interpolation - subhasish...

INTERPOLATION

öi¡¢no Q¾cÊ 1 | P a g e

Finite Differences

Let us have a table of values 𝑥𝑖 , 𝑦𝑖 , 𝑖 = 0, 1, 2, … . , 𝑛 of any function 𝑦 = 𝑓(𝑥). The values of 𝑥 are equally

spaced, i.e. 𝑥𝑖 = 𝑥0 + 𝑖𝑕, 𝑖 = 0, 1, 2, … . , 𝑛.

Suppose, we are required to recover the values of 𝑦 = 𝑓(𝑥) for certain intermediate value of 𝑥, then the method

of 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠 of the function is used.

Forward Differences

Forward differences can be expressed as,

∆𝑖𝑦𝑠 = 𝑦𝑠 , 𝑖 = 0

∆ ∆𝑖−1𝑦𝑠 = ∆𝑖−1𝑦𝑠+1 − ∆𝑖−1𝑦𝑠 , 𝑖 > 0

Hence for 𝑠 = 0,1,2,3, …,

∆𝑦0 = 𝑦1 − 𝑦0, ∆𝑦1 = 𝑦2 − 𝑦1, …… ., ∆𝑦𝑛−1 = 𝑦𝑛 − 𝑦𝑛−1

Here, ∆ is called as the 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟 and ∆𝑦0, 𝑦1, … are called as the

𝑓𝑖𝑟𝑠𝑡 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠. The differences of the first forward differences are called as the

𝑠𝑒𝑐𝑜𝑛𝑑 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠 and are denoted by, ∆2𝑦0, ∆2𝑦1 ….,. Similarly, 𝑡𝑕𝑖𝑟𝑑 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠,

∆3𝑦0, ∆3𝑦1, … and 𝑓𝑜𝑢𝑟𝑡𝑕 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠, ∆4𝑦0, ∆4𝑦1, … and higher orders can be obtained as given

below,

∆2𝑦0 = ∆𝑦1 − ∆𝑦0 = 𝑦2 − 𝑦1 − 𝑦1 − 𝑦0 = 𝑦2 − 2𝑦1 + 𝑦0

∆2𝑦1 = ∆𝑦2 − ∆𝑦1 = 𝑦3 − 𝑦2 − 𝑦2 − 𝑦1 = 𝑦3 − 2𝑦2 + 𝑦1

∆3𝑦0 = ∆2𝑦1 − ∆2𝑦0 = 𝑦3 − 2𝑦2 + 𝑦1 − 𝑦2 − 2𝑦1 + 𝑦0 = 𝑦3 − 3𝑦2 + 3𝑦1 − 𝑦0

A table can be formed showing the forward differences,

𝑥 𝑦 ∆ ∆2 ∆3 ∆4

𝑥0 𝑦0

∆𝑦0

𝑥1 𝑦1 ∆2𝑦0

∆𝑦1 ∆3𝑦0

𝑥2 𝑦2 ∆2𝑦1 ∆4𝑦0

∆𝑦2 ∆3𝑦1

𝑥3 𝑦3 ∆2𝑦2

∆𝑦3

𝑥4 𝑦4

Backward Differences

Backward differences can be expressed as,

∇𝑖𝑦𝑠 = 𝑦𝑠 , 𝑖 = 0

∇ ∇𝑖−1𝑦𝑠 = ∇𝑖−1𝑦𝑠 − ∇𝑖−1𝑦𝑠−1, 𝑖 > 0

Hence for 𝑠 = 1,2,3, ..,

∇𝑦1 = 𝑦1 − 𝑦0, ∇𝑦2 = 𝑦2 − 𝑦1, …… ., ∇𝑦𝑛 = 𝑦𝑛 − 𝑦𝑛−1

INTERPOLATION


Here, ∇ is called as the 𝑏𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟 and ∇𝑦0, ∇𝑦1, … are called as the

𝑓𝑖𝑟𝑠𝑡 𝑏𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠. The differences of the first backward differences are called as the

𝑠𝑒𝑐𝑜𝑛𝑑 𝑏𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠 and are denoted by, ∇2𝑦2, ∇2𝑦3, ... Similarly, 𝑡𝑕𝑖𝑟𝑑 𝑏𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠,

∇3𝑦3, ∇3𝑦4, … and 𝑓𝑜𝑢𝑟𝑡𝑕 𝑏𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠, ∇4𝑦4, ∇4𝑦5, … and higher orders can be obtained as given

below,

∇2𝑦1 = ∇𝑦1 − ∇𝑦0 = 𝑦2 − 𝑦1 − 𝑦1 − 𝑦0 = 𝑦2 − 2𝑦1 + 𝑦0

∇2𝑦2 = ∇𝑦2 − ∇𝑦1 = 𝑦3 − 𝑦2 − 𝑦2 − 𝑦1 = 𝑦3 − 2𝑦2 + 𝑦1

∇3𝑦1 = ∇2𝑦2 − ∇2𝑦1 = 𝑦3 − 2𝑦2 + 𝑦1 − 𝑦2 − 2𝑦1 + 𝑦0 = 𝑦3 − 3𝑦2 + 3𝑦1 − 𝑦0

A table can be formed showing the backward differences,

𝑥 𝑦 ∇ ∇2 ∇3 ∇4

𝑥0 𝑦0

∇𝑦1

𝑥1 𝑦1 ∇2𝑦2

∇𝑦2 ∇3𝑦3

𝑥2 𝑦2 ∇2𝑦3 ∇4𝑦4

∇𝑦3 ∇3𝑦4

𝑥3 𝑦3 ∇2𝑦4

∇𝑦4

𝑥4 𝑦4

Central Differences

Central differences can be expressed as,

𝛿𝑖𝑦𝑠 = 𝑦𝑠 , 𝑖 = 0

𝛿 𝛿𝑖−1𝑦𝑠 = 𝛿𝑖−1𝑦𝑠+1 2 − 𝛿𝑖−1𝑦𝑠−1/2, 𝑖 > 0

Hence for 𝑠 = 1/2, 3/2, ..,

𝛿𝑦1/2 = 𝑦1 − 𝑦0 , 𝛿𝑦3/2 = 𝑦2 − 𝑦1, …… ., 𝛿𝑦𝑛−1/2 = 𝑦𝑛 − 𝑦𝑛−1

Here, 𝛿 is called as the 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟 and 𝛿𝑦1/2, 𝛿𝑦3/2, … are called as the

𝑓𝑖𝑟𝑠𝑡 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠. The differences of the first central differences are called as the

𝑠𝑒𝑐𝑜𝑛𝑑 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠 and are denoted by, 𝛿2𝑦1, 𝛿2𝑦2, … Similarly, 𝑡𝑕𝑖𝑟𝑑 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠,

𝛿3𝑦3/2, 𝛿3𝑦5/2, … and 𝑓𝑜𝑢𝑟𝑡𝑕 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠, 𝛿4𝑦2, 𝛿2𝑦3, … and higher orders can be obtained as given

below,

𝛿2𝑦1 = 𝛿𝑦3/2 − 𝛿𝑦1/2 = 𝑦2 − 𝑦1 − 𝑦1 − 𝑦0 = 𝑦2 − 2𝑦1 + 𝑦0

𝛿2𝑦2 = 𝛿𝑦5/2 − 𝛿𝑦3/2 = 𝑦3 − 𝑦2 − 𝑦2 − 𝑦1 = 𝑦3 − 2𝑦2 + 𝑦1

𝛿3𝑦3/2 = 𝛿2𝑦2 − 𝛿2𝑦1 = 𝑦3 − 2𝑦2 + 𝑦1 − 𝑦2 − 2𝑦1 + 𝑦0 = 𝑦3 − 3𝑦2 + 3𝑦1 − 𝑦0

INTERPOLATION


A table can be formed showing the central differences,

𝑥 𝑦 𝛿 𝛿2 𝛿3 𝛿4

𝑥0 𝑦0

𝛿𝑦1/2

𝑥1 𝑦1 𝛿2𝑦1

𝛿𝑦3/2 𝛿3𝑦3/2

𝑥2 𝑦2 𝛿2𝑦2 𝛿4𝑦2

𝛿𝑦5/2 𝛿3𝑦5/2


𝛿𝑦7/2

𝑥4 𝑦4

It is clear from the three tables that in a definite numerical case, the same numbers occurs in the same positions.

