- 1.
When performing direct interpolation, values of y corresponding to various values of the argument x are given and we are required to find a value of the entry y corresponding to a value of x inter-mediate between the given values. If it is required to obtain an interpolated value of the argument corresponding to an intermediate value of the entry, the process adopted is called inverse interpolation. In other words, for direct interpolation we assume a curve y = ux passing through the points (x, y) and estimate the value of y corresponding to some intermediate value x': for inverse interpolation we have a similar curve but are required to find a value of x corresponding to a value y'.
For certain functions we may obtain the result easily. If y = sin x, then x = sin-1 y ; if y = x3, then x = yI ; if y = ax, then x = log y/log a. The required values of x can be calculated imin these examples.On the other hand, if the data are simply corresponding numerical values of x and y, all that we can write down is a formula such as Newton's or Stirling's : we must then endeavour to obtain a value for x by solving an equation. For example
y=ux= (I +0)xuo=u0+xz u0+x2t 2u0+x3A3u0+....
If second differences may be assumed constant we have a quadratic equation which can be solved at once. Should this assumption be inadmissible, then we are faced with an equation of higher degree than the second and the solution of such an equation may be very laborious. In these circumstances we resort to approximate methods of solution of the equation.
INVERSE INTERPOLATION 85
Example I.
Obtain a value for x when ux = 19, given the following values: x o I 2
Zlx o I 20
We may write down at once
y = ux = (1 +A ) x=u ,++x , A 2 u , ;
i.e. 19=0+x+18x2=x+9(x2—x)
so that 9x2 — 8x — 19 = 0,
from which x = 1.964 ... or — P075 ....
But since we have to find an interpolated value of x corresponding to a value of y we may treaty as the argument and x as the entry. Let us write the data in the form
Y
x=vy o I 2
and apply the Lagrange formula to calculate v19 as if for direct inter
We shall have
v19 I 2
19.18.(— I) 18.1.(— 19) + (— I).20.19
or v19=2.8.
It will be seen therefore that there are two distinct sets of results. By adopting the first method x has the values 1.964... or — 1'075... and by adopting the second method x has the unique value 2.8. It remains to examine the reasons for the difference and to ascertain which result is more likely to approximate to the true interpolated value or values.
In the first method it will be apparent that we have taken a curve of the form y = a + bx + cx2—a parabolic curve—and have obvalues of x corresponding to y = 19. This gives two values of x, one on each side of the vertex of the parabola. In applying the Lagrange formula inversely we have assumed that x is a parafunction of y and have given y a particular value (19) in the equation x = a + /3y + yy2. If we substitute the value of y corresponding to each value of x from the data, it is easily seen that a = o, = 398/380 and y = — 18/380. The Lagrange equais therefore 190x = I99y — 9y2.
86 FINITE DIFFERENCES
Now if the two curves y=9x2—8x
and x = 199 y — 9 y'
190 190
be plotted on the same graph, it will be seen that they take different shapes, thus:
Fig. 13.
On the curve y = 9x2 — 8x the abscissae of the points whose ordinates are 19 are 1.96... and — 1.07..., whereas on the other curve there is only one point for which the ordinate is 19, namely the point (2.8, 19). Unless therefore the two curves obtained from the data, (i) by treating x as the argument and (ii) by treating y as the argument, intersect at the required interpolated value, as for example at K in the above figure, the two methods are bound to give different results.
The question whether the Newton equation or the inverse interpolation method will give an answer more nearly to the true value can only be decided by a consideration of the data. For instance, if in this example there were reason to believe that for increasing positive values x and y increased together, it is more likely that the interpolated values of x when y = 19 would be 1.964... and — 1.075... than the value x = 2.8.
1
Y
-y=9 x2-8x
oy
/b
x
_9_ 19
INVERSE INTERPOLATION 87
3. The above example shows in a simple manner that the different processes of inverse interpolation may give different sets of results. We now proceed to a more general investigation of the problem.
