Chain rules for divided differences and the error
in parametric curve interpolation
Michael Floater
Centre of Mathematics for Applications,
Department of Informatics,
University of Oslo
Parametric curve interpolation
We can interpolate a sequence of points x0 , x1 , . . . , xn ∈ Rd , with
a (vector-valued) polynomial p : R → Rd , degree ≤ n, by choosing
any parameter values t0 < t1 < · · · < tn and demanding that
p(ti ) = xi , i = 0, 1, . . . , n.
Examples:
I
find the ti to reduce the degree of p, de Boor, Höllig & Sabin
(1987), Degen (1993), Scherer & Mørken (1997), Schaback
(1998), Grandine & Hogan (2004).
I
choose ‘uniform’ ti : let ti = i.
I
choose ‘chordal’ ti : let t0 = 0 and ti+1 = ti + kxi+1 − xi k.
Approximation error
Suppose the points xi are samples xi = f(si ) from some (often
unknown) curve f : R → Rd , where s0 < s1 < · · · < sn . What is
the error of approximation, i.e., what is the distance between
p|[t0 ,tn ] and f|[s0 ,sn ] ?
A natural way to measure the distance is to let φ : R → R be the
polynomial, degree ≤ n, such that φ(ti ) = si , i = 0, 1, . . . , n, and
look at the difference
f(φ(t)) − p(t),
t ∈ [t0 , tn ].
Since p interpolates f ◦ φ at t = ti , Newton’s error formula gives
f(φ(t)) − p(t) = (t − t0 )(t − t1 ) · · · (t − tn )[t0 , t1 , . . . , tn , t](f ◦ φ).
In this talk:
1. Derive formulas for divided differences of composite functions
2. Some resulting error analysis
3. Other divided difference formulas...
Derivatives of composite functions
Basic chain rule:
(g ◦ f )0 (x) = g 0 (f (x))f 0 (x).
Higher order chain rules:
(g ◦ f )00 (x) = g 00 (f (x))(f 0 (x))2 + g 0 (f (x))f 00 (x),
(g ◦ f )000 = g 000 (f 0 )3 + 3g 00 f 0 f 00 + g 0 f 000 ,
(g ◦ f )(4) = g (4) (f 0 )4 + 6g 000 (f 0 )2 f 00 + g 00 (3(f 00 )2 + 4f 0 f 000 ) + g 0 f (4) ,
(g ◦ f )(5) = · · · .
Faa di Bruno’s formula
Published in 1857:
(g ◦f )(n) (x) =
X
n!
g (k) (f (x))
b1 ! · · · bn !
f (1) (x)
1!
!b1
···
f (n) (x)
n!
The sum is over all k = 1, . . . , n and solutions b1 , . . . , bn ≥ 0 to
b1 + b2 2 + · · · + bn n = n,
b1 + · · · + bn = k,
i.e., partitions of the integer n.
Cavaliere Francesco Faa di Bruno, Note sur une nouvelle formule
de calcul differentiel, Quarterly J. Pure Appl. Math, 1 (1857),
359–360.
!bn
Examples
3 = b1 + b2 2 + b3 3
3=1+1+1=1+2=3
000
(g ◦ f ) = g 000 (f 0 )3 + 3g 00 f 0 f 00 + g 0 f 000
4 = b1 + b2 2 + b3 3 + b4 4
4=1+1+1+1=1+1+2=1+3=2+2=4
(g ◦ f )(4) = g (4) (f 0 )4 + 6g 000 (f 0 )2 f 00 + 4g 00 f 0 f 000 + 3g 00 (f 00 )2 + g 0 f (4)
Divided differences
[x0 ]f = f (x0 )
f (x1 ) − f (x0 )
[x0 , x1 ]f =
,
x − x0
1
f (x2 ) − f (x1 ) f (x1 ) − f (x0 ) .
−
(x2 − x0 ),
[x0 , x1 , x2 ]f =
x2 − x1
x1 − x0
...
