Question

Determine the spacing h in a table of equally spaced values for the function f(x)= (2+x)^4 , 1≤x≤2 so that the quadratic interpolation in this table satisfies | error|≤ 10^-6

Answer 1

Answer on Question #66785 – Math – Algorithms | Quantitative Methods

Question

Determine the spacing $h$ in a table of equally spaced values for the function

$$f(x) = (x + 2)^4, \quad 1 \leq x \leq 2$$

so that the quadratic interpolation in this table satisfies $|\text{error}| \leq 10^{-6}$.

Solution

$$f(x) - P_n(x) = \frac{(x - x_0)(x - x_1) \dots (x - x_n)}{(n + 1)!} f^{(n+1)}(c_x)$$

where $P_n(x)$ is the polynomial of degree less than or equal to $n$ interpolating the $f$ at the $n + 1$ data points $x_0, x_1, x_2, \ldots, x_n$ in $[a, b]$; $c_x$ is between the maximum and minimum of $x, x_0, x_1, x_2, \ldots, x_n$.

We have $n = 2$. Then:

$$f(x) - P_2(x) = \frac{(x - x_0)(x - x_1)(x - x_2)}{3!} f'''(c_x)$$ $$f'(x) = 4(x + 2)^3$$ $$f''(x) = 12(x + 2)^2$$ $$f'''(x) = 24(x + 2)$$ $$f(x) - P_2(x) = \frac{(x - x_0)(x - x_1)(x - x_2)}{6} \cdot 24(c_x + 2)$$

where $1 \leq x \leq 2$, $c_x$ is between the maximum and minimum of $x, x_0, x_1, x_2$.

If we assume that $x_0 \leq x \leq x_2$ and that $h = x_1 - x_0 = x_2 - x_1$, then $c_x$ is between $x_0$ and $x_2$, and we have:

$$|f(x) - P_2(x)| \leq \left| \frac{(x - x_0)(x - x_1)(x - x_2)}{6} \cdot 24(x_2 + 2) \right| \leq 96 \left| \frac{(x - x_0)(x - x_1)(x - x_2)}{6} \right|$$

If $h = x_1 - x_0 = x_2 - x_1$, then

$$\max_{x_0 \leq x \leq x_2} \left| \frac{(x - x_0)(x - x_1)(x - x_2)}{6} \right| = \frac{h^3}{9\sqrt{3}}.$$ $$| error | = | f (x) - P _ {2} (x) | \leq \frac {h ^ {3}}{9 \sqrt {3}} \cdot 9 6.$$

Thus,

$$\frac {h ^ {3}}{9 \sqrt {3}} \cdot 9 6 = 1 0 ^ {- 6},$$ $$h = \frac {\sqrt [ 3 ]{\frac {3 \sqrt {3}}{3 2}}}{1 0 0} \approx 0. 0 0 5 5.$$

Answer: $h = \frac{\sqrt[3]{\frac{3\sqrt{3}}{32}}}{100} \approx 0.0055$ .

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