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 XIII
MISCELLANEOUS THEOREMS
1. Indeterminate forms.
It has been demonstrated in Chapter zx (para. 8) that the limit of f (x) as x -~ a is frequently required although f (a) itself has no
meaning. Forms such as o 9 which result from the direct substitu
tion of a for x in f (x) are called indeterminate forms. To obtain Lt f (x), where f (a) is an indeterminate form, we may resort to
x-aa
algebraic methods as previously shown, or we may adapt the processes of the differential calculus to the solution of the problem.
Let I (x) and 0 (x) be two functions of x continuous as far as the value x = a and let 0 (a) = o = 0 (a), so that     Via) is of the
indeterminate form
'
o
Let f (x) _    (x) . Write a + h for x, so that Lt is the same as
x~a
Lt. h—0
Then f (x) _    (x)    (a + h)    (a) + kb' (a + Blh)
0 (x) 0 (a + h) 0 (a) + h/ (a + 02h)'
But 0 (a) and :/i (a) are each zero.
Therefore f (x) _        (a + 91 h) (dividing numerator and denomi

i/i' (a + 02h)
To obtain Lt f (x) when f (a) is the indeterminate form °- we
x--->a    o
therefore differentiate numerator and denominator separately and put x = a in the result.
If 0' (a) and 0' (a) are both zero (so that the form o- is again
0
nator by h).
Therefore Lta f (x) = hLto      (x) = 0,     (a) , since when h -* o, a+O1hand a+B2heach -+a.
INDETERMINATE FORMS 207 obtained) a further differentiation must be effected and a subfor x in 0,, (x) ; and so on if this be indeterminate.

Lt x6 -5x+4= Lt 6x5-5=I=1. x—>1 x3 — 2x + I x---1 3x2 — 2 I
Example 2.
Find Lt sin-10 - 0

Lt sin-1 B - 0    form a .
e ~o    ea    o
Differentiating numerator and denominator separately: I

2
    = Lt v — B    form o-.
o ~o    302    0
Differentiating again:    = Lt 0 (I — 02)-'I
    e-,o    60
- B2)
    = eLto    6
    0L, to        dividing through by B,
— 1 —s•
 B-*0    0'
I

208    DIFFERENTIAL CALCULUS
(For a proof of this theorem, see Gibson's Elementary Treatise on the Calculus, p. 420.)
 Lt -
ex x->w
Lt n(n— i)xn-2

...........................
= o. Example 4.    1
Find Lt ~ I — log + x } x->o lx x2 g l    )
x.* Lt o {
x — xIzlog(t+x)} is of the form co -00. Write it as    Lt x — log (i      + x)    form -°
x->0    x2    0
= Lt I -`I+x)—1
x-> 0
2X
Lt    +x) 2
x->0    2
00
form -
00
ex
x- a

ILLUSTRATIVE EXAMPLES    209
Example 5.
Lt (I — x2)log(1-x).
This takes the form o° when 1 is substituted for x.
y = (I— x2)1°g(1-x),


I
'NY = log (I — x) log (I — x2).
l    x2)
Lt logy = Lt °g I —    x)    form — 0o
x-)-1        x--1 log (I — x)    — CO

— 2X
x2
I
I — s
which, on removing the common factor        , is — x
Lt 2X = I. x->1 + x
Lt y=e.
x- 1
minate form when a is
0' (x) will not be rLtt- ~~) ,
(ii) the differentiation is performed on the numerator and


denominator separately.    (x) must not be differentiated as the (x)
quotient of two functions of x.
4. Partial differentiation.
If f (x, y) = o defines an implicit function of x and y, we may obtain dr by differentiating in the usual manner and then solving the resulting equation for dx.
F    14
Let Then
= Lt x-~1
Note. Care must he taken in applying the principles of the differential calculus to indeterminate forms to remember that
(i) the method holds only when the function takes an indeter-
(x)
substituted for x. Otherwise Lt
x—*a (x)
2I0 DIFFERENTIAL CALCULUS
For example, if
  
 
x2 + xy + y2 =O,
then 2x + y + xy' + 2yy' = 0,
and y' = — (zx + y)/(x + 2y).

