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Interpolation Introduction

In many practical situations, for a function which either may not be explicitly

specified or may be difficult to handle, we often have a tabulated data where

and for In such cases, it may be required to represent

or replace the given function by a simpler function, which coincides with the values of at the

tabular points This process is known as INTERPOLATION.

Interpolation is also used to estimate the value of the function at the non tabular points. Here, we shall consider only those functions which are sufficiently smooth, i.e., they are differentiable sufficient number of times. Many of the interpolation methods, where the tabular points are equally spaced, use difference operators. Hence, in the following we introduce various difference operators and study their properties before looking at the interpolation methods.

We shall assume here that the TABULAR POINTS are equally spaced, i.e.,

for each The real number is called the STEP LENGTH. This

gives us Further, gives the value of the function at the

tabular point. The points are known as NODES or NODAL VALUES.

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Difference Operator Forward Difference Operator

DEFINITION 11.2.1 (First Forward Difference Operator)

We define the FORWARD DIFFERENCE OPERATOR, denoted by as

The expression gives the FIRST FORWARD DIFFERENCE of and the operator is

called the FIRST FORWARD DIFFERENCE OPERATOR. Given the step size this formula uses the values at

and the point at the next step. As it is moving in the forward direction, it is called the forward difference operator.

DEFINITION 11.2.2 (Second Forward Difference Operator) The second forward difference operator,

is defined as

We note that

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In particular, for we get,

and

DEFINITION 11.2.3 ( Forward Difference Operator) The forward difference operator, is defined as

EXERCISE 11.2.4 Show that In general, show that for any positive

integers and with

EXAMPLE 11.2.5 For the tabulated values of find and

Solution: Here,

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Remark 11.2.6 Using mathematical induction, it can be shown that

Thus the forward difference at uses the values at

EXAMPLE 11.2.7 If where and are real constants, calculate

Solution: We first calculate as follows:

Now,

Thus, for all

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Remark 11.2.8 In general, if is a polynomial

of degree then it can be shown that

The reader is advised to prove the above statement.

Remark 11.2.9

1. For a set of tabular values, the horizontal forward difference table is written as:

……..

2. In many books, a diagonal form of the difference table is also used. This is written as:

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………

However, in the following, we shall mostly adhere to horizontal form only.

Backward Difference Operator

DEFINITION 11.2.10 (First Backward Difference Operator) The FIRST BACKWARD DIFFERENCE OPERATOR,

denoted by is defined as

Given the step size note that this formula uses the values at and the point at the previous step. As it moves in the backward direction, it is called the backward difference operator.

DEFINITION 11.2.11 ( Backward Difference Operator) The backward difference operator,

is defined as

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In particular, for we get

Note that

EXAMPLE 11.2.12 Using the tabulated values in Example 11.2.5, find and

Solution: We have and

EXAMPLE 11.2.13 If where and are real constants, calculate

Solution: We first calculate as follows:

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Now,

Thus, for all

Remark 11.2.14 For a set of tabular values, backward difference table in the horizontal form is written as:

. .. .

EXAMPLE 11.2.15 For the following set of tabular values write the forward and backward difference tables.

9 10 11 12 13 14

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5.0 5.4 6.0 6.8 7.5 8.7

Solution: The forward difference table is written as

9 5 0.4 = 5.4 - 5 0.2 = 0.6 - 0.4 0= 0.2-0.2 -.3 = -0.3 - 0.0 0.6 = 0.3 - (-0.3)

10 5.4 0.6 0.2 -0.3 0.3

11 6.0 0.8 -0.1 0.0

12 6.8 0.7 -0.1

13 7.5 0.6

14 8.1

In the similar manner, the backward difference table is written as follows:

9 5

10 5.4 0.4

11 6 0.6 0.2

12 6.8 0.8 0.2 0.0

13 7.5 0.7 -0.1 - 0.3 -0.3

14 8.1 0.6 -0.1 0.0 0.3 0.6

Observe from the above two tables that , etc.

