Problems from Lagrange's method

Development:

We have $n+1=3$ points, therefore $n=2$. The Lagrange polynomial is written as:

$\begin{array}{rl}{P}_2(x)=& f_0\cdot l_0(x)+f_1\cdot l_1(x)+f_2\cdot l_2(x)\\ =& 0\cdot l_0(x)+{\displaystyle \frac{1}{2}}\cdot l_1(x)+{\displaystyle \frac{1}{\cdot }}{l}_2(x)={\displaystyle \frac{1}{2}}\cdot l_1(x)+l_2(x)\end{array}$

Let's calculate $l_1(x)$ and $l_2(x)$:

$$\begin{array}{rl}{l}_1(x)=& {\displaystyle \frac{(x-x_0)\cdot (x-x_2)}{(x_1-x_0)\cdot (x_1-x_2)}}={\displaystyle \frac{(x-0)\cdot (x-{\displaystyle \frac{\pi }{2}})}{({\displaystyle \frac{\pi }{6}}-0)\cdot ({\displaystyle \frac{\pi }{6}}-{\displaystyle \frac{\pi }{2}})}}\\ =& {\displaystyle \frac{x^2-{\displaystyle \frac{\pi }{2}}x}{{\displaystyle \frac{\pi }{6}}\cdot ({\displaystyle \frac{-\pi }{3}})}}=-{\displaystyle \frac{18}{{\pi }^2}}{x}^2+{\displaystyle \frac{9}{\pi }}x\\ l_2(x)=& {\displaystyle \frac{(x-x_0)\cdot (x-x_1)}{(x_2-x_0)\cdot (x_2-x_1)}}={\displaystyle \frac{(x-0)\cdot (x-{\displaystyle \frac{\pi }{6}})}{({\displaystyle \frac{\pi }{2}}-0)\cdot ({\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{\pi }{6}})}}\\ =& {\displaystyle \frac{x^2-{\displaystyle \frac{\pi }{6}}x}{{\displaystyle \frac{\pi }{2}}\cdot ({\displaystyle \frac{\pi }{3}})}}={\displaystyle \frac{6}{{\pi }^2}}{x}^2-{\displaystyle \frac{1}{\pi }}x\end{array}$$

Replacing the obtained values:

$$\begin{array}{rl}{P}_2(x)=& {\displaystyle \frac{1}{2}}\cdot l_1(x)+l_2(x)={\displaystyle \frac{1}{2}}\cdot (-{\displaystyle \frac{18}{{\pi }^2}}{x}^2+{\displaystyle \frac{9}{\pi }}x)+{\displaystyle \frac{6}{{\pi }^2}}{x}^2-{\displaystyle \frac{1}{\pi }}x\\ =& -{\displaystyle \frac{9}{{\pi }^2}}{x}^2+{\displaystyle \frac{9}{2\pi }}x+{\displaystyle \frac{6}{{\pi }^2}}{x}^2-{\displaystyle \frac{1}{\pi }}x=-{\displaystyle \frac{3}{{\pi }^2}}{x}^2+{\displaystyle \frac{7}{2\pi }}x\end{array}$$

Thus:

$$\sin ({\displaystyle \frac{\pi }{3}})=f({\displaystyle \frac{\pi }{3}})\approx P_2({\displaystyle \frac{\pi }{3}})={\displaystyle \frac{5}{6}}=0.8333\dots$$

To give a bound for the error, it is necessary to know $f^{(n+1)}(x)=f^{(3)}(x)$. In our case,

$$f^{(3)}(x)=-\cos (x)\Rightarrow |f^{(3)}(x)|=|\cos (x)|⩽1$$

Therefore: $$|f({\displaystyle \frac{\pi }{3}})-P_2({\displaystyle \frac{\pi }{3}})|⩽{\displaystyle \frac{1}{6}}\cdot {\displaystyle \frac{\pi }{3}}\cdot {\displaystyle \frac{\pi }{6}}\cdot {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{\pi }{648}}=0.047⩽0.5\cdot {10}^{-1}$$

Solution:

$\sin ({\displaystyle \frac{\pi }{3}})=0.8333$ and $|\text{error}|⩽0.05$

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