Where there are two or more independent variables an alternamethod can be adopted which is often simpler in its applicaThis is the method known as partial differentiation.
Consider the function f (x, y) = x2 + xy + y2 where x and y are independent variables. Then if we differentiate f (x, y) with respect to x keeping y constant, the result is said to be the partial differential coefficient of f (x, y) with respect to x; similarly, on differentiating with respect to y keeping x constant, we obtain the partial differential coefficient of f (x, y) with respect to y.
The usual notation is ax f (x, y) ; y f (x, y) for partial derivatives with respect to x and y respectively.
In the above example of = 2x + y and of = x + zy.
Generally, if u be written for f (x, y),
au = Lt f    (x + Ox, y)    -f (x, y)
ax    ,x-,-o    Ax
au = Lt f    (x, _y + Ay) - f (x, y)
ay Av=o    Ay

5. To prove that, if x and y he functions of a third variable z, then
du au dx au dy dz=axdz+aydi
Let x, y, u become x + zx, y + Ay, u + Du respectively when z becomes z + Oz.
Now if we let x vary while y + Ay remains constant, we have, by the Mean Value Theorem,
f(x+Ox,y+Oy)=f(x,y+Ay)+Lxaxf(x+O4x,y+oy). Similarly,
a
f(x,y+Ay)=f(x,y)+Ayyf(x,y+02oy).
and
or
PARTIAL DIFFERENTIATION
Adding these two results and dividing by °z, we obtain
°u f (x + °x, y + °y) — f (x, y)
°z    °z
    °x a    a
f (x + B1 °x, y + °y) +
—    f (x, y + °2°Y).
    ax    z ay
Taking the limit °z —> o, so that Au, °x, °y also —> o,
°u °x °y become du, dx, dy respectively, °z' Oz' °z    dz dz dz
axf (x + BI°x, y + °y) becomes ijx f (x, y), i.e. and    - f (x, y + O2°y) becomes y f (x, y), i.e. oy. du _ au dx au dy
dz ax dz + ay dz'
Corollary. If z = x, so that y is a function of x, and u, although expressed in terms of x and y, is a function of the single variable x,
du = au + au - , since dx = I . dx ax ay dx    dx

Suppose now that u = f (x, y) = o; then au au dy ° - ax ay dx'
dy    au au
dx    ax/ ay'
This is a convenient formula for obtaining
implicit function of x.
    For example, if    x2 + xy + y2 = o,
ply 0 =(2x+y)+(x+2y)d.
2II
then
dx when y is an
dy    2x+y dx = x + 2y
as before.
6. A further investigation of the theory of partial differentiation involves detailed mathematical analysis. Two important theorems are, however, worthy of mention:
I4-2

2I2    DIFFERENTIAL CALCULUS
it can be shown that
a2u    a2u ax ay = ay ax
In other words, the operations of differentiating partially with respect to x and y are commutative.
(ii) If u = f (x, y) is a homogeneous function of the nth degree in x and y, then
x ~x + y y = nu (Euler's Theorem).
The proofs of these theorems are difficult, and it will be sufficient to verify that they are true by simple examples.
a2u    2
If u = yx, show that ax      ay ay     ax z
ay = xyx-1' ax ay = ax ay    yx-1 + xyx-1 log y,
au    log    a2u    a (au\l    110
ax =Yx Y'    —
oyax ay \ax/ — xYx-gY YxY
= xyx-1 log y + yx-1,
ax3 + by3 + cx2y + dxy2.
to prove that, if
u = ax3 + by3 + cx2y + dxy2,
au    au xax+ya = 311. Y
au = 3ax2 + 2cxy + dy2,
ax
au = 3by2 + cx2 + 2dxy.
ay
(i) Defining ax ay as the operationx (y) , i.e. the process of
partial differentiation of a—y with respect to x, keeping y constant, Y
Example 6.
which proves the proposition.
Example 7.
Show that Euler's Theorem for a homogeneous function of x
holds for
and y
It is required then

RELATION BETWEEN D AND 0    213
ax y au
x +y = 3ax3 + 2cx2y + dxy2 + 3by3 + cx2y + 2dxy2 = 3u.
Euler's Theorem can be extended to any number of variables, so that, if u is a homogeneous function of x, y, z, ... of degree n,
au    au    au
xax+yoy+zaz+... =nu. 7. Relation between the operators dx and A.
It has been shown that, if us be a rational integral function of x, we may express us in the form exDuo, where D denotes the operation of differentiation (Chap. xi, para. 13).
I.e.    Exuo = exD u0 ,
or    Ex = exD.
If now x = 1 we have
E eD or 1 + 0 = eD,
so that    Dlog(1+A)
02 03 04
-0— z + 3 — 4 +...,
A2 A3 A4    2
2
D2r0— +3—4+...
02 03 + 1204 — A5 + .... Similarly,    D3 A3 — A4 +.415 —
We have therefore a convenient method for expressing the differential coefficients of a function of x in terms of the differences of the function.
Example 8.
µx (the force of mortality) = — lx dxx , where lx is the number of
persons at exact age x in any year of time. Given the following table, find a value for µms.
Age ...    50    51    52    53
lx    73,499    72,724    71,753    70,599

