solution of equations by divided difference interpolaton

solution of equations by divided difference interpolaton

Divided diference interpolation can be used to obtain approximate
solutions to equations and to invert

functions

numerically. The
idea is that, given an equation

f

⁢

(

y

)

=

x

which we want to solve
for

y

, we first take several numbers

y

1

,

…

,

y

n

and
compute

x

1

⁢

…

⁢

x

n

as

x

i

=

f

⁢

(

y

i

)

. Then we compute the

divided differences

of the

y

i

’s regarded as functions of the

x

i

’s and form the divided difference series. Substituting

x

in this series provides an approximation to

y

.

To illustrate how this works, we will examine the

transcendental
equation

x

+

e

-

x

=

2

. We note that

2

+

e

-

2

=

2.13533

and

1.5

+

e

-

1.5

=

1.72313

, so there will be a solution
between

1.5

and

2

, likely closer to

2

than

1.5

. Therefore,
as our values of the

y

i

’s, we shall take

1.5

,

1.6

,

1.7

,

1.8

,

1.9

,

2.0

,

2.1

. We now tabulate

x

i

=

y

i

+

e

-

y

i

for those values:

y

i

x

i

1.5

1.72313

1.6

1.80190

1.7

1.88268

1.8

1.96530

1.9

2.04957

2.0

2.13533

2.1

2.22246

Next, we form a

divided difference table

of the

y

i

’s as a
function of the

x

i

’s:

1.72313

1.50000

1.26952

1.80190

1.60000

-

0.19799

1.23793

0.12082

1.88268

1.70000

-

0.16873

-

0.039609

1.21036

0.10789

0.091553

1.96530

1.80000

-

0.14201

-

0.077347

-

0.13457

1.18666

0.08210

0.024360

2.04957

1.90000

-

0.12127

-

0.067102

1.16604

0.05930

2.13533

2.00000

-

0.10602

1.14771

2.22246

2.10000

From this table, we form the series

1.50000

+

1.26952

⁢

(

x

-

1.72313

)

-

0.19799

⁢

(

x

-

1.72313

)

⁢

(

x

-

1.80190

)

+

0.12082

⁢

(

x

-

1.72313

)

⁢

(

x

-

1.80190

)

⁢

(

x

-

1.88268

)

-

0.039609

⁢

(

x

-

1.72313

)

⁢

(

x

-

1.80190

)

⁢

(

x

-

1.88268

)

⁢

(

x

-

1.96530

)

+

0.091553

⁢

(

x

-

1.72313

)

⁢

(

x

-

1.80190

)

⁢

(

x

-

1.88268

)

⁢

(

x

-

1.96530

)

⁢

(

x

-

2.04957

)

-

0.13457

⁢

(

x

-

1.72313

)

⁢

(

x

-

1.80190

)

⁢

(

x

-

1.88268

)

⁢

(

x

-

1.96530

)

⁢

(

x

-

2.04957

)

⁢

(

x

-

2.13533

)

Substituting

2.00000

for

x

, we obtain

1.84140

. Given that

1.84140

+

e

-

1.84140

<

2

<

1.84141

+

e

-

1.84141

,

this answer is correct to all

5

decimal places.

In the presentation above, we tacitly assumed that there was a
solution to our equation and focussed our attention on finding
that answer numerically. To

complete

the treatment we will
now show that there indeed exists a unique solution to the
equation

x

+

e

-

x

=

2

in the

interval

(

0

,

∞

)

.

Existence

follows from the

intermediate value theorem

. As
noted above,

1.5

+

e

-

1.5

<

2

<

3

+

e

-

2

.

Since

x

+

e

-

x

depends continuously on

x

, it follows
that there exists

x

∈

(

1.5

,

2

)

such that

x

+

e

-

x

=

2

.

As for uniqueness, note that the

derivative

of

x

+

e

-

x

is

1

-

e

-

x

. When

x

>

0

, we have

e

-

x

<

1

, or

1

-

e

-

x

>

0

. Hence,

x

+

e

-

x

is a strictly increaing
function of

x

, so there can be at most one

x

such that

x

+

e

-

x

=

2

.

Title

solution of equations by divided difference interpolaton

Canonical name

SolutionOfEquationsByDividedDifferenceInterpolaton

Date of creation

2013-03-22 16:49:22

Last modified on

2013-03-22 16:49:22

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

16

Author

rspuzio (6075)

Entry type

Application

Classification

msc 39A70

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