The American Term Life Insurance History Project

You are reading a page from An Elementary Treatise on Actuarial Mathematics by Harry Freeman (1932)
Part of the American Term Life Insurance History Project
Term Life Insurance

CHAPTER III
INTERPOLATION FOR EQUAL
INTERVALS

Interpolation may be defined as the operation of obtaining the value of a function for any intermediate value of the argument, being given the values of the functions for certain values of the argument. The process has been picturesquely described by Thiele as " the art of reading between the lines of a table.^" Where the form of the function y = u_s is known or can be deduced from the given values, the ordinary algebraic process of substitution can be used, and the required value obtained with little difficulty. In actuarial work the relation connecting the function and the independent variable is seldom evident, and it is then that recourse must be had to finite difference methods.

Before proceeding to examine the practical aspect of interthe question of negative and fractional values of

in the

expression

ⁿ

must be considered. The proof of the identity

x+nh =

)

ⁿ

by means of operators or by induction, as in the

previous chapter, ceases to have a meaning if

is negative or

fractional, since the reasoning assumes that the argument x advances

by steps of h at a time. The assumption that, so long as the ordialgebraic rules are followed, the expansion is true for all

values of the quantities involved and not only for certain specified

values, is not necessarily true, and an analogy can be drawn between

the application of the binomial theorem to algebraic quantities and

to operators. For example, the expansion

+ x

)

ⁿ

is only convergent,

i.e. is arithmetically intelligible, for negative values of

provided

that x is numerically less than unity.

Thus

(I —

x)-

+ 2x

+ 4x

+ ... +

+ ...

leads to an absurd result if we put x

= 2,

for then we should have

(—

^—2

= +

4+1

2+3

2+•

. •

which is impossible, since

(—

)

^—2

1/l—

= I.

    EXPANSION OF (I + 0)ⁿI1x    45
Similarly we must have regard to the possibility of expanding (I + 0)ⁿu_x by the use of the binomial theorem.
Consider the two following series of corresponding values of x and ux:
(i)    x    0    I    2    3    4    5
    u_s    I    4    9 16 25 36
Then to find the value of, say, u_i we shall have
u_l=E u₀=(1 +0)'u0
    ^I)
= uo + 011₀ + E -2    ) 0²uo + ....
2¹
The leading differences are Duo = 3 and A²u_o = 2, higher differbeing zero.
=I+3/2-=9/4,

which is otherwise evident, since u_s is (I + x)², and, for the value

^x= ^us = (3/²)²⁼ 9/4^.
(ii) x o	1	2	3	4	5
ux 1	5	25	125	625	3125

This is evidently a geometrical progression and the function from which the values are derived is y = u_s = 5^x. If we attempted to express u, as a rational integral function of x so that

u_s = a + bx + cx² + dx³— ...,
and then applied the relation Eⁿuo = (i + 0)ⁿ uo for the value n = we should obtain as above
u =E^kuo=(1+A)'ua

= uo + -i-duo + 0-- ¹⁾0² u₀

2! + ".
Here the leading differences are 1, 4, 16, 64, 256, ... and tend to become successively larger. But u_i = = 2^.24 approximately, and this cannot be the same as the divergent^. series found by ex(1 + A)ⁱ ^u0^.Hence unless the function is capable of being expressed as a rational integral function of x we cannot use the relation Eⁿ= (1 +A)ⁿ for the value n = ?,.

46 FINITE DIFFERENCES

Let us now consider the problem more generally. If fi_x be a rational integral function of degree k in x, we may write

fix

⁼

a + bx +

+ ... +

If we adopt the binomial expansion for expressing

_,x

in terms of uo and the leading differences of uo, namely,

⁼

uo+x

₁

Au,+x

₂

up+...+x,.O

up+...

we have two series for

which are equivalent for more than

values of the variable, since they are true for all positive integral values of x.

Hence by a well-known algebraic theorem, they are true for all values of x, positive or negative, integral or fractional.

Therefore so long as

Zlx

is a rational integral function of x the

binomial expansion is valid for all values of x. It is not necessarily

valid for other forms of function, and we are led to the conclusion

that we can expand

_ux+nh

in terms of

, A

u,, ... A

ux, ...

by the

binomial theorem for all forms of the function if

be a positive integer, but only for other values of

_h,

is a rational integral function of x.

