Interpolation Formula
Interpolation formula is a method to find new values of any function using the set of available values through interpolation. It is an important statistical tool used to calculate the value between two points on the curve of a function from the given points which also lie on the same curve.
In 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 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
- Lagrange Interpolation Formula
- Leading Difference Formula
- Quadratic Interpolation Formula
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.
[Tex]f(x) = f(x_0) + (x β x_0) \dfrac{f(x_1) β f(x_0)}{x_1 β x_0} [/Tex]
or
[Tex]y = y_0 + (x β x_0) \dfrac{y_1 β y_0}{x_1 β x_0} [/Tex]
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.
[Tex]f(x) = \dfrac{(x β x_0)(x-x_2)(x-x_3)(x-x_4)}{(x_1 β x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)} + \dfrac{(x β x_0)(x β x_1)(x β x_3)(x β x_4)}{(x_2 β x_0)(x_2 β x_1)(x_2 β x_3)(x_2 β x_4)}+\dfrac{(x β x_0)(x β x_1)(x β x_2)(x β x_4)}{(x_3 β x_0)(x_3 β x_1)(x_3 β x_2)(x_3 β x_4)}+β¦. [/Tex]
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:
β³n0 = (y β 1)n = yn β ayn-1 + byn-2 β cyn-3 +β¦.+ y0 = 0
Quadratic Interpolation Formula
It is a refined form of linear interpolation method and is given as:
[Tex]f(x_j + ΞΈh) β f_j + ΞΈΞf_j + \frac{1}{2}ΞΈ(ΞΈ-1)Ξ^2f_j [/Tex]
Interpolation Formula for unequal intervals
In numerical analysis, interpolation formulas are used to estimate values between known data points. For unequal intervals, one common method is the Lagrange Interpolation. This technique is especially useful when the intervals between your data points vary, as it doesnβt require the intervals to be equal.
[Tex]P(x) = \sum_{i=0}^{n} y_i \ell_i(x) \quad \text{where} \quad \ell_i(x) = \prod_{\substack{j=0 \\ j \neq i}}^{n} \frac{x β x_j}{x_i β x_j}[/Tex]
Key Points
- P(x): This is the polynomial that interpolates the points.
- ββ: The summation symbol, aggregating contributions from each term indexed by ?i.
- ??yiβ: The function values at each ??xiβ.
- β?(?)βiβ(x): The Lagrange basis polynomial for each point ??xiβ, which is crucial for ensuring that ?(?)P(x) passes through each data point exactly.
- ββ: The product symbol, representing the multiplication across all terms except when ?=?j=i.
- ?β????β??xiββxjβxβxjββ: The individual terms of the product that construct each basis polynomial β?(?)βiβ(x), ensuring it equals 1 at ??xiβ and 0 at all other ??xjβ.
Interpolation Formula for equal intervals
The interpolation formula for equal intervals, often used in methods like Newtonβs divided differences or polynomial interpolation, can be expressed in LaTeX as follows:
[Tex]P(x) = f(x_0) + \Delta x f[x_0, x_1] \cdot (x β x_0) + \Delta x^2 f[x_0, x_1, x_2] \cdot (x β x_0)(x β x_1) + \ldots [/Tex]
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 +β¦..+ 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, βinterβ-polation works within the scope of the given data. | Extrapolation works outside the scope of the given data. |
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Solved Examples on Interpolation Formula
Example 1: Interpolate the value of y at x = 3, if two of the ordered pairs are (β5, 5), (5, 7).
Solution:
Given
x = 3, x0 = β5, x1 = 5, y0 = 5, y1 = 7
Since, [Tex]y = y_0 + (x β x_0) \dfrac{y_1 β y_0}{x_1 β x_0} [/Tex]
[Tex]y = 5 + (3 β (β5)) \dfrac{7 β 5}{5 β (-5)} [/Tex]
= 5 + 8(.2)
=5 + 1.6
y = 6.6
Example 2: Interpolate the missing value in the following table:
X | Y |
---|---|
1990 | 100 |
1991 | 107 |
1992 | ? |
1993 | 157 |
1994 | 212 |
Solution:
Since the known values of Y are 4, the fourth leading difference will be zero. Thus,
β³40 = (y β 1)4 = y4 β 4y3 + 6y2 β 4y1 + y0 = 0
Substituting the values of yβs we have,
212 β 4(157) + 6y2 β 4y1 + y0 = 0
6y2 = 628 β 212 + 428 β 100
y2 = 124
Examples 3: The table below depicts insurance premiums payable to different age groups. Find the premium payable at 17.
Age | Premium Payable |
---|---|
15 | 11.1 |
25 | 12.6 |
35 | 14.3 |
45 | 16.1 |
55 | 18.3 |
Solution:
Age Premium(Y) β³1 β³2 β³3 β³4 15 = x0 11.1 = y0 +1.5 = β³10 25 = x1 12.6 = y1 +0.2 = β³20 +1.7 = β³11 β0.1 = β³30 35 = x2 14.3 = y2 +0.1 = β³21 +0.4 = β³40 +1.8 = β³13 +0.3 = β³31 45 = x3 16.1 = y3 +0.4 = β³22 +2.2 = β³14 55 = x4 18.3 = y4 Now, x = 17 β 15/25 β 15 = 2/10 = 0.2
Hence, Y17 = 11.1 + 0.3 β 0.016 β 0.0048 β 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:
x | y |
---|---|
2 | 1 |
3 | 5 |
4 | 13 |
6 | 61 |
7 | 125 |
Solution:
Applying Lagrange Formula,
[Tex]f(x) = \dfrac{(x β x_0)(x-x_2)(x-x_3)(x-x_4)}{(x_1 β x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)} + \dfrac{(x β x_0)(x β x_1)(x β x_3)(x β x_4)}{(x_2 β x_0)(x_2 β x_1)(x_2 β x_3)(x_2 β x_4)}+\dfrac{(x β x_0)(x β x_1)(x β x_2)(x β x_4)}{(x_3 β x_0)(x_3 β x_1)(x_3 β x_2)(x_3 β x_4)}+β¦. [/Tex]
f(x) = [Tex]1.\dfrac{(5β3)(5-4)(5-6)(5-7)}{(2β3)(4-3)(4-6)(4-7)}+5.\dfrac{(5β2)(5β4)(5β6)(5β7)}{(3β2)(3β4)(3β6)(3β7)}+13.\dfrac{(5β2)(5β3)(5β6)(5β7)}{(4β2)(4β3)(4β6)(4β7)}+β¦. [/Tex]
= 0.1 β 2.5 + 1.3 + 30.5 β 12.5
= 28.6
Example 5: Reconstruct the following data with the class intervals halved using interpolation:
Marks | Number of Students |
---|---|
0 β 10 | 50 |
10 β 20 | 65 |
20 β 30 | 85 |
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 = 5β10/10 = β0.5
y5 = y0 + x.β³10 + [x(xβ1).β³2]/2 = 25
Similarly,
Interpolation of less than 15 = 80
Interpolation of less than 25 = 155
Thus, the reconstructed table is as follows:
Marks Cumulative Frequency Marks Number of Students Less than 5 25 0 β 5 22 Less than 10 50 5 β 10 25 Less than 15 80 10 β 15 30 Less than 20 115 15 β 20 35 Less than 25 155 20 β 25 40 Less than 30 200 25 β 30 45
Interpolation Formula β FAQs
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.
Why is Interpolation Important?
Interpolation is needed to compute the value of a function for an intermediate value of the independent function.
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
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
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 +β¦..+ anenx
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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