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 VII
SUMMATION
If therefore f (x) is the function whose first difference is (x) we can find the sum of any number of terms of the series whose general term is q (x) in terms of values of f (x), for any given interval of differencing.
Generally, the interval of differencing is unity and the first term of the algebraic series under consideration is (I). By putting a and h each = I, the required relation then becomes
n

(x)=f(n+i)—f(I)•
0x2 = (x + 1)2 — x2 = 2x + I ;


0x2 — Lx = 2X,


or    ?A(x2—x)=x;
I (x2 — x), or 2x (x — I),
is the function whose first difference is x.
Again, it can be quite easily seen that, since Aax = (a — I) ax,
ax
O    —    a.
a—I

98    FINITE DIFFERENCES
a
Therefore    a a I is the function whose difference is ax. —
n
We can therefore find ux by the above method if ux is of the
form kx or kax.
The relation Ox(m) = mx('n-1) enables the sum of any series whose nth term can be expressed in the factorial notation to be summed immediately.


Example 1. -
Sum to n terms the series whose xth term is x (x — 1) (x — 2). Now    x (x — 1) (x — 2) = x(3),


and since    /,x(4) = 4x(3),
n
E x(3) = Lfx(4)n+~ 1

where [f(x)] 1+1 represents the process of substituting n + I and 1 for x successively in f (x) and deducting the second result from the first.
n
... E x(3) =    [(n + 1)(4) — 1(4)]
J
= (n + 1) n (n — 1) (n — 2),
since the product of four successive terms of which 1 is the first includes o and is therefore obviously zero.
In the above example the law of the series is known. It often happens that, owing to the fact that a limited number of terms is given, the law of the series must be assumed.
Example 2.
Find the sum of n terms of the series o, to, 33, 77, 150, ....
By taking out the differences it is seen that Du = 1o, O2u = 13, and that the only two values of A3u available are each 8. We must assume therefore that ,,3 u is constant.
We may adopt several different methods for the solution of the problem. Of the three following methods, the first is probably the best. (i)    = (1 + 0)n-1 u1 (if u1 be the first term)
= u1 + (n — 1) Au1 + (n — 1)2 z. 2u1 + (n — 1)3 L3n1 = 0 + Io (n — I) + 13 (n — 1)2 + 8 (n — 1)3.


E u = E {Io (n — I) + 13 (n — 1)2 + 8 (n — I)31. 1    1

SUMMATION

Since (n — I), = nr+1 — (n — I)r+1 = 0 (n — 1)r+1, it follows that
E (n — I )r = [(n — 1),,.+,T+1
= nr+1 n
E u = Ion2 + 13113 + 8n4. 1
     I00    FINITE DIFFERENCES
n    I    I    n+i
1
(x + I) (x + 2) (x + 3)    (x + I) (x + 2)Ji
— 2 [(n + 2) (n + 3) 2 . 3]'
  SUMMATION
For Dux=b+c(r—1)rx,
and
A2ux=c(r—1)2rx=krx,
where k = c (r — 1)2.

It follows that for nth differences
Annx= c(r— 1)nrx.
Example 4.
Sum to n terms the series 1, 6, u, 18, 31, 58, I15, ....
The difference table is
Third differences are in G.P. with common ux Aux A2ux A,ux ratio 2.
    1    Assume therefore that
    5    ux = a + bx(1) + cx(2) + dex,
    6    o    Dux=b+2cx(1) +d2x(2— 1)
    5    2    = b + 2cx(1) + d2x,
    II    2
    7    4    A2u, = 2c + dzx,
    18    6    A3ux = dzx.
    13    8    Inserting the differences for the value
    31    14
    27    16 x = 1, we have
    58    30    2=~13u,=2d; o=A2u,=2c+2d;
    57    5=Du,=b+2c+2d.
    I15    From these equations we find easily that
d=1; c=—1; b=5.
Putting x = i in ux = a + bx(1) + cx(2) + dex, we have
1 =u1=a+b+ 2d,
    whence    a=1—b—2d=—6;
ux=—6+5x(1) —x(2) +2x;
    .'. n E ux    [— 6x(1) + 5x(2) x(3) +
2x n+1 =    - -
    1    2    3    2—I =—6(n+1)a) +6+ 5(n? 1)(a) _(n 31)(s)
+2n+1—2,
since all the factorials except the first vanish for the lower limit. This simplifies to
2n+1 — 2 — 6 (2n2 — 15n + 19).
Alternatively, we may proceed thus: A3ux = d2x.

