LagrangePolynomials.lhs

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\begin{document}
\sffamily
\begin{center}
\begin{LARGE}
{\bf Direct Collocation}\\
\vspace*{0.3cm}
\end{LARGE}\end{center}
\begin{center}
\begin{Large}
Greg Horn
\end{Large}

\end{center}
\vspace*{1cm}

\section{Lagrange Interpolation}
The Lagrange interpolating polynomial of degree $D$, defined by nodes $x_0..x_D$ at timepoints $\tau_0..\tau_D$ is
\begin{equation}
\begin{aligned}
x(\tau) &= \sum_{j=0}^D \myl_j(\tau) x_j
\end{aligned}
\label{eq:lagrange_interp_poly}
\end{equation}
where
\begin{equation}
\myl_j(\tau) = \prod_{k=0,k \neq j}^D\frac{\tau-\tau_k}{\tau_j-\tau_k}
\end{equation}

\begin{code}
{-# OPTIONS_GHC -Wall #-}
{-# Language FlexibleContexts #-}
{-# Language GADTs #-}

module Dyno.LagrangePolynomials ( lagrangeDerivCoeffs, lagrangeXis, runComparison ) where

import qualified Data.Vector as V

import Casadi.SXFunction ( sxFunction )
import Casadi.Function ( evalDMatrix )
import Casadi.SharedObject ( soInit )
import Casadi.SX ( SX, ssym, sgradient )
import Casadi.DMatrix ( DMatrix, ddata, ddense )

import Dyno.TypeVecs

lagrangeXis :: Fractional a => [a] -> a -> Int -> a
lagrangeXis taus tau j =
  product [(tau - tk) / (tj - tk) | k <- [0..deg]
                                  , k /= j
                                  , let tk = taus !! k
                                        tj = taus !! j
                                  ]
  where
    deg = length taus - 1

\end{code}

The derivative of this polynomial on an intermediate point is given by
\begin{equation}
x'(\tau) = \sum_{j=0}^D \myldot_j(\tau) x_j
\label{eq:lagrange_interp_poly_deriv}
\end{equation}

Written out for $D=3$, this looks like

\begin{align}
x(\tau) = & x_0 \taus{}{1}{0} \taus{}{2}{0} \taus{}{3}{0} + \\
          & x_1 \taus{}{0}{1} \taus{}{2}{1} \taus{}{3}{1} + \\
          & x_2 \taus{}{0}{2} \taus{}{1}{2} \taus{}{3}{2} + \\
          & x_3 \taus{}{0}{3} \taus{}{1}{3} \taus{}{2}{3}
\end{align}

Evaluating this at an interpolation node $\tau_m$ eliminates all terms except $m=k$, giving
\begin{equation}
x(\tau_m) = x_m
\end{equation}

The derivative for $D=3$ is

\begin{align}
x'(\tau) = &x_0 (\taud{}{1}{0} \taus{}{2}{0} \taus{}{3}{0}
               + \taus{}{1}{0} \taud{}{2}{0} \taus{}{3}{0}
               + \taus{}{1}{0} \taus{}{2}{0} \taud{}{3}{0}) + \\
           &x_1 (\taud{}{0}{1} \taus{}{2}{1} \taus{}{3}{1}
               + \taus{}{0}{1} \taud{}{2}{1} \taus{}{3}{1}
               + \taus{}{0}{1} \taus{}{2}{1} \taud{}{3}{1}) + \\
           &x_2 (\taud{}{0}{2} \taus{}{1}{2} \taus{}{3}{2}
               + \taus{}{0}{2} \taud{}{1}{2} \taus{}{3}{2}
               + \taus{}{0}{2} \taus{}{1}{2} \taud{}{3}{2}) + \\
           &x_3 (\taud{}{0}{3} \taus{}{1}{3} \taus{}{2}{3}
               + \taus{}{0}{3} \taud{}{1}{3} \taus{}{2}{3}
               + \taus{}{0}{3} \taus{}{1}{3} \taud{}{2}{3})
\end{align}

evaluating this at $\tau_0$ gives:

\begin{align}
x'(\tau_0) = &x_0 (\taud{0}{1}{0} \taus{0}{2}{0} \taus{0}{3}{0}
                 + \taus{0}{1}{0} \taud{0}{2}{0} \taus{0}{3}{0}
                 + \taus{0}{1}{0} \taus{0}{2}{0} \taud{0}{3}{0}) +\\
             &x_1 (\taud{0}{0}{1} \taus{0}{2}{1} \taus{0}{3}{1}
                 + \taus{0}{0}{1} \taud{0}{2}{1} \taus{0}{3}{1}
                 + \taus{0}{0}{1} \taus{0}{2}{1} \taud{0}{3}{1}) +\\
             &x_2 (\taud{0}{0}{2} \taus{0}{1}{2} \taus{0}{3}{2}
                 + \taus{0}{0}{2} \taud{0}{1}{2} \taus{0}{3}{2}
                 + \taus{0}{0}{2} \taus{0}{1}{2} \taud{0}{3}{2}) +\\
             &x_3 (\taud{0}{0}{3} \taus{0}{1}{3} \taus{0}{2}{3}
                 + \taus{0}{0}{3} \taud{0}{1}{3} \taus{0}{2}{3}
                 + \taus{0}{0}{3} \taus{0}{1}{3} \taud{0}{2}{3})
\end{align}
which simplifies to
\begin{align}
x'(\tau_0) = &x_0 (\taud{0}{1}{0}
                 + \taud{0}{2}{0}
                 + \taud{0}{3}{0}) +\\
             &x_1 (\taud{0}{0}{1} \taus{0}{2}{1} \taus{0}{3}{1}) +\\
             &x_2 (\taud{0}{0}{2} \taus{0}{1}{2} \taus{0}{3}{2}) +\\
             &x_3 (\taud{0}{0}{3} \taus{0}{1}{3} \taus{0}{2}{3}) \\
= & x_0 C_{0,0} + x_1 C_{0,1} + x_2 C_{0,2} + x_3 C_{0,3}
\end{align}

