The linear interpolation is a method which originates general Newton interpolation and approximation to determine for an unknown value that is between two given numbers; that is, an intermediate value is found. It is also applied to approximate functions, where the values f (a) and f (b) are known and we want to know the intermediate of f (x).
There are different types of interpolation, such as linear, quadratic, cubic and of higher degrees, the simplest being the linear approximation. The price that must be paid with linear interpolation is that the result will not be as accurate as with approximations using functions of higher degrees.
Definition
Linear interpolation is a process that allows you to deduce a value between two well-defined values, which can be in a table or in a line graph.
For example, if you know that 3 liters of milk are worth $ 4 and that 5 liters are worth $ 7, but you want to know what the value of 4 liters of milk is, you interpolate to determine that intermediate value.
Method
To estimate an intermediate value of a function, the function f (x) is approximated by means of a line r (x), which means that the function varies linearly with «x» for a section «x = a» and «x = b "; that is, for a value "x" in the interval (x 0, x 1) and (y 0, y 1), the value of "y" is given by the line between the points and is expressed by the following relationship:
(y - y 0) ÷ (x - x 0) = (y 1 - y 0) ÷ (x 1 - x 0)
For an interpolation to be linear, the interpolation polynomial must be of degree one (n = 1), so that it fits the values of x 0 and x 1.
Linear interpolation is based on similarity of triangles, in such a way that, deriving geometrically from the previous expression, the value of "y" can be obtained, which represents the unknown value for "x".
That way you have to:
a = tan Ɵ = (opposite leg 1 ÷ adjacent leg 1) = (opposite leg 2 ÷ adjacent leg 2)
Expressed in another way, it is:
(y - y 0) ÷ (x - x 0) = (y 1 - y 0) ÷ (x 1 - x 0)
Solving for «and» from the expressions, we have:
(y - y 0) * (x 1 - x 0) = (x - x 0) * (y 1 - y 0)
(y - y 0) = (y 1 - y 0) *
Thus, the general equation for linear interpolation is obtained:
y = y 0 + (y 1 - y 0) *
In general, linear interpolation gives a small error on the real value of the true function, although the error is minimal compared to if you intuitively choose a number close to the one you want to find.
This error occurs when trying to approximate the value of a curve with a straight line; In these cases, the size of the interval must be reduced to make the approximation more precise.
For better results regarding the approximation, it is advisable to use functions of degree 2, 3 or even higher degrees to perform the interpolation. For these cases the Taylor theorem is a very useful tool.
Solved exercises
Exercise 1
The number of bacteria per unit volume existing in an incubation after x hours is presented in the following table. You want to know what is the volume of bacteria for the time of 3.5 hours.
Solution
The reference table does not establish a value that indicates the amount of bacteria for a time of 3.5 hours, but there are upper and lower values corresponding to a time of 3 and 4 hours, respectively. That way:
x 0 = 3 and 0 = 91
x = 3.5 y =?
x 1 = 4 and 1 = 135
Now, the mathematical equation is applied to find the interpolated value, which is the following:
y = y 0 + (y 1 - y 0) *.
Then the corresponding values are substituted:
y = 91 + (135 - 91) *
y = 91 + (44) *
y = 91 + 44 * 0.5
y = 113.
Thus, it is obtained that for a time of 3.5 hours, the number of bacteria is 113, which represents an intermediate level between the volume of bacteria existing in the times of 3 and 4 hours.
Exercise 2
Luis has an ice cream factory, and he wants to do a study to determine the income he had in August based on the expenses made. The administrator of the company makes a graph that expresses this relationship, but Luis wants to know:
What is the income for August, if an expense of $ 55,000 was incurred?
Solution
A graph is given with values of income and expenses. Luis wants to know what the August revenues are if the factory had an expense of $ 55,000. This value is not directly reflected in the graph, but the values are higher and lower than this.
First a table is made where to easily relate the values:
Now, the interpolation formula is used to thus determine the value of y
y = y 0 + (y 1 - y 0) *
Then the corresponding values are substituted:
y = 56,000 + (78,000 - 56,000) *
y = 56,000 + (22,000) *
y = 56,000 + (22,000) * (0.588)
y = 56,000 + 12,936
y = $ 68,936.
If an expense of $ 55,000 was made in August, the income was $ 68,936.
References
- Arthur Goodman, LH (1996). Algebra and trigonometry with analytical geometry. Pearson Education.
- Harpe, P. d. (2000). Topics in Geometric Group Theory. University of Chicago Press.
- Hazewinkel, M. (2001). Linear interpolation ", Encyclopedia of Mathematics.
- , JM (1998). Elements of numerical methods for Engineering. UASLP.
- , E. (2002). A chronology of interpolation: from ancient astronomy to modern signal and image processing. Proceedings of the IEEE.
- numerical, I. a. (2006). Xavier Tomàs, Jordi Cuadros, Lucinio González.