IO2    FINITE DIFFERENCES
Deduct d2x from ux and difference the function ux — dzx in the usual way.
By giving x the values I, 2, 3, 4 in succession it is easily seen that ux — dzx takes the values — I, 2, 3, 2 respectively. On forming a difference table we find that the leading differences are
I (ux — d2x) = 3, 02(ux — dzx) = — 2.
Then    ux=2x—I+3(x— I)1 -2 (x— I)2, so that
Eux= [ 2 x    1)1+3(x— I)2—2(x—I)3]1+1
= 271+1 — 2 — n1 + 3712 — 2713,
which, on simplification, gives the same result as that above.
6. The form uxvx. Summation by parts.
When the general term of the series is the product of two funcn
tions of x and when the value of each of the summations E ux and

It follows that when the function uxvx can be put in the form VxOUx the summation can be performed at once if E [Ux+10Vx] can be evaluated (but not otherwise).
Example 5.
Find the sum of the series a + za2 + 3a3 + 4a4 + ... to n terms. The terms are -successive values of the function y = xax, and since
as    x a — I
x
we may write x for Vx and    a a     I for Ux in the relation above.
n
E vs is known, a method known as "summation by parts " can be
adopted.

We have 0 (UxVx) = Ux+1 Vx+1 — UxVx
= Ux+1 (Vx+1 — Vx) + Vx (Ux+1 — Us) = Ux+10Vx + Vx DUx
VxLUx = t (UxVx) — Ux+10Vx;
.. E[VxzUx]= [UXVxJ"'+1—E [Ux+10Vx]•
I    1    1
1

11SUMMATION E [xax] = E [xA (a — ax I-)
[a      ax I x]1+1 — E ax+l
+      I Ax
ax    n+1 n ax+l
[     xi n+1
,since Lx=I, a—1    1    I a—II
[a axI xJl+l — [
(a as+l
an+1    a    an+2    a2 =(n+ i) a — I a — I (a — I)2 + (a — I)2
The above method is laborious if the rational integral function of x is of higher degree than the first. If, for example, the function were y = x3ax, the term corresponding to Ux+iO Vx would be
a— 0x3 or ax+l (3x2 + 3X + I), and the formula would have to
x+l
a —I    a
be applied separately to this expression, twice in succession, in order to obtain eventually a simple function of ax.
n
7. The result in para. I, namely, that E (x) = f (n + I) — f (I),
where Of (x) = (x), can be obtained by the use of the operator E.
n
We have    E(x)=(I)+ci(2)+...+(n)
= (I +E+E2+...+En-1)c (I)
En—1 E — I (I)
since
n
Eck (x)=(En—I)f(I)
=f(n+1)-f(i).
n    En_I
Thus the operator E is equivalent to the operator .   — i , and we may safely substitute E En n    II for in
any series of operations. 1

104    FINITE DIFFERENCES
En —
Again, since Enux = us+n, the identity E u,. = E — I us can be
expressed as
E us =    I (ur+n — us) 1
= 0 (ux+n — ux)
= 0-1 (ux+n — u,.).
The result E (x) = f (n + I) — f (I) is sometimes written as
E (~J (x) = [f (x)11 +1,



as in the numerical example given in para. 2.
8. The relation between the operators l and A. It has been seen that if
of (x) = (x)
f (x) = Ecb (x),
where the summation is performed between certain limits.
If therefore we omit the limits we may say that with certain reservations summation is the inverse process to differencing.
Consequently    (x) = zf (x) = DEq (x),
so that    of 1,
i.e.    E-0_A -1.
Now although DE - I it does not follow that EA = I, for we shall obtain the same result by differencing f (x) + c, where c is a constant, as by differencing f (x) alone.
Thus
so that
0 [ f (x) + c] =
f (x) + c = 0—1() (x) =
EA    I.
 (x) ;
(x) = EAf (x),

The symbol E may be treated as the equivalent of the inverse symbol A-1 provided that it be remembered that E and 0 are not commutative, and that where we are not summing between limits an arbitrary constant must be inserted.
or
Eof(x)= [f(x)
1
n+1
I
then

