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Interpolation Formula

Last Updated : 04 Sep, 2023
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In the course of statistical analysis and interpretation, sometimes it is found that a given series happens to be incomplete rather than complete, i.e., some values in the series remain unknown. But to derive correct results, it becomes essential to find the missing or unknown values in the series. The statistical technique that is used to estimate the unknown values on the basis of available data is called interpolation.

What Is Interpolation?

Interpolation is an important statistical tool. It is the process of calculating a value between two points on the curve of a function from the given points which also lie on the same curve. In other words, interpolation involves the calculation of new values from the already available set of values. Using interpolation, the diverse data can be converted into a concise function, such that each point in the data passes through the curve of such function. It is generally used in geography to predict data points such as noise level, rainfall, elevation, and so on.

Interpolation Formula

There are various formulae used to find the Interpolation and some of them are,

Linear Interpolation Formula

This method is used where the given data set does not consist of more than two ordered pairs. The value of the function at a certain value of the abscissa is obtained using the given ordered pairs.

f(x) = f(x_0) + (x − x_0) \dfrac{f(x_1) − f(x_0)}{x_1 − x_0}

or

y = y_0 + (x - x_0) \dfrac{y_1 - y_0}{x_1 - x_0}

Lagrange Interpolation Formula

This method is employed where the given data set consists of more than two ordered pairs and can be used to derive a polynomial, which further assumes certain arbitrary values on certain points.

f(x) = \dfrac{(x − x_0)(x-x_2)(x-x_3)(x-x_4)}{(x_1 − x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)} + \dfrac{(x − x_0)(x − x_1)(x − x_3)(x − x_4)}{(x_2 − x_0)(x_2 − x_1)(x_2 − x_3)(x_2 − x_4)}+\dfrac{(x − x_0)(x − x_1)(x − x_2)(x − x_4)}{(x_3 − x_0)(x_3 − x_1)(x_3 − x_2)(x_3 − x_4)}+....

Leading Difference Formula

This method is used when the data set itself is kind of incomplete, i.e., where the frequency of one or more data sets is missing. In this case, the leading difference up to the number of known items is always zero and is given as:

â–³n0 = (y − 1)n = yn − ayn-1 + byn-2 − cyn-3 +….+ y0 = 0

Quadratic Interpolation Formula

It is a refined form of linear interpolation method and is given as:

f(x_j + θh) ≈ f_j + θΔf_j + \frac{1}{2}θ(θ-1)Δ^2f_j

Uses of Interpolation

Various uses of Interpolation are given below,

Deriving a Function From a Data Set

The diverse and scattered data points can be turned into a compact function using interpolation so that each point in the data travels through the curve of such a function. This makes it so much easier to understand the whole data in just one glance. Data can be converted into the following form:

p(x) = a0 + a1ex + a2e2x +…..+ anenx

Obtaining Piecewise Polynomials

With the help of interpolation, one can also approximate functions into polynomials, i.e., simpler forms, which would make it easier to integrate and differentiate such functions to facilitate the calculations pertaining to areas under curves. Many a time, the problems involving integration are very difficult to be solved analytically. In such cases, interpolation is used to alter the integrands and polynomials to make calculations easier.

Difference Between Interpolation and Extrapolation

The difference between Interpolation and Extrapolation are discussed in the table given below,

Interpolation

Extrapolation

It is the process of calculating a value between two points on the curve of a function from the given points which also lie on the same curve. It is the way of predicting the magnitude of the reliant factor for an individual entity that is beyond the scope of the dataset.
As the name suggests, “inter”-polation works within the scope of the given data.Extrapolation works outside the scope of the given data.

Read, More

Solved Examples Interpolation Formula

Example 1: Interpolate the value of y at x = 3, if two of the ordered pairs are (−5, 5), (5, 7).

Solution:

Given

x = 3, x0 = −5, x1 = 5, y0 = 5, y1 = 7 

Since, y = y_0 + (x - x_0) \dfrac{y_1 - y_0}{x_1 - x_0}

y = 5 + (3 − (−5)) \dfrac{7 − 5}{5 − (-5)}

   = 5 + 8(.2)

=5 + 1.6

y = 6.6

Example 2: Interpolate the missing value in the following table:

XY
1990100
1991107
1992?
1993157
1994212

Solution:

Since the known values of Y are 4, the fourth leading difference will be zero. Thus, 

△40 = (y − 1)4 = y4 − 4y3 + 6y2 − 4y1 + y0 = 0

Substituting the values of y’s we have,

212 − 4(157) + 6y2 − 4y1 + y0 = 0 

6y2 = 628 − 212 + 428 − 100

y2 = 124

Examples 3: The table below depicts insurance premiums payable to different age groups. Find the premium payable at 17.

