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CHAPTER IV
INTERPOLATION FOR UNEQUAL
INTERVALS

In the previous chapter formulae have been developed on the assumption that the argument proceeded by equal intervals. Although in actuarial problems the data are generally given at equidistant intervals of the independent variable, it sometimes happens that we are required to interpolate when values of the function are known for unequal intervals. In other words, instead of values of u_x for arguments x + h, x + zh, x + 3h, ... being given, the known values are x + a, x + b, x + c, ..., where a — b, b — c, ... are not necessarily equal.

Divided differences.

Since we cannot take out the differences as hitherto defined, we

adopt a process of differencing involving the argument as well as the entry. The differences obtained by this process are called " divided " differences, and are found in the following manner.

Let

11x

be given for the values

a, x = b, x = c, x = d, ...,

where the intervals need not be equal. Then we have

First divided differences :

lib

—

— Zl

_b.

11d —

b—a ' c—b ' d—c '

which may be denoted by

11a;

11,;

21;

....

Second divided differences :

Zlb

—

uc — O

11b

—

c—a'd—b 'e—c

;

^'2

11b;

^'2

11~; ....

Third divided differences :

^D,2u

^-01211

d — a

oru

; ...

and so on.

5⁸    FINITE DIFFERENCES A numerical example will make this clear.
Example 1.
Take out the divided differences of u_x given the following table:
    x    I    2    4    7    12
    u_x22    30    82    1o6    206

The table is
x    ux    0^'1[x    O^'2u_x    A^'3ux    ^a'Qux
I    22
30-22₌₈ 2 - I
z    30
26-8₌₆ 4-I
^82-30=26 4-2
-3^.6-⁶_=-16 7-1
4    82
8-26
⁼-36 7 -2

•5¹ - (- I'6)
I2-I

I^.5-(-3^.6)

=•₅₁
I2-2

Io6-82_₈ 7-4
7
Io6

20-8 I2-4
206 - Io6 ⁼20 I2-7

12 2o6
It is essential to arrange the work systematically if error is to be avoided. The numerators and denominators must be set out, either in parallel columns or in the form of fractions. Where there is ample space the columnar arrangement is better, especially where the divisors are cumbrous. It should be noted that while the numerators are the first differences of the preceding divided differences, the denominators are all formed directly from the arguments, differencing first in the ordinary way, then in pairs, then in triplets, ... and so on. It will also be seen that the divisor is always the difference between the values of x for the last and first u_xinvolved in the difference.
3. Newton's divided difference formula.
An alternative notation for divided differences is as follows. Let the function be f (x) and the arguments x₀, x_i, x2, ... x_T. Then the successive divided differences are

NEWTON^'S DIVIDED DIFFERENCE FORMULA
f(^xo,^xl)=^f(xo)-f(x1) xo—x1 '
f(xo,x1,x2)=f (4,^x1)^—f(^x1,^x2) -
X_i)— x₂

...............
where the arguments are all different.
If we replace x_o, x₁, ... x, by x, a_o, a₁, ... a_n, we obtain a system that may be written
(x — ao) f (^x, ^ao) = f (^x) — f (^ao),
(x — ^a1)f (^x, ^ao, ^a1) = f (^x, ^ao) ^—.f (^ao, ^a1),
(x—an)f(x,ao,...an)=f(x,a0,...an_1)—f(a°,a1, ... a_n). From this we obtain in succession
f (^x) = f (^ao) + (x — ao) f (^x, ^ao)
= f (^ao) + (x — ^ao) f (^ao , ^a1) + (x — ao) (x — ^a1) f (^x, ^ao, ^a1)
and generally .f(x)=f(ao)+(x—ao)f(ao,a1)+(x—ao)(x—a1)f(ao,a1,a2)+...
+ (x — a0) (x — a₁) ... (x — an_1) f (^ao, a1, ... an) + R,
where

R= (x — ao) (x—a₁)... (x—a_n)f (^x,^ao,^a1, ... an).

This is Newton^'s formula with divided differences.

It should be noted that the term preceding R in the expansion is (x — a_o) (x — a₁) ... (x — a_n_₁) f (ao, a₁, ... a_n), there being n factors in the coefficient. In R, however, the coefficient has n + 1 factors, the new factor (x — a_n) being introduced.

Reverting to the notation
~~^n—11!    — ~~^n—1Zf
,^gnu =    arrl    ar
dr
^an+r — a,.

