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In [[numerical analysis]], '''Hermite interpolation''', named after [[Charles Hermite]], is a method of [[interpolation|interpolating data points]] as a [[polynomial function]]. The generated '''[[Hermite polynomial]]''' is closely related to the [[Newton polynomial]], in that both are derived from the calculation of [[divided differences]].
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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''&nbsp;+&nbsp;1) values
:<math>
\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}
</math>
must be known, rather than just the first ''n'' values required for Newton interpolation. The resulting polynomial may have degree at most ''n''(''m''&nbsp;+&nbsp;1)&nbsp;&minus;&nbsp;1, whereas the Newton polynomial has maximum degree ''n''&nbsp;&minus;&nbsp;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''&nbsp;&minus;&nbsp;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 <math>m = 1</math> for all points.) Therefore, given <math>n + 1</math> data points <math>x_0, x_1, x_2, \ldots, x_n</math>, and values <math>f(x_0), f(x_1), \ldots, f(x_n)</math> and <math>f'(x_0), f'(x_1), \ldots, f'(x_n)</math> for a function <math>f</math> that we want to interpolate, we create a new dataset
:<math>z_0, z_1, \ldots, z_{2n+1}</math>
such that
:<math>z_{2i}=z_{2i+1}=x_i.</math>
 
Now, we create a [[Divided differences|divided differences table]] for the points <math>z_0, z_1, \ldots, z_{2n+1}</math>. However, for some divided differences,
:<math>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}</math>
which is undefined!
In this case, we replace the divided difference by <math>f'(z_i)</math>. All others are calculated normally.
 
=== General case ===
In the general case, suppose a given point <math>x_i</math> has ''k'' derivatives. Then the dataset <math>z_0, z_1, \ldots, z_{N}</math> contains ''k'' identical copies of <math>x_i</math>. When creating the table, [[divided differences]] of <math>j = 2, 3, \ldots, k</math> identical values will be calculated as
 
:<math>\frac{f^{(j)}(x_i)}{j!}.</math>
 
For example,
:<math>f[x_i, x_i, x_i]=\frac{f''(x_i)}{2}</math>
:<math>f[x_i, x_i, x_i, x_i]=\frac{f^{(3)}(x_i)}{6}</math>
etc.
 
=== Example ===
Consider the function <math>f(x) = x^8 + 1</math>. Evaluating the function and its first two derivatives at <math>x \in \{-1, 0, 1\}</math>, we obtain the following data:
:{| class="wikitable" style="text-align: center; padding: 1em;"
|-
| ''x'' || ''&fnof;''(''x'') || ''&fnof;''<nowiki>'</nowiki>(''x'')  || ''&fnof;''<nowiki>''</nowiki>(''x'')
|-
| &minus;1 ||  2  ||  &minus;8    || 56
|-
| 0  ||  1  ||  0    || 0
|-
| 1  ||  2  ||  8    || 56
|}
 
Since we have two derivatives to work with, we construct the set <math>\{z_i\} = \{-1, -1, -1, 0, 0, 0, 1, 1, 1\}</math>. Our divided difference table is then:
:<math>
\begin{matrix}
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_7)}{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_8)}{1} = 8  &                        &                          &      &    &  &    & \\
z_8 =  1  &  f[z_8] = 2  &                          &                        &                          &      &    &  &    & \\
\end{matrix}
</math>
and the generated polynomial is
:<math>
\begin{align}
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{align}
</math>
by taking the coefficients from the diagonal of the divided difference table, and multiplying the ''k''th coefficient by <math>\prod_{i=0}^{k-1} (x - z_i)</math>, as we would when generating a Newton polynomial.
 
==Error==
Call the calculated polynomial ''H'' and original function ''f''. Evaluating a point <math>x \in [x_0, x_n]</math>, the error function is
:<math>f(x) - H(x) = \frac{f^{(K)}(c)}{K!}\prod_{i}(x - x_i)^{k_i}</math>
where ''c'' is an unknown within the range <math>[x_0, x_N]</math>, ''K'' is the total number of data-points plus one, and <math>k_i</math> is the number of derivatives known at each <math>x_i</math> plus one.
 
==See also==
*[[Cubic Hermite spline]]
*[[Newton series]], also known as [[finite differences]]
*[[Neville's schema]]
*[[Polynomial interpolation]]
*[[Lagrange polynomial|Lagrange form]] of the interpolation polynomial
*[[Bernstein polynomial|Bernstein form]] of the interpolation polynomial
*[[Chinese remainder theorem#Applications | Chinese remainder theorem - Applications]]
 
==References==
* {{ cite book|last1=Burden|first1=Richard L.|first2= J. Douglas |last2=Faires|title=Numerical Analysis|publisher= Belmont: Brooks/Cole|year= 2004}}
* {{Citation |last=Spitzbart |first=A. |title=A Generalization of Hermite's Interpolation Formula |journal=[[American Mathematical Monthly]] |volume=67 |issue=1 |pages=42&ndash;46 |date=January 1960 |jstor=2308924 |doi= }}
 
==External links==
*[http://mathworld.wolfram.com/HermitesInterpolatingPolynomial.html Hermites Interpolating Polynomial] at Mathworld
 
[[Category:Interpolation]]
[[Category:Finite differences]]
[[Category:Factorial and binomial topics]]

Revision as of 21:38, 20 February 2014

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