Question

Given:f(x) = sin x , a =π/6, n=4, 0

Answer 1

See below

Explanation:

Remark:

This is the complete question

Given f(x) = sin x , a =π/6, n=4, 0<x< π/3

a. approximate f by a Taylor polynomial with degree n at the number a

b. use Taylor's inequality to estimate the accuracy of the approximation f(x) ~ Tn(x) when x lies in the given interval .

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Answer:

Recall Taylor's approximation with a polynomial for a function f if we know the value of f at a point a and there exists the derivatives involved:

`\bf f(x)=f(a)+f'(a)(x-a)+\displaystyle(f''(x))/(2!)(x-a)^2+...+\displaystyle(f^((n))(x))/(n!)(x-a)^n+R_(n+1)(x)`

where

`\bf R_(n+1)(x)=\displaystyle(f^((n+1))(c))/((n+1)!)(x-a)^(n+1)`

for some c such that c < | a-x |

Now, if f(x) = sin(x) and a =π/6 we have

`\bf f(a)=sin(\pi/6)=1/2\\\\f'(a)=cos(\pi/6)=√(3)/2\\\\f''(a)=-sin(\pi/6)=-1/2\\\\f^((3))(a)=-cos(\pi/6)=-√(3)/2\\\\f^((4))(a)=sin(\pi/6)=1/2\\\\f^((5))(c)=cos(c)`

and we can approximate sin(x) with the polynomial

`\bf sin(x)=\displaystyle(1)/(2)+\displaystyle(√(3))/(2)(x-\pi/6)-\displaystyle(1)/(4)(x-\pi/6)^2-\displaystyle(√(3))/(12)(x-\pi/6)^3+\displaystyle(1)/(48)(x-\pi/6)^4`

The error of the approximation when x lies in the interval 0<x< π/3 can be bounded by

`\bf \left|R_(5)(x)\right|\leq\left|f^((5))(c)(x-\pi/3)^5\right|\leq\left|cos(c)(x-\pi/3)^5\right|\leq\left|x-\pi/3\right|^5`