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EXERCISE 11.2.16

1. Show that

2. Prove that

3. Obtain in terms of Hence show that

Remark 11.2.17 In general it can be shown that or

Remark 11.2.18 In view of the remarks (11.2.8) and (11.2.17) it is obvious that, if is a

polynomial function of degree then is constant and for

Central Difference Operator

DEFINITION 11.2.19 (Central Difference Operator) The FIRST CENTRAL DIFFERENCE OPERATOR, denoted

by is defined by

and the #MATH5164# MATHEND000# CENTRAL DIFFERENCE OPERATOR is defined as

Thus,

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In particular, for define and then

Thus, uses the table of It is easy to see that only the even central differences use the

tabular point values

Shift Operator

DEFINITION 11.2.20 (Shift Operator) A SHIFT OPERATOR, denoted by is the operator which shifts the

value at the next point with step i.e.,

Thus,

Averaging Operator

DEFINITION 11.2.21 (Averaging Operator) The AVERAGING OPERATOR, denoted by gives the average value between two central points, i.e.,

Thus and

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Relations between Difference operators 1. We note that

Thus,

2. Further, Thus,

Thus gives us

So we write,

Similarly,

3. Let us denote by Then, we see that

Thus,

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Recall,

So, we have,

That is, the action of is same as that of

4. We further note that,

5. and

6. Thus,

7. 8. i.e.,

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9. 10. In view of the above discussion, we have the following table showing the relations

between various difference operators:

E

E E

EXERCISE 11.3.1

1. Verify the validity of the above table. 2. Obtain the relations between the averaging operator and other difference operators.

3. Find and for the following tabular values:

0 1 2 3 4

93.0 96.5 100.0 103.5 107.0

11.3 12.5 14.0 15.2 16.0

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Newton's Interpolation Formulae As stated earlier, interpolation is the process of approximating a given function, whose values are

known at tabular points, by a suitable polynomial, of degree which takes the

values at for Note that if the given data has errors, it will also be reflected in the polynomial so obtained.

In the following, we shall use forward and backward differences to obtain polynomial function

approximating when the tabular points 's are equally spaced. Let

where the polynomial is given in the following form:

(11.4.1)

for some constants to be determined using the fact that for

So, for substitute in (11.4.1) to get This gives us Next,

So, For or equivalently

Thus, Now, using mathematical induction, we get

Thus,

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As this uses the forward differences, it is called NEWTON'S FORWARD DIFFERENCE FORMULA for interpolation, or simply, forward interpolation formula. EXERCISE 11.4.1 Show that

and and in general,

For the sake of numerical calculations, we give below a convenient form of the forward interpolation formula.

Let then

With this transformation the above forward interpolation formula is simplified to the following form:

(11.4.2)

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If =1, we have a linear interpolation given by

(11.4.3)

For we get a quadratic interpolating polynomial:

(11.4.4)

and so on.

It may be pointed out here that if is a polynomial function of degree then

coincides with on the given interval. Otherwise, this gives only an approximation to the

true values of

If we are given additional point also, then the error, denoted by

is estimated by

Similarly, if we assume, is of the form

then using the fact that we have

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Thus, using backward differences and the transformation we obtain the Newton's backward interpolation formula as follows:

(11.4.5)

EXERCISE 11.4.2 Derive the Newton's backward interpolation formula (11.4.5) for Remark 11.4.3 If the interpolating point lies closer to the beginning of the interval then one uses the Newton's forward formula and if it lies towards the end of the interval then Newton's backward formula is used. Remark 11.4.4 For a given set of n tabular points, in general, all the n points need not be used for interpolating polynomial. In fact N is so chosen that forward/backward difference almost remains constant. Thus N is less than or equal to n. EXAMPLE 11.4.5

1. Obtain the Newton's forward interpolating polynomial, for the following tabular data and interpolate the value of the function at

x 0 0.001 0.002 0.003 0.004 0.005

y 1.121 1.123 1.1255 1.127 1.128 1.1285

2. Solution: For this data, we have the Forward difference difference table

0 1.121 0.002 0.0005 -0.0015 0.002 -.0025

.001 1.123 0.0025 -0.0010 0.0005 -0.0005

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.002 1.1255 0.0015 -0.0005 0.0

.003 1.127 0.001 -0.0005

.004 1.128 0.0005

.005 1.1285

3. Thus, for where and we get

4. Thus,

5.

6. Using the following table for approximate its value at Also, find an error

estimate (Note ).

0.70 72 0.74 0.76 0.78

0.84229 0.87707 0.91309 0.95045 0.98926

7. Solution: As the point lies towards the initial tabular values, we shall use Newton's Forward formula. The forward difference table is:

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0.70 0.84229 0.03478 0.00124 0.0001 0.00001

0.72 0.87707 0.03602 0.00134 0.00011

0.74 0.91309 0.03736 0.00145

0.76 0.95045 0.03881

0.78 0.98926

8. In the above table, we note that is almost constant, so we shall attempt degree polynomial interpolation.

9. Note that gives Thus, using forward

interpolating polynomial of degree we get

10. 11.

12. An error estimate for the approximate value is

13.

14. Note that exact value of (upto decimal place) is and the approximate value, obtained using the Newton's interpolating polynomial is very close to this value. This is also reflected by the error estimate given above.