evaluating this at $\tau_1$ gives:
\begin{align}
x'(\tau_0) = &x_0 (\taud{1}{1}{0} \taus{1}{2}{0} \taus{1}{3}{0}) +\\
             &x_1 (\taud{1}{0}{1}
                 + \taud{1}{2}{1}
                 + \taud{1}{3}{1}) +\\
             &x_2 (\taus{1}{0}{2} \taud{1}{1}{2} \taus{1}{3}{2}) +\\
             &x_3 ( \taus{1}{0}{3} \taud{1}{1}{3} \taus{1}{2}{3}) \\
= & x_0 C_{1,0} + x_1 C_{1,1} + x_2 C_{1,2} + x_3 C_{1,3}
\end{align}

evaluating this at $\tau_2$ gives:
\begin{align}
x'(\tau_2) = &x_0 (\taus{2}{1}{0} \taud{2}{2}{0} \taus{2}{3}{0}) + \\
             &x_1 (\taus{2}{0}{1} \taud{2}{2}{1} \taus{2}{3}{1}) + \\
             &x_2 (\taud{2}{0}{2}
                 + \taud{2}{1}{2}
                 + \taud{2}{3}{2}) + \\
             &x_3 (\taus{2}{0}{3} \taus{2}{1}{3} \taud{2}{2}{3}) \\
= & x_0 C_{2,0} + x_1 C_{2,1} + x_2 C_{2,2} + x_3 C_{2,3}
\end{align}

evaluating this at $\tau_3$ gives:
\begin{align}
x'(\tau_3) = &x_0 (\taus{3}{1}{0} \taus{3}{2}{0} \taud{3}{3}{0}) + \\
             &x_1 (\taus{3}{0}{1} \taus{3}{2}{1} \taud{3}{3}{1}) + \\
             &x_2 (\taus{3}{0}{2} \taus{3}{1}{2} \taud{3}{3}{2}) + \\
             &x_3 (\taud{3}{0}{3} 
                 + \taud{3}{1}{3} 
                 + \taud{3}{2}{3}) \\
= & x_0 C_{3,0} + x_1 C_{3,1} + x_2 C_{3,2} + x_3 C_{3,3}
\end{align}

The general formula for $C_{j,k}$ is
\begin{equation}
C_{j,k} =
\begin{cases}
\sum_{i=0,i\ne k}^D{\frac{1}{\tau_k-\tau_i}} & j=k \\
\frac{1}{\tau_k-\tau_j} \prod_{i=0,i\ne j,i\ne k}^D{\frac{\tau_j-\tau_i}{\tau_k-\tau_i}}  & j \ne k
\end{cases}
\end{equation}

\begin{code}
lagrangeDerivCoeffs :: (Dim deg, Fractional a) => Vec deg a -> Vec deg (Vec deg a)
lagrangeDerivCoeffs taus' = mkVec' [mkVec' [cjk j k | k <- [0..deg]] | j <- [0..deg]]
  where
    taus = unVec taus'
    deg = V.length taus - 1

    cjk j k
      | j == k = sum [ 1/(tau_k - taus V.! i) | i <- [0..deg], i /= k ]
      | otherwise =
        1 / (tau_k - tau_j) *
        product [ (tau_j - tau_i)/(tau_k - tau_i)
                | i <- [0..deg], i /= j, i /= k, let tau_i = taus V.! i
                ]
      where
        tau_k = taus V.! k
        tau_j = taus V.! j
\end{code}

Testing code:
\begin{code}
ssyms :: String -> Int -> IO [SX]
ssyms name k = mapM (ssym . (name ++) . show) $ take k [(0::Int)..]

runComparison :: IO ()
runComparison = do
  let deg = 6
      sampleTaus = take (deg+1) [0.1,0.2..]
      sampleTaus' = map realToFrac sampleTaus

  tau <- ssym "t"
  taus <- ssyms "t" (deg + 1)

  let zs :: [SX]
      zs = map (lagrangeXis taus tau) [0..deg]
      inputs = tau : taus
      zdot = map (`sgradient` tau) zs
  zdotAlg <- sxFunction (V.fromList inputs) (V.fromList zdot)
  soInit zdotAlg

  --mapM_ print zdot'
  
  putStrLn "numeric:"
  vals' <- V.mapM (\tau_i -> evalDMatrix zdotAlg (V.fromList (tau_i : sampleTaus'))) (V.fromList sampleTaus')
  let d2d :: DMatrix -> Double
      d2d x = case V.toList (ddata (ddense x)) of
        [y] -> y
        ys -> error $ "d2d: need length 1, got length " ++ show (length ys)

      vals = fmap (fmap d2d) vals'
  V.mapM_ print vals
  
  putStrLn "\nnumeric difference:"
  let cmp :: V.Vector Double -> V.Vector Double -> IO ()
      cmp v1s v2s = print $ V.zipWith (-) v1s v2s
      f :: Dim n => Vec n Double -> IO ()
      f st = V.zipWithM_ cmp vals $ fmap unVec $ unVec $ lagrangeDerivCoeffs st
  reifyVector (V.fromList sampleTaus) f
  return ()

\end{code}

\end{document}