How to find intermediate value

To determine the unknown intermediate values of a function or a table of data in mathematics we use the technique of interpolation. A discrete set of known parameters can be specified with the arguments x0, x1 . . . xn and function values yj=f(xj) (where j=0, 1, . . . , n). In a simple particular case of the task of finding intermediate values of a given row can be solved using the linear interpolation.

Instruction

1. The essence of linear interpolation can be described by the following assumption: in the interval between the known neighbouring tabular values of the argument xi and xj consider the function y=f(x) can be approximately regarded as linear. In other words, in this range the value function is proportional to the change in the argument.
2. More clearly this assumption can be shown in graphic form in the Cartesian coordinate system. Consider the cut functions u and u appears to be a continuous straight line with known coordinates. When searching for intermediate values of the function Y, unknown argument X is between adjacent values XI and xj. Thus, we can write the following inequalities XI < X < xj, yi < Y < yj.
3. Express written terms of proportions of the following: (yj – yi)/(xj – XI) = (Y – yi)/(X – XI). Here yj and xj is the target value, yi, XI – initial value of the segment, X and Y – the desired intermediate values.
4. As can be seen from the proportions for a given increment of the argument X - XI it is easy to find the corresponding change of the function Y – yi. Express the increment: Y – yi = ((yj – yi)/(xj – XI))*(X – XI).
5. Thus, the intermediate function values can be determined, knowing only the increment on which there is a change of argument. Calculate the difference yj – yi xj – XI at a given step of the argument X – XI. Substituting these values in the formula increment, find the rate of change of a function.
6. Find the intermediate value Y. For this to the increment value add to the initial figure o functions on the considered segment. Likewise, is any intermediate value with a specified step increment.
7. If there is a problem in defining X for given values of the function y=f(x) is the inverse linear interpolation. Its essence is finding values of X using the same proportion, but now as a known parameter is the increment of the function Y – u. Using a similar transformation is unknown intermediate value of the argument X = ((yj – yi)/(xj – XI))/(Y – u) + XI.