Thus we obtain,

∆𝑦0 = ∇𝑦1 = 𝛿𝑦1/2

∆3𝑦1 = ∇3𝑦4 = 𝛿3𝑦5/2

Symbolic Relations & Separations of Symbols

1. The averaging operator 𝜇 is defined as,

𝜇𝑦𝑠 =1

2 𝑦𝑠+1/2 + 𝑦𝑠−1/2

2. The shift operator 𝐸 is defined as,

𝐸𝑦𝑠 = 𝑦𝑠+1

which shows that the effect of 𝐸 is to shift the functional value 𝑦𝑠 to the next higher value 𝑦𝑠+1. In

general,

𝐸𝑛𝑦𝑠 = 𝑦𝑠+𝑛

3. The differentiation operator 𝐷 is defined, such that

𝐷𝑦 𝑥 =𝑑

𝑑𝑥𝑦 𝑥

We can expand 𝑦(𝑥 + 𝑕) about 𝑕 as,

𝑦 𝑥 + 𝑕 = 𝑦 𝑥 + 𝑕𝑦′ 𝑥 +𝑕2

2!𝑦′′ 𝑥 +

𝑕3

3!𝑦′′′ 𝑥 + ⋯

𝐸𝑦 𝑥 = 1 + 𝑕𝐷 +𝑕2𝐷2

2!+

𝑕3𝐷3

3!+ ⋯ 𝑦 𝑥

𝐸𝑦 𝑥 = 𝑒𝑕𝐷𝑦 𝑥

𝐸 = 𝑒𝑕𝐷

INTERPOLATION


Relationship Corner:

A. We know that,

∆𝑦0 = 𝑦1 − 𝑦0 = 𝐸𝑦0 − 𝑦0 = 𝐸 − 1 𝑦0

Hence,

∆= 𝐸 − 1 ⟹ 𝐸 = 1 + ∆

It implies that the effect of the operator 𝐸 on 𝑦0 is the same as that of 1 + ∆ on 𝑦0.

B. We know that,

∇𝑦1 = 𝑦1 − 𝑦0 = 𝑦1 −1

𝐸𝑦1 = 1 −

1

𝐸 𝑦1 = 1 − 𝐸−1 𝑦1

Hence,

∇= 1 − 𝐸−1

It implies that the effect of the operator ∇ on 𝑦1 is the same as that of 1 − 𝐸−1 on 𝑦1.

C. We know that,

δ𝑦1/2 = 𝑦1 − 𝑦0 = 𝐸1/2𝑦1/2 −1

𝐸1/2𝑦1/2 = 𝐸1/2 −

1

𝐸1/2 𝑦1/2 = 𝐸1/2 − 𝐸−1/2 𝑦1/2

Hence,

δ = 𝐸1/2 − 𝐸−1/2

It implies that the effect of the operator δ on 𝑦1/2 is the same as that of 𝐸1/2 − 𝐸−1/2 on 𝑦1/2.

D. We know that,

𝜇𝑦𝑠 =1

2 𝑦𝑠+1/2 + 𝑦𝑠−1/2 =

1

2 𝐸1/2𝑦𝑠 +

1

𝐸1/2𝑦𝑠 =

1

2 𝐸1/2 + 𝐸−1/2 𝑦𝑠

Hence,

𝜇 =1

2 𝐸1/2 + 𝐸−1/2

E. On squaring, we get

𝜇2 =1

4 𝐸1/2 + 𝐸−1/2

2

𝜇2 =1

4 𝐸 + 𝐸−1 + 2

𝜇2 =1

4 𝐸1/2 − 𝐸−1/2

2+ 4 =

1

4 𝛿2 + 4

Hence,

𝜇 = 1 +𝛿2

4

INTERPOLATION


F. We have,

∆= 𝐸 − 1

∆= 𝐸 1 − 𝐸−1 = ∇𝐸

∆= 𝐸1/2 𝐸1/2 − 𝐸−1/2 = 𝛿𝐸1/2

G. We have,

𝜇 =1

2 𝐸1/2 + 𝐸−1/2

δ = 𝐸1/2 − 𝐸−1/2

Adding we get,

2𝜇 + 𝛿 = 2𝐸1/2

⟹ 𝐸1/2 = 𝜇 +𝛿

2

Subtracting we get,

2𝜇 − 𝛿 = 2𝐸−1/2

⟹ 𝐸−1/2 = 𝜇 −𝛿

2

Theorem: Prove that, ∆𝑛𝑢𝑥−𝑛 = 𝑢𝑥 − 𝑛𝑢𝑥−1 +𝑛(𝑛−1)

2𝑢𝑥−2 + ⋯ + (−1)𝑛𝑢𝑥−𝑛

Proof: The RHS is given as,

𝑢𝑥 − 𝑛𝑢𝑥−1 +𝑛(𝑛 − 1)

2𝑢𝑥−2 + ⋯ + (−1)𝑛𝑢𝑥−𝑛

= 𝑢𝑥 − 𝑛𝐸−1𝑢𝑥 +𝑛(𝑛 − 1)

2𝐸−2𝑢𝑥 + ⋯ + (−1)𝑛𝐸−𝑛𝑢𝑥

= 1 − 𝑛𝐸−1 +𝑛 𝑛 − 1

2𝐸−2 + ⋯ + −1 𝑛𝐸−𝑛 𝑢𝑥

= 1 − 𝐸−1 𝑛𝑢𝑥

= 1 −1

𝐸

𝑛

𝑢𝑥 = 𝐸 − 1

𝐸 𝑛

𝑢𝑥 =∆𝑛

𝐸𝑛𝑢𝑥

= ∆𝑛𝐸−𝑛𝑢𝑥

= ∆𝑛𝑢𝑥−𝑛

= LHS

QED

INTERPOLATION


Theorem: Prove that, 𝑒𝑥 𝑢0 + 𝑥∆𝑢0 +𝑥2

2!∆2𝑢0 + ⋯ = 𝑢0 + 𝑢1𝑥 + 𝑢2

𝑥2

2!+ ⋯

Proof: The LHS is given as,

= 𝑒𝑥 𝑢0 + 𝑥∆𝑢0 +𝑥2

2!∆2𝑢0 + ⋯

= 𝑒𝑥 1 + 𝑥∆ +𝑥2

2!∆2 + ⋯ 𝑢0

= 𝑒𝑥𝑒𝑥∆𝑢0 = 𝑒𝑥 1+∆ 𝑢0 = 𝑒𝑥𝐸𝑢0

= 1 + 𝑥𝐸 +𝑥2𝐸2

2!+ ⋯ 𝑢0

= 𝑢0 + 𝑢1𝑥 + 𝑢2𝑥2

2!+ ⋯

= RHS

QED

Differences of a Polynomial

Let 𝑦(𝑥) be a polynomial of the 𝑛𝑡𝑕 degree given as,

𝑦 𝑥 = 𝑎0𝑥𝑛 + 𝑎1𝑥

𝑛−1 + 𝑎2𝑥𝑛−2 + ⋯ + 𝑎𝑛

Hence, we get

𝑦 𝑥 + 𝑕 − 𝑦 𝑥 = 𝑎0 𝑥 + 𝑕 𝑛 − 𝑥𝑛 + 𝑎1 𝑥 + 𝑕 𝑛−1 − 𝑥𝑛−1 + ⋯

= 𝑎0 . 𝑛𝑕 . 𝑥𝑛−1 + 𝑎1′ . 𝑥𝑛−2 + ⋯ + 𝑎𝑛

′

where, 𝑎1′ , 𝑎2

′ , … , 𝑎𝑛′ are the new coefficients.

The above equation is the 𝑓𝑖𝑟𝑠𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 of a polynomial of 𝑛𝑡𝑕 degree. The degree of the new polynomial

is (𝑛 − 1). It is represented as,

∆𝑦(𝑥) = 𝑎0 . 𝑛𝑕 . 𝑥𝑛−1 + 𝑎1′ . 𝑥𝑛−2 + ⋯ + 𝑎𝑛

′

The 𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒, ∆2𝑦(𝑥), will be a polynomial of degree (𝑛 − 2). The coefficient of 𝑥𝑛−2 will be

𝑎0 𝑛(𝑛 − 1)𝑕2. Progressing on the same lines we find the 𝑛𝑡𝑕 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒, ∆𝑛𝑦(𝑥), to be 𝑎0 𝑛! 𝑕𝑛 which is a

constant. Hence, the (𝑛 + 1)𝑡𝑕 and the higher differences of a polynomial of 𝑛𝑡𝑕 degree will be zero.

Conversely, we can say that if for a give set of equally spaced values of 𝑥, the 𝑛𝑡𝑕 differences of a tabulated

function are constant and the (𝑛 + 1)𝑡𝑕 and the higher differences vanish then the tabulated function represents

a polynomial of degree 𝑛.

INTERPOLATION


Newton’s Formulae for Interpolation

Newton’s Forward Difference Formula

Let us have a set of (𝑛 + 1) values, 𝑥0, 𝑦0 , 𝑥1, 𝑦1 , … , 𝑥𝑛 , 𝑦𝑛 of 𝑥 and 𝑦. We are required to find a

polynomial of 𝑛𝑡𝑕 degree such that 𝑦 and 𝑦𝑛(𝑥) agree at the tabulated points. Let points 𝑥𝑖 be evenly spaced,

𝑥𝑖 = 𝑥0 + 𝑖𝑕, 𝑖 = 0, 1, 2, … , 𝑛

The 𝑛𝑡𝑕 degree polynomial 𝑦𝑛(𝑥) can be written as,

𝑦𝑛 𝑥 = 𝑎0 + 𝑎1 𝑥 − 𝑥0 + 𝑎2 𝑥 − 𝑥0 𝑥 − 𝑥1 + ⋯ + 𝑎𝑛 𝑥 − 𝑥0 𝑥 − 𝑥1 𝑥 − 𝑥2 … 𝑥 − 𝑥𝑛−1

Since, 𝑦 and 𝑦𝑛(𝑥) agree at the tabulated points, we get

𝑎0 = 𝑦0; 𝑎1 =𝑦1 − 𝑦0

𝑥1 − 𝑥0=

∆𝑦0

𝑕; 𝑎2 =

∆2𝑦0

𝑕22!; ………… ; 𝑎𝑛 =

∆𝑛𝑦0

𝑕𝑛𝑛!