15. Apply degree interpolation polynomial for the set of values given in Example

11.2.15, to estimate the value of by taking

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Also, find approximate value of Solution: Note that is closer to the values lying in the beginning of tabular values, while is towards the end of tabular values. Therefore, we shall use forward difference formula for and the backward difference formula for

Recall that the interpolating polynomial of degree is given by

Therefore,

1. for and we have This gives,

2.

3. for and we have This gives,

4.

Note: as is closer to we may expect estimate calculated using

to be a better approximation.

5. for we use the backward interpolating polynomial, which gives,

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Therefore, taking and we have

This gives,

EXERCISE 11.4.6

1. Following data is available for a function

x 0 0.2 0.4 0.6 0.8 1.0

y 1.0 0.808 0.664 0.616 0.712 1.0

2. Compute the value of the function at and 3. The speed of a train, running between two station is measured at different distances from

the starting station. If is the distance in from the starting station, then the

speed (in ) of the train at the distance is given by the following table:

x 0 50 100 150 200 250

v(x) 0 60 80 110 90 0

4. Find the approximate speed of the train at the mid point between the two stations.

5. Following table gives the values of the function at the different

values of the tabular points

x 0 0.04 0.08 0.12 0.16 0.20

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S(x) 0 0.00003 0.00026 0.00090 0.00214 0.00419

6. Obtain a fifth degree interpolating polynomial for Compute and also find an error estimate for it.

7. Following data gives the temperatures (in ) between 8.00 am to 8.00 pm. on May 10, 2005 in Kanpur:

Time 8 am 12 noon 4 pm 8pm

Temperature 30 37 43 38

8. Obtain Newton's backward interpolating polynomial of degree to compute the temperature in Kanpur on that day at 5.00 pm.

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Lagrange's Interpolation Formula Introduction

In the previous chapter, we derived the interpolation formula when the values of the function are

given at equidistant tabular points However, it is not always possible to obtain

values of the function, at equidistant interval points, In view of this, it is desirable to derive an interpolating formula, which is applicable even for unequally distant points. Lagrange's Formula is one such interpolating formula. Unlike the previous interpolating formulas, it does not use the notion of differences, however we shall introduce the concept of divided differences before coming to it.

Divided Differences DEFINITION 12.2.1 (First Divided Difference) The ratio

for any two points and is called the FIRST DIVIDED DIFFERENCE of relative to

and It is denoted by

Let us assume that the function is linear. Then is constant for any two tabular

points and i.e., it is independent of and Hence,

Thus, for a linear function if we take the points and then, i.e.,

Thus,

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So, if is approximated with a linear polynomial, then the value of the function at any point

can be calculated by using where is the

first divided difference of relative to and

DEFINITION 12.2.2 (Second Divided Difference) The ratio

is defined as SECOND DIVIDED DIFFERENCE of relative to and

If is a second degree polynomial then is a linear function of Hence,

In view of the above, for a polynomial function of degree 2, we have

Thus,

This gives,

From this we obtain,

So, whenever is approximated with a second degree polynomial, the value of at any point can be computed using the above polynomial, which uses the values at three points

and

EXAMPLE 12.2.3 Using the following tabular values for a function obtain its second degree polynomial approximation.

0 1 2

0.1 0.16 0.2

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1.12 1.24 1.40

Also, find the approximate value of the function at Solution: We shall first calculate the desired divided differences.

Thus,

Therefore

EXERCISE 12.2.4

1. Using the following table, which gives values of corresponding to certain values

of find approximate value of with the help of a second degree polynomial.

322.8 324.2 325

2.50893 2.51081 2.5118

2. Show that

So,

That is, the second divided difference remains unchanged regardless of how its arguments are interchanged.

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3. Show that for equidistant points and where

and 4. Show that for a linear function, the second divided difference with respect to any three

points, and is always zero.

Now, we define the divided difference.

DEFINITION 12.2.5 ( Divided Difference) The #MATH5307# MATHEND000# DIVIDED

DIFFERENCE of relative to the tabular points is defined recursively as

It can be shown by mathematical induction that for equidistant points,

(12.2.1)

where, and

In general,

where and is the length of the interval for Remark 12.2.6 In view of the remark (11.2.18) and (12.2.1

EXAMPLE 12.2.7 Show that can be written as

), it is easily seen that for a

polynomial function of degree the divided difference is constant and the divided difference is zero.

Solution:By definition, we have

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so, Now since,

we get the desired result.

EXERCISE 12.2.8 Show that can be written in the following form:

where,

and

Further show that for

Remark 12.2.9 In general it can be shown that where,

and

Here, is called the remainder term.

