Hugh MacColl.

Symbolic logic and its applications online

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53_67
Y 9~

= t r 1 (477-469) p + t r 2 (469-477) p ,forQ p = (63Q) p
= x 1 e + aw = x 1 (see § 11, Formula? 22, 23).

Thus we get AB = ,i' a - 12 = .r .i- From tne aata AB there-
fore we infer that x lies between x a and ,i\ ; that is.

53
between positive infinity and — .

53 ,_4
greater than — - or 7-.



,67 53V



In other words, x is



122



SYMBOLIC LOGIC



[§§ 139, 140

Now, here evidently the formula of § 135 was not
wanted ; for it is evident by mere inspection that u\ is
greater than ,r 2 , so that a\ being therefore the nearest
inferior limit, the limit ,r 2 is superseded and may be left
out of account. In fact A implies B, so that we get
AB = A = ,r aU .

140. Given that 7x — 53 is negative and 07 — 9*
positive ; required the limits of x.

Let A denote the first datum, and B the second. We



get—



A = (7£-53) N = (x-



53






x - —



■ x,





53


x 1 -


7




67


x 2 -


9



2'./3-

Hence, we get

AxS = Xy^ £#?2'. p ~~ ^'l'. 2' . 0.

By Formula 2 of § 1 3 5 we get —

53 67\ N , / 7 53

= %(477 - 4G9) N +,%(469 - 477) N

= Xyij + x 2 ,e = x% (see §11, Formulae 22, 23).

This shows that the nearer superior limit x 2 super-
sedes the more distant superior limit x\ ; so that we get






A-b — ,Vy 2 '. — ®%.
07



Thus x lies between the superior



limit x 2 (or — ] and negative infinity.



141]



CALCULUS OF LIMITS



123



CHAPTER XVI

141. We will now consider the limits of two variables,
and first with only numerical constants (see § 156).

Suppose we have given that the variables x and y are
both positive, while the expressions 2y — 3# — 2 and
3^ + 2^ — 6 are both negative; and that from these data
we are required to find the limits of y and x in the order
y, x. Table op Limits.

Let A denote our whole
data. We have

A = y r x p (2y - 3x - 2) N (3?/ + 2x
-6) N .

Beginning with the first
bracket factor, we get*

(2y - 3x -2Y = (y-^x-lJ = (y- ? A ) N = Vv
Then, taking the second bracket factor, we get



o

2/i = ^ + l


_ 6


2

y 2 =2 - X
o


2

X 2 — o




a? 3 ~ o





(3?/ + 2x - 6) N = ( y + - x - 2 I =



Also 2/V = 2/ a '. a- a '.o ( seG §§ 137 ' 139), so that

A = y a '. o»a'. <#i#2' = Va.: v. 2'. o ;v v. o = Vi>. 2'. a' «'. o ;
for the nearer superior limits y 1 and y 2 supersede the
more distant limit y a . Applying Formula (2) of § 135 to
the statement y v v , we get

/13

Vv. * = vvivi - VzY + y-Ay-2 - ?a) n = yA ^ - 1



+ y 2 ^x-lj=y r (x-^



+ !h\ x ~



13



* The limits are registered in the table, one after another, as they are
found, so that the table grows as the process proceeds.



124 SYMBOLIC LOGIC [§141

Substituting this alternative for y v 2 . in the expression for
A, we get

A = (y r x v + y^\)y^. o = (yv. 0% + feiftK-.o

= VV. V C a. V . + y<H. ti C a'. 1. == 2/l'. 0^1'. ' 2/2'. O^a'. 1 '1

omitting in the first term the superior limit x a because it
is superseded by the nearer superior limit x x ; and omit-
ting in the second term the limit x , because it is super-
seded by the nearer limit x v The next step is to apply
Formula (3) of § 135 to the ^-factors y vo and y z . We
get



yv. = Vv. 0(2/1 - y<>Y = yv. 0(2/1)* = yv. d -® + l

= yv.o(3x + 2) 1 ' = y v Jx + ^ J = yi'. (* - x ^f

— 2/1'. 0^2 !

y%. = 2/2'. 0(2/2 - ?7o) P = 2/2'. o(2/ 2 ) P = 2/2'. o( 2 - -x J
= y 2 , (6 - 2^ = ^,0(3 -xf = y^ Q (x- 3) N

= 2/2'.0<%-

Substituting these equivalents of ?y r and ?/ 2 . in A, we get

A = }Jx. cftv. 2.0 "J~ 2/2'. O^a'. 3'. 1 = 2/l'. 0^1'. 1 2/2'. O'^V. 1 >

for evidently x is a nearer inferior limit than ,r 2 , and
therefore supersedes ,v 2 ; while x 3 is a nearer superior
limit than x a (which denotes positive infinity), and there-
fore supersedes x a . We have now done with the ?/-state-
ments, and it only remains to apply Formula (3) of § 135
to the ^'-statements x vo and x si . It is evident, however,
by mere inspection of the table, that this is needless, as
it would introduce no new factor, nor discover any incon-
sistency, since x x is evidently greater than x , that is, than
zero, and x 3 is evidently greater than x x . The process
therefore here terminates, and the limits are fully deter-



§141]



CALCULUS OF LIMITS



1 25



mined. We have found that either x varies between x x
and zero, and y between y 1 and zero ; or else x varies
between x B and x v and y between y 2 and zero.