Setting 𝑥 = 𝑥0 + 𝑝𝑕, and substituting for 𝑎0, 𝑎1, … , 𝑎𝑛 , we get

𝑦𝑛 𝑥 = 𝑦0 + 𝑝∆𝑦0 +𝑝(𝑝 − 1)

2!∆2𝑦0 + ⋯ +

𝑝 𝑝 − 1 𝑝 − 2 … (𝑝 − 𝑛 + 1)

𝑛!∆𝑛𝑦0

This is the 𝑁𝑒𝑤𝑡𝑜𝑛’𝑠 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛𝑡𝑒𝑟𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 and is useful for interpolation near the

𝑏𝑒𝑔𝑖𝑛𝑛𝑖𝑛𝑔 of a set of tabular values.

The error committed in replacing the function 𝑦(𝑥) by means of a polynomial 𝑦𝑛(𝑥) is given as,

𝑦 𝑥 − 𝑦𝑛 𝑥 = 𝑥 − 𝑥0 𝑥 − 𝑥1 𝑥 − 𝑥2 … 𝑥 − 𝑥𝑛

𝑛 + 1 !𝑦𝑛+1 휀 , 𝑥0 < 휀 < 𝑥𝑛

Since, 𝑝 =𝑥−𝑥0

𝑕, we can write the above equation as,

𝑦 𝑥 − 𝑦𝑛 𝑥 =𝑝 𝑝 − 1 𝑝 − 2 … 𝑝 − 𝑛

𝑛 + 1 !𝑕𝑛+1𝑦𝑛+1 휀 , 𝑥0 < 휀 < 𝑥𝑛

As we don’t have any information concerning 𝑦𝑛+1 𝑥 , the above error equation is not useful in practice.

Expanding 𝑦(𝑥 + 𝑕) about 𝑕, we get

𝑦 𝑥 + 𝑕 = 𝑦 𝑥 + 𝑕𝑦′ 𝑥 +𝑕2

2!𝑦′′ 𝑥 +

𝑕3

3!𝑦′′′ 𝑥 + ⋯

Neglecting the terms containing higher powers of 𝑕, we get

𝑦′ 𝑥 =1

𝑕 𝑦 𝑥 + 𝑕 − 𝑦 𝑥 =

1

𝑕∆𝑦 𝑥

We can write 𝑦′ 𝑥 as 𝐷𝑦(𝑥) where, 𝐷 ≡𝑑

𝑑𝑥. The above equation gives,

𝐷 ≡1

𝑕∆

Hence, we get

𝐷𝑛+1 ≡1

𝑕𝑛+1∆𝑛+1

INTERPOLATION


Thus, we obtain

𝑦𝑛+1 𝑥 =1

𝑕𝑛+1∆𝑛+1 𝑦 𝑥

Hence, the error equation can be written as,

𝑦 𝑥 − 𝑦𝑛 𝑥 =𝑝 𝑝 − 1 𝑝 − 2 … 𝑝 − 𝑛

𝑛 + 1 !∆𝑛+1𝑦 휀 , 𝑥0 < 휀 < 𝑥𝑛

This form is suitable for computation.

Newton’s Backward Difference Formula

Let us have a set of (𝑛 + 1) values, 𝑥0, 𝑦0 , 𝑥1, 𝑦1 , … , 𝑥𝑛 , 𝑦𝑛 of 𝑥 and 𝑦. We are required to find a

polynomial of 𝑛𝑡𝑕 degree such that 𝑦 and 𝑦𝑛(𝑥) agree at the tabulated points. Let points 𝑥𝑖 be evenly spaced,

𝑥𝑖 = 𝑥0 + 𝑖𝑕, 𝑖 = 0,1,2, … , 𝑛

The 𝑛𝑡𝑕 degree polynomial 𝑦𝑛(𝑥) can be written as,

𝑦𝑛 𝑥 = 𝑎0 + 𝑎1 𝑥 − 𝑥𝑛 + 𝑎2 𝑥 − 𝑥𝑛 𝑥 − 𝑥𝑛−1 + ⋯ + 𝑎𝑛 𝑥 − 𝑥𝑛 𝑥 − 𝑥𝑛−1 𝑥 − 𝑥𝑛−2 … 𝑥 − 𝑥1

Since, 𝑦 and 𝑦𝑛(𝑥) agree at the tabulated points, we get

𝑎0 = 𝑦𝑛 ; 𝑎1 =𝑦𝑛 − 𝑦𝑛−1

𝑥𝑛 − 𝑥𝑛−1=

∇𝑦𝑛

𝑕; 𝑎2 =

∇2𝑦𝑛

𝑕22!; ………… ; 𝑎𝑛 =

∇𝑛𝑦0

𝑕𝑛𝑛!

Setting 𝑥 = 𝑥𝑛 + 𝑝𝑕, and substituting for 𝑎0, 𝑎1, … , 𝑎𝑛 , we get

𝑦𝑛 𝑥 = 𝑦𝑛 + 𝑝∇𝑦𝑛 +𝑝(𝑝 + 1)

2!∇2𝑦𝑛 + ⋯ +

𝑝 𝑝 + 1 𝑝 + 2 … (𝑝 + 𝑛 − 1)

𝑛!∇𝑛𝑦0

This is the 𝑁𝑒𝑤𝑡𝑜𝑛’𝑠 𝑏𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛𝑡𝑒𝑟𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 and is useful for interpolation near the

𝑒𝑛𝑑 of a set of tabular values.

The error committed in replacing the function 𝑦(𝑥) by means of a polynomial 𝑦𝑛(𝑥) is given as,

𝑦 𝑥 − 𝑦𝑛 𝑥 = 𝑥 − 𝑥𝑛 𝑥 − 𝑥𝑛−1 𝑥 − 𝑥𝑛−2 … 𝑥 − 𝑥0

𝑛 + 1 !𝑦𝑛+1 휀 , 𝑥0 < 휀 < 𝑥𝑛

Since, 𝑝 =𝑥−𝑥𝑛

𝑕, we can write the above equation as,

𝑦 𝑥 − 𝑦𝑛 𝑥 =𝑝 𝑝 + 1 𝑝 + 2 … 𝑝 + 𝑛

𝑛 + 1 !𝑕𝑛+1𝑦𝑛+1 휀 , 𝑥0 < 휀 < 𝑥𝑛

As we don’t have any information concerning 𝑦𝑛+1 𝑥 , the above error equation is not useful in practice.

Expanding 𝑦(𝑥 + 𝑕) about 𝑕, we get

𝑦 𝑥 + 𝑕 = 𝑦 𝑥 + 𝑕𝑦′ 𝑥 +𝑕2

2!𝑦′′ 𝑥 +

𝑕3

3!𝑦′′′ 𝑥 + ⋯

Neglecting the terms containing higher powers of 𝑕, we get

𝑦′ 𝑥 =1

𝑕 𝑦 𝑥 + 𝑕 − 𝑦 𝑥 =

1

𝑕∆𝑦 𝑥

INTERPOLATION


We can write 𝑦′ 𝑥 as 𝐷𝑦(𝑥) where, 𝐷 ≡𝑑

𝑑𝑥. The above equation gives,

𝐷 ≡1

𝑕∆

Hence, we get

𝐷𝑛+1 ≡1

𝑕𝑛+1∆𝑛+1

Thus, we obtain

𝑦𝑛+1 𝑥 =1

𝑕𝑛+1∆𝑛+1 𝑦 𝑥

We know that, ∆= 𝐸 − 1 and ∇= 1 − 𝐸−1 . Hence, we get, ∆=∇

1−∇. The above equation becomes

𝑦𝑛+1 𝑥 =1

𝑕𝑛+1

∇

1 − ∇ 𝑛+1

𝑦 𝑥

Hence, the error equation can be written as,

𝑦 𝑥 − 𝑦𝑛 𝑥 =𝑝 𝑝 + 1 𝑝 + 2 … 𝑝 + 𝑛

𝑛 + 1 !

∇

1 − ∇ 𝑛+1

𝑦 휀 , 𝑥0 < 휀 < 𝑥𝑛

This form is suitable for computation.

Central Difference Interpolation Formulae

Gauss’s Central Difference Formulae

Gauss’s Forward Formula: The difference table can be written with taking the central ordinate to be 𝑦0

corresponding to 𝑥0. Those above are labeled as 𝑦−1, 𝑦−2, … and those below are labeled 𝑦1, 𝑦2, … .

