INTERPOLATION

Introduction

• The process of estimating intermediate values between given data points is called Interpolation .
• The most common method used for this purpose is Polynomial Interpolation.

Find the functional value for 𝑥 = 2.5 , i.e. 𝑓 2.5 = ? Is this a problem of Interpolation or extrapolation??

𝑥	1	3	4	6	9
𝑓(𝑥)	4	16	43	67	23

Solution:
Step 1: Obtain the polynomial equation using tabular data points.
Suppose we got our polynomial as,
𝑓 𝑥 = 4𝑥
4 + 3𝑥
2 + 2
Step 2 : Now, put 𝑥 = 2.5 in the above expression of 𝑓 𝑥 which gives the desired answer

Note: The number of data points minus one defines the order of interpolation. If n number of data points are given, the order of the polynomial will be n-1.

Linear Interpolation

• Connecting two data points with a straight line.
• Suppose we are given two data points (𝑥₁, 𝑓 𝑥₁ ) and (𝑥₂, 𝑓 𝑥₂ ) .
• We are going to derive the formula which helps to calculate 𝑓 𝑥 for the given 𝑥 where 𝑥₁ ≤ 𝑥 ≤ 𝑥₂ .

Let us find the equation of straight line from the given two points (𝑥₁, 𝑓 𝑥₁ ) and (𝑥₂, 𝑓 𝑥₂ ) .

Demerits of Linear Interpolation

• Linear Interpolation uses first order polynomial (i.e. straight line) to approximate the function.
• To reduce this error, we need to use higher order polynomial.

Lagrange Interpolation Polynomial

• Let 𝑥₀, 𝑓₀ , 𝑥₁, 𝑓₁ , and 𝑥₂, 𝑓₂ are the given data points.

x	𝑥0	𝑥1	𝑥2
f	𝑓0	𝑓1	𝑓2

Here, number of data points = 3 , so the polynomial will be of the order 2.

• Let us consider a second order polynomial of the form 𝑝₂ 𝑥 = 𝑏₁ 𝑥 − 𝑥₀ 𝑥 − 𝑥₁ + 𝑏₂ 𝑥 − 𝑥₁ 𝑥 − 𝑥₂ + 𝑏₃ 𝑥 − 𝑥₂ 𝑥 − 𝑥₀ ………………..[1] where 𝑏₁, 𝑏₂ 𝑎𝑛𝑑 𝑏₃ 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑐𝑜𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑜𝑓 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙.
• Given points 𝑥₀, 𝑓₀ , 𝑥₁, 𝑓₁ , and 𝑥₂, 𝑓₂ should satisfy the equation 1.
• At 𝑥 = 𝑥_𝑜, 𝑝₂ 𝑥₀ = 𝑓_𝑜,

Similarly, for 𝑥 = 𝑥₁, 𝑝₂ 𝑥₁ = 𝑓₁  

Example 2.1
For the following data ,use Lagrange interpolation polynomial to find 𝑦 𝑎𝑡 𝑥 = 3.2.

x	1	2	4	5
y	4	9	61	120

Solution: Here, 𝑥₀ = 1, 𝑥₁ = 2, 𝑥2 = 4, 𝑥₃ = 5 , 𝑓₀ = 4, 𝑓₁ = 9, 𝑓₂ = 61, 𝑓₃ = 120

Now,

Newton Interpolation Polynomial

• The Newton form of polynomial is 𝒑_𝒏 (𝒙) = 𝒂_𝟎+ 𝒂_𝟏𝒙 − 𝒙_𝟎 + 𝒂_𝟐 𝒙 − 𝒙_𝟎 𝒙 − 𝒙_𝟏 + 𝒂_𝟑 𝒙 − 𝒙_𝟎 𝒙 − 𝒙_𝟏 𝒙 − 𝒙_𝟐… … … … … . . +𝒂_𝒏 𝒙 − 𝒙_𝟎 𝒙 − 𝒙_𝟏 … . . (𝒙 − 𝒙_𝒏−𝟏)
• For given 𝑥₀, 𝑓₀ , 𝑥₁, 𝑓₁ , 𝑥₂, 𝑓₂ … … … (𝑥_𝑛, 𝑓_𝑛), we need to find the value of coefficients 𝑎₀, 𝑎₁, 𝑎₂, 𝑎₃ … … … . . 𝑎_𝑛.