6o    FINITE DIFFERENCES

For a further investigation of the remainder term R see the article ^"On certain Formulae of Approximate Summation, etc.^" in j.I.A. vol. LIII, by Prof. Steffensen, to whom the above proof is due.

Theorems similar to those proved for the ordinary advancing difference formula can be shown to be true for divided differences. The following important proposition holds equally for divided differences as for differences at equal intervals.

be a rational integral function of the nth degree in

then the nth divided difference is constant.

It will be sufficient to consider the function

ⁿ

Then, if the values of the argument

a, b, c, ...,

the first

divided difference is

(bn —

^an

)/(

—

a) = b

ⁿ

+ b

ⁿ

+ b

^n-3

+ ... + a

^n-1

a homogeneous function of

and

of degree

(n —

1).

Again, the second divided difference is

^n—1

^n—2

b +c

^n—3

+...

^n—1

)

—

^n—1

^n—2

^n—3

+ ... + a

^n—1

)

c—a'

(0—1

+ c

^n—2

^—3

+...

^n—1

)

—

^n—1

^n—2

^n—3

+...

^bn—1

)

c—a

^—1

—

^n—1

^—2

—

^n—2

C —

- +b

+...+b

^n_2

c—ac—ac — a'

and since

c — a

is a factor of all the numerators, this is a homofunction of

a, b, c

of the

(n —

z)th degree.

Generally, the pth divided difference is a homogeneous function

+ r) values of the argument of the

(n —

p)th degree.Therefore the nth divided difference will be of the (n — n)th degree, i.e. will be constant.

If nth divided differences are constant, higher divided differences

will be zero. In that event the expansion for

in para. 3 will stop at the term just preceding R, and we shall have identically

11a

(x — a

₀

)

(x — a

₀

) (x — a

₁

)

^'2

1La

+ ...

+ (x —

(x — a

₁

) ... (x —

n_1)

ao•

Newton^'s divided difference formula can be applied quite easily in practice, as the following example will show.

NEWTON^'S ADVANCING DIFFERENCE FORMULA 61 Example 2.
From the data in Ex. I, find u₈.
Assuming fourth divided differences constant, the formula gives u₈=u₁+(8—I)A'u₁+(8^—I)(8^—2)ZS.²u₁
+(8— I)(8—2)(8—4)A'3u1+(8— ^I)(^8—2)(^8—4)(^8—7)^A4uj = 11₁ + 7A'u₁ + 42A'²u₁ + 168A'³u₁ + 168A^'4u₁
= 22 + 56 + 252 — 268^.8 + 32'3
= 93 to the nearest integer.
6. Development of Newton's advancing difference formula. If the interval b — a = h, we have
,^'Zla = (u_b — ua)/(b — a) = (ua±h — ^ua)/^h = ^Iua/^h; also if the interval c — b = the interval b — a = h,
0^'211. _ _/(1 (1'u, — A'ua)/(C — a)
(^ua+2h — ua+h)/h — (^ua+h — ^Ua)lh

2h
(ⁿa+2h — ^2ua+h + ua)/2h2 = A²z1a/2h². Similarly, if the interval d — c is also h,

A'3ua = (z^'2u_b — A'2ua)/(d — a)
(^ua+3h — ^2ua+2h + ^ua+h)/2h² — (^ua42h — ^2ua+h + ^ua)1^2h2
3^h

= (^ua+3h — 3¹¹a+2h + 3^t1a+h — ua)/3.2h3 = A³ua/3 h³, and so on.
The formula for u_s in terms of u_a and its leading divided differences is
ux=u_a+ (x—a)+ (x — a) (x — b)I'²u_a
+ (x—a) (x—b) (x—c)A^'3u_a+ •.•
+(x—a) (x—b) (.x—c)... (x—k)A^'fdu_a. This becomes, on putting x — a = nh,
ua+nh = ua + tlhA'u_a + nh [nh — (b — a)] ,'²11₀
+ nh [nh — (b — a)] [nh — (c — a)]A'3ua+.... If now we have equal intervals, so that
b — a = h, c — a= 2h,..., and so on,

62    FINITE DIFFERENCES nh0u_a nh (nh — h) 0²u_a
ua+nh=ua+ h +    2! h²from the relations proved above,
i.e.    ^ua+nh = ua + n_iAua + n₂L²ua + n₃L 3ua + ...,
which is Newton^'s formula for advancing differences. 7. Lagrange's interpolation formula.

