Question

determine a unique polynomial f(x) of degree<=3 such that
f(x_0)=1, f'(x_0)=2, f(x_1)=2, f'(x_1)=3 where x_1 - x_0 = h

Answer 1

Answer on Question #66677 – Math – Algorithms | Quantitative Methods

Question

determine a unique polynomial $f(x)$ of degree $<= 3$ such that $f(x_0) = 1$, $f'(x_0) = 2$, $f(x_1) = 2$, $f'(x_1) = 3$ where $x_1 - x_0 = h$

Solution

$$f(x) = ax^3 + bx^2 + cx + d. \quad f'(x) = 3ax^2 + 2bx + c$$ $$\left\{ \begin{array}{l} f(x_0) = 1 \\ f(x_0 + h) = 2 \\ f'(x_0) = 2 \\ f'(x_0 + h) = 3 \end{array} \right. \Rightarrow \left\{ \begin{array}{c} ax_0^3 + bx_0^2 + cx_0 + d = 1 \\ a(x_0 + h)^3 + b(x_0 + h)^2 + c(x_0 + h) + d = 2 \\ 3ax_0^2 + 2bx_0 + c = 2 \\ 3a(x_0 + h)^2 + 2b(x_0 + h) + c = 3 \end{array} \right.$$

We can solve this system using Cramer’s rule and obtain the coefficients $a, b, c$ and $d$.

In particular case $(x_0 = 0)$:

$$\begin{array}{l} \left\{ \begin{array}{l} d = 1 \\ ah^3 + bh^2 + ch + d = 2 \\ c = 2 \\ 3ah^2 + 2bh + c = 3 \end{array} \right. \Rightarrow d = 1, \quad c = 2 \rightarrow \left\{ \begin{array}{l} ah^3 + bh^2 = 1 - 2h \\ 3ah^2 + 2bh = 1 \end{array} \right. \Rightarrow \\ a = \frac{5h - 2}{h^3}, \quad b = \frac{3 - 7h}{h^2}. \end{array}$$

So $f(x) = \frac{5h - 2}{h^3} x^3 + \frac{3 - 7h}{h^2} x^2 + 2x + 1$.

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