Lagrange Interpolation B305F2

1. **State the problem:** We want to find the Lagrange interpolation polynomial for the points $(0,3)$, $(1,3.19)$, $(2,3.16)$, and $(2.57,2.95)$. This polynomial will pass exactly through these points. 2. **Formula for Lagrange interpolation:** The polynomial is given by $$P(x) = \sum_{i=0}^n y_i L_i(x)$$ where each basis polynomial is $$L_i(x) = \prod_{j=0,j\neq i}^n \frac{x - x_j}{x_i - x_j}$$ 3. **Calculate each $L_i(x)$:** - For $L_0(x)$ with $x_0=0$: $$L_0(x) = \frac{x-1}{0-1} \cdot \frac{x-2}{0-2} \cdot \frac{x-2.57}{0-2.57} = \frac{x-1}{-1} \cdot \frac{x-2}{-2} \cdot \frac{x-2.57}{-2.57}$$ - For $L_1(x)$ with $x_1=1$: $$L_1(x) = \frac{x-0}{1-0} \cdot \frac{x-2}{1-2} \cdot \frac{x-2.57}{1-2.57} = x \cdot \frac{x-2}{-1} \cdot \frac{x-2.57}{-1.57}$$ - For $L_2(x)$ with $x_2=2$: $$L_2(x) = \frac{x-0}{2-0} \cdot \frac{x-1}{2-1} \cdot \frac{x-2.57}{2-2.57} = \frac{x}{2} \cdot (x-1) \cdot \frac{x-2.57}{-0.57}$$ - For $L_3(x)$ with $x_3=2.57$: $$L_3(x) = \frac{x-0}{2.57-0} \cdot \frac{x-1}{2.57-1} \cdot \frac{x-2}{2.57-2} = \frac{x}{2.57} \cdot \frac{x-1}{1.57} \cdot \frac{x-2}{0.57}$$ 4. **Substitute $y_i$ values:** $$P(x) = 3L_0(x) + 3.19L_1(x) + 3.16L_2(x) + 2.95L_3(x)$$ 5. **Simplify each term:** - Simplify $L_0(x)$: $$L_0(x) = \frac{(x-1)(x-2)(x-2.57)}{-1 \cdot -2 \cdot -2.57} = \frac{(x-1)(x-2)(x-2.57)}{-5.14}$$ - Simplify $L_1(x)$: $$L_1(x) = x \cdot \frac{x-2}{-1} \cdot \frac{x-2.57}{-1.57} = x \cdot (2 - x) \cdot \frac{2.57 - x}{1.57}$$ - Simplify $L_2(x)$: $$L_2(x) = \frac{x}{2} \cdot (x-1) \cdot \frac{x-2.57}{-0.57} = -\frac{x}{2} (x-1) \cdot \frac{x-2.57}{0.57}$$ - Simplify $L_3(x)$: $$L_3(x) = \frac{x}{2.57} \cdot \frac{x-1}{1.57} \cdot \frac{x-2}{0.57}$$ 6. **Final polynomial:** $$P(x) = 3 \cdot \frac{(x-1)(x-2)(x-2.57)}{-5.14} + 3.19 \cdot x (2 - x) \cdot \frac{2.57 - x}{1.57} - 3.16 \cdot \frac{x}{2} (x-1) \cdot \frac{x-2.57}{0.57} + 2.95 \cdot \frac{x}{2.57} \cdot \frac{x-1}{1.57} \cdot \frac{x-2}{0.57}$$ This polynomial can be expanded further if needed, but this form clearly shows the Lagrange interpolation polynomial passing through the four points.