# Hermite interpolation

In numerical analysis, Hermite interpolation, named after Charles Hermite, is a method of interpolating data points as a polynomial function. The generated Hermite interpolating polynomial is closely related to the Newton polynomial, in that both are derived from the calculation of divided differences. However, the Hermite interpolating polynomial may also be computed without using divided differences, see Chinese remainder theorem § Hermite interpolation.

Unlike Newton interpolation, Hermite interpolation matches an unknown function both in observed value, and the observed value of its first m derivatives. This means that n(m + 1) values

${\begin{matrix}(x_{0},y_{0}),&(x_{1},y_{1}),&\ldots ,&(x_{n-1},y_{n-1}),\\(x_{0},y_{0}'),&(x_{1},y_{1}'),&\ldots ,&(x_{n-1},y_{n-1}'),\\\vdots &\vdots &&\vdots \\(x_{0},y_{0}^{(m)}),&(x_{1},y_{1}^{(m)}),&\ldots ,&(x_{n-1},y_{n-1}^{(m)})\end{matrix}}$ must be known, rather than just the first n values required for Newton interpolation. The resulting polynomial may have degree at most n(m + 1) − 1, whereas the Newton polynomial has maximum degree n − 1. (In the general case, there is no need for m to be a fixed value; that is, some points may have more known derivatives than others. In this case the resulting polynomial may have degree N − 1, with N the number of data points.)

## Usage

### Simple case

When using divided differences to calculate the Hermite polynomial of a function f, the first step is to copy each point m times. (Here we will consider the simplest case $m=1$  for all points.) Therefore, given $n+1$  data points $x_{0},x_{1},x_{2},\ldots ,x_{n}$ , and values $f(x_{0}),f(x_{1}),\ldots ,f(x_{n})$  and $f'(x_{0}),f'(x_{1}),\ldots ,f'(x_{n})$  for a function $f$  that we want to interpolate, we create a new dataset

$z_{0},z_{1},\ldots ,z_{2n+1}$

such that

$z_{2i}=z_{2i+1}=x_{i}.$

Now, we create a divided differences table for the points $z_{0},z_{1},\ldots ,z_{2n+1}$ . However, for some divided differences,

$z_{i}=z_{i+1}\implies f[z_{i},z_{i+1}]={\frac {f(z_{i+1})-f(z_{i})}{z_{i+1}-z_{i}}}={\frac {0}{0}}$

which is undefined. In this case, the divided difference is replaced by $f'(z_{i})$ . All others are calculated normally.

### General case

In the general case, suppose a given point $x_{i}$  has k derivatives. Then the dataset $z_{0},z_{1},\ldots ,z_{N}$  contains k identical copies of $x_{i}$ . When creating the table, divided differences of $j=2,3,\ldots ,k$  identical values will be calculated as

${\frac {f^{(j)}(x_{i})}{j!}}.$

For example,

$f[x_{i},x_{i},x_{i}]={\frac {f''(x_{i})}{2}}$
$f[x_{i},x_{i},x_{i},x_{i}]={\frac {f^{(3)}(x_{i})}{6}}$

etc.

### Example

Consider the function $f(x)=x^{8}+1$ . Evaluating the function and its first two derivatives at $x\in \{-1,0,1\}$ , we obtain the following data:

x ƒ(x) ƒ'(x) ƒ''(x)
−1 2 −8 56
0 1 0 0
1 2 8 56

Since we have two derivatives to work with, we construct the set $\{z_{i}\}=\{-1,-1,-1,0,0,0,1,1,1\}$ . Our divided difference table is then:

${\begin{array}{llcclrrrrr}z_{0}=-1&f[z_{0}]=2&&&&&&&&\\&&{\frac {f'(z_{0})}{1}}=-8&&&&&&&\\z_{1}=-1&f[z_{1}]=2&&{\frac {f''(z_{1})}{2}}=28&&&&&&\\&&{\frac {f'(z_{1})}{1}}=-8&&f[z_{3},z_{2},z_{1},z_{0}]=-21&&&&&\\z_{2}=-1&f[z_{2}]=2&&f[z_{3},z_{2},z_{1}]=7&&15&&&&\\&&f[z_{3},z_{2}]=-1&&f[z_{4},z_{3},z_{2},z_{1}]=-6&&-10&&&\\z_{3}=0&f[z_{3}]=1&&f[z_{4},z_{3},z_{2}]=1&&5&&4&&\\&&{\frac {f'(z_{3})}{1}}=0&&f[z_{5},z_{4},z_{3},z_{2}]=-1&&-2&&-1&\\z_{4}=0&f[z_{4}]=1&&{\frac {f''(z_{4})}{2}}=0&&1&&2&&1\\&&{\frac {f'(z_{4})}{1}}=0&&f[z_{6},z_{5},z_{4},z_{3}]=1&&2&&1&\\z_{5}=0&f[z_{5}]=1&&f[z_{6},z_{5},z_{4}]=1&&5&&4&&\\&&f[z_{6},z_{5}]=1&&f[z_{7},z_{6},z_{5},z_{4}]=6&&10&&&\\z_{6}=1&f[z_{6}]=2&&f[z_{7},z_{6},z_{5}]=7&&15&&&&\\&&{\frac {f'(z_{6})}{1}}=8&&f[z_{8},z_{7},z_{6},z_{5}]=21&&&&&\\z_{7}=1&f[z_{7}]=2&&{\frac {f''(z_{7})}{2}}=28&&&&&&\\&&{\frac {f'(z_{7})}{1}}=8&&&&&&&\\z_{8}=1&f[z_{8}]=2&&&&&&&&\\\end{array}}$

and the generated polynomial is

{\begin{aligned}P(x)&=2-8(x+1)+28(x+1)^{2}-21(x+1)^{3}+15x(x+1)^{3}-10x^{2}(x+1)^{3}\\&\quad {}+4x^{3}(x+1)^{3}-1x^{3}(x+1)^{3}(x-1)+x^{3}(x+1)^{3}(x-1)^{2}\\&=2-8+28-21-8x+56x-63x+15x+28x^{2}-63x^{2}+45x^{2}-10x^{2}-21x^{3}\\&\quad {}+45x^{3}-30x^{3}+4x^{3}+x^{3}+x^{3}+15x^{4}-30x^{4}+12x^{4}+2x^{4}+x^{4}\\&\quad {}-10x^{5}+12x^{5}-2x^{5}+4x^{5}-2x^{5}-2x^{5}-x^{6}+x^{6}-x^{7}+x^{7}+x^{8}\\&=x^{8}+1.\end{aligned}}

by taking the coefficients from the diagonal of the divided difference table, and multiplying the kth coefficient by $\prod _{i=0}^{k-1}(x-z_{i})$ , as we would when generating a Newton polynomial.

## Error

Call the calculated polynomial H and original function f. Evaluating a point $x\in [x_{0},x_{n}]$ , the error function is

$f(x)-H(x)={\frac {f^{(K)}(c)}{K!}}\prod _{i}(x-x_{i})^{k_{i}}$

where c is an unknown within the range $[x_{0},x_{N}]$ , K is the total number of data-points, and $k_{i}$  is the number of derivatives known at each $x_{i}$  plus one.