Question

5. (10 points) What is the order of the following quadrature formula called Neuton's rule given by the coefficients (bo.bi,b, bs) (1/8,3/8,3/8, 1/8) and (o,,2)-

Answer 1

is the rule

Integral^1_0 f (x) dx = 1/8 (f (0) + 3f (1/3) + 3f (2/3) + f (1) is the rule For f (x) = 1 we get Integral^1_0 f (x) dx = 1, 1/8 (f (0) + 3 f (1/3) + 3f (2/3) + f (1)) = 1 For f (x) = x we get Integral^1_0 f (x) dx = 1/2, 1/8 (f (0) + 3f (1/3) + 3f (2/3) + f (1)) = 1/2 For f (x) = x^2 we get Integral^1_0 f (x) dx = 1/3, 1/8 (f (0) + 3f (1/3) + 3f (2/3) + f (1)) = 1/3 For f (x) = x^3 we get Integral^1_0 f (x) dx = 1/4, 1/8 (f (0) + 3f (1/3) + 3f (2/3) + f (1) = 1/4 but For f (x) = x^4 we get Integral^1_0 f (x) = dx = 1/5, 1/8 (f (0) + 3f (1/3) + 3f (2/3) + f (1)) = 11/54 So

For we get $\int_0^1 f(x)\,dx=1,\,\frac{1}{8}\left(f(0)+3f\left(\frac{1}{3} \right )+3f\left(\frac{2}{3} \right )+f\left(1 \right ) \right )=1$

For we get

For we get $\int_0^1 f(x)\,dx=\frac{1}{3},\,\frac{1}{8}\left(f(0)+3f\left(\frac{1}{3} \right )+3f\left(\frac{2}{3} \right )+f\left(1 \right ) \right )=\frac{1}{3}$

For we get but

For we get $\int_0^1 f(x)\,dx=\frac{1}{5},\,\frac{1}{8}\left(f(0)+3f\left(\frac{1}{3} \right )+3f\left(\frac{2}{3} \right )+f\left(1 \right ) \right )=\frac{11}{54}$

So the given approximation formula is correct for all polynomial up to degree but not higher

Hence, the order of the approximation is $h=4$

$\blacksquare$

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