Note on osculatory rational interpolation
Authors: Salzer Herbert E. 
Published: June 2010

In n-point osculatory interpolation of order

r
i

−

1

{r_i}\, - \,1

at points

x
i

{x_i}

,

i

=

1
,

2
,

⋯

,

n
,

i\, = \,1,\,2,\, \cdots \,,\,n,

, by a rational expression

N
(
x
)

/

D
(
x
)

{N(x)}/{D(x)}

, where

N
(
x
)

N(x)

and

D
(
x
)

D(x)

are polynomials

∑

a
j

x
j

\sum {{a_j}{x^j}}

and

∑

b
j

x
j

\sum {{b_j}{x^j}}

, We use the lemma that the system (1)
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="left-brace upper N left-parenthesis x Subscript i Baseline right-parenthesis slash upper D left-parenthesis x Subscript i Baseline right-parenthesis right-brace Superscript left-parenthesis m right-parenthesis Baseline equals f Superscript left-parenthesis m right-parenthesis Baseline left-parenthesis x Subscript i Baseline right-parenthesis comma m equals 0 comma 1 comma midline-horizontal-ellipsis comma r Subscript i Baseline minus 1">

{
N
(

x

i

)

/

D
(

x

i

)
}

(
m
)

=

f

(
m
)

(

x

i

)
,

m
=
0
,
1
,
⋯
,

r

i

−
1

{{\{N({{x}_{i}})/D({{x}_{i}})\}}^{(m)}}\,=\,{{f}^{(m)}}({{x}_{i}}),\,m=0,1,\cdots ,{{r}_{i}}-1

, is equivalent to (2)
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="upper N Superscript left-parenthesis m right-parenthesis Baseline left-parenthesis x Subscript i Baseline right-parenthesis equals left-brace f left-parenthesis x Subscript i Baseline right-parenthesis upper D left-parenthesis x Subscript i Baseline right-parenthesis right-brace Superscript left-parenthesis m right-parenthesis Baseline comma m equals 0 comma 1 comma midline-horizontal-ellipsis comma r Subscript i Baseline minus 1 comma upper D left-parenthesis x Subscript i Baseline right-parenthesis not-equals 0">

N

(
m
)

(

x
i

)

=

{

f
(

x
i

)
D
(

x
i

)

}

(
m
)

,

m

=

0
,

1
,

⋯

,

r
i

−

1
,

D
(

x
i

)

≠

0

{N^{(m)}}({{x_i}})\, = \,{\{{f({{x_i}})D({{x_i}})}\}^{(m)}},\,m\, = \,0,\,1,\, \cdots \,,\,{r_i}\, - \,1,\,D({{x_i}})\, \ne \,0

. This equivalence does not require

N
(
x
)

N(x)

or

D
(
x
)

D(x)

to be a polynomial or even a linear combination of given functions. The lemma implies that (1), superficially non-linear in

a
j

{a_j}

and

b
j

{b_j}

, being the same as (2), is actually linear. For the n-point interpolation problem, the linear system, of order

∑

i
=
1

n

r
i

\sum \limits _{i = 1}^n {{r_i}}

, which might be large, is replaceable by separate linear Systems of orders

r
i

{r_i}

(or even

r
i

+

r

i
+
1

+

⋯

+

r

i
+
j

{r_i}\, + \,{r_{i + 1}}\, + \, \cdots \, + \,{r_{i + j}}

when conveniently small) by applying the lemma to the continued fraction (3)
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="upper N left-parenthesis x right-parenthesis slash upper D left-parenthesis x right-parenthesis equals a Subscript 1 comma 0 Baseline plus StartFraction x minus x 1 vertical-bar Over vertical-bar a Subscript 1 comma 1 Baseline EndFraction plus StartFraction x minus x 1 vertical-bar Over vertical-bar a Subscript 1 comma 2 Baseline EndFraction plus midline-horizontal-ellipsis plus StartFraction x minus x 1 vertical-bar Over vertical-bar a Subscript 1 comma r 1 minus 1 Baseline EndFraction plus StartFraction x minus x 1 vertical-bar Over vertical-bar a Subscript 2 comma 0 Baseline EndFraction plus StartFraction x minus x 1 vertical-bar Over vertical-bar a Subscript 2 comma 1 Baseline EndFraction plus midline-horizontal-ellipsis plus StartFraction x minus x 1 vertical-bar Over vertical-bar a Subscript 2 comma r 2 minus 1 Baseline EndFraction plus StartFraction x minus x 1 vertical-bar Over vertical-bar a Subscript 3 comma 0 Baseline EndFraction plus midline-horizontal-ellipsis plus StartFraction x minus x Subscript n minus 1 Baseline vertical-bar Over vertical-bar a Subscript n comma 0 Baseline EndFraction plus StartFraction x minus x Subscript n Baseline vertical-bar Over vertical-bar a Subscript n comma 1 Baseline EndFraction plus midline-horizontal-ellipsis plus StartFraction x minus x Subscript n Baseline vertical-bar Over vertical-bar a Subscript n comma r Sub Subscript n Subscript minus 1 Baseline EndFraction">

N
(
x
)

/

D
(
x
)

=

a

1
,
0

+

x
−

x

1

|

|

a

1
,
1

+

x
−

x

1

|

|

a

1
,
2

+

⋯

+

x
−

x

1

|

|

a

1
,

r

1

−
1

+

x
−

x

1

|

|

a

2
,
0

+

x
−

x

1

|

|

a

2
,
1

+

⋯

+

x
−

x

1

|

|

a

2
,

r

2

−
1

+

x
−

x

1

|

|

a

3
,
0

+

⋯

+

x
−

x

n
−
1

|

|

a

n
,
0

+

x
−
 

x

n

|

|

a

n
,
1

+

⋯

+

x
−

x

n

|

|

a

n
,

r

n

−
1

{N(x)}/{D(x)\,=\,{{a}_{1,0}}\,+\,\tfrac {x- {{x}_{1}}|}{| {{a}_{1,1}} }\,+\,\tfrac {x- {{x}_{1}} |}{| {{a}_{1,2}} }\,+\,\cdots \,+\,\tfrac {x- {{x}_{1}} |}{| {{a}_{1,{{r}_{1}}-1}} }\,+\,\tfrac {x- {{x}_{1}} |}{| {{a}_{2,0}} }\,+\,\tfrac {x- {{x}_{1}} |}{| {{a}_{2,1}} }\,+\,\cdots \,+\,\tfrac {x- {{x}_{1}} |}{| {{a}_{2,{{r}_{2}}-1}} }\,+\,\tfrac {x- {{x}_{1}} |}{| {{a}_{3,0}} }\,+\,\cdots \,+\,\tfrac {x- {{x}_{n-1}} |}{| {{a}_{n,0}} }\,+\,\tfrac {x-\ {{x}_{n}} |}{| {{a}_{n,1}} }\,+\,\cdots \,+\tfrac {x- {{x}_{n}} |}{| {{a}_{n,{{r}_{n}}-1}} }}\;

. In (3), which has the property (proven in two ways) that the determination of

a

i
,
m

{a_{i,m}}

is independent of all a’s that follow, we find

a

i
,
m

{a_{i,m}}

stepwise, but several at a time (instead of singly which is more tedious), retrieving them readily from the solutions of those lower-order linear systems.