All finite difference formulae which are employed for the purpose of interpolation are based on the hypothesis that the functions in question can be represented by polynomials, i.e. by rational integral functions, with sufficient accuracy for the purpose in hand. This assumption is the justification for using integral formulae for fractional intervals : the processes to be explained in this and subsequent chapters are simply various methods of carrying out the calculations based on this assumption. One special advantage of these methods is that it is unnecessary to fix in advance the degree (n) of the polynomial; the interpolation formulae will be in such a form that to increase n will merely involve the introof a fresh term without affecting the other terms. The introduction of additional terms however will not necessarily imthe approximation or justify the assumption. Fortunately, in actuarial work the functions with which we deal are usually such that the assumption is sufficiently accurate for our purpose.

In applying the formula in para. 3 to a given set of data the following points should be noted :

NEWTON^'S FORMULA 47

(a)
If the basic curve is y = a + bx + cx² + ... + kx"-¹ there will be n constants, and in order to determine these constants n equations are necessary. For there to be n equations, values of y corresponding to n values of x must be given. Therefore either n points on the curve, or n other corresponding relations between x and y must be known. In that event the curve will be of degree n — I, and nth and higher differences may be assumed zero.

Our investigation has been confined to equidistant values of the argument. If the given values are not equidistant a formula slightly different in form from the expansion

A)`

can be developed with a modified method of differencing (see Chapter

Iv).

With regard to the statement (a) above that for a curve of degree n — 1 there must be n facts given, it is not essential that zz points on the assumed curve must be known. We may have given, for example, three

points and two values of the differential coefficient _dxHere we have

five facts ; we assume therefore a fourth degree curve, so that fifth and

higher differences are zero.

6. Newton's formula.The formula ^ux+,,h = u_s + n₁Au_s. + n₂A²u_s + ... is known as Newton^'s formula, and is the fundamental formula for interwhen the given values are at equidistant intervals. The expansion can be applied to solve many forms of the problem of interpolation.The following variations of the problem may arise :

(i)

Where there are

equidistant terms and it is required to find an intermediate term.

equidistant terms of which

n — i

are known and it is required to find the missing term.Where there are

equidistant terms of which

n — r

are known and it is required to find the r missing terms.

Note. Some modern writers have adopted the name ^"Newton-Gregory formula" for the above expansion, as the first publication appears to have occurred in a letter from James Gregory to John Collins on 23 Nov. 167o. The letter is given and the question of Newton^'s priority is fully discussed by D. C. Fraser in J.I.A. vol. LIi, pp. ^I17-35^.

48 FINITE DIFFERENCES
7. Examples of the different variations referred to above are given below :
Example i.
The values of annuities by a certain table are given for the following ages:
Age x 25 26 27 28 29 Annuity-value a_x 15•oo6 15^.326 15^.630 15^.919 16^.195
Determine the value of the annuity at age 27 _ti.
Five values are given : we must therefore assume that fourth differences are constant. The difference table is

x	a_x	'a_x	^Azax	°^3ax	0⁴a_x
25	15^.006
		'320
26	15^.326		— •oi6
		^.3⁰4		+ •001
27	15^.630		— •015		+ •001
		'289		+ •002
28	15'919		— '013
		•276
29	16^.195

The leading differences correspond to the argument x = 25 and we require the entry for age 27;. Our formula is therefore
^a27i = E'²ⁱ a₂, = (1 + ^A)^{=1 a}2s
— [I+25A+2.5 X 1^.5 _A.,₊2^.5 X 1^.5 X .5 _A3 2 6
₊2^.5 X ^1.5_^x'5 X (—•5)A4] ^a2s 24
= ^a2s + ²'5^Aa2s + ¹'⁸75^j2a23 + '3¹²5^A3a,a — '⁰39^06A4a,,
= 15•oo6 + .8000 — •0300 + •0003 — '00004
= ¹5'77^6.

Note. Since the data are given to three places of decimals, sufficient figures have been used to give three places only in the result. Since our interpolation is based on an assumption, namely, that fourth differences are constant, a result to more than this number of decimal places would be unjustifiable.

Example 2.
From the following data find the value of 11₄₇:
¹¹4,9 = 19.2884; ^U49 = ¹9'535⁶; ^U49 = 19^.6513; U₅₀= 19'7620.

CHANGE OF ORIGIN 49

We cannot form a difference table, since the given terms are not equidistant. As however four terms are available we may assume that third differences are constant, and that as a consequence fourth differences are zero.