✓ If only two data points 𝑥0, 𝑓0 , 𝑥1, 𝑓1 are given, 𝒑𝟏 𝒙 = 𝒂𝟎+ 𝒂𝟏 𝒙 − 𝒙𝟎 → 𝟏𝒔𝒕 𝒐𝒓𝒅𝒆𝒓 𝒑𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍 ✓ If three data points 𝑥0, 𝑓0 , 𝑥1, 𝑓1 , 𝑥2, 𝑓2 are given 𝒑𝟐 𝒙 = 𝒂𝟎 + 𝒂𝟏 𝒙 − 𝒙𝟎 + 𝒂𝟐 𝒙 − 𝒙𝟎 𝒙 − 𝒙𝟏 → 𝟐𝒏𝒅 𝒐𝒓𝒅𝒆𝒓 𝒑𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍 ✓ Similarly, for four data points 𝑥0, 𝑓0 , 𝑥1, 𝑓1 , 𝑥2, 𝑓2 𝑥3, 𝑓3 , 𝒑𝟑 𝒙 = 𝒂𝟎 + 𝒂𝟏 𝒙 − 𝒙𝟎 + 𝒂𝟐 𝒙 − 𝒙𝟎 𝒙 − 𝒙𝟏 + 𝒂𝟑(𝒙 − 𝒙𝟎)(𝒙 − 𝒙𝟏)(𝒙 − 𝒙𝟐) → 𝟑𝒓𝒅 𝒐𝒓𝒅𝒆𝒓 𝒑𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍

• For 𝑥 = 𝑥₀, 𝑝_𝑛 (𝑥₀ )= 𝑓₀ , we will get 𝑓₀ = 𝑎₀ . 𝑎₀ = 𝑓₀
• For 𝑥 = 𝑥₁, 𝑝_𝑛 (𝑥₁ ) = 𝑓₁ 

or,  𝑓₁ = 𝑎₀ + 𝑎₁ 𝑥₁ − 𝑥₀

or, 𝑓1 = 𝑓₀ + 𝑎₁ 𝑥₁ − 𝑥₀ [since 𝑎₀ = 𝑓₀]

𝑓₂ = 𝑎₀ + 𝑎₁ 𝑥₂ − 𝑥₀ + 𝑎₂(𝑥₂ − 𝑥₀)(𝑥₂ − 𝑥₁) which gives

Newton Divide Difference Table

The alternative way of finding the coefficients (𝑎0, 𝑎1, 𝑎2 𝑎𝑛𝑑 𝑠𝑜 𝑜𝑛) value is to use Newton divided difference table. For given 𝑥₀, 𝑓₀ , 𝑥₁, 𝑓₁ , 𝑥₂, 𝑓₂ , 𝑥₃, 𝑓₃ 𝑎𝑛𝑑 (𝑥₄, 𝑓₄)

Example 2.2

Find the functional value for 𝑥 = 7 using Newton interpolation polynomial.

x	5	6	9	11
𝑓(𝑥)	12	13	14	16

Solution: Since 4 data points are given, the required polynomial will be of the order 3.

 𝒑_𝟑 (𝒙) = 𝒂_𝟎 + 𝒂_𝟏 𝒙 − 𝒙_𝟎 + 𝒂_𝟐 𝒙 − 𝒙_𝟎 𝒙 − 𝒙_𝟏 + 𝒂_𝟑 𝒙 − 𝒙_𝟎𝒙 − 𝒙_𝟏 𝒙 − 𝒙_𝟐

To get the value of 𝑎₀, 𝑎₁, 𝑎₂, 𝑎₃ 𝑎𝑛𝑑 𝑎₄ , we are going to use Newton divided difference table.

Newton Forward Difference /Newton Backward Difference [Interpolation with equidistant points]

• Particularly used when the functional values are given at a sequence of equally spaced points.
• Also called as Newton Gregory technique.