𝑥 𝑦 ∆ ∆2 ∆3 ∆4

𝑥−2 𝑦−2

∆𝑦−2

𝑥−1 𝑦−1 ∆2𝑦−2

∆𝑦−1 ∆3𝑦−2

𝑥0 𝑦0 ∆2𝑦−1 ∆4𝑦−2

∆𝑦0 ∆3𝑦−1

𝑥1 𝑦1 ∆2𝑦0

∆𝑦1

𝑥2 𝑦2

𝑦𝑝 can be expressed as,

𝑦𝑝 = 𝑦0 + 𝐺1∆𝑦0 + 𝐺2∆2𝑦−1 + 𝐺3∆

3𝑦−1 + 𝐺4∆4𝑦−2 + …

where, 𝐺1, 𝐺2 , 𝐺3, 𝐺4, … have to be determined. 𝑦𝑝 can be expressed in terms of 𝑦0, ∆𝑦0 and higher order

differences as,

𝑦𝑝 = 𝐸𝑝𝑦0

= (1 + ∆)𝑝𝑦0

= 𝑦0 + 𝑝∆𝑦0 +𝑝(𝑝 − 1)

2!∆2𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2

3!∆3𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)

4!∆4𝑦0 …

INTERPOLATION


The right hand side can be also expressed in terms of 𝑦0, ∆𝑦0 and higher order differences as,

∆2𝑦−1 = ∆2𝐸−1𝑦0

= ∆2(1 + ∆)−1𝑦0

= ∆2(1 − ∆ + ∆2 − ∆3 + ⋯ )𝑦0

= ∆2𝑦0 − ∆3𝑦0 + ∆4𝑦0 − ∆5𝑦0 + ⋯

∆3𝑦−1 = ∆ ∆2𝑦−1

= ∆ ∆2𝑦0 − ∆3𝑦0 + ∆4𝑦0 − ∆5𝑦0 + ⋯

= ∆3𝑦0 − ∆4𝑦0 + ∆5𝑦0 − ∆6𝑦0 + ⋯

∆4𝑦−2 = ∆4𝐸−2𝑦0

= ∆4(1 + ∆)−2𝑦0

= ∆4(1 − 2∆ + 3∆2 − 4∆3 + ⋯ )𝑦0

= ∆4𝑦0 − 2∆5𝑦0 + 3∆6𝑦0 − 4∆7𝑦0 + ⋯

Hence, we have

𝑦0 + 𝑝∆𝑦0 +𝑝(𝑝 − 1)

2!∆2𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2

3!∆3𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)

4!∆4𝑦0 + ⋯

= 𝑦0 + 𝐺1∆𝑦0 + 𝐺2(∆2𝑦0 − ∆3𝑦0 + ∆4𝑦0 − ∆5𝑦0 + ⋯ ) + 𝐺3(∆3𝑦0 − ∆4𝑦0 + ∆5𝑦0 −

∆6𝑦0 + ⋯ ) + 𝐺4(∆4𝑦0 − 2∆5𝑦0 + 3∆6𝑦0 − 4∆7𝑦0 + ⋯ ) + …

Equating the coefficients of ∆𝑦0, ∆2𝑦0, ∆3𝑦0, ∆4𝑦0, … on both sides, we get

𝐺1 = 𝑝

𝐺2 =𝑝(𝑝 − 1)

2!

−𝐺2 + 𝐺3 =𝑝 𝑝 − 1 𝑝 − 2

3! ⟹ 𝐺3 =

𝑝 𝑝 + 1 𝑝 − 1

3!

𝐺2 − 𝐺3 + 𝐺4 =𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)

4! ⟹ 𝐺4 =

𝑝 𝑝 + 1 𝑝 − 1 (𝑝 − 2)

4! 𝑒𝑡𝑐.

Hence, we get

𝑦𝑝 = 𝑦0 + 𝑝∆𝑦0 +𝑝(𝑝 − 1)

2!∆2𝑦−1 +

𝑝 𝑝 + 1 𝑝 − 1

3!∆3𝑦−1 +

𝑝 𝑝 + 1 𝑝 − 1 (𝑝 − 2)

4!∆4𝑦−2 + …

INTERPOLATION


Gauss’s Backward Formula: We can also analyze the difference table as follows,

𝑥 𝑦 ∆ ∆2 ∆3 ∆4

𝑥−2 𝑦−2

∆𝑦−2

𝑥−1 𝑦−1 ∆2𝑦−2

∆𝑦−1 ∆3𝑦−2

𝑥0 𝑦0 ∆2𝑦−1 ∆4𝑦−2

∆𝑦0 ∆3𝑦−1

𝑥1 𝑦1 ∆2𝑦0

∆𝑦1

𝑥2 𝑦2

Hence, 𝑦𝑝 can be expressed as,

𝑦𝑝 = 𝑦0 + 𝐻1∆𝑦−1 + 𝐻2∆2𝑦−1 + 𝐻3∆

3𝑦−2 + 𝐻4∆4𝑦−2 + …

where, 𝐻1, 𝐻2, 𝐻3, 𝐻4, … have to be determined. 𝑦𝑝 can be expressed in terms of 𝑦0, ∆𝑦0 and higher order

differences as,

𝑦𝑝 = 𝐸𝑝𝑦0 = (1 + ∆)𝑝𝑦0

= 𝑦0 + 𝑝∆𝑦0 +𝑝(𝑝 − 1)

2!∆2𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2

3!∆3𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)

4!∆4𝑦0 …


∆𝑦−1 = ∆𝐸−1𝑦0

= ∆(1 + ∆)−1𝑦0

= ∆(1 − ∆ + ∆2 − ∆3 + ⋯ )𝑦0

= ∆𝑦0 − ∆2𝑦0 + ∆3𝑦0 − ∆4𝑦0 + ⋯

∆2𝑦−1 = ∆ ∆𝑦−1

= ∆(∆𝑦0 − ∆2𝑦0 + ∆3𝑦0 − ∆4𝑦0 + ⋯ )

= ∆2𝑦0 − ∆3𝑦0 + ∆4𝑦0 − ∆5𝑦0 + ⋯

∆3𝑦−2 = ∆3𝐸−2𝑦0

= ∆3(1 + ∆)−2𝑦0

= ∆3(1 − 2∆ + 3∆2 − 4∆3 + ⋯ )𝑦0

= ∆3𝑦0 − 2∆4𝑦0 + 3∆5𝑦0 − 4∆6𝑦0 + ⋯

∆4𝑦−2 = ∆ ∆3𝑦−1

= ∆(∆3𝑦0 − 2∆4𝑦0 + 3∆5𝑦0 − 4∆6𝑦0 + ⋯ )

= ∆4𝑦0 − 2∆5𝑦0 + 3∆6𝑦0 − 4∆7𝑦0 + ⋯

INTERPOLATION


Hence, we have

𝑦0 + 𝑝∆𝑦0 +𝑝(𝑝 − 1)

2!∆2𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2

3!∆3𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)

4!∆4𝑦0 + ⋯

= 𝑦0 + 𝐻1(∆𝑦0 − ∆2𝑦0 + ∆3𝑦0 − ∆4𝑦0 + ⋯ ) + 𝐻2(∆2𝑦0 − ∆3𝑦0 + ∆4𝑦0 − ∆5𝑦0 + ⋯ ) +

𝐻3(∆3𝑦0 − 2∆4𝑦0 + 3∆5𝑦0 − 4∆6𝑦0 + ⋯ ) + 𝐻4(∆4𝑦0 − 2∆5𝑦0 + 3∆6𝑦0 − 4∆7𝑦0 +

⋯ ) + …


𝐻1 = 𝑝

−𝐻1 + 𝐻2 =𝑝(𝑝 − 1)

2! ⟹ 𝐻2 =

𝑝 𝑝 + 1

2!

𝐻1 − 𝐻2 + 𝐻3 =𝑝 𝑝 − 1 𝑝 − 2

3! ⟹ 𝐻3 =

𝑝 𝑝 + 1 𝑝 − 1

3!

−𝐻1 + 𝐻2 + 2𝐻3 + 𝐻4 =𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)

4! ⟹ 𝐻4 =

𝑝 𝑝 + 1 𝑝 + 2 (𝑝 − 1)

4! 𝑒𝑡𝑐.