If the function is y = u_s we assume therefore that O⁴u_x = o whatever
the value of x. We may write
11₄₆= o,
i.e.    (E — ¹)^{4 u}46 = ^o,
or    (E⁴ — 4E³ + 6E² — 4E + I) 16₄₆= 0,
i.e.    ¹⁶50^—4¹¹49+⁶¹¹48^—4¹¹47+¹⁶46⁼⁰,
so that 19'7620 — 78^.6052 + I17^.2136 — 411₄₇ + 19'2884 = 0, from which    ¹⁶47 = ¹9'4¹47^.
Example 3.
Complete the following table:
X    2^.0    2^.1    2'2    2'3    2'4    2'5    2'6
u_x    '135    •111    ^.loo    'o82    •074
This is similar to Ex. 2. Instead of using the assumption once that L u = o, we write down two equations of the same form, thus
O⁵u₂.₀ = o, so that (E — 1)⁵ 11₂.₀= o, and    L1°u₂.₁ = o,    ,,    (E — 1)⁵16₂.₁= 0.
Our two equations then become
16₂.₆— 516₂.₅+ I0U2.4    10112.3 + 516₂.₂^—16₂.₁=0,
since the interval of differencing is o•1.
Inserting the known values of u_s and solving, the required values are easily found to be u₂.₁= • 123 and 16₂.₄= •090.

Note. The function in this question is y = e-^x and the tabular value of 14₂.₄ is •091 correct to three decimal places. This difference is due to the fact that our assumption that the curve y = u_s is a rational integral function of the fourth degree in x is only approximately true.

8. Change of origin.

If we had plotted the curve y = u_s on which the values of u_s in, say, Example 2 were assumed to lie, we should have had values of y corresponding to values of x at 4⁶, 4⁸, 49, 50. Precisely the same curve would result, however, if we changed the origin of our co-ordinates so that 46 was represented by the value x = o, 48 by

F ₄

5⁰ FINITE DIFFERENCES
x = 2, 49 by x = 3 and so on, the unit of x being unaltered. This process of changing the origin simplifies our notation considerably. In the three examples above we could have changed the origin and could have altered the questions to read :

Ex. 1.	Origin at age 25	Given u_o, u_1iu₂, u₃, u₄
	Unit of differencing 1 year of age	Required u₂1 .
Ex. 2.	Origin at x = 46	Given uo , u₂ , u₃, u₄
	Unit of differencing x = i	Required u₁.
Ex. 3.	Origin at x = 2^.0	Given uo, u₂, u_s
	Unit of differencing x = o^. l	Required u₁ and u₄ .

It is of interest at this stage to note the difference that exists between the ordinary accepted notation in algebra and that in finite differences for successive terms of a series. For example, in an algebraic series such as 1, 2, 4, 8, 16, ..., where each term is twice the preceding one, the terms are generally denoted by u₁, u₂, u₃ , u₄, ... , the nth term being un. This is because the terms follow strictly the order of the natural numbers : we cannot imagine, say, the 2 ₂th term of such a series. Graphically, the series is represented by a succession of isolated points. On the other hand, if the same series of values of the function y = u_x, 1, 2, 4, 8, 16, ..., corresponding to equidistant values of x, were given and we were required to interpolate for a value of the independent variable between two of the given terms, we should assume that a smooth curve could be drawn to pass through the points. Having chosen one of the values of x for our origin, we should then proceed to apply a suitable finite difference formula. For the above data the value of x most convenient for our origin would be the value for which y = 1, and our (imaginary) curve would pass through the point whose coordinates are (o, 1). In that event the terms would be denoted by uo , u₁, u₂ , u₃ , ... , the nth term being u_n_₁. This difference between the two sets of notation is of importance when dealing with summation of series.

9. If in Newton's formula
^us^.fnh = u_x + n₁0ux + n₂ O² Ua + n3 A3 u„ + ...
we put h = 1, x = o, and replace n by x, we obtain the series u_s = uo + x₁Ou_o + x₂0²up + x₃L³uo + ... .

SUBDIVISION OF INTERVALS 51
This is generally called the advancing difference formula, and gives u_s in terms of uo and its leading differences.
If, however, we wish to obtain u_s in terms of u__m and its leading differences, we may write the formula
u_s = ^U_m+(m+x) = Em+xu_m = (I + 0)m+x ^u—m =u_m+(m+x)10u-m+(m+ ^x)2^LI2u_m+^...
+ (m + x)r ^Lru_m + ... .

It is often more convenient to use this formula than to obtain u_x in terms of uo and differences of uo, the advantage being that thereby we can make use of values of the argument on either side of the origin.