On the same assumption as has been made hitherto, namely that the function concerned is a rational integral function of x, an interpolation formula can be evolved which is equivalent to the process of splitting up an algebraic fraction into its partial fractions.
Let n values of the function y = u_x be given, so that u, is sup-posed to be a rational integral function of the (n — I)th degree in x, and let the given values of x be a, b, c, ... k.

Then we may write
ux=A(x—b)(x—c)...(x—k)+B(x—a)(x—c)...(x—k)+... +K(x—a) (x—b)... (x
where there are n terms each of degree n — i in x.
This is true for all values of x involved. Put therefore x = a.
Then u_a= A (a — b) (a — c) ... (a — k),
ua
^`q(a—b) (a-c) ... (a—k)^'Similarly, by putting x = b,
u_b B _
(b—a)(b—c)...(b—k)'
In like manner all the coefficients can be found.
(x — b) (x — c) (x — k) (x — a) (x — c) (x — k)

.'. ^ux=ua(a—b) (a—c)... (a—k)+ub (b — a) (b—c)... (b—k) +.
(x—a)(x—b)(x—c)... +uk(k—a)(k—b)(k—c)...'

₊nh(nh—h)(nh—2h)4³u_a+...

₃!_h3

LAGRANGE^'S FORMULA 63
or otherwise
u_x _ ua I
(x—a)(x—b)... (x—k) (a — b) (a — c) ... (a — k)x — a
u_b
⁺(b—a) (b—c)... (b—k)x—b⁺
It is evident that this is exactly the same as splitting the fraction

(x—a)(x—b)... (x — k) into partial fractions.
This alternative expression is due to Euler and was given earlier than Lagrange^'s formula.
It is interesting to note that Euler's form, when written as
^{f (x)}— - ^{f (a)} + .f (^b) + ...
(x—a)(x—b) ... (a — x) (a—b)... (b — x) (b — a) ...

is an expression for the divided difference f (x, a, b, c, ... k). It follows therefore that Euler's formula (and consequently Lagrange's) can be evolved from the divided difference formula by equating the nth divided difference to zero.
8. Lagrange's formula is usually laborious to apply in practice and requires close attention to sign; it is generally simpler to employ other finite difference methods. Where the intervals are equal an advancing difference formula may be used, and for unequal intervals it is preferable to use divided differences.
The principles on which this formula has been developed are the same as those assumed for the difference formulae, namely that n values of the function being given, nth differences are assumed zero. The following examples show the application of the formula.

Example 3.
Given the data in Ex. I, obtain u₈ by the use of Lagrange's formula.
⁺(4^—^I)(4^—2)(4^— 7)(4^— ¹²) 0 1 ₄)^+...,

64
FINITE DIFFERENCES
22 30

7.6.4.1.(—4) (^—1)(^—3)(^—6)(^—11)^.7⁺I.(—2)(—5)(—Io).6

82 1o6
+ 3.2.(— 3) (— 8).4 + 6.5.3.(— 5).1

206

+ II.I0.8.5.(— 4)'
.', u₈ = — io•666 ... + 33^.6 — 95^.666 ... + 158^.293 ... + 7^.865 = 93 (to the nearest integer) as in Ex. 2.
Example 4.
Find the form of the function y = u_s given that
uo = 8, u_l = II, u₄ = 68, u₅ = 123. By Lagrange^'s formula :
4^x(^x—5) 4(x—1)(x—4)'

... u_x = [(²3^x+ 8) (x² — 5x + 4) — (19x — 8) (^x2— 5^x)]

=x³—x²+3x+8.

It is instructive to work out this example by divided differences, using two different orders for the values of x, thus illustrating the important principle that the order is indifferent.

(^a) x    us    ^O'ux    ^A'2ux    ^O'3ux
    0    8
3 + I = 3
    I    I1    ¹⁶-4⁼4
    4    68    57±3=19    36—4_9    5=5⁼¹

55⁼¹⁼55
    5    123

u_R
x(x— 1) (x
u_s

8    I    II    I
-4)(^x—5) (—1)(—4)(—5) x+1(—3)(—4) x—1
    68    I    123    I
+ 4.3(— ¹)^.^x—4 + 5^.4^.1.x—5
2 I    11    I    68    1    123
5 X^T 12x- I 12X—4⁺ 20 x—5
1 1 1 5 X + 4 0 1    57X — 24
    20 x(x—5)    I2(X— 1)(x—4)
23X+8    19x'—8

ux⁼8+3x+4x(x^—I)+Ix(x^—I)(x^—4)⁼x³-x²+3x+8.