The formulae of direct interpolation are based on the properties of rational integral functions of the variable, and any formula which proceeds to nth differences gives exact results when applied to a rational integral function of the nth degree. By stopping short of nth differences the formula can, of course, be used to obtain approximate results, and the success of the interpolation depends on the magnitude of the terms omitted. Thus, if we use rth differences for a polynomial of the nth degree in x, the result is the exact value of terms up to and including the term in x'. The terms beyond xr are disregarded, and this can only be done legitimately if they are relatively unimportant.
In questions of direct interpolation there is only one value of y, i.e. of ux, for a given value of x. There may be, however, more than one value of x for a given value of y. In fact, if y is a rational infunction of x, x is a rational function of y only when both functions are of the first degree. In other cases the inverse function may be an infinite series or an irrational function. For example, in the HMI Table of Mortality there is only one value of dx for a given value of lx (where lx represents the number of persons attaining exact age x in any year of time, and (Ix is the number of these who die before reaching age x + I). In the neighbourhood of the peak of the death curve, however, there will be two values of 1,. within a short range of interpolation for a given value of dx.
For these reasons the subject of inverse interpolation is more troublesome than that of direct interpolation, although it should always be remembered that the conditions for good interpolation are the same for inverse interpolation as for direct. One principal condition is that within the range of interpolation there should be only one value of x corresponding to a given value of y.
Let us consider the equation in Example I, namely
y = — 8x + 9x2,
where the range of interpolation is from o to 2. The first point to note is that the function is not a good subject for direct interpolaexcept when the formula is applied to its fullest extent—the
88 FINITE DIFFERENCES
second degree. The reason is that the last term is the predominating term throughout the greater part of the range.
In most instances, by altering the interval and reducing the range of interpolation, a function can be reduced to a good form for direct interpolation. Such a question as the following might be put:
Given the function y = ux = — 8x + 9x2, for what intervals of x should ux be tabulated so that in any interval an interpolated value of y can be obtained by first difference interpolation with an error less than, say, •ooi ?
| Put |
|
x = z/a; |
|
| |
|
8z |
922 |
| then |
|
ux = vz = — +
a a2 |
| i.e. |
a2v2 = — 8az + 9z2, |
| |
a2Av, = |
— 8a + 18z + 9, |
| and |
a2I 2vz = |
18, |
| so that |
02vz = |
18/a2. |
Suppose vs to be tabulated at unit intervals for values of z. Then, for an interpolated value vs+t between vs and vz+1,
vz+t = vs + tOvs + it (t — 1) 02`112 .
If we take vs + tOvs as the interpolated value, there is an error it (t — i) A2v2 and the maximum numerical value of it (t — 1) is i-, being the value when t is
Therefore, by the conditions,
it (t 1)A2v2<•oo1,
$02v2 < .001.
But A2v2 = i8/a2, .'. 18/8a2 < •ooI ;
i.e. a2 > 18000/8,
i.e. > 2250,
or a> 1/2250, and the most convenient value for a is 5o.
We must therefore tabulate us at intervals of A of unity, i.e. at intervals of •o2.
SUCCESSIVE APPROXIMATION 89 For example,
|
1.12 |
56 |
2.3296 |
|
1'14 |
57 |
2'5764 |
|
1.16 |
58 |
2.8304 |
|
I.18 |
59 |
3.0916 |
|
1.20 |
6o |
3.3600 |
We may use this table for finding values of x corresponding to given values of ux, and the interpolation is as legitimate as direct interpolation. While, however, direct interpolation to second differences is exact, inverse interpolation to second differences, while nearer the truth than first difference interpolation, is not exact.
- 4.
Practical methods of inverse interpolation.
It is evident that the problem of inverse interpolation is the same as that of direct interpolation for unequal intervals. The methods of Lagrange or of divided differences could therefore be employed to obtain any intermediate value of x corresponding to a value of y, given a table of y = zrx, by the use of the inverse rex = vv. The labour involved in applying either of these methods is often prohibitive, and the methods usually adopted in practice are given below.
5.
Successive approximation.
