C++ Vector Interpolation of Points with Derivatives

Visual Comparision of Zero Order and First Order Interpolation

Zero order interpolation is straight line interpolation. First order interpolation corresponds to the derivatives being satisfied for a polynomial.

This test image is interpolating two points on a circle. Since the circles derivative is known exactly this makes a good test case because we have perfect information.

The red points are the two interpolated points. The blue is the linear interpolation, and the green is the first order interpolation.

The orange line between the two is the circle that each is interpolating. Now you can see that the blue is further away from the true solution than the green. As the order is increased the interpolation converge to the true function.

I have deliberately used points spaced apart so you can see the interpolation effects. Though in real life there are so many situations to see this anyway.

Mathematica Script

The 3D version of the problem and I will be blowed if I am going to solve it by hand. Well not that I have not done that sort of thing before, but if a computer can do it without error it should do it. To make my life easier I have done the coefficient calculation for the two point interpolation of a curve in the first order.

f8[f0_,df0_,f1_,df1_] := Module[ {equ1,equ2,t1},
  equ1 = Sum[ a[i]*t^i,{i,0,3}];
  equ2 = D[equ1,t];
  t1 = Solve[ 
  {
    (equ1 /. t->0) == f0,
    (equ1 /. t->1) == f1,
    (equ2 /. t->0) == df0,
    (equ2 /. t->1) == df1
  }, 
  {a[0],a[1],a[2],a[3]} 
  ];
(*
  Print["t1=",t1];

  Print[ t1[[1]][[1]][[2]] ];
  Print[ t1[[1]][[2]][[2]] ];
  Print[ t1[[1]][[3]][[2]] ];
  Print[ t1[[1]][[4]][[2]] ]; 
*)
  Return[ equ1 /. t1 ];
]

f8[1,0,0,0]

fverify[a_,da_,b_,db_] := Module[ {e1,e2,c0,c1,c2,c3},
  e1 = f8[a,da,b,db];
Print["e1=",e1];
  e2 = D[e1,t];
Print["e2=",e2];
  c0 = (e1 /. t->0)[[1]];
  c1 = (e1 /. t->1)[[1]];
  c2 = (e2 /. t->0)[[1]];
  c3 = (e2 /. t->1)[[1]];
Print["c0=",c0];
Print["c1=",c1];
Print["c2=",c2];
Print["c3=",c3];
  Print[ c0 == a ];
  Print[ c1 == b ];
  Print[ c2 == da ];
  Print[ c3 == db ];
]

fverify[0,0,0,1]

Source

Makefile

Three Point Problem

The two point interpolation produced a curve. The three point interpolation produces a surface. n dimensional point interpolation generates a n-1 dimensional surface.

A linear or zero order interpolation between the points in 3D is called Barycentric Coordinates. I really object to silly names like this because it detracts from the maths, 0th order interpolation actually says something.

The aim is to apply the same method as was done in 2D successfully for 3D space. Consider the first derivative problem.

$p_{0} (u, v) = p_{0} f_{0} + p_{0u} f_{0u} + p_{0v} f_{0v}$

This is $p_{0}$ contribution over the surface.

Lets consider just one of those f polynomial functions and how many conditions that it needs to satisfy.

$f_{0} (1, 0) = 1$
$f_{0} (0, 1) = 0$
$f_{0} (0, 0) = 0$

$\frac{\partial}{\partial u} f_{0} (1, 0) = 0$
$\frac{\partial}{\partial u} f_{0} (0, 1) = 0$
$\frac{\partial}{\partial u} f_{0} (0, 0) = 0$

$\frac{\partial}{\partial v} f_{0} (1, 0) = 0$
$\frac{\partial}{\partial v} f_{0} (0, 1) = 0$
$\frac{\partial}{\partial v} f_{0} (0, 0) = 0$

This is just one of the functions. The other two must also be satisfied. Now a cubic and lower expansion of two variables has 10 coefficients so there is one extra coefficient. This is where the element of risk lies. Assuming symmetry is a good thing then either the constant coefficient or the u*v coefficient is to be ignored.

Why do I consider this risky? For a surface in 3D it really needs to be continuous. This means that given any two neighboring triangles which share a boundary of two points, the interpolated polynomial on one must match the interpolated polynomial of the other. Else the surface would be broken where they meet.

<TODO>

Move test code to its own class. ie vecInterptest.h
Document all members - even p0 and p1.
Extend to 3D space
Integration of a vector function about a point.
*** This is really important. Do any really elegant solutions exist? Or are numerical approaches the only way.
Here I believe is an acceptable approach.
sum(i=0..n-1,1/2*(r[i]+r[i+1]))*dtheta
(1/2*(r[0]+r[n])+sum(r[i],i=1..n-1))*dtheta for 2D case.
r[i] = f(i/n) where f(t) is the vector function.
dtheta = (theta(n)-theta(0))/n
The determination of partial derivatives numerically from a mesh of points.
General coefficient solver scripts in Mathematica would be nice.

points.

General coefficient solver scripts in Mathematica would be nice.

vecinterp

Intro

Visual Comparision of Zero Order and First Order Interpolation

Mathematica Script

Source

Three Point Problem

<TODO>