CHAPTER VI
FINITE DIFFERENCES. INVERSE INTERPOLATION
- In direct interpolation a series of values of the function is given and the problem is to find the value of the function correto some intermediate value of the argument.
In Inverse Interpolation the problem is reversed and it is required to find the value of the argument which corresponds to some value of the function, intermediate between two tabulated values.
- In certain cases of mathematical functions the desired result can be obtained by direct calculation. Thus if
y =.f (x) = az,log y — loga'and the value of x can be found equivalent to any given value of y.Where this is not the case various methods can be adopted. These will be examined in order.3. Let y = f (x) be the given value. Then(x — I)°2.f(0)+....If it be assumed that the higher orders of differences vanish, and that the values of A, A', etc. are obtained from the given terms of the series, then we have an equation in x which can be solved by the usual methods.The disadvantages of this plan are firstly that an equation of higher degree than the second is troublesome to solve, and secondly that for certain functions the degree of approximation may not be very close. Since a quadratic equation employs only three terms of the series, it often happens that no close approximation can be obtained. In all cases the intervals between terms should be as narrow as possible, so that accuracy may be increased and the use of higher orders of differences obviated as far as possible.
42 FINITE DIFFERENCES
A further approximation is obtained by taking second differences into account and writing x, in place of x in the equation, thus giving
x2— f(x)—f(0) ~).
Al(0) + (x1—1) A (0)
When third differences are taken into account x2 is written for x, giving
f(x)-f(0) (3).
x3 _ °f(0)+(x—1)02.1(0)+(x2—1)(x2—2)(0)
These processes can be repeated until the desired degree of approximation is reached. The method has the disadvantage of being somewhat laborious. On the other hand it has the advantage that an error of calculation at an early stage does not vitiate the result, being rectified by the further approximations.
- A different method of procedure is to treat x as a function of f (x). Thus since
y =f (x),we may write= (y).We therefore treat x as a function of y and, since the given values of y (i.e. f (0), f (1), etc.) will usually represent unequal intervals of the variable y, we must resort to interpolation by such a method as Lagrange's or Divided Differences, in order to obtain our value of x (i.e. (y)) corresponding to the given value off (x).
- The following example is worked out in each of the above ways. Example. Find the number of which the log is 2',. , having given
log 200 = 2.30103 210 = 2.32222 220 = 2.34242 230 = 2.36173.Method I.
| .f (x) |
002 |
A, |
| 2.30103 |
•02119— •00099 |
•00010 |
| 2.32222 |
•02020— •00089 |
|
| 2.34242 |
•01931 |
|
| 2.36173 |
|
|
INVERSE INTERPOLATION 43
f (x) = 2.33333 = 2.30103 + .02119x — •00099 x (x 1)
2
+'00010x(v—1)(x—2)
6
Or, by reduction,
x3 -32.7x2+1303.1x—1938=0.
Whence, solving the cubic,
x =1.5443.
Since the initial value of x is 200 and the unit of measurement is 10, the result of the calculation is to give 215'443 as the required value of x.
- 8.Method II.
f(x)=2.30103+•02119x—•00099x(x—1)2'also, f (x) = 2.32222 + •02020 (x -1) — •00089 (x -1) (x — 2) 2
The first approximation to the value of x is x' =1.5, so that 3 — x' =1.5. Since the values of 3 — x' and x' are approximately equal, we may take for our "weighted" equation the arithmetic mean of the above equations, giving
f (x) = 2.33333 = 2.30108 + .0216l x — •00047x2.Whence, solving the quadratic, x =1.5443 as before.
9.Method III.
1st approximation_ 2.33333 — 2.30103x'02119=1 5243,2nd approximation2.33333 — 2.30103_
x2 •02119 — 4 x •5243 x •00099 =1.5432,
3rd approximation2.33333 — 2.30103_•02119 — 4 x •5432 x •00099 + a x •5432 x (•5432 -1) x •00010=1.5442,which differs only slightly from the value obtained by Methods I and II.