^{2n + 1}}} \]
Hence we have determined the nth Taylor polynomials for the given function.

Note: In Taylor’s series expansion we have the derivatives, so we have to apply the differentiation to the function. If the function is a trigonometric function then we use the table of trigonometry ratios for the standard angles. While solving the above function we use simple arithmetic operations.