It may be observed here that the expression is a polynomial of degree and

for

Further, if is a polynomial of degree then in view of the Remark 12.2.6

, the remainder

term, as it is a multiple of the divided difference, which is 0 .

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Lagrange's Interpolation formula In this section, we shall obtain an interpolating polynomial when the given data has unequal tabular points. However, before going to that, we see below an important result.

THEOREM 12.3.1 The divided difference can be written as:

Proof. We will prove the result by induction on The result is trivially true for For

Let us assume that the result is true for i.e.,

Consider then the divided difference is

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which on rearranging the terms gives the desired result. Therefore, by mathematical induction, the proof of the theorem is complete. height6pt width 6pt depth 0pt

Remark 12.3.2 In view of the theorem 12.3.1

Now, if a function is approximated by a polynomial of degree then , its divided

difference relative to will be zero,(Remark

the divided difference of a function remains unchanged regardless of how its arguments are interchanged, i.e., it is independent of the order of its arguments.

12.2.6

) i.e.,

Using this result, Theorem 12.3.1

gives

or,

which gives ,

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Note that the expression on the right is a polynomial of degree and takes the value at

for

This polynomial approximation is called LAGRANGE'S INTERPOLATION FORMULA.

Remark 12.3.3 In view of the Remark (12.2.9

Remark 12.3.4 We have seen earlier that the divided differences are independent of the order of its arguments. As the Lagrange's formula has been derived using the divided differences, it is not necessary here to have the tabular points in the increasing order. Thus one can use

Lagrange's formula even when the points are in any order, which was not possible in the case of Newton's Difference formulae.

), we can observe that is another form of

Lagrange's Interpolation polynomial formula as obtained above. Also the remainder term gives an estimate of error between the true value and the interpolated value of the function.

Remark 12.3.5 One can also use the Lagrange's Interpolating Formula to compute the value of

for a given value of This is done by interchanging the roles of and i.e. while

using the table of values, we take tabular points as and nodal points are taken as

EXAMPLE 12.3.6 Using the following data, find by Lagrange's formula, the value of at

0 1 2 3 4

9.3 9.6 10.2 10.4 10.8

11.40 12.80 14.70 17.00 19.80

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Also find the value of where

Solution: To compute we first calculate the following products:

Thus,

Now to find the value of such that we interchange the roles of and and calculate the following products:

Thus,the required value of is obtained as:

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EXERCISE 12.3.7 The following table gives the data for steam pressure vs temperature :

360 365 373 383 390

154.0 165.0 190.0 210.0 240.0

Compute the pressure at

EXERCISE 12.3.8 Compute from following table the value of for :

5.60 5.90 6.50 6.90 7.20

2.30 1.80 1.35 1.95 2.00

Also find the value of where

Gauss's and Stirling's Formulas In case of equidistant tabular points a convenient form for interpolating polynomial can be derived from Lagrange's interpolating polynomial. The process involves renaming or re-designating the tabular points. We illustrate it by deriving the interpolating formula for 6 tabular points. This can be generalized for more number of points. Let the given tabular points be

These six points in the given order are not equidistant. We re-designate them for the sake of convenience as follows:

These re-designated tabular points in their given order are equidistant. Now recall from remark (12.3.3

) that Lagrange's interpolating polynomial can also be written as :

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which on using the re-designated points give:

Now note that the points and are equidistant and the divided difference are independent of the order of their arguments. Thus, we have

where for Now using the above relations and the transformation

we get

Thus we get the following form of interpolating polynomial

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(12.4.1)

Similarly using the tabular points

and the re-

designating them, as and we get another form of interpolating polynomial as follows:

(12.4.2)

Now taking the average of the two interpoating polynomials (12.4.1) and (12.4.2

) (called GAUSS'S FIRST AND SECOND INTERPOLATING FORMULAS, respectively), we obtain Sterling's Formula of interpolation:

(12.4.3)

These are very helpful when, the point of interpolation lies near the middle of the interpolating interval. In this case one usually writes the diagonal form of the difference table. EXAMPLE 12.4.1 Using the following data, find by Sterling's formula, the value of

at

0.20 0.21 0.22 0.23 0.24

1.37638 1.28919 1.20879 1.13427 1.06489

Here the point lies near the central tabular point Thus , we define

to get the difference table in diagonal form as:

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(here,

and etc.).

Using the Sterling's formula with we get as follows:

Note that tabulated value of at is 1.1708496.

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EXERCISE 12.4.2 Compute from the following table the value of for :

0.00 0.02 0.04 0.06 0.08

0.00000 0.02256 0.04511 0.06762 0.09007

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