explicit formula for divided differences

explicit formula for divided differences

Theorem 1

.

The

n

-th

divided difference

of a

function

f

can be written explicitly as

Δ

n

⁢

f

⁢

[

x

0

,

…

,

x

n

]

=

∑

i

=

0

n

f

⁢

(

x

i

)

∏

0

≤

j

≤

n

j

≠

i

(

x

i

-

x

j

)

Proof.

We will proceed by recursion on

n

. When

n

=

1

, the formula to be proven
reduces to

Δ

1

⁢

f

⁢

[

x

0

,

x

1

]

=

f

⁢

(

x

0

)

x

0

-

x

1

+

f

⁢

(

x

1

)

x

1

-

x

0

,

which agrees with the definition of

Δ

1

⁢

f

.

To prove that this is correct when

n

>

1

, one needs to check that it the

recurrence relation

for divided differences.

∑

0

≤

i

≤

n

+

1

i

≠

0

f

⁢

(

x

i

)

∏

0

≤

j

≤

n

+

1

j

≠

i

j

≠

0

(

x

i

-

x

j

)

-

∑

0

≤

i

≤

n

+

1

i

≠

1

f

⁢

(

x

i

)

∏

0

≤

j

≤

n

+

1

j

≠

i

j

≠

1

(

x

i

-

x

j

)

=

f

⁢

(

x

1

)

∏

0

≤

j

≤

n

+

1

j

≠

i

j

≠

0

(

x

1

-

x

j

)

-

f

⁢

(

x

0

)

∏

0

≤

j

≤

n

+

1

j

≠

i

j

≠

1

(

x

0

-

x

j

)

+

⁢

∑

2

≤

i

≤

n

+

1

f

⁢

(

x

i

)

⁢

(

1

∏

0

≤

j

≤

n

+

1

j

≠

i

j

≠

0

(

x

i

-

x

j

)

-

1

∏

0

≤

j

≤

n

+

1

j

≠

i

j

≠

1

(

x

i

-

x

j

)

)

=

-

(

x

0

-

x

1

)

⁢

f

⁢

(

x

0

)

∏

0

≤

j

≤

n

+

1

j

≠

i

(

x

0

-

x

j

)

+

(

x

1

-

x

0

)

⁢

f

⁢

(

x

1

)

∏

0

≤

j

≤

n

+

1

j

≠

i

(

x

1

-

x

j

)

+

⁢

∑

2

≤

i

≤

n

+

1

f

⁢

(

x

i

)

⁢

(

x

i

-

x

0

∏

0

≤

j

≤

n

+

1

j

≠

i

(

x

i

-

x

j

)

-

x

i

-

x

1

∏

0

≤

j

≤

n

+

1

j

≠

i

(

x

i

-

x

j

)

)

=

(

x

1

-

x

0

)

⁢

f

⁢

(

x

0

)

∏

0

≤

j

≤

n

+

1

j

≠

i

(

x

0

-

x

j

)

+

(

x

1

-

x

0

)

⁢

f

⁢

(

x

1

)

∏

0

≤

j

≤

n

+

1

j

≠

i

(

x

1

-

x

j

)

+

(

x

1

-

x

0

)

⁢

∑

2

≤

i

≤

n

+

1

(

x

1

-

x

0

)

⁢

f

⁢

(

x

i

)

∏

0

≤

j

≤

n

+

1

j

≠

i

(

x

i

-

x

j

)

=

(

x

1

-

x

0

)

⁢

∑

0

≤

i

≤

n

+

1

f

⁢

(

x

i

)

∏

0

≤

j

≤

n

+

1

j

≠

i

(

x

i

-

x

j

)

Thus, we see that, if

Δ

n

⁢

f

⁢

[

x

0

,

…

,

x

n

]

=

∑

i

=

0

n

f

⁢

(

x

i

)

∏

0

≤

j

≤

n

j

≠

i

(

x

i

-

x

j

)

,

then

Δ

n

+

1

⁢

f

⁢

[

x

0

,

…

,

x

n

+

1

]

=

∑

i

=

0

n

+

1

f

⁢

(

x

i

)

∏

0

≤

j

≤

n

+

1

j

≠

i

(

x

i

-

x

j

)

.

Hence, by induction, the formula holds for all

n

.
∎

This formula may be phrased another way by introducing the

polynomials

p

n

defined as

p

n

⁢

(

x

)

=

∏

i

=

0

n

(

x

-

x

i

)

.

We may write

Δ

n

⁢

f

⁢

[

x

0

,

…

,

x

n

]

=

∑

i

=

0

n

f

⁢

(

x

i

)

p

′

⁢

(

x

i

)

.

Either form of the

explicit formula

makes it obvious that divided differences
are

symmetric functions

of

x

0

,

x

1

,

…

.

Title

explicit formula for divided differences

Canonical name

ExplicitFormulaForDividedDifferences

Date of creation

2013-03-22 14:41:16

Last modified on

2013-03-22 14:41:16

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

21

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 39A70

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