In the first place we obtain either by inspection or by a rough graph two values of x lying on either side of the required intervalue. (For example, a value for x when y is zero in the function y = x2 — 4x + 2—i.e. a solution of the equation x2 — 4X + 2 = o—lies between the values x = 1 and x = 2.) We then choose a suitable origin and unit of differencing so that if x be the interpolated value and lies between two successive
Avz'2396 •2468'2540 •2612•2684
90 FINITE DIFFERENCES
values of the argument, the interval will be small and x will be as near to the origin as possible.
Suppose that the required value lies between o and I. The method proceeds as follows :
1/x = uo + xAuo + ix (x — I) A2uo + ix (x — I) (x — 2) 03uo + .... Since x is small, a first approximation (a1) will be obtained by neglecting terms involving second and higher differences of uo.
us = uo + a1zuo approximately,
i.e. a1 = (ux — u„)wu0, first approximation.
Again, neglecting third and higher differences, we may write
us = uo + a2,uo + a2 (al — I) 1 2 uo ,
where a2 is a second approximation and is therefore not very different from a1. This gives
a2 =Duo + 2 (al u0 i) A211~ second approximation. Similarly
_ us — uo
a3 Duo + (a2 — I) A2 uo + s (a2 — 1) (a2 — 2) 03 ua'
third approximation,
6. Elimination of third differences.
We have, as far as third differences, by expressing us in terms of uo , 4uo , ... ,
us = + xAuo + ix (x — I),2uo + ix (x — 1) (x — 2) A3uo. Also, in terms of u1 i Au1, ... ,
ux=u1+(x—I) !u1 + (x — I) (x—2)A2u1
+ (x— 1)(x—z)(x—3)A3u1.
If now we ignore the terms containing third differences and multiply both sides of the first equation by 3 — a and both sides of the second equation by a (where a is an approximation to the required value, found by inspection or otherwise) and add, a new quadratic equation in x will be formed. The error involved in ignoring the third differences will be small, since
x (x — I) (x — 2) (3 — a) A3u, + a (x — I) (x — 2) (x — 3) A3u1 will be small provided that! 3u0 and A3u1 are not very different.
and so on.
x y
|
0 |
19,231 |
|
| |
|
— 363 |
|
I |
18,868 |
|
| |
|
— 349 |
|
2 |
18,519 |
|
| |
|
— 337 |
|
3 |
18,182 |
|
| |
|
— 327 |
|
4 |
17855 |
|
By the ordinary advancing difference formula
18,600 = 19,231 — 363x -{- 14x (x — 1) — 2x (x — 1) (x — 2)' 2! 3!
where the value of x is required corresponding to the value 18,600 of y.
- (i)Successive approximation.
Since x is small, the first approximation will be19,231 — 18,600al =363 or I1383....631or I.7634... .a2 = 363—7(a1—1)Similarly,a3 = 63or 1.7646....363 — 7 (a2 — I) + 3 (a2 — I) (a2 — 2)The required result is therefore 52 + 1.7646... = 53.7646....
- (ii)Elimination of third differences.
We have18,600=ux=(1 +J)xuo= 19,231 -363x+7x(x—I)—3x(x—I)(x—2) and18,600 = ux = (I + 0)x—Iu, = 18,868 — 349 (x — I) + 6 (x — I) (x — 2)—*(x—I)(x—z)(x—3).By inspection a rough value of the interpolated value is I.75, allowing for the change of origin.ILLUSTRATIVE EXAMPLE7. The following question is solved by both these methods. Example 2.Find the value of x for which y is 18,600, givenx5253545556y19,23118,86818,51918,18217855
Changing the origin, the difference table is
14 I2 I0
92 FINITE DIFFERENCES
If, therefore, we multiply the two equations by (3 — P75) and I.75 respectively, and add, we may neglect the fourth term. The factors being P25 and 1.75 we can use 5 and 7: we thus obtain
12 X 18,600 = 5 [19,231 — 363X + 7X (x — 1)]
+7[18,868—349(x—I)+6(x—1)(x—2)]
| or |
223,200 = 230i758 |
— 4419x + 77x2, |
| i.e. |
77x2 — 4419X |
+ 7558 = o. |
Solving the quadratic, the value required is x = 1.7646..., which agrees with the value of a3 in method (i) to four decimal places.