214    DIFFERENTIAL CALCULUS The difference table is
x    lx    11x    A21x    ~3lx
50    73,499

3•
x (x +    1'x    A3u-1 + ... .
RELATION BETWEEN D AND A    215 A first approximation will evidently be dux ux+l — u,r_1
ax =    2 or, if the unit of differencing be h,
ux+n — u=-n
2h
9. Another formula, giving the differential coefficient in terms of differences of ux, can be obtained from Bessel's formula.
tlx = uD + ul (x — 2) Au, x (x — I) A2u-1 + A2u0 2    21    2
+(x— x(x— I)A3a_l+....
3 Changing x to x +
u0+ul { x<1u } (x+(x—2)A2u    +A2ito
x+} =    2    2!    2
Differentiating :    ll
dxux+=Duo+xA2ula    (x
3 -I+
Ifx=o,    dx
d ul=Du„—O-u-    ....
4
Changing the origin to x — we have the approximation
d    03 ux-11
dx ux = Du _„ — 24 + ....
An interesting discussion on the calculation of the values of differential coefficients of a function by means of selected values of the variable will be found in T.F.A. vol. Ix, pp. 238 et seq. (G. J. Lidstone). By means of prepared tables of coefficients Mr Lidstone evolves formulae for the values of the successive differential coefficients, using both advancing differences and central differences. In addition, alternative processes are given for the values of the derivatives when the intervals are un

216    DIFFERENTIAL CALCULUS
10. Osculatory interpolation.
It may happen that we know the values of u,r at intervals of a unit, and that we wish to calculate a complete table of values with smaller intervals. For example it is a common practice to calculate every fifth value in a life-table, and to complete the table by interpolation : here the unit interval for the preliminary calculations is five years.
If we decide to use a third difference formula then every interinvolves four of the given values. For the interval o to I the best course is to base the formula on the four values u_1, 110, uI, u2i thus giving equal weight to values on either side of the interval. An appropriate formula is Bessel's formula, namely,
ux = (tip + u1) + (x - i) Otlp x (x — I) ,2110 + 0211_1 +(x-x(x-I)0311 I    (i).
    
31 - This is the formula for the interval o to 1. For the interval Ito 2 we should use the corresponding formula based on the four given values uo, u1, u2, u3.
Thus for the two intervals o to I and 1 to 2, the two interpolation curves have a common ordinate when x = 1; they may not necessarily have a common tangent. Two neighbouring interpolacurves will usually cut one another at the point of junction, and while there will be a smooth run of values within each unit interval the values will not run smoothly with those in the next interval.
We are led therefore to enquire whether we can find a series of curves of interpolation which shall have common tangents as well as common ordinates at the points of junction. Such curves are said to have contact of the first order at the points where they join, and the necessary condition for this contact is evidently that ux
and dx must have the same values at these points on the one curve
as on the other.
The interpolation curve given by Bessel's formula (i) for the interval o to I was based on the conditions that it should have u_I, uo, u1, 112 as ordinates. We retain the condition that no and u,

OSCULATORY INTERPOLATION    217
should be ordinates and abandon the condition that u_1 and u2 should be ordinates. Instead, we shall stipulate that the differcoefficients of us when x = o and when x = I have known values. To fix these values we shall proceed as follows :
The interpolation curves for the two intervals — I to o and o to I are to have the same tangent when x = o as the curve of second degree which has u_1, uo and u1 for ordinates. The equation to this curve may be written
Fz(— I, o, I)= (u0+u1)+ (x — .) Au0+ 2x(x— I)A2u_1 ...(ii).
The interpolation curves for the two intervals o to i and I to 2 are to have the same tangent when x = I as the curve of the second degree which has uo, u1 and u2 for ordinates. In a similar manner to the above the equation to this curve may be written
Fz(o,I,2)— (u0+u1)+(x)Duo+x(x—I)A'uo...(111). It will be noticed that the forms that we have chosen for the equations (ii) and (iii) differ only in the last term.
Differentiating (ii),
Fs' (— 1, o, 1) = Duo + (x — ) O2u_1,
so that    F„ (— I, o, 1) = Duo — 2,2u_1         (iv). Differentiating (iii),
Fs' (o, I, 2) = Duo + (x — ) A2u0i
and    F1' (o, I, 2) = Duo + 2\2uo     (v).
The values of Fo' (— I, o, I) and F1' (o, I, 2) in (iv) and (v) are to be values of the differential coefficients of the required intercurve for the interval o to 1.
Let vx be that curve, so that its ordinates when x = — I, o, z, 2 form the basis for an ordinary interpolation formula of the third
degree:

vx = 2 (v0 + v1) + (x    Ov0 + x    (~    I) O2?10 2 A2v—1
+(--__3(x— 1)Q3v_1     (vi).
The tangents to this curve are given by the equation
vz' _ ~v0 + (x — 2) A2v0 2 A2v_1 + 3x2 6 x + 2 03v_1 ... (vii).