Hence, we get

𝑦𝑝 = 𝑦0 + 𝑝∆𝑦−1 +𝑝 𝑝 + 1

2!∆2𝑦−1 +

𝑝 𝑝 − 1 𝑝 − 2

3!∆3𝑦−2 +

𝑝 𝑝 + 1 𝑝 + 2 (𝑝 − 1)

4!∆4𝑦−2 + …

Stirling’s Formula

Taking mean of Gauss’ forward and backward formulae, we obtain the Stirling’s formula as

𝑦𝑝 = 𝑦0 + 𝑝 ∆𝑦−1 + ∆𝑦0

2 +

𝑝2

2!∆2𝑦−1 +

𝑝 𝑝2 − 1

3! ∆3𝑦−2 + ∆3𝑦−1

2 +

𝑝2(𝑝2 − 1)

4!∆4𝑦−2 + ⋯

The above difference table can be written in terms of central differences as,

𝑥 𝑦 𝛿 𝛿2 𝛿3 𝛿4

𝑥2 𝑦2

𝛿𝑦−3/2

𝑥−1 𝑦−1 𝛿2𝑦−1

𝛿𝑦−1/2 𝛿3𝑦−1/2

𝑥0 𝑦0 𝛿2𝑦0 𝛿4𝑦0

𝛿𝑦1/2 𝛿3𝑦1/2


𝛿𝑦3/2

𝑥2 𝑦2

Hence, the Stirling’s formula can now be expressed as,

𝑦𝑝 = 𝑦0 + 𝑝 𝛿𝑦−1/2 + 𝛿𝑦1/2

2 +

𝑝2

2!𝛿2𝑦0 +

𝑝 𝑝2 − 1

3! 𝛿3𝑦−1/2 + 𝛿3𝑦1/2

2 +

𝑝2(𝑝2 − 1)

4!𝛿4𝑦0 + ⋯

INTERPOLATION


Bessel’s Formula

This is one of the most useful formula for practical interpolation. The difference table now can be represented

as,

𝑥 𝑦 ∆ ∆2 ∆3 ∆4

𝑥−2 𝑦−2

∆𝑦−2

𝑥−1 𝑦−1 ∆2𝑦−2

∆𝑦−1 ∆3𝑦−2

𝑥0

𝑦0

𝑦1

∆2𝑦−1

∆2𝑦0

∆4𝑦−2

∆4𝑦−1

∆𝑦0 ∆3𝑦−1

𝑥1

∆𝑦1 ∆3𝑦0

𝑥2 𝑦2 ∆2𝑦1

∆𝑦2

𝑥3 𝑦3

The brackets mean that the average of the values enclosed in the brackets has to be taken.


𝑦𝑝 =𝑦0 + 𝑦1

2+ 𝐵1∆𝑦0 + 𝐵2

∆2𝑦−1 + ∆2𝑦0

2+ 𝐵3∆

3𝑦−1 + 𝐵4

∆4𝑦−2 + ∆4𝑦−1

2+ …

𝑦𝑝 =𝑦0 + (𝑦0 + ∆𝑦0)

2+ 𝐵1∆𝑦0 + 𝐵2

∆2𝑦−1 + ∆2𝑦0

2+ 𝐵3∆

3𝑦−1 + 𝐵4

∆4𝑦−2 + ∆4𝑦−1

2+ …

𝑦𝑝 = 𝑦0 + 𝐵1 +1

2 ∆𝑦0 + 𝐵2

∆2𝑦−1 + ∆2𝑦0

2+ 𝐵3∆

3𝑦−1 + 𝐵4

∆4𝑦−2 + ∆4𝑦−1

2+ …

where, 𝐵1, 𝐵2, 𝐵3, 𝐵4, … have to be determined. 𝑦𝑝 can be expressed in terms of 𝑦0, ∆𝑦0 and higher order

differences as,


= (1 + ∆)𝑝𝑦0

= 𝑦0 + 𝑝∆𝑦0 +𝑝(𝑝 − 1)

2!∆2𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2

3!∆3𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)

4!∆4𝑦0 …


∆2𝑦−1 = ∆2𝐸−1𝑦0

= ∆2(1 + ∆)−1𝑦0

= ∆2(1 − ∆ + ∆2 − ∆3 + ⋯ )𝑦0

= ∆2𝑦0 − ∆3𝑦0 + ∆4𝑦0 − ∆5𝑦0 + ⋯

INTERPOLATION


∆3𝑦−1 = ∆ ∆2𝑦−1

= ∆ ∆2𝑦0 − ∆3𝑦0 + ∆4𝑦0 − ∆5𝑦0 + ⋯

= ∆3𝑦0 − ∆4𝑦0 + ∆5𝑦0 − ∆6𝑦0 + ⋯

∆4𝑦−1 = ∆ ∆3𝑦−1

= ∆ ∆3𝑦0 − ∆4𝑦0 + ∆5𝑦0 − ∆6𝑦0 + ⋯

= ∆4𝑦0 − ∆5𝑦0 + ∆6𝑦0 − ∆7𝑦0 + ⋯

∆4𝑦−2 = ∆4𝐸−2𝑦0

= ∆4(1 + ∆)−2𝑦0

= ∆4(1 − 2∆ + 3∆2 − 4∆3 + ⋯ )𝑦0

= ∆4𝑦0 − 2∆5𝑦0 + 3∆6𝑦0 − 4∆7𝑦0 + ⋯

Hence, we have

𝑦0 + 𝑝∆𝑦0 +𝑝(𝑝 − 1)

2!∆2𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2

3!∆3𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)

4!∆4𝑦0 + ⋯

= 𝑦0 + 𝐵1 +1

2 ∆𝑦0 + 𝐵2

2∆2𝑦0−∆3𝑦0+∆4𝑦0−∆5𝑦0+⋯

2 + 𝐵3(∆3𝑦0 − ∆4𝑦0 + ∆5𝑦0 − ∆6𝑦0 +

⋯ ) + 𝐵4 2∆4𝑦0−3∆5𝑦0+4∆6𝑦0−5∆7𝑦0+⋯

2 + …


𝐵1 +1

2= 𝑝

𝐵2 =𝑝(𝑝 − 1)

2!

−𝐵2

2+ 𝐵3 =

𝑝 𝑝 − 1 𝑝 − 2

3! ⟹ 𝐵3 =

𝑝 𝑝 − 1 𝑝 − 1/2

3!

𝐵2

2− 𝐵3 + 𝐵4 =

𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)

4! ⟹ 𝐵4 =

𝑝 𝑝 + 1 𝑝 − 1 (𝑝 − 2)

4! 𝑒𝑡𝑐.

Hence, we get

𝑦𝑝 = 𝑦0 + 𝑝∆𝑦0 +𝑝 𝑝−1

2! ∆2𝑦−1+∆2𝑦0

2 +

𝑝 𝑝−1 𝑝−1/2

3!∆3𝑦−1 +

𝑝 𝑝+1 𝑝−1 𝑝−2

4! ∆4𝑦−2+∆4𝑦−1

2 + …

INTERPOLATION


Everett’s Formula

This is an extensively used interpolation method and uses only even order differences as shown below,

𝑥 𝑦 ∆ ∆2 ∆3 ∆4

𝑥−2 𝑦−2

∆𝑦−2

𝑥−1 𝑦−1 ∆2𝑦−2

∆𝑦−1 ∆3𝑦−2

𝑥0 𝑦0 ∆2𝑦−1 ∆4𝑦−2

∆𝑦0 ∆3𝑦−1

𝑥1 𝑦1 ∆2𝑦0 ∆4𝑦−1

∆𝑦1 ∆3𝑦0

𝑥2 𝑦2 ∆2𝑦1

∆𝑦2

𝑥3 𝑦3


𝑦𝑝 = 𝐸0𝑦0 + 𝐸2∆2𝑦−1 + 𝐸4∆

4𝑦−2 + …

+𝐹0𝑦1 + 𝐹2∆2𝑦0 + 𝐹4∆

4𝑦−1 + …

where, 𝐸0 , 𝐹0, 𝐸2, 𝐹2, 𝐸4, 𝐹4 , … have to be determined. 𝑦𝑝 can be expressed in terms of 𝑦0, ∆𝑦0 and higher order

differences as,


= (1 + ∆)𝑝𝑦0

= 𝑦0 + 𝑝∆𝑦0 +𝑝(𝑝 − 1)

2!∆2𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2

3!∆3𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)

4!∆4𝑦0 …


𝑦1 = 𝑦0 + ∆𝑦0

∆2𝑦−1 = ∆2𝐸−1𝑦0

= ∆2(1 + ∆)−1𝑦0

= ∆2(1 − ∆ + ∆2 − ∆3 + ⋯ )𝑦0

= ∆2𝑦0 − ∆3𝑦0 + ∆4𝑦0 − ∆5𝑦0 + ⋯

∆4𝑦−1 = ∆2 ∆2𝑦−1

= ∆. ∆(∆2𝑦0 − ∆3𝑦0 + ∆4𝑦0 − ∆5𝑦0 + ⋯ )

= ∆(∆3𝑦0 − ∆4𝑦0 + ∆5𝑦0 − ∆6𝑦0 + ⋯ )

= ∆4𝑦0 − ∆5𝑦0 + ∆6𝑦0 − ∆7𝑦0 + ⋯

INTERPOLATION


∆4𝑦−2 = ∆4𝐸−2𝑦0

= ∆4(1 + ∆)−2𝑦0

= ∆4(1 − 2∆ + 3∆2 − 4∆3 + ⋯ )𝑦0

= ∆4𝑦0 − 2∆5𝑦0 + 3∆6𝑦0 − 4∆7𝑦0 + ⋯

Hence, we have

𝑦0 + 𝑝∆𝑦0 +𝑝(𝑝 − 1)