OTHER USES OF THE SYMBOL E    I05
The process of summation in finite differences is similar to the corresponding process in integral calculus and the relations between the symbols are analogous to those existing between the symbols of differentiation and integration. As a result, finite difference summation is often referred to as integration. Eux is said to be the
n
indefinite integral of ux; E ux the definite integral ; and a function that can be integrated, such as ax, to be "immediately integrable." 9. Other uses of the symbol E.
One of the commonest functions in the theory of life contingencies is the expression obtained by multiplying lx (the number attaining age x) by the interest factor vx. This product is denoted by Dx, and the connection between certain functions dependent on Dx is indicated thus :
N=D;
Sx=EIYx.
Here E denotes summation from age x to the end of the mortality table, it being understood that values beyond the end of the table are zero.
When E is used in this special sense,
Eux = ux + ux+1 + ux+2 + ... to the end of the table ;
DEux = (ux+l + ux+2 + ux+3 + ...) — (ux + ux-F-1 + ux+2 + ...)
= — ux,
    so that here    DE — I.
In point of fact, the correct way of showing the relation between
the functions D and N, etc., is
=
Nx = E Dx+t ,
t=0
where x is fixed and t is the variable.
In modern mathematical works it is now usual to use the
    co    n
notations E , E etc., where the variable t is specified for the lower
t=0 t=a
b
limit only. The still shorter form E is often used where there is
a
no doubt about the variable.

1o6    FINITE DIFFERENCES
Again, in algebraic series, E is often used loosely to indicate the sum of the first x terms of a series, thus :


Eux=ul+u2+...+ux,
.. DEux = (u1 + u2 + ... + ux+I) — (u1 + u2 + ... + ux) = ux+1
= Eux,
whence 0E - E.
On the other hand, if E is specially defined so that Eux indicates a, summation beginning with u1 and ending with the last term preceding ux, then


Eux=ul+u2+u3+...+ux-I,
AEux= (161+u2+u3+... +ux) — (u1+u2+u3+...+ux_1)
= ux.
In these circumstances therefore AE I.
These illustrations serve to show that great care must be exin introducing E into any formula. The sense in which it is to be used should be clearly defined in every instance : the safest course is always to state the limits where possible.
10. Application of the relation between E and A.
By treating the operator E as equivalent to A-1, the method of separation of symbols can be employed for the solution of problems. For example, a convenient formula for the evaluation of Eaxux can be evolved by which the necessity for the continued application of summation by parts can be obviated.
Example 6.
Prove that


Eaxux —     ax f — a + a2A2 —     a3,,3      +    ux { a constant
a    l    a    I    (a    I)2    (a    1)3    }
Now    EPaxux = ax+Pux+p = as (aE)vux
and    Eaxux = A-1 axux + c
(E — I)-1 axux + c.
Let    (E)Ao+AiE+A2E2+...+ATET+... Then the (r + I)th term in (E) axux is ATETaxux.

ILLUSTRATIVE EXAMPLES I.e.    Arax+rux+r = axArarux+r = axArarErux
= axAr(aE)rux;
:.    (E) ax ux = axe (aE) ux ,
so that if q (E) is the operation (E — 0-1
(E — I)_laxux = as (aE — I)-lux;
Eaxux = ax (aE — I)-lux + c
ax(a+aA— I)-lux+c
1

    = ax(a — I)-1 [1+    a0    ux +c
a—      ~
ax [    aA    a2t~2 —

    I— a — + (a —    ]ux +c.
a— I    I    I)2
Example 7.
Apply the above formula to evaluate E3x (x3 + x2 + x + 1). Aux = A (x3 + x2 + x + I) = 3x2+ 5X+3, A2ux=A2(x3+x2+x+ I)=6x+8,
and    A3ux =A3 (x3 + x2 + x + I)=6;
E3x(x3+x3+x+ I)
3      3x lux — (3 3 I) Dux + (3 321)2 A2ux — (3 331) A3ux} + c
= 2x[x3+x2+x+ 1 —(3x2+5x+3)+4(6x+8)—=8'.6]+c = g [4x3 — 14X2 + 28X — 23] + C.
It is often of advantage to set out the rational integral function in factorial notation; the successive differences can then be obwith little difficulty.
Thus, by the method given in Chapter II, para. 23, x3 + x2 + x + 1 = x(3) + 4x(2) + 3x(1) + I,
so that        Dux = 3x12 + 8x(1) + 3,
A2ux = 6x(1) + 8,
A3 ux = 6,
and, by adopting the formula for Eaxux, we have easily that
E3x (x3 + x2 + x + 1) = 8 (4x(3) — 2X'2) + 18x(' — 23) + c.