[x1 , . . . , xn ]f − [x0 , . . . , xn−1 ]f
[x0 , . . . , xn ]f =
xn − x0
f (n) (x)
x0 ,...,xn →x
n!
(n)
f (ξ)
[x0 , . . . , xn ]f =
,
n!
lim
[x0 , . . . , xn ]f =
some ξ ∈ [x0 , xn ].
Chain rule for divided differences
[x0 , . . . , xn ](g ◦ f ) =
n
X
X
k
Y
[f (xi0 ), . . . , f (xik )]g
[xij−1 , . . . , xij ]f .
k=1 0=i0 <···<ik =n
j=1
I
The second sum is over all sequences i1 < i2 < · · · < ik−1 of
integers from the set {1, 2, . . . , n − 1}.
n−1
.
Number of terms in second sum = k−1
I
Total number of terms = 2n−1 .
I
Unlike Faa di Bruno’s formula there are no factorials!
I
Examples
[x0 , x1 ](g ◦ f ) = [f (x0 ), f (x1 )]g [x0 , x1 ]f ,
[x0 , x1 , x2 ](g ◦ f ) = [f (x0 ), f (x2 )]g [x0 , x1 , x2 ]f
+ [f (x0 ), f (x1 ), f (x2 )]g [x0 , x1 ]f [x1 , x2 ]f ,
[x0 , x1 , x2 , x3 ](g ◦ f ) = [f 0 , f 3 ]g [x0 , x1 , x2 , x3 ]f
+ [f 0 , f 1 , f 3 ]g [x0 , x1 ]f [x1 , x2 , x3 ]f
+ [f 0 , f 2 , f 3 ]g [x0 , x1 , x2 ]f [x2 , x3 ]f
+ [f 0 , f 1 , f 2 , f 3 ]g [x0 , x1 ]f [x1 , x2 ]f [x2 , x3 ]f .
Second chain rule for divided differences
[x0 , . . . , xn ](g ◦ f ) =
n
X
[f (x0 ), . . . , f (xk )]g [xk , . . . , xn ]([x0 , ·]f · · · [xk−1 , ·]f ).
k=1
I
The last part can be expanded using the product rule.
n−1
, the same as in the
Number of terms of order k = k−1
previous formula.
I
Total number of terms = 2n−1 , as before.
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Again, no factorials!
I
Examples
[x0 , x1 ](g ◦ f ) = [f (x0 ), f (x1 )]g [x 0 , x1 ]f ,
[x0 , x1 , x2 ](g ◦ f ) = [f (x0 ), f (x1 )]g [x 0 , x1 , x2 ]f
+ [f (x0 ), f (x1 ), f (x2 )]g [x 0 , x2 ]f [x 1 , x2 ]f ,
[x0 , x1 , x2 , x3 ](g ◦ f ) = [f 0 , f 1 ]g [x 0 , x1 , x2 , x3 ]f
+ [f 0 , f 1 , f 2 ]g [x 0 , x2 ]f [x 1 , x2 , x3 ]f
+ [f 0 , f 1 , f 2 ]g [x 0 , x2 , x3 ]f [x 1 , x3 ]f
+ [f 0 , f 1 , f 2 , f 3 ]g [x 0 , x3 ]f [x 1 , x3 ]f [x 2 , x3 ]f .
Polynomial interpolation, uniform parameterization
Theorem. If ti = i, i = 0, 1, . . . , n then
kf ◦ φ − pk[t0 ,tn ] ≤ Cn (sn − s0 )2 kf 00 kφ([t0 ,tn ]) .
Why? Example: n = 2:
f(φ(t)) − p(t) = t(t − 1)(t − 2)[0, 1, 2, t](f ◦ φ),
and, with s = φ(t), and if 0 ≤ t ≤ 1 < 2,
[0, t, 1, 2](f ◦ φ) = [s0 , s, s2 ]f [0, t]φ [t, 1, 2]φ
+ [s0 , s1 , s2 ]f [0, t, 1]φ [1, 2]φ
+ [s0 , s, s1 , s2 ]f [0, t]φ [t, 1]φ [1, 2]φ.