2!∆2𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2

3!∆3𝑦0 +

𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)

4!∆4𝑦0 + ⋯

= 𝐸0𝑦0 + 𝐸2(∆2𝑦0 − ∆3𝑦0 + ∆4𝑦0 − ∆5𝑦0 + ⋯ ) + 𝐸4(∆4𝑦0 − 2∆5𝑦0 + 3∆6𝑦0 − 4∆7𝑦0 + ⋯ )+ …

+𝐹0(𝑦0 + ∆𝑦0) + 𝐹2∆2𝑦0 + 𝐹4(∆4𝑦0 − ∆5𝑦0 + ∆6𝑦0 − ∆7𝑦0 + ⋯ ) + …

Equating the coefficients of 𝑦0, ∆𝑦0, ∆2𝑦0, ∆3𝑦0, ∆4𝑦0, … on both sides, we get

𝐹0 = 𝑝

𝐹0 + 𝐸0 = 1 ⟹ 𝐸0 = 1 − 𝐹0 = 1 − 𝑝 = 𝑞

−𝐸2 =𝑝 𝑝 − 1 𝑝 − 2

3! ⟹ 𝐸2 = −

(1 − 𝑞) (1 − 𝑞) − 1 (1 − 𝑞) − 2

3!=

𝑞(𝑞2 − 12)

3!

𝐸2 + 𝐹2 =𝑝(𝑝 − 1)

2! ⟹ 𝐹2 =

𝑝(𝑝 − 1)

2!+

𝑝 𝑝 − 1 𝑝 − 2

3!=

𝑝(𝑝2 − 12)

3!

𝐸4 =𝑞 𝑞2 − 12 𝑞2 − 22

5!

𝐹4 =𝑝 𝑝2 − 12 𝑝2 − 22

5! 𝑒𝑡𝑐.

Hence, we get

𝑦𝑝 = 𝑞𝑦0 +𝑞(𝑞2 − 12)

3!∆2𝑦−1 +

𝑞 𝑞2 − 12 𝑞2 − 22

5!∆4𝑦−2 + …

+𝑝𝑦1 +𝑝(𝑝2 − 12)

3!∆2𝑦0 +

𝑝 𝑝2 − 12 𝑝2 − 22

5!∆4𝑦−1 + …

Relationship between Bessel’s and Everett’s Formula

Bessel’s formula is given as,

𝑦𝑝 = 𝑦0 + 𝑝∆𝑦0 +𝑝 𝑝−1

2! ∆2𝑦−1+∆2𝑦0

2 +

𝑝 𝑝−1 𝑝−1/2

3!∆3𝑦−1 +

𝑝 𝑝+1 𝑝−1 𝑝−2

4! ∆4𝑦−2+∆4𝑦−1

2 + …

= 𝑦0 + 𝑝(𝑦1 − 𝑦0) +𝑝 𝑝 − 1

2! ∆2𝑦−1 + ∆2𝑦0

2 +

𝑝 𝑝 − 1 𝑝 − 1/2

3! ∆2𝑦0 − ∆2𝑦−1

+𝑝 𝑝 + 1 𝑝 − 1 𝑝 − 2

4! ∆4𝑦−2 + ∆4𝑦−1

2 + …

INTERPOLATION


𝑦𝑝 = 1 − 𝑝 𝑦0 + 𝑝 𝑝 − 1

4−

𝑝 𝑝 − 1 𝑝 −12

6 ∆2𝑦−1 + ⋯

+𝑝𝑦1 + 𝑝 𝑝 − 1

4+

𝑝 𝑝 − 1 𝑝 −12

6 ∆2𝑦0 + ⋯

𝑦𝑝 = 𝑞𝑦0 +𝑞(𝑞2 − 12)

3!∆2𝑦−1 + ⋯ + 𝑝𝑦1 +

𝑝(𝑝2 − 12)

3!∆2𝑦0 + …

This is Everett’s formula truncated after second differences. Hence, Everett’s formula truncated after second

differences is equivalent to Bessel’s formula truncated after third differences.

Lagrange’s Interpolation Formula

The above formulae have been derived with the requirement that the values of the independent variables be

evenly spaced. It is hence desirable to have formulae with unevenly spaced independent variables. Lagrange’s

interpolation method is one such formula.

Let 𝑦(𝑥) be continuous and differentiable (𝑛 + 1) times in the interval (𝑎, 𝑏). Let 𝑥0, 𝑦0 , 𝑥1, 𝑦1 , … , (𝑥𝑛 , 𝑦𝑛)

be the (𝑛 + 1) points where the values of 𝑥 need not be necessarily equally spaced. We wish to find a

polynomial of degree 𝑛, such that

𝐿𝑛 𝑥𝑖 = 𝑦 𝑥𝑖 = 𝑦𝑖 , 𝑖 = 0, 1, … , 𝑛

Let us consider the simplest case i.e. the equation of a straight line (a linear polynomial) passing through two

points 𝑥0, 𝑦0 and 𝑥1, 𝑦1 . Such a polynomial is given as,

𝐿𝑛 𝑥 =𝑥 − 𝑥1

𝑥0 − 𝑥1𝑦0 +

𝑥 − 𝑥0

𝑥1 − 𝑥0𝑦1

= 𝑙0(𝑥)𝑦0 + 𝑙1(𝑥)𝑦1

= 𝑙𝑖(𝑥)𝑦𝑖

1

𝑖=0

where, 𝑙0 𝑥 =𝑥−𝑥1

𝑥0−𝑥1 and 𝑙1 𝑥 =

𝑥−𝑥0

𝑥1−𝑥0

It can be deduced from the above equations that,

𝑙0 𝑥0 = 1 , 𝑙0 𝑥1 = 0

𝑙1 𝑥0 = 0 , 𝑙1 𝑥1 = 1

The above relations can be expressed as,

𝑙𝑖 𝑥𝑗 = 1, 𝑖 = 𝑗0, 𝑖 ≠ 𝑗

To derive the general formula let the desired 𝑛𝑡𝑕 degree polynomial be,

𝐿𝑛 𝑥 = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + ⋯ + 𝑎𝑛𝑥𝑛 … (𝐴)

INTERPOLATION


We know that, 𝐿𝑛 𝑥𝑖 = 𝑦 𝑥𝑖 = 𝑦𝑖 , 𝑖 = 0, 1, … , 𝑛. Hence, we get

𝑦0 = 𝑎0 + 𝑎1𝑥0 + 𝑎2𝑥02 + ⋯ + 𝑎𝑛𝑥0

𝑛

𝑦1 = 𝑎1 + 𝑎1𝑥1 + 𝑎2𝑥12 + ⋯ + 𝑎𝑛𝑥1

𝑛 … (𝐵)

⋮ ⋮ ⋮ ⋮ ⋮

𝑦𝑛 = 𝑎0 + 𝑎1𝑥𝑛 + 𝑎2𝑥𝑛2 + ⋯ + 𝑎𝑛𝑥𝑛

𝑛

These equations will have a solution if the following determinant is not zero.

1 𝑥0 𝑥02 ⋯ 𝑥0

𝑛

1 𝑥1 𝑥12 ⋯ 𝑥1

𝑛

⋮ ⋮ ⋮ ⋮ ⋮1 𝑥𝑛 𝑥𝑛

2 ⋯ 𝑥𝑛𝑛

≠ 0

The value of the determinant, called Vandermonde’s determinant is

𝑥0 − 𝑥1 𝑥0 − 𝑥2 … 𝑥0 − 𝑥𝑛 𝑥1 − 𝑥2 … 𝑥1 − 𝑥𝑛 … 𝑥𝑛−1 − 𝑥𝑛

Eliminating 𝑎0, 𝑎1, 𝑎2, … , 𝑎𝑛 from equations 𝐴 and 𝐵, we get

𝐿𝑛 𝑥 1 𝑥 𝑥2 ⋯ 𝑥𝑛

𝑦0 1 𝑥0 𝑥02 ⋯ 𝑥0

𝑛

𝑦1 1 𝑥1 𝑥12 ⋯ 𝑥1

𝑛

⋮ ⋮ ⋮ ⋮ ⋮ ⋮𝑦𝑛 1 𝑥𝑛 𝑥𝑛

2 ⋯ 𝑥𝑛𝑛

= 0

This shows that, 𝐿𝑛 𝑥 is a linear combination of 𝑦0 , 𝑦1, 𝑦2, … , 𝑦𝑛 . Hence, we can write

𝐿𝑛 𝑥 = 𝑙𝑖(𝑥)𝑦𝑖

𝑛

𝑖=0

where 𝑙𝑖(𝑥) are polynomials of 𝑥 of degree 𝑛. Since, 𝐿𝑛 𝑥𝑗 = 𝑦𝑗 for 𝑗 = 0,1,2,3, … , 𝑛, we get

