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spline interpolation

For the friends of curves I have below the source code for the spline interpolation added. The function can also be like the Linear Interpolation in Excel "installed" and is used for interpolation of measured values from existing data pairs.

The program dates back to the source code of the freeware program xNumber 5.6 - Multi Precision Floating Point Computing and Numerical Methods for EXCEL.
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• 'Interpolate one point from sorted X,Y data pairs using Cubic Spline Interpolation
• Function Interpol_cspline(Xin As Range, Yin As Range, Xtarget As Double)
• Dim n%, ny% 'Xin and Yin counts
• Dim i% 'loop counter
• Dim ihi%, m%
• Dim a#, b#, p#, qn#, sig#, un#, h0#, h1#, h2#, Yout#
• ny = Yin.Cells.Count
• n = Xin.Cells.Count
• ' Next check to be sure that there are the came counts of Xin and Yin
• If n <> ny Then
• Interpol_cspline = CVErr(xlErrRef) 'Uneven counts of Xin and Yin"
• GoTo err_ret
• End If
• 
  
• ReDim x(1 To n) As Double
• ReDim Y(1 To n) As Double
• 'ReDim u(1 To n - 1) As Single
• ReDim U(1 To n - 1) As Double 'mod. LV 31-8-02
• ReDim ypp(1 To n) As Double 'these are the 2nd derivative values
• 'populate the input arrays
• For i = 1 To n
• x(i) = Xin(i)
• If i > 1 Then
• If x(i) <= x(i - 1) Then 'Check if X values are not repeating and are sorted in ascending order
• Interpol_cspline = CVErr(xlErrValue) 'Error: Not sorted or repeating
• GoTo err_ret
• End If
• End If
• Y(i) = Yin(i)
• Next i
• ypp (1) = 0 'First knot boundary condition
• U (1) = 0' First knot boundary condition
• For i = 2 To n - 1
• h0 = x (i + 0) - x (i - 1)
• h1 = x (i + 1) - x (i - 0)
• h2 = x (i + 1) - x (i - 1)
• 
  
• themselves = h0 / h2
• p = themselves * ypp (i - 1) + 2 #
• ypp (i) = (to - 1) / p
• U (i) = (Y (i + 1) - Y (i)) / h1 - (Y (i) - Y (i - 1)) / h0
• U (i) = (6 # * U (i) / h2 - say * U (i - 1)) / p
• 
  
• Next in
• 
  
• qn = 0
• un = 0
• ypp(n) = (un - qn * U(n - 1)) / (qn * ypp(n - 1) + 1) 'Last knot boundary condition
• For i = n - 1 To 1 Step -1
• ypp(i) = ypp(i) * ypp(i + 1) + U(i) 'Backfill the 2nd derivatives
• Next i
• ''''''''''''''''''''''''''''''''''''''''
• 'Spline evaluation at target X point
• '''''''''''''''''''''''''''''''''''''''''
• 'Find correct interval using halving binary search
• i = 1
• ihi = n
• Do While (i <>
• m = (i + ihi) \ 2 'Calculate the midpoint
• If (Xtarget < ihi =" m" i =" m">
• Loop
• ' i = beginning of the correct interval
• h0 = x(i + 1) - x(i) 'Calc the width of the X interval
• a = (x(i + 1) - Xtarget) / h0
• b = (Xtarget - x(i)) / h0
• Yout = a * Y(i) + b * Y(i + 1) + ((a ^ 3 - a) * ypp(i) + (b ^ 3 - b) * ypp(i + 1)) * (h0 ^ 2) / 6#
• Interpol_cspline = Yout
• err_ret:
• End Function

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