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Following conditions are applied binomial interpolation method: 

• The X-variable (independent variable) advances by equal intervals say 15, 20, 25, 30 or say 2, 4, 6, 8, 10 etc. 
• The value of X for which the value of Y is to be estimated must be one of the values of X.

Interpolation is the technique of estimating the value dependent variable( Y) for any intermediate value of the independent variable(X).

The equation (y – 1 )⁵ = 0 is: 

y₅ – 5y_-4 + 10y₃ -10y₂ + 5y₁ -y₀ = 0

Since 4 known values of y are given, then 4th leading difference may be zero.

Δ⁴₀ = (y – 1 )⁴ = y₄ – 4y₃ + 6y₂ – 4y₁ + y₀ = 0 (i) 

and the second equation can be obtained by increasing the suffixes of each term of ‘Y’ by one, keeping the coefficients same; we get:

Δ⁴₁ = (y – 1)⁴ = y₅ – 4y₄ + 6y₃ – 4y₂ + y₁ = 0 …………. (ii)

From equation (i) y₄ – 4y₃ + 6y₂ – 4y₁ + y₀ = 0

ie., 38 – 4(33) + 6y₂ – 4(20) + 13 = 0

So, by simplifying, 38 – 132 + 6y₂ – 80 + 13 = 0 ;

6y₂ – 161

We get y₂ =26.83

From (ii) y₅ – 4y₄ + 6y₃ – 4y₂ + y = 0 = 0

y₅ – 4(38) + 6(33) -4(26.83) + 20 = 0

y₅ – 152 + 198 – 107.32 + 20 = 0

∴ y = 41.32.

Since the known values are 5, the estimation is based on the expansion of Δ⁵ 

= (y – 1)⁵ = 0

∴ Δ⁵= (y – 1)⁵ 

= y₅ – 5y₄ + 10y₃ – 10y₂ + 5y₁ – y₀ = 0

We have to determine the value of y_3;

20.1 – 5(23.1) + 10y₃ – 10(29.1) + 5(32.2) – 35.4 = 0

So, by simplifying, 10y₃ – 260.8 = 0

∴ y₃ = 26.08 years

Hence, the probable expectation of life at the age 25 is 26.08 years.

Now expand (y – 1)⁵ = 0 with change of subscript, keeping coefficients as it is.

y₆ – 5y₅ + 10y₄ – 10y₃ + 5y₂ – y₁ = 0

y₆ – 5(20.1) + 10(23.1) – 10(26.08) + 5(29.1)-32.2 = 0

y₆– 17 = 0, y₆ = 17 years.

Extrapolation is a procedure of estimating the unknown value of dependent variable for a given value of independent variable which is outside the limits or the range of the independent variable’.

In making use of the techniques of interpolation the following assumptions are made

• There are no sudden jumps in the values of independent variable from one period to another. 
• The rate of change of figures from one period to another is uniform.

The procedure of estimating the missing value of y for a given value of x, where x is within the limits x₀ and x_n we use the technique Interpolation.

Since the known values are 5, the fifth leading differences will be 

zero, i.e. Δ⁵ = 0

Δ⁵ = (y – 1 )⁵ 

= y₅ – 5y₄ +10y₃ – 10y₂ + 5y₁ – y₀ = 0 …….(i)

And the second equation can be obtained by, increasing the suffixes of each term of’y’ by one, keeping the coefficients same;

ie. (y- 1)⁵ = y₆ – 5y₅ + 10y₄ – 10y₃+ 5y₂ – y, = 0 ………. (ii) 

determine the value of Y₂ from equation (i)

162 – 5(142) +10 (128) – 10y₂ + 5(112) – 100 = 0

by simplifying, -10y₂ = -1192

y₂ = 119.2 

Hence, the missing Index number for 1990 is 119.2

From (ii) y₆ – 5(162) + 10(142) – 10(128) + 5(119.2) – 112 = 0 

Here y₂ =119.2

y₆ – 810 + 1420 – 1280 + 596- 112 = 0.

∴ y₆ = 186

Hence the cost of living index number for the year 2010 is 186.

• Binomial expansion method and 
• Newton’s advancing difference method

Interpolation is the technique of estimating the value dependent variable(Y) for any intermediate ) value of the independent variable(X). I Extrapolation is the technique of estimating the value of dependent variable (Y) any value of independent variable (X) which is outside the given series.

The procedure of estimating the missing value of y for a given value of x, where x is within the limits x₀ and x_n we use Interpolation. Here “Interpolation is a procedure of estimating the unknown value of dependent variable for a given value of independent variable which is within the limits or the range of the independent variable”.

But if the value of y is to be estimated for a value of x which is outside the limits x₀ and x_n then procedure Extrapolation is used. “Extrapolation is a procedure of estimating the unknown value of dependent variable for a given value of independent variable which is outside the limits or the range of the independent variable”.

Extrapolation is the technique of estimating the value of dependent variable (Y) any value of independent variable (X) which is outside the given series.

The procedure of estimating the missing value of y for a given value of x, where x is within the limits x₀ and x_n we use Interpolation. Here “Interpolation is a procedure of estimating the unknown value of dependent variable for a given value of independent variable which is within the limits or the range of the independent variable”.

But if the value of y is to be estimated for a value of x which is outside the limits x, and x, then procedure Extrapolation is used. Here “Extrapolation is a procedure of estimating the unknown value of dependent variable for a given value of independent variable which is outside the limits or the range of the independent variable”.