Linear interpolation is a method used to estimate intermediate values between two known points. This method considers a linear relationship between the known points to determine the value of interest. This process is widely used in various fields, such as mathematics, statistics, engineering, and others.
In this article, we'll cover the linear interpolation method, explaining step-by-step how to perform this type of calculation. We'll also present examples of solved exercises to facilitate understanding and practical application of this method. With this, we hope to contribute to the understanding and efficient use of linear interpolation in various contexts.
Step by step guide to perform an interpolation calculation efficiently.
To perform an interpolation calculation efficiently, follow these steps:
Step 1: Gather the data needed for the interpolation problem. This includes the known values of the reference points and the x-value for which you want to interpolate.
Step 2: Calculate the difference between the known values of x closest to the desired value of x. This will help determine the interpolation interval.
Step 3: Calculate the difference between the y-values corresponding to the nearest reference points. This will determine the variation of y in the interpolation interval.
Step 4: Using the linear interpolation formula, calculate the value of y for the desired value of x. The formula is given by:
y = y1 + [(x – x1)/(x2 – x1)] * (y2 – y1)
Step 5: Substitute the known values into the formula and solve to obtain the value of y that corresponds to the desired value of x.
By following these steps, you will be able to perform an interpolation calculation efficiently and obtain accurate results for the problem at hand.
When is it appropriate to use linear interpolation in mathematical analysis or modeling?
Linear interpolation is a method used in mathematical analysis or modeling when it is desired to estimate intermediate values from known values. This method is suitable when the data are distributed in a linear and continuous manner.
For example, when we have a table with values of a function at discrete points and we want to estimate the value of the function for an intermediate point, linear interpolation can be a good choice. It consists of drawing a straight line between two known points and using that line to estimate the desired value.
It is important to emphasize that linear interpolation may not be suitable in cases where the data exhibits non-linear behavior, such as curves or oscillations. In these cases, other more complex interpolation methods may be more appropriate.
How to efficiently fill gaps in a table with intermediate values.
Linear interpolation is a method used to efficiently fill gaps in a table with intermediate values. This method consists of calculating the missing values based on the known values in the table, using a simple formula.
To apply linear interpolation, you need at least two known points in the table. From these points, you can calculate the intermediate values using the following formula:
Y = Y1 + (X – X1) * (Y2 – Y1) / (X2 – X1)
Where Y is the intermediate value we want to find, Y1 and Y2 are the known values from the table, X1 and X2 are the corresponding values of X and X is the value of X for which we want to find the intermediate value Y.
To illustrate, let's solve a practical exercise using linear interpolation. Suppose we have the following table:
| X | Y |
|---|---|
| 1 | 3 |
| 3 | 7 |
Now, let's use the linear interpolation formula to find the value of Y when X is equal to 2:
Y = 3 + (2 – 1) * (7 – 3) / (3 – 1)
Y = 3 + 1 * 4 / 2
Y = 3 + 2
Y = 5
Therefore, the value of Y when X is equal to 2 is 5.
As we can see, linear interpolation is a simple and efficient method for filling gaps in a table with intermediate values. Using this technique, we can obtain accurate results quickly and easily.
Practical guide to performing quadratic interpolation efficiently.
To perform quadratic interpolation efficiently, it's important to follow a few simple steps. First, you need to have the known points that will be used for the interpolation. Then, you need to calculate the coefficients of the quadratic function that passes through these points.
To calculate the coefficients of a quadratic function, we can use the method of least squares. This method consists of finding the quadratic function that minimizes the sum of the squares of the differences between the calculated values and the actual values of the known points.
Once the coefficients of the quadratic function have been calculated, we can use it to interpolate values between known points. To do this, simply plug the desired value into the quadratic function and obtain the corresponding interpolated value.
It's important to note that quadratic interpolation is more accurate than linear interpolation because it takes into account the curvature of the data. However, it's important to note that quadratic interpolation can lead to less accurate results in regions far from known points.
Therefore, when performing quadratic interpolation, it's crucial to ensure that the known points are evenly distributed and that the quadratic function is calculated correctly. This allows for more accurate and reliable interpolation results.