Io8    FINITE DIFFERENCES
To express 4x(3) — 2x(2) + 18x(1) — 23 in the form
ax3 + bx2 + cx + d,
we use the method of detached coefficients applied inversely:
    2 4    -2    18    — 23
    o    8    — 10
4 Hence
E3x(x3+x2+x+I)= (4x3—14x2+28x—23)+c, as before.
11. Series in general.
Certain series can be summed by purely algebraic methods only : others can be summed by algebra or by finite differences. For example, binomial, exponential and logarithmic series depend upon algebraic theorems, and algebraic methods should therefore be employed to find the sums of such series. As has been shown above, series whose nth terms can be expressed as rational integral functions of n can be summed by finite difference methods. Arithmetical and geometrical progressions (and series dependent thereon) can also be summed by finite differences, but for these algebraic processes are preferable.
There remain a few other types of series to which either algebra or finite difference methods can be applied.
Example 8.
Evaluate
I 0
4
0

4
0
28

Example 9.
Show that the general term of the recurring series uo+u,x+u2x2+...+urx''+...,
for which the scale of relation is I — px — qx2, is Aan + Bbn, where a, b are functions of p, q and A, B are constants.
Since I — px — qx2 is the scale of relation,
un — pun-1 — qun-2 = o
i.e.    un — pE-1un — qE 2un = o,
or    (I — pE-1 — qE-2) un = o.
Therefore    (I aE-1) (I — bE-1) un = o
if    a+b=p and ab=—q.
This will be true if either
(I — aE-') un or (1 — bE-1) un = o,
i.e. if    un — aun_1 or un — bun_, =    o.
Now if    un — aun_1 = o,
then    un = aun_,,
and the series is a geometrical progression with common ratio a. The general term of the "a" series is therefore Aan.
Similarly the general term of the "b" series is Bbn, where A and B are constants. But if a new series be formed by the addition of these two progressions the relationship will hold good for this new series. In other words the most general solution is
un = Aan + Bb",
where we may give A and B any values, but a and b must satisfy the equations    a + b = p and ab = — q.
12. If we expand us by a finite difference formula, we can find the sum of a number of consecutive values of us by integrating both sides of the identity.
For example
ux=uo+thuo+x2A2uo+x3A3uo+...
x~2~    x(3)
z~
=uo +x1) Duo+      A2uo+    03uo+..., 3•
as in Example z (ii).

II0    C    FINITE DIFFERENCES

Eux = c + x(1)u0 + 2x(2)Au0 + Sx(3)A2u0 + .... If the limits be i and n, we have
n
Eux = I x(i) uo+ 4x(2) t uo + sx(3) A2uo + ... ~ n4 1' which can be evaluated in the usual way.
Example 10.
If fourth and higher differences are ignored, prove that the sum of n successive terms of a function, of which uo is the central term, is

nun +    (n3 — n) A2u_i
where n is odd.
Since u0 is the central term, it will be convenient to use a central difference formula.
Gauss's forward formula gives
ux= u0+xAuo+ 2x (x I)A2u_1+ (x + I)x(x — I)A3u_1 = u0 + x(1) Au0 + x(2) A2u-1 + s (x + 1)(3) A3u_1


Eux = C + xu0 + 2x(2) Au0 + .ix(3) A2u 1 + 2T (x + 1)(4) A3u_I .
On summation between the limits — (n — i) and (n — 1) the coof Auo and A3 u_3 will cancel, and we shall have
1(n-1)    1k(n+1)
Eux = LC+xuo+2x(2)Au0+2x(3)A2u_1+ (x+I)(4)A3u_1J -+(n-1)    -}(n-1)
— n+I—/—n1
2    l    2
n+I n—1 n—3 ( n -1.n + 1n+311A2u-1 +g    2    2    2 —\    2    2    2 ~J

= nuo + (n3 — n) A2 u_1


EXAMPLES 7

Sum the following series :
I. 7, 14, 19, 22, 23, 22, ... to n terms.

EXAMPLES
I I. (x + 3) (x + 4) (x + 5). 12. X (x + 2) (x + 4).
13.
 (3x—2)(3x+ 1)(3x+4). 14. x (x + I) (x + 3).
15. x(x+I)(x+3)(x+4). 16. (2x+3)(zx+5)(2x+7)(2x+9).
  (x + 3) (X + 4). i8.    (x + I) (X + 3)'

x
19. (x + z) (x + 3) (x + 4)'    zo. (3x — 2) (3x + 1) (3x + 4)*
    x+3     21. x(x+I)(x+2)•
 112    FINITE DIFFERENCES
3o. Sum ton terms 1.32+3.52+5.72+7.92+..••
    EXAMPLES    I13