Answer:
See below
Step-by-step 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:
[tex]\bf f(x)=f(a)+f'(a)(x-a)+\displaystyle\frac{f''(x)}{2!}(x-a)^2+...+\displaystyle\frac{f^{(n)}(x)}{n!}(x-a)^n+R_{n+1}(x)[/tex]
where
[tex]\bf R_{n+1}(x)=\displaystyle\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}[/tex]
for some c such that c < | a-x |
Now, if f(x) = sin(x) and a =π/6 we have
[tex]\bf f(a)=sin(\pi/6)=1/2\\\\f'(a)=cos(\pi/6)=\sqrt{3}/2\\\\f''(a)=-sin(\pi/6)=-1/2\\\\f^{(3)}(a)=-cos(\pi/6)=-\sqrt{3}/2\\\\f^{(4)}(a)=sin(\pi/6)=1/2\\\\f^{(5)}(c)=cos(c)[/tex]
and we can approximate sin(x) with the polynomial
[tex]\bf sin(x)=\displaystyle\frac{1}{2}+\displaystyle\frac{\sqrt{3}}{2}(x-\pi/6)-\displaystyle\frac{1}{4}(x-\pi/6)^2-\displaystyle\frac{\sqrt{3}}{12}(x-\pi/6)^3+\displaystyle\frac{1}{48}(x-\pi/6)^4[/tex]
The error of the approximation when x lies in the interval 0<x< π/3 can be bounded by
[tex]\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[/tex]
