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

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

Explanation

Solution

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)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+...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⁡xf(x) = \sin x. The formula for the Taylor’s series expansion is given by f(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+...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)=∑n=0∞f(n)(a)n!(x−a)nf(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(a)}}{{n!}}} {(x - a)^n}

Now consider the function f(x)=sin⁡xf(x) = \sin x, the value of f at a=0 is f(0)=0f(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⁡xf'(x) = \cos x. The value of the first derivative at a = 0 is f′(0)=cos⁡(0)=1f'(0) = \cos (0) = 1.

The second derivative of the function is f′′(x)=−sin⁡xf''(x) = - \sin x, the value of second derivative at a = 0 is f′′(0)=−sin⁡(0)=0f''(0) = - \sin (0) = 0
The third derivative of the function is f′′′(x)=−cos⁡xf'''(x) = - \cos x, the value of third derivative at a = 0 is f′′′(0)=−cos⁡(0)=−1f'''(0) = - \cos (0) = - 1
The fourth derivative of the function is f′′′′(x)=sin⁡xf''''(x) = \sin x, the value of third derivative at a = 0 is f′′′′(0)=sin⁡(0)=0f''''(0) = \sin (0) = 0
The fifth derivative of the function is fv(x)=cos⁡x{f^v}(x) = \cos x, the value of third derivative at a = 0 is fv(0)=cos⁡(0)=1{f^v}(0) = \cos (0) = 1
Therefore the function is written as,
sin⁡x=0+1(x−0)+02!(x−0)2+−13!(x−0)3+04!(x−0)4+15!(x−0)5\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−x33!+x55!±...\sin x = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} \pm ...
In generally it is written as
∴sin⁡x=∑n=0∞(−1)n(2n+1)!x2n+1\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.