◆ Polcof()

Sub Polcof	(	XX As	Double,
		N As	Long,
		X() As	Double,
		C() As	Double,
		D() As	Double,
		Info As	Long
	)

Coefficients of polynomial interpolation

Purpose: This routine computes the coefficients of the polynomial fit produced by a previous call to Polint.

Parameters

[in]	XX	The point about which the Taylor expansion is to be made.
[in]	N	The number of data points (must remain unchanged between the call to Polint and the call to Polcof).
[in]	X()	Array X(LX - 1) (LX >= N) The array of abscissas (must remain unchanged between the call to Polint and the call to Polcof).
[out]	C()	Array C(LC - 1) (LC >= N) An array of information produced by subroutines Polint (must remain unchanged between the call to Polint and the call to Polcof).
[out]	D()	Array D(LD - 1) (LD >= N) The array of coefficients for the Taylor expansion about XX. P(Z) = D(0) + D(1)(Z-XX) + D(2)((Z-XX)^2) + ... + D(N-1)*((Z-XX)^(N-1))
[out]	Info	= 0: Successful exit = -2: The argument N had an illegal value (N < 1) = -3: The argument X() is invalid = -4: The argument C() is invalid = -5: The argument D() is invalid

Reference: SLATEC

Example Program

Compute the coefficients of the polynomial fit by interpolating the following natural logarithm table. Then compute ln(0.115) and ln(0.125) using obtained polynomial.

  x       ln(x)
------ ---------
 0.10   -2.3026
 0.11   -2.2073
 0.12   -2.1203
 0.13   -2.0402
------ ---------

Sub Ex_Polcof()
    Const N As Long = 4
    Dim X(N - 1) As Double, Y(N - 1) As Double, C(N - 1) As Double, D(N - 1) As Double
    Dim Xe As Double, XX As Double, Info As Long
    '-- Data
    X(0) = 0.1: Y(0) = -2.3026
    X(1) = 0.11: Y(1) = -2.2073
    X(2) = 0.12: Y(2) = -2.1203
    X(3) = 0.13: Y(3) = -2.0402
    '-- Polynomial interpolation
    Call Polint(N, X(), Y(), C(), Info)
    If Info <> 0 Then
        Debug.Print "Error in Polint: Info =", Info
        Exit Sub
    End If
    '-- Compute coefficients of polynomial
    XX = 0.12
    Call Polcof(XX, N, X(), C(), D(), Info)
    Debug.Print D(0), D(1), D(2), D(3)
    '-- Compute ln(0.115) and ln(0.125)
    Xe = 0.115
    Debug.Print "ln(" + CStr(Xe) + ") =", D(0) + D(1) * (Xe - XX) + D(2) * ((Xe - XX) ^ 2) + D(3) * ((Xe - XX) ^ 3)
    Xe = 0.125
    Debug.Print "ln(" + CStr(Xe) + ") =", D(0) + D(1) * (Xe - XX) + D(2) * ((Xe - XX) ^ 2) + D(3) * ((Xe - XX) ^ 3)
    Debug.Print "Info =", Info
End Sub

Example Results: -2.1203 8.33166666666668 -34.5000000000022 233.333333333222

ln(0.115) = -2.16285

ln(0.125) = -2.079475

Info = 0