𝑙𝑖 𝑥𝑗 = 1, 𝑖 = 𝑗0, 𝑖 ≠ 𝑗

Hence, 𝑙𝑖(𝑥) can be written as,

𝑙𝑖 𝑥 = 𝑥 − 𝑥0 𝑥 − 𝑥1 … 𝑥 − 𝑥𝑖−1 𝑥 − 𝑥𝑖+1 … 𝑥 − 𝑥𝑛

𝑥𝑖 − 𝑥0 𝑥𝑖 − 𝑥1 … 𝑥𝑖 − 𝑥𝑖−1 𝑥𝑖 − 𝑥𝑖+1 … 𝑥𝑖 − 𝑥𝑛

𝑙𝑖 𝑥 =

𝑥 − 𝑥𝑗

𝑛

𝑗 =0𝑖≠𝑗

/[ 𝑥𝑖 − 𝑥𝑗

𝑛

𝑗=0𝑖≠𝑗

]

Now we get,

(𝑥)𝑛+1

= 𝑥 − 𝑥0 𝑥 − 𝑥1 … 𝑥 − 𝑥𝑖−1 𝑥 − 𝑥𝑖 𝑥 − 𝑥𝑖+1 … 𝑥 − 𝑥𝑛

INTERPOLATION


Then,

(𝑥𝑖)′

𝑛+1=

𝑑

𝑑𝑥 𝑥

𝑛+1 𝑥=𝑥𝑖

Hence,

𝑙𝑖 𝑥 = (𝑥)𝑛+1

𝑥 − 𝑥𝑖 (𝑥𝑖)′𝑛+1

𝐿𝑛 𝑥 = (𝑥)𝑛+1

𝑥 − 𝑥𝑖 (𝑥𝑖)′𝑛+1

𝑦𝑖

𝑛

𝑖=0

This is called the Lagrange’s interpolation formula. The coefficients, 𝑙𝑖 𝑥 , are called as Lagrange interpolation

coefficients. Interchanging 𝑥 and 𝑦, we obtain the formula,

𝐿𝑛 𝑦 = 𝑦 𝑛+1

𝑦 − 𝑦𝑖 𝑦𝑖 ′𝑛+1

𝑥𝑖

𝑛

𝑖=0

This is the inverse Lagrange’s interpolation formula.

Divided Differences

The Lagrange interpolation formula has the disadvantage that if another interpolation point is added, the

interpolation coefficients 𝑙𝑖 𝑥 have to be computed again. Newton’s general interpolation formula overcomes

this hurdle. This formula works on the principle of divided differences.

Let 𝑥0, 𝑦0 , 𝑥1, 𝑦1 , … , (𝑥𝑛 , 𝑦𝑛) be the (𝑛 + 1) points. Then the divided differences of orders 1, 2, 3, … , 𝑛 are

defined by the relations,

𝑥0, 𝑥1 =𝑦1 − 𝑦0

𝑥1 − 𝑥0

=𝑦1

𝑥1 − 𝑥0−

𝑦0

𝑥1 − 𝑥0

Also,

𝑥1, 𝑥0 =𝑦0 − 𝑦1

𝑥0 − 𝑥1

=𝑦0

𝑥0 − 𝑥1−

𝑦1

𝑥0 − 𝑥1

=𝑦1

𝑥1 − 𝑥0−

𝑦0

𝑥1 − 𝑥0

= 𝑥0 , 𝑥1

𝑥0, 𝑥1, 𝑥2 = 𝑥1, 𝑥2 − 𝑥0 , 𝑥1

𝑥2 − 𝑥0

=1

𝑥2 − 𝑥0 𝑦2 − 𝑦1

𝑥2 − 𝑥1−

𝑦1 − 𝑦0

𝑥1 − 𝑥0

INTERPOLATION


=1

𝑥2 − 𝑥0

𝑦2

𝑥2 − 𝑥1− 𝑦1

1

𝑥2 − 𝑥1+

1

𝑥1 − 𝑥0 +

𝑦0

𝑥1 − 𝑥0

=𝑦0

𝑥1 − 𝑥0 (𝑥2 − 𝑥0)−

𝑦1

𝑥1 − 𝑥0 (𝑥2 − 𝑥1)+

𝑦2

(𝑥2 − 𝑥0)(𝑥2 − 𝑥1)

=𝑦0

𝑥0 − 𝑥1 (𝑥0 − 𝑥2)+

𝑦1

𝑥1 − 𝑥0 (𝑥1 − 𝑥2)+

𝑦2

(𝑥2 − 𝑥0)(𝑥2 − 𝑥1)

Hence for (𝑛 + 1) points, we get

𝑥0, 𝑥1, 𝑥2 , … , 𝑥𝑛 = 𝑥1, 𝑥2, … , 𝑥𝑛 − 𝑥0 , 𝑥1, … , 𝑥𝑛−1

𝑥𝑛 − 𝑥0

=𝑦0

𝑥0 − 𝑥1 … (𝑥0 − 𝑥𝑛)+

𝑦1

𝑥1 − 𝑥0 … (𝑥1 − 𝑥𝑛)+ ⋯ +

𝑦𝑛

𝑥𝑛 − 𝑥0 … (𝑥𝑛 − 𝑥𝑛−1)

Hence the divided differences are symmetrical in their arguments.

A divided difference table is given as,

𝑥 𝑦 𝑦[. , . ] 𝑦[. , . , . ] 𝑦[. , . , . , . ] 𝑦[. , . , . , . , . ]

𝑥0 𝑦0

𝑦[𝑥0, 𝑥1]

𝑥1 𝑦1 𝑦[𝑥0, 𝑥1, 𝑥2]

𝑦[𝑥1, 𝑥2] 𝑦[𝑥0, 𝑥1, 𝑥2, 𝑥3]

𝑥2 𝑦2 𝑦[𝑥1, 𝑥2 , 𝑥3] 𝑦[𝑥0, 𝑥1, 𝑥2, 𝑥3 , 𝑥4] 𝑦[𝑥2, 𝑥3] 𝑦[𝑥1, 𝑥2, 𝑥3, 𝑥4]

𝑥3 𝑦3 𝑦[𝑥2, 𝑥3 , 𝑥4]

𝑦[𝑥3, 𝑥4]

𝑥4 𝑦4

Let the arguments be equally spaced given by,

𝑥1 − 𝑥0 = 𝑥2 − 𝑥1 = ⋯ = 𝑥𝑛 − 𝑥𝑛−1 = 𝑕 Hence,

𝑥0, 𝑥1 =𝑦1 − 𝑦0

𝑥1 − 𝑥0=

1

𝑕∆𝑦0

𝑥0, 𝑥1, 𝑥2 = 𝑥1, 𝑥2 − 𝑥0 , 𝑥1

𝑥2 − 𝑥0

=1

2𝑕 ∆𝑦1

𝑕−

∆𝑦0

𝑕 =

1

2𝑕2∆2𝑦0

=1

2! 𝑕2∆2𝑦0

𝑥0, 𝑥1, 𝑥2 , … , 𝑥𝑛 = 𝑥1, 𝑥2, … , 𝑥𝑛 − 𝑥0 , 𝑥1, … , 𝑥𝑛−1

𝑥𝑛 − 𝑥0

=1

𝑛! 𝑕2∆𝑛𝑦0

INTERPOLATION


Newton’s General Interpolation Formula

By definition of divided differences we get,

𝑥, 𝑥0 =𝑦 − 𝑦0

𝑥 − 𝑥0

Hence,

𝑦 = 𝑦0 + 𝑥 − 𝑥0 𝑥, 𝑥0

Again,

𝑥, 𝑥0, 𝑥1 = 𝑥, 𝑥0 − 𝑥0, 𝑥1

𝑥 − 𝑥1

𝑥, 𝑥0 = 𝑥0, 𝑥1 + 𝑥 − 𝑥1 𝑥, 𝑥0, 𝑥1

Hence, the first equation becomes,

𝑦 = 𝑦0 + 𝑥 − 𝑥0 𝑥0, 𝑥1 + 𝑥 − 𝑥1 𝑥, 𝑥0, 𝑥1

𝑦 = 𝑦0 + 𝑥 − 𝑥0 𝑥0, 𝑥1 + 𝑥 − 𝑥0 𝑥 − 𝑥1 𝑥, 𝑥0, 𝑥1

Again,

𝑥, 𝑥0 , 𝑥1, 𝑥2 = 𝑥, 𝑥0 , 𝑥1 − 𝑥0, 𝑥1, 𝑥2

𝑥 − 𝑥2

Hence,

𝑦 = 𝑦0 + 𝑥 − 𝑥0 𝑥0, 𝑥1 + 𝑥 − 𝑥0 𝑥 − 𝑥1 𝑥0, 𝑥1, 𝑥2 + 𝑥 − 𝑥0 𝑥 − 𝑥1 (𝑥 − 𝑥2) 𝑥, 𝑥0 , 𝑥1, 𝑥2

Proceeding further, we can obtain the general formula to be

𝑦 = 𝑦0 + 𝑥 − 𝑥0 𝑥0, 𝑥1 + 𝑥 − 𝑥0 𝑥 − 𝑥1 𝑥0, 𝑥1, 𝑥2 + 𝑥 − 𝑥0 𝑥 − 𝑥1 𝑥 − 𝑥2 𝑥0, 𝑥1, 𝑥2, 𝑥3 + ⋯+ 𝑥 − 𝑥0 𝑥 − 𝑥1 … (𝑥 − 𝑥𝑛) 𝑥, 𝑥0, 𝑥1, … , 𝑥𝑛

This is the Newton’s general interpolation formula with divided differences, the last term being the remainder

term after (𝑛 + 1) terms.