AgePremium Payable
1511.1
2512.6
3514.3
4516.1
5518.3

Solution:

AgePremium(Y)â–³1â–³2â–³3â–³4
15 = x011.1 = y0    
  +1.5 = â–³10   
25 = x112.6 = y1 +0.2 = â–³20  
  +1.7 = â–³11 âˆ’0.1 = â–³30 
35 = x214.3 = y2 +0.1 = â–³21 +0.4 = â–³40
  +1.8 = â–³13 +0.3 = â–³31 
45 = x316.1 = y3 +0.4 = â–³22  
  +2.2 = â–³14   
55 = x418.3 = y4    

Now, x = 17 – 15/25 – 15 = 2/10 = 0.2

Hence, Y17 = 11.1 + 0.3 − 0.016 − 0.0048 − 0.0134 = 11.4

Thus premium payable at 17 is 11.4

Example 4: Find the value of y when x = 5 from the following data:

xy
21
35
413
661
7125

Solution:

Applying Lagrange Formula, 

f(x) = \dfrac{(x − x_0)(x-x_2)(x-x_3)(x-x_4)}{(x_1 − x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)} + \dfrac{(x − x_0)(x − x_1)(x − x_3)(x − x_4)}{(x_2 − x_0)(x_2 − x_1)(x_2 − x_3)(x_2 − x_4)}+\dfrac{(x − x_0)(x − x_1)(x − x_2)(x − x_4)}{(x_3 − x_0)(x_3 − x_1)(x_3 − x_2)(x_3 − x_4)}+....

f(x) = 1.\dfrac{(5−3)(5-4)(5-6)(5-7)}{(2−3)(4-3)(4-6)(4-7)}+5.\dfrac{(5−2)(5−4)(5−6)(5−7)}{(3−2)(3−4)(3−6)(3−7)}+13.\dfrac{(5−2)(5−3)(5−6)(5−7)}{(4−2)(4−3)(4−6)(4−7)}+....

       = 0.1 − 2.5 + 1.3 + 30.5 − 12.5

      = 28.6

Example 5: Reconstruct the following data with the class intervals halved using interpolation:

MarksNumber of Students
0 – 1050
10 – 2065
20 – 3085

Solution:

For estimating the marks in the class interval 0-5, 10-15, and 20-25, we will have to interpolate the values of less than 5, less than 15 and less than 25 respectively. 

Interpolation of less than 5

x = 5−10/10 = −0.5

y5 = y0 + x.△10 + [x(x−1).△2]/2 = 25

Similarly, 

Interpolation of less than 15 = 80

Interpolation of less than 25 = 155

Thus, the reconstructed table is as follows:

MarksCumulative FrequencyMarksNumber of Students
Less than 5250 – 522
Less than 10505 – 1025 
Less than 158010 – 1530
Less than 2011515 – 2035
Less than 2515520 – 2540
Less than 3020025 – 3045

FAQs on Interpolation Formula

1. What is Interpolation?

Interpolation is an important statistical tool. It is the process of calculating a value between two points on the curve of a function from the given points which also lie on the same curve.

2. Why is Interpolation Important?

Interpolation is needed to compute the value of a function for an intermediate value of the independent function.

3. What are Piecewise Polynomials?

Approximated functions converted into polynomials, i.e., simpler forms, which would to make it easier to integrate and differentiate such functions to facilitate the calculations pertaining to areas under curves are called piecewise polynomials

4. What are the Various Methods of Interpolation?

The various methods of interpolation are,

  • Linear Interpolation Formula
  • Nearest Neighbor Method
  • Quadratic Interpolation
  • Lagrange Formula
  • Newton Formula

5. What is the Use of Interpolation?

Deriving a Function From a Data Set: The diverse and scattered data points can be turned into a compact function using interpolation, so that each point in the data travels through the curve of such function. This makes it so much easier to understand the whole data in just one glance. Data can be converted into the following form:

p(x) = a0 + a1ex + a2e2x +…..+ anenx

6. What is the Difference between Interpolation and Extrapolation?

Interpolation is the process of calculating a value between two points on the curve of a function from the given points which also lie on the same curve, whereas extrapolation is the way of predicting the magnitude of the reliant factor for an individual entity that is beyond the scope of the dataset.



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