For determination of unknown intermediate values of any function or tabular data in calculus mathematics the interpolation device is used. A discrete set of the known parameters can be set by x0 arguments, h1... xn and values of the yj=f (xj) function (where j=0, 1..., n). In a simple special case the problem of search of intermediate values of the specified row can be solved by means of carrying out linear interpolation.
Instruction
1. The essence of linear interpolation can be described the following assumption: in an interval between the known next tabular values of an argument xi and xj the considered y=f (x) function can be considered approximately linear. In other words, on this interval the value of function changes in proportion to change of an argument.
2. More visually this assumption can be displayed in a graphic look in the Cartesian system of coordinates. The considered piece of the yi and yj function is represented to a continuous straight line with the known coordinates. By search of intermediate value of function Y, the unknown argument of X is between the next xi and xj values. Thus, it is possible to write down the following inequalities of xi
Express the written-down conditions as a proportion of the following look: (yj – yi) / (xj – xi) = (Y – yi) / (X – xi). Here yj and xj – final values, yi, xi – initial values of a piece, Y and H – required intermediate values.
Apparently from a proportion at the set increment of an argument of X - xi is easy to find corresponding change of function Y – yi. Express increment: Y-yi = ((yj – yi) / (xj – xi)) * (X – xi).
Thus, intermediate values of function can be defined, knowing only increment on which there was a change of an argument. Calculate the differences of yj – yi and xj – xi at the set step of an argument of X – xi. Substituting the received values in an increment formula, find a function change indicator.
Find intermediate value Y. For this purpose to the received value of increment add an initial indicator of the yi function on the considered piece. The same way there is any intermediate value with the set increment step.
If there is a task in definition of an argument of the X in preset values y=f (x) function, the return linear interpolation is carried out. Its essence consists in search of value X by means of the same proportion, only now function Y increment – yi acts as the known parameter. By means of similar transformations there is an unknown intermediate value of an argument of X = ((yj – yi) / (xj – xi)) / (Y – yi) + xi.
3. Express the written-down conditions as a proportion of the following look: (yj – yi) / (xj – xi) = (Y – yi) / (X – xi). Here yj and xj – final values, yi, xi – initial values of a piece, Y and H – required intermediate values.
4. Apparently from a proportion at the set increment of an argument of X - xi is easy to find corresponding change of function Y – yi. Express increment: Y-yi = ((yj – yi) / (xj – xi)) * (X – xi).
5. Thus, intermediate values of function can be defined, knowing only increment on which there was a change of an argument. Calculate the differences of yj – yi and xj – xi at the set step of an argument of X – xi. Substituting the received values in an increment formula, find a function change indicator.
6. Find intermediate value Y. For this purpose to the received value of increment add an initial indicator of the yi function on the considered piece. The same way there is any intermediate value with the set increment step.
7. If there is a task in definition of an argument of the X in preset values y=f (x) function, the return linear interpolation is carried out. Its essence consists in search of value X by means of the same proportion, only now function Y increment – yi acts as the known parameter. By means of similar transformations there is an unknown intermediate value of an argument of X = ((yj – yi) / (xj – xi)) / (Y – yi) + xi.