1.md

Chapter 7: Exercise 1

a

For $x \le \xi$, $f_1(x)$ has coefficients $a_1 = \beta_0, b_1 = \beta_1, c_1 = \beta_2, d_1 = \beta_3$.

b

For $x \gt \xi$, $f(x)$ has the form of: $$ \beta_0 + \beta_1 x + \beta_2 x^2 + \beta_3 x^3 + \beta_4 (x - \xi)^3 \ = \beta_0 + \beta_1 x + \beta_2 x^2 + \beta_3 x^3 + \beta_4 (x^3 - 3 x^2 \xi + 3 x \xi^2 - \xi^3) \ = (\beta_0 - \beta_4 \xi^3) + (\beta_1 + 3 \beta_4 \xi^2) x + (\beta_2 - 3 \beta_4 \xi) x^2 + (\beta_3 + \beta_4) x^3 $$

Thus, $a_2 = \beta_0 - \beta_4 \xi^3, b_2 = \beta_1 + 3 \beta_4 \xi^2, c_2 = \beta_2 - 3 \beta_4 \xi, d_2 = \beta_3 + \beta_4$.

c

$$ f_1(\xi) = \beta_0 + \beta_1 \xi + \beta_2 \xi^2 + \beta_3 \xi^3 \\ f_2(\xi) = (\beta_0 - \beta_4 \xi^3) + (\beta_1 + 3 \beta_4 \xi^2) \xi + (\beta_2 - 3 \beta_4 \xi) \xi^2 + (\beta_3 + \beta_4) \xi^3 \\ = \beta_0 - \beta_4 \xi^3 + \beta_1 \xi + 3 \beta_4 \xi^3 + \beta_2 \xi^2 - 3 \beta_4 \xi^3 + \beta_3 \xi^3 + \beta_4 \xi^3 \\ = \beta_0 + \beta_1 \xi + \beta_2 \xi^2 + 3 \beta_4 \xi^3 - 3 \beta_4 \xi^3 + \beta_3 \xi^3 + \beta_4 \xi^3 - \beta_4 \xi^3 \\ = \beta_0 + \beta_1 \xi + \beta_2 \xi^2 + \beta_3 \xi^3 $$

d

$$ f'(x) = b_1 + 2 c_1 x + 3 d_1 x^2 \\ f_1'(\xi) = \beta_1 + 2 \beta_2 \xi + 3 \beta_3 \xi^2 \\ f_2'(\xi) = \beta_1 + 3 \beta_4 \xi^2 + 2 (\beta_2 - 3 \beta_4 \xi) \xi + 3 (\beta_3 + \beta_4) \xi^2 \\ = \beta_1 + 3 \beta_4 \xi^2 + 2 \beta_2 \xi - 6 \beta_4 \xi^2 + 3 \beta_3 \xi^2 + 3 \beta_4 \xi^2 \\ = \beta_1 + 2 \beta_2 \xi + 3 \beta_3 \xi^2 + 3 \beta_4 \xi^2 + 3 \beta_4 \xi^2 - 6 \beta_4 \xi^2 \\ = \beta_1 + 2 \beta_2 \xi + 3 \beta_3 \xi^2 $$

e

$$ f''(x) = 2 c_1 + 6 d_1 x \\ f_1''(\xi) = 2 \beta_2 + 6 \beta_3 \xi \\ f_2''(\xi) = 2 (\beta_2 - 3 \beta_4 \xi) + 6 (\beta_3 + \beta_4) \xi \\ = 2 \beta_2 + 6 \beta_3 \xi $$