10. Subdivision of intervals.

A frequent problem in actuarial work is the interpolation for values of u_s at individual points, given every fifth or tenth value of the function. For example, the problem may be to complete the series uo, u_1iu₂, u_3f... from the known values uo, u₅, ^u1o, ^u15, ... or from ^U0, ^u10, ^u20, ^u30, • • •

A simple method for obtaining the individual values where quinquennial values are known is given below.
Let 8u_x denote the difference for unit intervals of x and Du_x denote the difference for quinquennial intervals.
Then _ux+5may be expressed as either (I + 8) ⁵u_s or as (I + 0) u_s .
Symbolically    (I + 8)5    + A,
i.e.    (I + 8) _ (I + 0)*,
or    8- (I+0)¹—I.
From this relation we can find easily that
8u_x = (•20 — •o80² + •048.\³ — ...) us. Hence    8²u_x = (•2 — .o8A² + .o48 A³ — ...)²u_s = (•04zX² — •o32 ³+ ...) u_x.
Similarly 8³u_x = (•oo80³ — ...) u_x.

The same principle can be adopted if decennial values are known. In that event Du_x, L ^tux, ... will represent differences for decennial intervals, and the individual differences will be found

from the identity 8 (i + z )^gib— I.

52 FINITE DIFFERENCES
An example will show the application of the method.
Example 4.

From the following table of yearly premiums for policies maturing at quinquennial ages, estimate the premium for policies maturing at all
ages from 45 to 65 inclusive :

Age x    ₄₅50
The leading differences for quinquennial intervals are
Au_x    ^Au_x    Mux    ,4u_x
— .₄6₇+ .₁₄6    — '046    + '017
The formulae required are
8u_x = ('zA — 'o8A² + 'o48A³ — '0336A⁴) u_x= — '1078592, 8²u_x = (^.o4A² — •03zA³ + .0256A⁴) u_x = + '⁰⁰7747², 8³u_x = (^.0o8A³ — .0096A⁴) u_x = — '0005312,
S⁴ux = '0016A⁴u_x = + '0000272,
assuming fourth differences constant.
We have therefore by completing the table of differences,

Age	u_x	Su_x	ni_x	S³ux	8⁴u_x
45	2^.871
		— •10786
4⁶	²'7⁶3		+ '⁰⁰7747
		— •10011		— *000J312
47	z'663		+ '007216		+ 0000272
		— 09290		— 0005040
4⁸	²'57⁰		+ '006712		+ 0000272
		— o86i8		— 0004768
49	²'4⁸4		+ '006235		+ 0000272
		— ⁰7995		— ⁰⁰⁰449⁶
5⁰	²'4⁰4		+ '005786
		— '⁰74¹7
51	²'33⁰

and so on.
Note that since we require the value of the premium to the nearest penny, three decimal places will be required in the u_x column. To obtain results correct to three figures, four decimal places will be needed : 8u_x must therefore be given to five decimal places, S²u_x to six and S³u_x to seven.

55
Premium 2^.871 2^.404 2^.083 1'862 1'712
6o 65

EXAMPLES

EXAMPLES 3

Given the following data (a), find the missing term or terms (b):

1.(a) uo = 58o, 14₁= 556, U2 = 520, U₄ = 385; (^b) ^u3•

2.(a) u₁ = 386, u₃ = 53⁰, u₅ = 810; (b) u₂i u₄.

^(a)

= 150, 14

₁

= 192,

₂

= 241, U

₄

= 374;

(

)

3•

(

)

¹⁴

⁼

94,

⁼

⁶

¹⁴

₅

⁼

(

)

¹⁴

5.(a) 14₉= 6021, 14₁= 5229, U2 = 4559, U3 = 3979; (b) u .

6.(a) ¹⁴50 = 9²345, ¹⁴51⁼ 9¹55⁶, ¹⁴52⁼ 9⁰74⁸, ^u55⁼ ^88204;_(b)^U53;^u54^.

(a) u_

₁

202,

₀

175,

₁

= 82,

₂

55; (b) u

(a)

₉

₁

3, u

₂

= 10,

₃

= 34,

₅

= 209, 14

₈

= 1002;

(b)

₄

;

9.(a) uo = 192'I, u₁ = 187'5, u₂ = 184^.7, u3 = 184'6, 144 = _194'6,U5 = 199'4, U₇ = 212^.7, U₉ = 224'3 ; (^b) us ; u₈. lo. (a) uo = 98203, 14₁= 97843, U2 = 97459, U3 = 97⁰34; (^b) ^u2.25^.

The numbers of members of a certain Society are as given in the following table :

YearNumber

1910

⁸

451911867

1912Make the best estimate

⁸

⁶

you can of the

1914821numbers in 1912 and

577

1916.

1916

1917757

1918761

979

⁶

Findp₅₃ ifp_;,o = '98428,p51= '9⁸335,^p54 = '9⁸⁰⁰8,p₅₅ = *97⁸77^.

uo,

₁

₂

, ... u

be consecutive terms of a series, prove that

₃

= •05

₀

— •31

₁

+ •7

¹⁴

2 +

•

¹⁴

—

•3145 + '051

₈

Supply the missing term:

uo = 72795

U4 =

⁶

₁

=71651u

₅

=66566

₂

= 7

⁰

⁸

= 65152.