Since [s0 , s, s1 , s2 ]f = ([s, s1 , s2 ]f − [s0 , s, s1 ]f)/(s2 − s0 ), and
|[u, v ]φ|, |[u, v , w ]φ| ≤ K (s2 − s0 ), result follows.
Polynomial interpolation, chordal parameterization
Theorem. If t0 = 0 and ti+1 = ti + kxi+1 − xi k, i = 0, 1, . . . , n,
then
kf ◦ φ − pk[t0 ,tn ] ≤ Cn γs (sn − s0 )4 kf (4) kφ([t0 ,tn ]) .
Why? Because chord lengths approximate arc lengths. If
kf 0 (s)k = 1, then
0 < (si+1 − si ) − (ti+1 − ti ) ≤
Use divided difference chain rule...
1
(si+1 − si )3 kf 00 k2 .
12
Proof of chain rule
Use product rule (Popoviciu, 1933, and Steffensen, 1939). For
product of r functions,
[x0 , . . . , xn ](f1 f2 · · · fr ) =
X
r
Y
[xij−1 , . . . , xij ]fj .
0=i0 ≤i1 ≤···≤ir =n j=1
Use divided difference of a monomial. If n ≤ r ,
X
[x0 , . . . , xn ](·)r =
x0µ0 x1µ1 · · · xnµn .
µ0 +···+µn =r −n
µi ≥0
By interpolation, enough to prove the formula when g is a
polynomial of degree ≤ n. By linearity, enough to prove it when
g (y ) = y r , r ≤ n. Then (g ◦ f )(x) = f r (x) and by product rule,
X
[x0 , . . . , xn ](f r ) =
r
Y
[xαβ−1 , . . . , xαβ ]f .
0=α0 ≤α1 ≤···≤αr =n β=1
Take d.d.’s of order 0 outside the product: if
(α0 , α1 , . . . , αr ) = (i0 , . . . , i0 , i1 , . . . , i1 , . . . , ik , . . . , ik )
| {z } | {z }
| {z }
1+µ0
1+µ1
1+µk
for some k, 1 ≤ k ≤ r , and 0 = i0 < i1 < · · · < ik = n,
µ0 + · · · + µk = r − k, and µi ≥ 0 then
r
Y
β=1
[xαβ−1 , . . . , xαβ ]f =
fi0µ0 fi1µ1
µk
· · · fik
k
Y
[xij−1 , . . . , xij ]f .
j=1
It follows that
[x0 , . . . , xn ](f r )
=
r
X
X
X
µ0
µk
fi0 · · · fik
k=1 0=i0 <···<ik =n µ0 +···+µk =r −k
=
r
X
X
j=1
[fi0 , . . . , fik ](·)r
k=1 0=i0 <···<ik =n
=
n
X
X
k=1 0=i0 <···<ik =n
QED.
k
Y
[xij−1 , . . . , xij ]f
k
Y
[xij−1 , . . . , xij ]f
j=1
k
Y
[fi0 , . . . , fik ](·)
[xij−1 , . . . , xij ]f .
r
j=1
Chain rule for multivariate divided differences
Divided differences of a bivariate function f (x, y ): let
[x0 ; y0 ]f = f (x0 , y0 ) and
[x0 , . . . , xm ;y0 , . . . , yn ]f
[x1 , . . . , xm ; y0 , . . . , yn ]f − [x0 , . . . , xm−1 ; y0 , . . . , yn ]f
=
xm − x0
[x0 , . . . , xm ; y1 , . . . , yn ]f − [x0 , . . . , xm ; y0 , . . . , yn−1 ]f
.
=
yn − y0
lim [x , . . . , xm ; y0 , . . . , yn ]f =
,...,x →x 0
x0
m
y0 ,...,yn →y
D (m,n) f (x, y )
m!n!