8. It is often convenient to employ a central difference formula as the basic equation, as in the following example.
Example 3.
Find the root of the equation x3 — 9X — 14 = 0 which lies between 3 and 4.
| Let y = x3 — 9X — 14. Then we have, by actual calculation, |
|
x 3.0 3.2
y — 14.000 — 10.032
The difference table is |
3.4
— 5.296
Ay |
3.6
•256
A2y |
3.8 4.0
6.67z 14.000
0'y |
| x |
y |
|
| 3.0 |
— 14.000 |
|
|
|
| |
|
3.968 |
|
|
| 3.2 |
— 10•032 |
|
•768 |
|
| |
|
4'736 |
|
•048 |
| 3'4 |
— 5'296 |
|
•816 |
|
| |
|
5'552 |
|
'048 |
| 3.6 |
•256 |
|
•864 |
|
| |
|
6.416 |
|
•048 |
| 3.8 |
6.672 |
|
•912 |
|
| |
|
7.328 |
|
|
| 4.0 |
14.000 |
|
|
|
Taking the origin at 3.6 and using Stirling's formula :
u, = + x Au, + Au-1 + x2 02u-1 + x (x2 — I) A'u_1 + L.3 u_2 2 2 6 2
the interval of differencing being 0'2;
i.e. o = '256 + 5'984X + '432x2 + 'oo8x (x2 — I).
The cubic equation can be solved by successive approximation, or we can repeat Stirling's formula for the next value of us and adopt the alternative method outlined above.
INVERSE INTERPOLATION 93
If the first of these methods be adopted, it will be found that successive approximations to the value of x are — •o4278, — '042913, — •042971. From the last we obtain as the required solution
3.6 — ('042971 X '2) or 3.5914058,
which is correct to six decimal places, the seventh being nearer to 7.
If we choose our origin at the point (x, y) and the value of the intervalue is x + a, then, when a lies between — and , it is advanto use Stirling's formula. If a lies between 4 and i Bessel's formula should be applied. (Whittaker and Robinson: Interpolation, p. 6o.)
- The method of successive approximation is probably the best method for ordinary use. If we want a result that can be obtained to the required degree of accuracy by taking out differences as far as 3u, or if the curve is a cubic, the elimination method will give a satisfactory answer. The disadvantage of this process is that we obtain thereby one result only—a root of a quadratic—and that we cannot approach our interpolated value by steps as is done in the method of successive approximation. Moreover when fourth and higher differences are not negligible, the elimination method breaks down.
Another method of inverse interpolation can be employed when the function is of the form Ax3 + Bx2 + Cx + D. This method depends upon divided differences with "repeated arguments." It can be shown that if, in the general equation for ux in terms of uo and the divided differences of uo, we let all the arguments coincide, the rth divideddifference becomes r ~ !-x ux1 . This enables a table of values to bex=rbuilt up by repetition of certain values of x. As the arithmetical work is certainly no less than that involved in other methods, and as this method is restricted in its application to comparatively simple rational integral functions, it would seem to be of little practical value. A concise account of the method will be found in Steffensen's Interpolation, pp. 84 et seq.
- The general investigation of the accuracy of finite difference methods of approximation is a problem in direct interpolation, and has been dealt with previously. In dealing with the subject of successive approximation, however, it is of interest to include in the present chapter an elementary demonstration of the fact
94 FINITE DIFFERENCES
that in certain circumstances a better interpolation can be obtained by neglecting higher differences than by retaining them. For example, if we have a third difference curve, then
ux = uo + x.Auo + x202uo + x3,3uo exactly. The error (a) in taking two terms is x2,,2uo + x3..V uo,
and (/3) in taking three terms is x3A3u0. (a) may be expressed as
x (x — 1) (x — 2) [ 3 2u0 + 1 A3110
6 x— 2. u J
and (/3) as x (x — 1) (x — 2) 03u
6 0.
Then, ignoring the sign of x (x (x — 2) 03uo, which will
be the same for both (a) and (/3), (a) will be less than (/3) if
3 O2uo
x—2A3uo
is negative and less than 2.
