- Inicio
- Interpolation
- Inverse interpolation
- Ejercicios
Inverse interpolation
We know the following points of the function $f (x)$:
| $x$ | $0$ | $1$ | $2$ | $3$ |
| $f(x)$ | $0$ | $1$ | $4$ | $9$ |
Find the value of x $x$ such that $f (x) = 2$.
We have a case of inverse interpolation. Therefore we write the points in the table but exchanging the columns.
| $0$ | $0$ | |||
| $\dfrac{1-0}{1-0}=1$ | ||||
| $1$ | $1$ | $\dfrac{\dfrac{1}{3}-1}{4-0}=\dfrac{\dfrac{-2}{3}}{4}=-\dfrac{1}{6}$ | ||
| $\dfrac{2-1}{4-1}=\dfrac{1}{3}$ | $\dfrac{-\dfrac{1}{60}+\dfrac{1}{6}}{9-0}=\dfrac{\dfrac{3}{20}}{9}= \dfrac{1}{60}$ | |||
| $4$ | $2$ | $\dfrac{\dfrac{1}{5}-\dfrac{1}{3}}{9-1}=\dfrac{\dfrac{-2}{15}}{8}= -\dfrac{1}{60}$ | ||
| $\dfrac{3-2}{9-4}=\dfrac{1}{5}$ | ||||
| $9$ | $3$ |
$$\begin{array}{rl} P_3(y)=& 0+1\cdot (y-0)-\dfrac{1}{6}(y-0)(y-1)+\dfrac{1}{60}(y-0)(y-1)(y-4)\\ =&y-\dfrac{1}{6}y^2+\dfrac{1}{6}y+\dfrac{1}{60}y^3-\dfrac{1}{60}y^2-\dfrac{1}{15}y\\ =& \dfrac{1}{60}y^3-\dfrac{1}{4}y^2+\dfrac{37}{30}y \end{array}$$
We evaluate in $2$:
$$x\approx P_3(2)=\dfrac{1}{60}\cdot8-\dfrac{1}{4}\cdot4+\dfrac{37}{30}\cdot2= \dfrac{8}{5}=1.6$$
$$x=1.6$$