6.4 Hermite interpolation
If derivatives are known as well as values, a polynomial can match both. The Hermite polynomial \(H_{2n+1}\) of degree at most \(2n+1\) satisfies \(H(x_i) = f(x_i)\) and \(H'(x_i) = f'(x_i)\) at \(n+1\) distinct nodes.
Theorem 6.4.1. If \(f\in C^1[a,b]\) and \(x_0,\dots ,x_n\in [a,b]\) are distinct, there is a unique polynomial of degree at most \(2n+1\) agreeing with \(f\) and \(f'\) at every node. If \(f\in C^{2n+2}[a,b]\), its error is \[f(x)-H_{2n+1}(x) = \frac {f^{(2n+2)}(\xi (x))}{(2n+2)!}(x-x_0)^2\cdots (x-x_n)^2 .\]
The proof of the error formula repeats that of Theorem 6.1.3 with \(g(t) = f(t)-H(t)-(f(x)-H(x))\prod \frac {(t-x_i)^2}{(x-x_i)^2}\), which has double zeros at the nodes. The easiest way to construct \(H\) is by divided differences with each node repeated: set \(z_{2i} = z_{2i+1} = x_i\), and wherever a first divided difference would divide by zero, use the derivative instead, \[f[z_{2i},z_{2i+1}] = f'(x_i),\] which is the limit of \(f[x_i,x_i+\delta ]\) as \(\delta \to 0\) and agrees with (6.1). Then Newton’s formula on \(z_0,\dots ,z_{2n+1}\) gives \(H_{2n+1}\).
Example 6.4.2. Using \(f\) and \(f' = -J_1\) at \(1.3\), \(1.6\), \(1.9\),
| \(x\) | \(1.3\) | \(1.6\) | \(1.9\) |
| \(f(x)\) | \(0.6200860\) | \(0.4554022\) | \(0.2818186\) |
| \(f'(x)\) | \(-0.5220232\) | \(-0.5698959\) | \(-0.5811571\) |
find the Hermite polynomial \(H_5\) and approximate \(f(1.5)\).
Solution. With the repeated nodes \(z_0,\dots ,z_5 = 1.3, 1.3, 1.6, 1.6, 1.9, 1.9\) (one row each) the divided-difference table is
| \(f[z]\) | 1st | 2nd | 3rd | 4th | 5th |
| \(0.620086\) | |||||
| \(0.620086\) | \(-0.522023\) | ||||
| \(0.455402\) | \(-0.548946\) | \(-0.089743\) | |||
| \(0.455402\) | \(-0.569896\) | \(-0.069833\) | \(0.066366\) | ||
| \(0.281819\) | \(-0.578612\) | \(-0.029054\) | \(0.067966\) | \(0.002667\) | |
| \(0.281819\) | \(-0.581157\) | \(-0.008484\) | \(0.068567\) | \(0.001002\) | \(-0.002775\) |
The entries \(-0.522023\), \(-0.569896\), \(-0.581157\) in the first column are the given derivatives. Newton’s formula gives \begin {align*} H_5(x) &= 0.6200860-0.5220232(x-1.3)-0.0897427(x-1.3)^2+0.0663656(x-1.3)^2(x-1.6)\\ &\quad +0.0026667(x-1.3)^2(x-1.6)^2\\ &\quad -0.0027747(x-1.3)^2(x-1.6)^2(x-1.9), \end {align*}
and \(H_5(1.5) = 0.5118277\), correct to all seven digits: three values and three slopes do better than five values.
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