Scenario 1:

x	5	6	9	11
𝑓(𝑥)	12	13	14	16

𝐻𝑒𝑟𝑒, 𝑥₁ − 𝑥₀ = 6 − 5 = 1 𝑤ℎ𝑒𝑟𝑒𝑎𝑠, 𝑥₂ − 𝑥₁ = 9 − 6 = 3 Not equally spaced. We can’t use Newton Forward/Backward Difference. Rather, we need to rely on Lagrange or Newton Interpolation polynomial.

Scenario 2:

x	10	20	30	40	50
𝑓(𝑥)	0	7	26	63	124

Here, 𝒙 values are equally spaced. In this case we can use Newton forward difference or Newton backward difference.
𝑠𝑡𝑒𝑝 𝑠𝑖𝑧𝑒 ℎ = 10

• If required point is close to the start of the table, use Newton forward difference. 𝐴𝑡 𝑥 = 15, 𝑓 (𝑥) = ? 𝑜𝑟 𝐴𝑡 𝑥 = 25, 𝑓 𝑥 = ? [ First value of 𝒙 (𝐢. 𝐞. 𝑥₀) acts as a reference point ]
• If required point is close to the end of the table , use Newton backward difference. 𝐴𝑡 𝑥 = 48, 𝑓 𝑥 =? 𝑜𝑟 𝐴𝑡 𝑥 = 35, 𝑥 = ? [ Last value of 𝒙 (𝐢. 𝐞. 𝑥_𝑛) acts as a reference point ]
• Simple difference is used here rather than divided difference which was used in Newton divided difference table. 𝑓₁ − 𝑓₀ = 1𝑠𝑡 𝑜𝑟𝑑𝑒𝑟 𝑠𝑖𝑚𝑝𝑙𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒

Newton Forward Difference

Let us create a forward difference table for given 𝑥₀, 𝑓₀ , 𝑥₁, 𝑓₁ , 𝑥₂, 𝑓₂ , 𝑥₃, 𝑓₃ 𝑎𝑛𝑑 (𝑥₄, 𝑓₄)

Then, the Newton forward difference formula is given by,

Example 2.3:

Estimate the value of 𝑠𝑖𝑛𝜃 𝑎𝑡 𝜃 = 25⁰ using Newton-Gregory forward difference formula.

𝜃	10	20	30	40	50
sin(𝜃)	0.1736	0.3420	0.5000	0.6428	0.7660

Solution: Let us draw a Newton forward difference table to find the value of coefficients

𝒇_𝟎 = 𝟎. 𝟏𝟕𝟑𝟔 , ∆𝒇_𝟎 = 𝟎. 𝟏𝟔𝟖𝟒, ∆ ^𝟐𝒇_𝟎 = −𝟎. 𝟎𝟏𝟎𝟒 , ∆ ^𝟑𝒇_𝟎 = −𝟎. 𝟎𝟎𝟒𝟖, ∆ ^𝟒𝒇_𝟎 = 𝟎. 𝟎𝟎𝟎4

Newton forward difference formula is given by,

= 0.422665625

Newton Backward Difference

Let us create a forward difference table for given 𝑥₀, 𝑓₀ , 𝑥₁, 𝑓₁ , 𝑥₂, 𝑓₂ , 𝑥₃, 𝑓₃ 𝑎𝑛𝑑 (𝑥₄, 𝑓₄)

Then, the Newton backward difference formula is given by,

Example 2.4
Estimate the value of 𝑠𝑖𝑛𝜃 𝑎𝑡 𝜃 = 45 ⁰using Newton-Gregory backward difference formula.

𝜃	10	20	30	40	50
sin(𝜃)	0.1736	0.3420	0.5000	0.6428	0.7660

Solution: Let us draw a Newton backward difference table to find the value of coefficients

From table, 𝑓₄ = 0. 7660, 𝛻𝑓₄ = 0.1232, 𝛻 ² 𝑓₄ = −0. 0196, 𝛻 ³ 𝑓₄ = −0.0044, 𝑎𝑛𝑑 𝛻 ⁴ 𝑓₄ = 0.0004

Newton backward difference formula is given by

= 0.707509