Introduce shorthand
Univariate divided difference:
[x; i, j]f := [xi , . . . , xj ]f ,
i ≤ j.
p-variate divided difference:
[x; i, j]f = [xi11 , . . . , xj11 ; . . . ; xipp , . . . , xjpp ]f
where i = (i 1 , . . . , i p ), j = (j 1 , . . . , j p ) and i ≤ j.
1
p
D1n · · · Dpn f (x)
D n f (x)
lim [x; 0, n]f =
=
.
xk →x
n1 ! · · · np !
n!
p-variate product rule
Two functions:
[x; 0, n](fg ) =
X
[x; 0, i]f [x; i, n]g .
0≤i≤n
r functions:


r
Y
[x; 0, n]  fj  =
j=1
r
Y
[x; ij−1 , ij ]fj .
X
0=i0 ≤i1 ≤···≤ir =n j=1
n
n
i2
i
i1
0
0
i3
(p, q)-variate chain rule
With f = (f1 , . . . , fq ) : Rp → Rq , g : Rq → R, and fr ,j := fr (xj ),
[x; 0, n](g ◦ f) =
|n|
X
X
[f1,ij0 , . . . , f1,ij1 ; . . . ; fq,ijq−1 , . . . , fq,ijq ]g ×
k=1 0=i0 <···<ik =n
0=j0 ≤···≤jq =k
q
Y
jr
Y
r =1 j=jr −1 +1
[x; ij−1 , ij ]fr .
Example 1: p = 2, q = 1, n = (2, 1)
The formula is a sum over eight terms, one for each of the paths:
k =1:
1 path
k =2:
4 paths
k =3:
3 paths
[x; 00, 21](g ◦ f ) = [f00 , f21 ]g [x; 00, 21]f
+ [f00 , f10 , f21 ]g [x; 00, 10]f [x; 10, 21]f
+ [f00 , f20 , f21 ]g [x; 00, 20]f [x; 20, 21]f
+ [f00 , f01 , f21 ]g [x; 00, 01]f [x; 01, 21]f
+ [f00 , f11 , f21 ]g [x; 00, 11]f [x; 11, 21]f
+ [f00 , f10 , f20 , f21 ]g [x; 00, 10]f [x; 10, 20]f [x; 20, 21]f
+ [f00 , f10 , f11 , f21 ]g [x; 00, 10]f [x; 10, 11]f [x; 11, 21]f
+ [f00 , f01 , f11 , f21 ]g [x; 00, 01]f [x; 01, 11]f [x; 11, 21]f .
Example 2: p = 2, q = 2, n = (1, 1)
The formula is a sum over the three paths:
k =1:
1 path
k =2:
2 paths: left path, right path
A path of length k gives rise to k + 1 terms: the number of ways
to split a path of length k into two sub-paths. So there are
2 × 1 + 3 × 2 = 8 terms.
With a = f1 and b = f2 ,
[x; 00, 11](g ◦ f) = [a00 , a11 ; b11 ]g [x; 00, 11]a
+ [a00 ; b00 , b11 ]g [x; 00, 11]b
+ [a00 , a10 , a11 ; b11 ]g [x; 00, 10]a [x; 10, 11]a
+ [a00 , a10 ; b10 , b11 ]g [x; 00, 10]a [x; 10, 11]b
+ [a00 ; b00 , b10 , b11 ]g [x; 00, 10]b [x; 10, 11]b
+ [a00 , a01 , a11 ; b11 ]g [x; 00, 01]a [x; 01, 11]a
+ [a00 , a01 ; b01 , b11 ]g [x; 00, 01]a [x; 01, 11]b
+ [a00 ; b00 , b01 , b11 ]g [x; 00, 01]b [x; 01, 11]b.
Further work
If g (f (x)) = x, express divided differences of f in terms of divided
differences of g .
If f (x, y (x)) = 0, express divided differences of y in terms of
(bivariate) divided differences of f .