Question

For the given functions f (x), let x0 = 1, x1 = 1.25, and x2 = 1.6. Construct interpolation polynomials of degree at most one and at most two to approximate f (1.4), and find the absolute error.

a. f (x) = sin πx

b. f (x) =cube root of (x − 1)

c. f (x) = log₁₀ (3x − 1)

d. f (x) = e^2x − x

Answer 1

The Lagrange polynomial

are $L_{1,0}(x) = \frac{x-x_1}{x_0-x_1} = \frac{x-1.25}{1-1.25}=5-4x$ and $L_{1,1}(x) = \frac{x-x_0}{x_1-x_0} = \frac{x-1}{1.25-1}=4x-4$ for the first degree. Substituting x = 1.4 get

$L_{1,0}(1.4) = -0.6$ and $L_{1,1}(1.4) = 1.6$

For second degree

$L_{2,0}(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} = \frac{(x-1.25)(x-1.6)}{(1-1.25)(1-1.6)}=\frac{20}{3}(1.25-x)(1.6-x)$ and $L_{2,0}(x) =-0.2$

$L_{2,1}(x) = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} = \frac{(x-1)(x-1.6)}{(1.25-1)(1.25-1.6)}=\frac{80}{7}(x-1)(1.6-x)$ and $L_{2,1}(1.4) \approx 0.914286$

$L_{2,2}(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} = \frac{(x-1)(x-1.25)}{(1.6-1)(1.6-1.25)}=\frac{100}{21}(x-1)(x-1.25)$ and $L_{2,2}(1.4) \approx 0.285714$

a) $f(x)=sin(\pi x)$ :

$f(x_0)=sin(\pi) = 0$ ;

$f(x_1)=sin(1.25 \pi) = \frac{-1}{\sqrt{2}}\approx -0.707107$ ;

$f(x_2) = sin(1.6 \pi) \approx -0.951057$

$f(1.4) = f(x_0) L_{1,0}(x) + f(x_1)L_{1,1}(x) = 0(-0.6) -0.707107 \cdot 1.6 = -1.13137$

Precise value is -0.951057, and error is (-0.951057) - (-1.13137) =0.180315

For second degree

$f(1.4) = f(x_0) L_{2,0}(x) + f(x_1)L_{2,1}(x) + + f(x_1)L_{2,1}(x) = 0(-0.2) -0.707107 \cdot 0.914286 -0.951057 \cdot 0.285714 = -0.918228$

Error is -0.951057-(-0.918228) = -0.0328285

b) $f(x) = \sqrt[3]{x-1}$

$f(1)=0$

$f(1.25) \approx 0.629961$

$f(1.6) \approx 0.843433$

$f(1.4) = f(x_0) L_{1,0}(x) + f(x_1)L_{1,1}(x) \approx$ 1.00794

Precise value is 0.736806, and error is -0.271131

Second degree:

$f(1.4) = f(x_0) L_{2,0}(x) + f(x_1)L_{2,1}(x) + + f(x_1)L_{2,1}(x) \approx 0.816945$

error -0.0801384

c)

$f(x) = log_{10}(3x-1)$

$f(x_0) \approx 0.30103$

$f(x_1) \approx 0.439333$

$f(x_2) \approx 0.579784$

$f(1.4) = f(x_0) L_{1,0}(x) + f(x_1)L_{1,1}(x) \approx 0.522314$

Precise value is 0.50515, and error is -0.0171643

Second degree:

$f(1.4) = f(x_0) L_{2,0}(x) + f(x_1)L_{2,1}(x) + + f(x_1)L_{2,1}(x) \approx 0.507122$

error is -0.00197208

d)

$f(x)= e^{2x}-x$

$f(x_0)\approx 6.38906$

$f(x_1)\approx 10.9325$

$f(x_2) \approx 22.9325$

$f(1.4) = f(x_0) L_{1,0}(x) + f(x_1)L_{1,1}(x) \approx 13.6586$

Precise value is 15.0446, and error is 1.38609

Second degree:

$f(1.4) = f(x_0) L_{2,0}(x) + f(x_1)L_{2,1}(x) + + f(x_1)L_{2,1}(x) \approx 15.2698$

error is -0.225117