FINITE DIFFERENCES

^u235 = ²'37¹⁰7 ^U237 = ²'37474
^u236 = ^2.37²9^1 u238 = ^2'37⁶5^8.Find ^u235^.63 •

Given uo = — •5, u₁ = — '484, U₅ = 0, u_E= •256 find the missing terms.

Given the following data:

= o,

u10

15, use = 50; estimate

If you were given in addition

₅

= 35 how would your estimate be revised? Illustrate your answer by a diagram.

Find the value of an annuity at 5 per cent. given the following table :

54 '4

Annuity value

29203 16

28889

5'37

4'53375

76483

Obtain approximations to the missing values:

⁰

5354555

⁶

f (x)

684

_3'756

780 3

803 3

826

The area A of a circle diameter d is given for the following values :

d8o859

⁰

100

A5026 5674 6362 7088 7

⁸

Find approximate values for the areas of circles of diameters 82 and 91 respectively.

Calculate a value for sin 33⁰ 13' 30" from the following table of sines:

angle x°3

⁰

3334

sin x°•5000 •5

⁰

99 '544

⁶

'559

U75 = ²459; u80 = 2018; U85 = 1180; u9U = 402. Calculate the values of u₈₂and u₇₉.

From the

data in

Qu.

complete the table for values of

corresponding to individual values of x from 75 to 85.

Four values of a function at decennial points are given. Express

, 8

(the differences for unit intervals) in terms of the differences

of the function for decennial intervals.

Find the values u

₁

to u

₉

inclusive, given uo = o,

11,

₀

74,

2o =

'347, use = •518:

EXAMPLES 55

u_o = 23^.1234; U6 = ²3'7²34; U12 = 24'6834; U₁₈= 26'1330. Comthe series uo to u₆.

If you were asked at very short notice to obtain approximate values for the complete series

f (o), f (I), f (2), ... f (20),

being given that f (o) = •013,

(lo)

= •

248,

(15) = '57

⁸

, and

(20) = •983, what methods would you adopt, and what value would you obtain for

f (9)?

U₀+ u₈ = P9243U₂+ U₆= P9823

+ U7 =

P9590

₃

P9956. Find

₄

Tables are available giving premiums at age ₄₀at the following rates per cent :

Rate per cent.3314415

⁶

₄₀

'025891 '024654 •023517 '022470 '021509 '019811

It is desired to obtain P

₄₀

at 51 per cent. Obtain this, using

(a) two of the above values;

(P)

four of the above values; (y) six of the above values.

Given

(x)

= 500426,

1010

(

)

= 3

⁰

(x)

= 175212 and

(lo)

40356,

147

find

f (I).

29.

ul=l;

⁼

5'4

;

^=18'

47,

₇

₈

U9 +

+ ull

U12

= 90

36.

Find the value of

for all values of

from

to 12 inclusive.

x=10

If you were given u₀, u_l , u₂ and E u_x how would you complete the

x=1

table of

up to

Given up = I17^.7; u₂ = 110^.5 ; u₄ = 102^.7; ^u10 = 75'4, obtain the values of u_x for all integral values of x from o to lo.

Obtain the following relation between nine terms of the series represented by

₂

, ... u

₉

₅

= s (

6) —

(

+ 83

(

8) —

°7i

(

9),

and find

₅

given

⁼

'7455

⁶

;

₂

'55938; u

₃

= •

42796; u

₄

'32788;

₆

'1843z; u

₇

•

13165; u

= 'o8828;

₉

33• It is asserted that a quantity, which varies from day to day, is a rational and integral function of the day of the month, of less than the fifth degree, and that its values on the first seven days of the month are

30,30,28,25,22,20,20.

56 FINITE DIFFERENCES
Examine whether these assertions are consistent. If so, assume them to be true, and find (r) the degree of the function, (2) its value on the sixteenth of the month.

Extrapolation may be defined as the process of obtaining further terms of a series as opposed to interpolation, which is the process of finding intermediate terms.

The values of a certain function, corresponding to the values 4, 6, 8, to of the argument, are 914, 742, 605, 500 respectively; extrapolate to calculate the value of the function corresponding to the value

of the argument.

Given u_o = 1876, u1 = 777, u₃ = 19, U₆ = — 218, interpolate the values of u₂, u₄ and u₅, and find the form of the function, assuming it to be a rational integral function.