Linear Interpolation: Method, Solved Exercises
Linear interpolation is a method derived from general Newton interpolation and approximation to determine an unknown value that lies between two 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 you want to know the intermediate of f (x).
There are different types of interpolation, including linear, quadratic, cubic, and higher-order interpolation, with the simplest being linear approximation. The tradeoff with linear interpolation is that the result won't be as accurate as with approximations using higher-order functions.
Definition of
Linear interpolation is a process that allows you to deduce a value between two well-defined values, which can be in a table or a linear graph.
For example, if you know that 3 liters of milk are worth $4 and 5 liters are worth $7, but you want to know what the value of 4 liters of milk is, you interpolate to find that intermediate value.
Method
To estimate an intermediate value of a function, the function f (x) is approximated by means of the 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 (x0 , x1 ) y( y0 , y1 ) , the value of "y" is given by the line between the points and expressed by the following relation:
(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 x0 and x1.
Linear interpolation is based on the similarity of triangles, so that by geometrically deriving the previous expression, it is possible to obtain the value of «y», which represents the unknown value for «x».
Therefore, you must:
a = tan Ɵ = (opposite leg 1 ÷ adjacent leg 1 ) = (opposite leg 2 ÷ adjacent leg 2 )
Expressed another way, it is:
(y – y 0 ) ÷ (x – x 0 ) = (y 1 – y 0 ) ÷ (x 1 – x 0 )
Removing “and” from the expressions, you have:
(y – y 0 ) * (x 1 – x 0 ) = (x – x 0 ) * (y 1 – y 0 )
(y – y 0 ) = (y 1 – y 0 ) * [(x – x 0 ) ÷ (x 1 – x 0 )]
Thus, the general equation for linear interpolation is obtained:
y = y 0 + (y 1 – y 0 ) * [(x – x 0 ) ÷ (x 1 – x 0 )]
In general, linear interpolation gives a small error on the actual value of the true function, although the error is minimal compared to the intuitive number of a number close to the one found.
This error occurs when trying to approximate the value of a curve with a straight line; in such cases, the interval size must be reduced to make the approximation more accurate.
For best approximation results, it's advisable to use functions of degree 2, 3, or even higher degrees to perform the interpolation. For these cases, Taylor's theorem is a very useful tool.
Solved exercises
1 Exercise
The number of bacteria per unit volume in an incubation after x hours is shown in the following table. You want to know the volume of bacteria over the 3,5-hour period.
Solution
The reference table doesn't establish a value that indicates the number of bacteria over a 3,5-hour period, but there are higher and lower values corresponding to a period of 3 and 4 hours, respectively. Thus:
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 as follows:
y = y 0 + (y 1 – y 0 ) * [(x – x 0 ) ÷ (x 1 – x 0 )].
Then the corresponding values are substituted:
y = 91 + (135 – 91) * [(3,5 – 3) ÷ (4-3)]
y = 91 + (44) * [(0,5) ÷ (1)]
y = 91 + 44 * 0,5
y=113.
Thus, it is obtained that, for a period of 3,5 hours, the amount of bacteria is 113, which represents an intermediate level between the volume of bacteria existing at times of 3 and 4 hours.
2 Exercise
Luis owns an ice cream factory and wants to conduct a study to determine his August income based on his expenses. The company administrator creates a graph that reflects this relationship, but Luis wants to know:
What is the revenue for August if an expense of $55.000 was made?
Solution
A graph showing revenue and expense figures is provided. Luis wants to know what the August revenue will be if the factory has an expense of $55.000. This figure isn't directly reflected in the graph, but the figures are higher and lower than that.
First, a table is made to easily relate the values:
Now the interpolation formula is used to determine the value of y
y = y 0 + (y 1 – y 0 ) * [(x – x 0 ) ÷ (x 1 – x 0 )]
Then the corresponding values are substituted:
y = 56.000 + (78.000 – 56.000) * [(55.000 – 45.000) ÷ (62.000 – 45.000)]
y = 56.000 + (22.000) * [(10.000) ÷ (17.000)]
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 revenue was $68.936.
References
- Arthur Goodman, L.H. (1996). Algebra and Trigonometry with Analytic 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.
- numeric, I. a. (2006). Xavier Tomàs, Jordi Cuadros, Lucinio González.