Bibliography

1) Introductory Methods of Numerical Analysis – S. S. Sastry

2) IGNOU BDP Mathematics MTE – 10

3) Computer Oriented Numerical Methods – V Rajaraman

INTERPOLATION


DIY (Do It Yourself)

1. From the following table find the value of 𝑦 0.23 and 𝑦(0.29).

𝑥 0.20 0.22 0.24 0.26 0.28 0.30

𝑦 1.6596 1.6698 1.6804 1.6912 1.7024 1.7139

2. The area A of a circle of diameter d is given in the following table. Find the area of the circle when the

diameter is 82 units and for 98 units.

𝑑 80 85 90 95 100

𝐴 5026 5674 6362 7088 7854

3. From the following table, find the number of students who obtained less than 45 marks and who

obtained more than 75 marks.

𝑀𝑎𝑟𝑘𝑠 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80

𝑁𝑜. 𝑜𝑓 𝑆𝑡𝑢𝑑𝑒𝑛𝑡𝑠 31 42 51 35 31

4. Find a polynomial which fits the data

𝑥 3 5 7 9 11

𝑦 6 24 58 108 174

5. The population of a town was as under. Estimate the population for the year 1955 and 1923.

𝑌𝑒𝑎𝑟 1921 1931 1941 1951 1961

𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 (𝑖𝑛 𝑡𝑕𝑜𝑢𝑠𝑎𝑛𝑑𝑠) 46 66 81 93 101

6. The following table gives the amount of a chemical dissolved in water at different temperatures. Find

the amount dissolved at 120 and at 310.

𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 100 150 200 250 300 350

𝐴𝑚𝑜𝑢𝑛𝑡 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑒𝑑 19.97 21.51 22.47 23.52 24.65 25.89

7. Find the missing term in the table:

𝑥 0 1 2 3 4

𝑦 1 3 9 ? 81

8. The values of 𝑦 are consecutive terms of a series. Find the first and tenth terms of the series. Find also

the polynomial that approximates these values.

𝑥 3 4 5 6 7 8 9

𝑦 13 21 31 43 57 73 91

9. Use Gauss’s forward formula for central interpolation to find the value of 𝑦 when 𝑥 = 32.

𝑥 25 30 35 40

𝑦 0.2707 0.3027 0.3386 0.3794

10. Use Gauss’s backward formula for central interpolation to find the value of 𝑦 when 𝑥 = 12516.

𝑥 12500 12510 12520 12530

𝑦 = 𝑥 111.803399 111.848111 111.892806 111.937483

11. Use Stirling’s formula for central interpolation to find the value of 𝑦 when 𝑥 = 1.91.

𝑥 1.7 1.8 1.9 2.0 2.1 2.2

𝑦 = 𝑒𝑥 5.4739 6.0496 6.6859 7.3891 8.1662 9.0250

INTERPOLATION


12. Use Stirling’s formula for central interpolation to find the value of 𝑦 when 𝑥 = 4.325.

𝑥 4.1 4.2 4.3 4.4 4.5

𝑦 30.1784 33.3507 36.8567 40.7316 45.0141

13. Use Stirling’s formula for central interpolation to find the value of sin(0.197).

𝑥 0.15 0.17 0.19 0.21 0.23

sin 𝑥 0.14944 0.16918 0.18886 0.20846 0.22798

14. Use Bessels’s formula for central interpolation to find the value of 𝐾(32).

𝑥 20 25 30 35 40 45 𝐾(𝑥) 14.035 13.674 13.257 12.734 12.089 11.309

15. Use Bessel’s formula for central interpolation to find the value of 𝑦(2.15).

𝑥 0 1 2 3 4 𝑦(𝑥) 6.9897 7.4036 7.7815 8.1281 8.4510

16. Use Everett’s formula for central interpolation to find the value of 𝑦(25).

𝑥 20 24 28 32 𝑦(𝑥) 2854 3162 3544 3992

17. From the following table find the value of 𝑦(0.543) using Everett’s formula for central interpolation.

𝑥 0.1 0.2 0.3 0.4 0.5 0.6 0.7

𝑦 2.631 3.328 4.097 4.944 5.875 6.896 8.013

18. Given the table of values, evaluate 155 using Lagrange’s interpolation formula.

𝑥 150 152 154 156

𝑦 = 𝑥 12.247 12.329 12.410 12.490

19. Using Lagrange’s interpolation formula, find the value of 𝑓(𝑥) when 𝑥 = 0 from the following table:

𝑥 3 2 1 −1

𝑦 3 12 15 −21

20. Using Lagrange’s interpolation formula, find the value of 𝑓(𝑥) when 𝑥 = 5 from the following table:

𝑥 0 1 3 4 7

𝑦 4 1 43 112 655

21. Using Lagrange’s formula, find the interpolating polynomial which approximates the following data:

𝑥 −2 −1 2 3

𝑦 −12 −8 3 5

22. Given the table of values, use Lagrange’s formula to find 𝑥 when 𝑥3

= 3.756.

𝑥 50 52 54 56

𝑦 = 𝑥3

3.684 3.732 3.779 3.825

23. Given the table of values, use Lagrange’s formula to find 𝑥 when 𝑦 = 5.

𝑥 1.8 2.0 2.2 2.4 2.6

𝑦 2.9422 3.6269 4.4571 5.4662 6.6947

INTERPOLATION


24. Using Lagrange’s interpolation formula, express the following functions as sum of partial functions

a. 𝑥2+6𝑥+1

𝑥−1 𝑥+1 𝑥−4 𝑥−6 b.

𝑥2+𝑥−3

𝑥3−2𝑥2−𝑥+2

25. Using Newton’s general interpolating formula, find the values of 𝑓 8 and 𝑓 15

𝑥 4 5 7 10 11 13

𝑓(𝑥) 48 100 294 900 1210 2028

26. Using Newton’s general interpolating formula, find the interpolating polynomial which approximates

the following data:

𝑥 0 1 4 5

𝑦 8 11 68 123

27. From the table of values given below, obtain the value of 𝑦 when 𝑥 = 1.5, using Newton’s general

interpolating formula

𝑥 0 1 2 4 5

𝑦 5 14 41 98 122

28. Show that 𝑙0 𝑥 + 𝑙1 𝑥 + 𝑙2 𝑥 + 𝑙3 𝑥 = 1 for all 𝑥.

29. From Bessel’s formula, derive the following formula,

𝑦1/2 =1

2 𝑦0 + 𝑦1 −

1

16 ∆𝑦−1

2 + ∆𝑦02 +

3

256 ∆𝑦−2

4 + ∆𝑦−14 − ⋯

30. Prove the following relations:

a. 𝛿2𝐸 = ∆2 b. 1 + 𝛿2𝜇2 = 1 +1

2𝛿2

2

c. ∇= 𝛿𝐸−1/2

d. ∆ − ∇= 𝛿2 e. 𝜇 = cosh 𝑢

2 , 𝑢 = 𝑕𝐷 f. 𝜇𝐸 = 𝐸𝜇

g. 𝑓 ′ 𝑥 = 𝜇𝛿𝑓 𝑥 −1

6𝜇𝛿2𝑓 𝑥 +

1

30𝜇𝛿5𝑓(𝑥)

Answer Key

1. 1.6751, 1.7081 2. 5280.1056, 7542.5056 3. 48, 17 4. 2𝑥2 − 7𝑥 + 9

5. 50.5968, 96.8368 6. 20.7483, 24.8794 7. 31 8. 3, 111, 𝑥2 + 𝑥 + 1

9. 0.3165 10. 111.87493 11. 6.7531 12. 37.7894

13. 0.1957 14. 13.0618 15. 7.8353 16. 3250.875

17. 6.3027 18. 12.4501 19. 6 20. 229

22. 53.0159 23. 2.314 25. 448, 3150 27. 26.3516

21. −

1

15𝑥3 −

3

20𝑥2

+241

60𝑥 − 3.9

24.

𝑖) −1

2

1

𝑥 + 1 +

1

2

1

𝑥 − 1 +

1

𝑥 − 2

𝑖𝑖) 2

35

1

𝑥 + 1 +

4

15

1

𝑥 − 1 −

41

30

1

𝑥 − 4 +

73

70

1

𝑥 − 6

26. 𝑥3 − 𝑥2 + 3𝑥 + 8

interpolation - subhasish...

Documents