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- Interpolation of Hermite
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Interpolation of Hermite
We know that: $f (0) = 3$, $f'(0) = 1$, $f(1) = 2$ and $f'(1)=-2$. Calculate the polynomial of Hermite that interpolates these points.
In this case we have $n+1=2$ points, therefore the degree of the polynomial of Hermite will be $2n+1 = 3$. We proceed as we explained, we write in a table the points repeating those in which we know the derivative:
| $0$ | $3$ | |||
| ${f'}_0=1$ | ||||
| $0$ | $3$ | $\dfrac{-1-1}{1-0}=-2$ | ||
| $\dfrac{2-3}{1}=-1$ | $\dfrac{-1+2}{1-0}=1$ | |||
| $1$ | $2$ | $\dfrac{-2+1}{1-0}=-1$ | ||
| ${f'}_1=-2$ | ||||
| $1$ | $2$ |
Then, the polynomial is written in the same way, taking the first element of every column (starting from the second).
$$\begin{array}{rl} P_3(x)=& 3+1(x-0)-2(x-0)^2+1(x-0)^2(x-1)\\ =& 3+x-2x^2+x^3-x^2= x^3-3x^2+x+3 \end{array}$$
$$P_3(x)= x^3-3x^2+x+3 $$