In these circumstances the error made by retaining first differ
ences only is less than that made in continuing to second differences. As an illustration consider the function x3 + 5X + 50.
It is easily seen that uo = 50; Duo = 6; i2uo = 6; A3u,, = 6. If
2
x is ±, for example,
x 3 2 D03Zro = — 3/11, which is negative and less
uo
than 2.
(a) is therefore less than (33).
This can be otherwise shown by finding the values of the errors. = 5I/4 exactly.
Also u4 = (I + 0)I uo
= u0 + 4t,1r0 — 02uo + ... ,
up + jjL.U0 = 511 = 51~ ,
uo + 4Auo — 02uo = SOH b = 50H,
51i 51g k4 (a),
and 51 — 50} _ ?4 (3).
(a) is less than (/3), so that the approximation to first differences is better than that to second differences.
EXAMPLES
EXAMPLES 6
I. Given u1 = o, u2 = I12, u3 = 287, U5 = 612, find u4. Using u1 , u,, u3 and u4, find a value for x when us = 270.
2. The following values of u, are given: u10 = 544, u15 = 1227, u20 = 1775. Find, correct to one decimal place, the value of x for which us = I000.
3. Having given log 1 = o, log 2 = •30103, log 3 = '47712 and log 4 = '60206, find the number whose logarithm is •30500:
- (i)by expressing log x in terms of log I and its differences and solving for x;
by using Lagrange's formula applied inversely.Explain the nature of the assumptions in each case.4. Apply Lagrange's formula inversely to find to one decimal place the age for which the annuity-value is 13.6, given the following table :Age ............X3035404550Annuity-value at 41 per cent.as 15'9 14'9 14'113'3 12'55. f (o) = 16.35, f (5) = 14'88, f (lo) = 13'59, 1(15) = 12.46. Find x when f (x) = 14.00.6. Given the following table of f (x) :f (o) = 217, 1(1) = 140, f (2) = 23, f (3) = — 6, find approximately the value of x for which the function is zero.7. The following values of f (x) are given:
f (Io) = 1754, f (15) = 2648, f (20) = 3564.Find, correct to one decimal place, the value of x for which f (x) = 3000.
8. Given four values of a function uo , u1, U2, U3, show how to calculate an approximate value for x from the equationx (x — 1) 2x (x — 1) (x — 2) 3us = uo + xhuo + 2, uo +uo 3•by obtaining a quadratic equation in place of a cubic.Use the method to find x when us = P05, given u1b = 1.0000, u1.1 = 1'0323, U1.2 = 1'0627, U1.3 = P0914.
9. Given that (I.2o)3 = 1.728, (1'21)3 = I•772, (I'22)3 = 1.816, (1'23)3 = I'861, (1.24)3 = I'907, explain carefully how to find the real root of the equation x3 + x — 3 = o by a method of inverse interpolaWhat method would you adopt in practice? Obtain a value for the root to four decimal places.
96 FINITE DIFFERENCES lo. The following table is available:
Age x ... ... 44 45 46 47
ax at 4per cent. 13'40 13.16 12'93 12.68
Find, to two decimal places, the age corresponding to an annuity of 13.00.
I1. Find, to two decimal places, the real root of the equation x3+x—5=0
by means of divided differences applied inversely, using the values of the expression when x = o, 1, z and 3.
What is the reason for the poor result obtained in this case? (The true solution is x = 1.516 approximately.)
- The equation x3 — 6x — I1 = o has a root between 3 and 4. Obtain it by inverse interpolation correct to three places of decimals.
The formula for the value of an annuity-certain for n years at rate per cent. i is given byn a,l = I — v n+ i)—1. iGiven the following table, obtain to three decimal places the rate per cent. for which a20 is 14:Rate per cent.33442a~14'877514'212413'J90313'0079
- Solve the equation x = lo log10x, given the following data:
Argument x ... 1'35
log x ... '1303
1'37 I'38
'1367 '1399
1.36 '1335
15. Apply Lagrange's formula (inversely) to find a root of the equation u, = o, when u30 = — 30, u34 = — 13, u38 = 3, U42 = 18.