218    DIFFERENTIAL CALCULUS
The conditions to be satisfied are:
vo = 110,
v1 = u1,
vo' =F0' (— I, o, I) _!1u0 — 2A2u—1,
    
v1' = F1' (o, I, 2) = tuo + 2J2uo     (viii). We have at once (vo + v1) = 2 (u0 + u1), and L vo = Duo; these determine the first and second terms in (vi).
0'    I .~2    A27,
1
    From (vu), v ' - -,vo —    °-     1 + 03v_
2    2    12
A22'0 + A2v_I    I
v1    OZa+2    2    + I2,2—1,
and therefore    vo' + v1' = 2Avo + i.,3V_1, v1' — vo' = > (,2vo + L 2v_I). But, from (viii),
Vol + v1' = 2:,u0 + ?,3
vl/ — v0' = (
2 (A2110 + L\2u_1)•
Comparing the two expressions for vo' + v1' we have at once A3v_I = 3.. 3u_1 (since L vo = Duo), and from the two expressions for v1' — vo'
2 (A2vo + A2v—1) _ 2 (02u0 + A2 u-1).
We have now found the values of all four terms of the formula (vi), and we can write the formula in terms of u's as follows :
vx = 2 (uo + u1) + (x — Au, x (x — I) 02u0 + 02u_1

+ x (x — I) (x — 2) 0311_1     (ix).
2
This result is a formula of osculatory interpolation, and differs from the ordinary central difference formula (i) only in the last term.
The difference is
vx — ux = x (x — 13 (x — 2),3u_i — x (x — I 6(2x — I) 03 u_1
2
OSCULATORY INTERPOLATION 219 11. The problem of osculatory interpolation has been discussed by many eminent actuarial authorities in the past. The method was
devised by Dr Sprague (see J.I.A. vol. xx11, p. 270) and was subdeveloped by Mr George King and Dr Buchanan. An elementary demonstration of the method, depending upon addifferences, is given by Mr King in the Supplement to the 75th Annual Report of the Registrar-General. Dr Buchanan has given an alternative method based on Everett's formula (j.I.A. vol. xLII, pp. 369—94) and in the appendix to his paper there appears a simple demonstration of the formulae by Mr G. J. Lidstone.
The proof in para. 10 above is due to Mr D. C. Fraser, and deon Bessel's formula. The ordinary interpolation formula ending with the term ,3u_1 can be written in many different forms, all giving identical results, and the addition of the expression


x (x - I) (zx — 1)A3u_1 produces an osculatory formula. 6
Suppose, for example, that we take the descending difference formula
ux = u_1 + (x + I) ,u 1 + x (x2 1) A2u 1 + x (x   6    I) A3u-1'

Adding the term (x — I) (zx — I)

Z'x = u_1 (x + I) Au_1 + x (x + I) A'u_,
+x(x2— I)Tx(x — I) (2x- I)l A3u 1 2    6

= 1!p -4- xAu_1 + x (x2    2
I) A2u_1 + x2    Ii A3u_i

which is the form obtained by Mr Lidstone in the appendix to Dr Buchanan's paper.
An interesting Note on the application of a graphic method to formulae of osculatory interpolation appears in the Actuarial Students' Magazine, No. 3 (Edinburgh, 1930). By treating osculatory interpolation as a particular case of divided differences, Mr Fraser shows that a diagram similar to the hexagon diagram for ordinary differences (Chap. viii, para. II) can be employed to obtain the various forms of osculatory interpolation formulae.
6     A3u_1 we have

220    DIFFERENTIAL CALCULUS
EXAMPLES 13
n
 x-*0    x9
21. Determine a, b, c so that, as 0 tends to zero, the function
0 (a + b cos 0) - c sin 0 05 shall tend to the limit unity.
4.
Lt ax 1
x,O bx -
6. Lt [xi - al + (x - a)I] (x2 - a2)- . x-> a
8. Lt x-a+
x->a    V~/xn - an
1 Io. Lt (1 + x3)' .
x-*0
1
xl tan 2a    2
19. Prove that the limit of (2 - - I    as x tends to a is err . a
20. a and p are positive integers. Find (a T
I _ p)x - ax
Lt \


EXAMPLES
    au    au    au
xax+ya +Zaz=3u. Y
Find the value of d x , when x = 8, by using divided differences.
xe + yi
26. U = =    .
x'+y'

222    DIFFERENTIAL CALCULUS