A156744 Coefficients for estimation of derivative from equally spaced numerical data using the Lagrange interpolating polynomial.

-1, 0, 1, 1, -8, 0, 8, -1, -1, 9, -45, 0, 45, -9, 1, 3, -32, 168, -672, 0, 672, -168, 32, -3

Offset

1,5

Comments

An effective way to approximate the derivative of equally spaced numerical data is to differentiate its Lagrange interpolating polynomial. If y[x] is equally spaced data from x = -n to +n, its Lagrange interpolating polynomial P(x) has degree 2*n+1. Then P'(0) may be expressed as a weighted sum over the y[x]. This is the triangle of coefficients C[n,m] such that P'(0) = (1/d[n]) * Sum_{m=-n}^n C[n,m] y[m]. The denominator d[n] is given by sequence A099996. This is very useful in numerical analysis. For example, when n=1, this gives the centered difference approximation to the derivative.

Links

Table of n, a(n) for n=1..24.

Example

Irregular triangle begins:
  -1, 0, 1;
  1, -8, 0, 8, -1;
  -1, 9, -45, 0, 45, -9, 1;
  3, -32, 168, -672, 0, 672, -168, 32, -3;
  -2, 25, -150, 600, -2100, 0, 2100, -600, 150, -25, 2;  ...

Mathematica

facs[x_, j_, n_] := Product[If[k == j, 1, x - k], {k, -n, n}]
coefs[x_, n_] := Table[facs[x, j, n]/facs[j, j, n], {j, -n, n}]
d[n_] := Apply[LCM, Table[j, {j, 1, 2*n}]]
dCoefs[x_, n_] := d[n] D[coefs[y, n], y] /. {y -> x}
MatrixForm[Table[dCoefs[0, n], {n, 1, 5}]]

Crossrefs

When divided by sequence A099996, this triangle gives the coefficients needed to estimate derivatives from equally spaced numerical data using Lagrange interpolation. The first and last entry of each row of the triangle has absolute value lcm{1, 2, ..., 2*n}/n*binomial(2n, n), as seen in A068553.

Sequence in context: A347157 A186981 A347159 * A228438 A021557 A242943

Adjacent sequences: A156741 A156742 A156743 * A156745 A156746 A156747

Keyword

sign,tabf,more

Author

Bruce Boghosian, Feb 14 2009

Status

approved