EXAMPLES    65
(^b) ^{x    ux    A'ux    A'3ux    A'3ux}5 ¹²3

9. Newton^'s formula for divided differences may be considered as the basic formula in finite differences. It has been shown that, by making the intervals equal, the ordinary advancing difference formula follows, and that Lagrange^'s formula can be evolved from the divided difference formula by equating the nth divided difference to zero. Moreover, by taking the limiting values when the intervals tend to zero, an important theorem known as Taylor^'s theorem follows. (See Chapter xi.)

EXAMPLES 4

Given terms at unequal intervals, explain how to apply the method of divided differences to find an interpolated value : illustrate your answer by finding u₅ given

345,

55 =

⁰

2010, u

₄

₉₀

= 3815,

U5.15=

10965.

^u40 = 43833, U42 = 46568, ^u44 = 49431, U45 = 50912. Use divided differences to find u₄₃.

3. Given the following table, find log 656.

No.654658659, 661

Log 2

81562'81822

81892.8202

^u50 = ¹'⁶99⁰, ^U52 = P7160, ^u54 = P7324, ^u55 = P7404. Find ^u53 ^by

divided differences.

^u35^.0 = ^I175, ^u35^.5 = 1280, ^U39^.5 = 2180, U₄₀.₅= 2420. Obtain a value for u₄₀ (i) by advancing differences, (ii) by divided differences.

35,

⁶

= 436,

₃

₁

= 484. Find u

₈

₀

66 FINITE DIFFERENCES
7. Find the first three divided differences of the function y = x-² for the arguments x = 1, m, n, p.
Find by Lagrange^'s formula the value of

8.u₄₈given u4o = 15'22, ^U45 = ¹3'99, ^U50 = 12^.62, U₅₅= I I' 13.

given uo = 17'378,

⁸

94,

⁰

¹²

12'412.

1o.

₂₂

given u

₁₀

40, u

₁₅

66, u

₂₀

20.82,

u25

85.

II.

given

•400, u

₂

•

128, u

₃

= •

224, u

₄

= •

376.

12. Use Lagrang

s formula to find the form of the function

y = f (x)

givenx

₀

_{a_l}

the above

formula.

14.By means of Lagrange^'s formula, prove that, approximately,

(

)

(

—

^U_

(12_3 —

^24_

5),

(

)

= 2

^u_

11 —

₃

(123 —

l) —

(

—1 —

₃

)].

Four equidistant values u__l, u_o, u_l, and u₂being given, a value is interpolated by Lagrange^'s formula. Show that it may be written in the form

= yu0

+ y (.~''? _

I ). Q

(x2 ,

il222o ,

where x+y

⁼

16. If

, a

)

(

l) —

(ao_) ,

(

l) =

(

2) —

(a1)

etc. be

— ao

₂

—a

divided differences of the first order;

₂

, ao)

(a2,

^—

(

₂

—a

₀

etc. divided differences of the second order and so on, find

(2, 4, 9,

Io),

where

f (x) = (i) x

—

2x,

(ii)

⁴

Prove that if u_x be a rational integral function of x of the nth degree, and if values u_a , u_b , u~ , ... of u_x be given, then the expression for u_x in terms of its divided differences is the same whatever the order of arrangement of the u^'s.

)

⁶

597

⁰

⁸⁰

13. Values of

are given for all integral values of

from o to

Show that u

is capable of expression in the form

_{n_1}

n-2

(x—72)~(n—I)![x—n+I—(n 1)

_ _

^—

n+2

F(nI)

_2x

n+3—...f(n—

Find u

₄

given

= 4, ul = 7,

₂

12,

₃

= 20,

by using

EXAMPLES 67

18.Apply Lagrange^'s formula to find f (5) and f (6), given that

f(I)=2,f(2)=4,f(3)=8,f(4)= 16 _andf(7)=128;

and explain why the results differ from those obtained by completing the series of powers of 2.

19._{U_30=30;u-13=34;u3=38;u18=42.}Find u₀.

u70 = 7'69; u

₇₂

= 7'07;

u73

= 6

78 ;

75 = 6

18. Interpolate to find u

₇₁

by divided differences, using the following orders of the argument:

(

ⁱ

)

⁰

, 73, 75, 7

;(

ⁱⁱ

)

, 75, 7

⁰

, 73

By means of divided differences, find the value of u₁₉from the following table :

II17

2123

⁶

⁸

3,5

²⁶

^194,486

279,84

⁶

3,5

²⁶