How can i solve this Lagrange Polynomial problem?

Question

It's probably one of the easiest Lagrange Polynomial problem but I don't know much about this topic. So here is the question.

Consider the graph of function $f(x)=\sin(⁡x)$ over interval $[0, \frac{\pi}{2}]$.

Use two points $(0,0)$ and $(\frac{\pi}{2}, 1)$ to construct a Lagrange linear interpolation polynomial $P_1(x)$. Use this polynomial to approximate $\sin⁡(\frac{2\pi}{9})$ and find the upper bound for the error in this approximation.

Answer 1

In this case you can apply the liner interpolation, but using the Lagrange's formula you must see the definition.

Considering the points $\{(x_0,y_0), \dots, (x_N, y_N)\}$ you can define the interpolation polynomial in the Lagrange form as: $$ p_N(x) = \sum_{k=0}^N y_kL_k(x) $$ where the Lagrange polynomial are: $$ L_k(x) = \prod_{j=0, j \neq k}^N \frac{x-x_j}{x_k-x_j} $$

In your case $N=1$ and

$$ L_0(x) = \frac{x -x_1}{x_1-x_0} = \frac{x-\frac{\pi}{2}}{ 0 - \frac{\pi}{2}} = \frac{\pi - 2x}{\pi} $$ $$ L_1(x) = \frac{x - x_0}{x_1 - x_0} = \frac{x - 0}{\frac{\pi}{2} - 0} = \frac{2x}{\pi} $$

Note that, according with the theory $L_k(x_j) = \delta_{kj}$, $L_0(x_0) = 1 \; L_0(x_1) = 0$ same for $L_1(x)$. With these polynomials you obtain, $$ p_1(x) = y_0L_0(x) + y_1L_1(x) = 0 + 1 \cdot \frac{2x}{\pi} = \frac{2x}{\pi} $$

This imply that the $$ f(\frac{2\pi}{9}) = \sin(\frac{2\pi}{9}) \approx 0.64278760968653925 $$ $$ p_1(\frac{2\pi}{9}) = \frac{2}{\pi} \cdot \frac{2\pi}{9} = \frac{4}{9} = 0.\bar{4} $$

Now always from the theory you know that the error $E_N$ can be express as:

$$ E_n(x) = \frac{f^{(n+1)}(\zeta_x)}{(n+1)!} \omega_n(x) \quad \omega_N(x) = \prod_{k=0}^n (x - x_k) $$

in your case $$ E_1(x) = \frac{f^{(2)}(\zeta_x)}{(2)!}(x-x_0)(x-x_1) = \frac{f^{(2)}(\zeta_x)}{(2)!} \cdot x(x-\frac{\pi}{2}) $$

Now note that:

• $f(x) \, , x(x-\frac{\pi}{2})$ are analytic function, so they are continuous, and also their derivate.
• $f(x) \,, f^{(2)}(x)$ are bounded function, $|f^{(2)}(x)| \leq 1$
• you are workin on $\Omega = [0, \frac{\pi}{2}]$ it is compact and for Weierstrass's theoreme continuous functions have max and min

for the above reason you can estimate the error as: $$ |E_1(x)| \leq \frac{C_1C_2}{2!} \quad C_1 = \max_{\zeta \in \Omega} |f^{(2)}(\zeta)| \quad C_2 = \max_{\zeta \in \Omega} |\zeta(\zeta-\frac{\pi}{2})| \quad $$

you can assume $C_1 \leq 1$ and $C_2 \leq \frac{\pi^2}{16}$. This comes from a simple study of function, critical point in $\frac{\pi}{4}$, it is a max for $|\zeta(\zeta-\frac{\pi}{2})|$ so

$$ |E_1(x)| \leq 1 \cdot \frac{\pi^2}{16} \cdot \frac{1}{2!} = \frac{\pi^2}{32} \approx 0.30842513753404244 $$

Note that the case of $\frac{2\pi}{9}$ is according with this estimate.