The figure below will illustrate the preceding process
and table of reference. The symbol x denotes the
distance of any point P (taken at random out of those in
the shaded figure) from the line x , and the symbol y
denotes the distance of the point P from the line y .
The first equivalent of the data A is the statement



x z x o x r




llv 2- o x o> which asserts that y 1 and y 2 are superior limits
of y, that y (or zero) is an inferior limit of y, and that
x (or zero) is an inferior limit of x. It is evident that
this compound statement A is true for every point P in
the shaded portion of the figure, and that it is not true
for any point outside the shaded portion. The final
equivalent of the data A is the alternative y v% x r _
+ Vv. o tl V. i> the first term of which is true for every point
P in the quadrilateral contained by the lines y v y , x v x Q ;
and the second term of which is true for the triangle
contained by the lines y 2 , y 0) x v



126



SYMBOLIC LOGIC



[§142



Table of Limits.




142. Given tliat y 2 — 4./.' is negative and y + 2x — 4
positive ; required the limits of y and x.
Let A denote our data. We get

A = (v/-4 t r)-\y + 2,,;-4) p

= (7/ 2 -4 tt -) N (y-yi)";
tf - ± x y = {(y-2 JxXy + 2 Jx)Y
= {y-2 s /xr(y+2 s fxy

for (y — 2 s/^YiV + 2 x/^) N * s impossible. We therefore
get

a = 2/2'. 3(2/ - 2/i) p = 2/2'. s2/i = y-2.3. i

By Formula (1) of § 135 we get

2/ 3 . i = 2/3(2/3 - Vif + y/yi - 7hY

= y 3 (2tf - 2 ^ - 4) p + Vl {2x -2jx- 4)*
= y 3 (# - x/« - 2 ) p + y^a? - s/x - 2 ) N (see §§ 126,
127)

slx-l



^ X ~~l) ~i*



Y



-"((•"-D-lM^-i)-!}'

= ? / 3 (.j-4) p +2/i(*-4) n
= y 3 (# ~ «i) P + ^ - ^'i) N = 2/3^i + 2/r*r-
Therefore

A = 2/2'.3^1+//2'.l^l'-

We now apply Formula (3) of § 135, thus

!h. 3 = 2/2'. 3(2/2 - VsY = Ik. s( 2 */« + 2 xA')'' = y*. 3 e
2/2'. 1 = 2/2'. 1(2/2 - 2/i) r = 2/2'. i(2# + 2 V'/' - 4) P



= yr. i(* + «/* - -)" = V*. i{( V* + 2J - (2) }

= 2/2'. i(« - 1 ) r = h: M' ~ x -zY = 2/2'. i#2-



§§ 142, 143] CALCULUS OF LIMITS 127

Thus the application of Formula (3) of § 135 to y 2 , 3
introduces no new factor, but its application to the other
compound statement y 2 , 1 introduces the new statement
x 2 , and at the same time the new limit x 2 . Hence we
finally get (since Form 3 of § 135 applied to x a . a and
Xy 2 makes no change)

A^y.,.3^+^1%.2 (see §§137, 138).

This result informs us that " either x lies between x a
(positive infinity) and x ., and y between the superior




oc jc z



limit y 2 and the inferior limit y 3 ; or else x lies be-
tween £&, and x 2 , and y between y 2 and y v The above
figure will show the position of the limits. With this
geometrical interpretation of the symbols x, y, &c., all
the points marked will satisfy the conditions expressed
by the statement A, and so will all other points
bounded by the upper and lower branches of the para-
bole, with the exception of the blank area cut off by the
line y v

143. Given that y 2 — ±x is negative, and y + 2x — 4
also negative ; required the limits of y and x.

Here the required limits (though they may be found



128 SYMBOLIC LOGIC [§§ 143-145

independently as before) may be obtained at once from
the diagram in § 142. The only difference between
this problem and that of § 142 is that in the present
case y + 2x — 4 is negative, instead of being, as before,
positive. Since y 2 — 4a; is, as before, negative, y. 2 will be,
as before, a superior limit, and y 3 an inferior limit of y ;
so that, as before, all the points will be restricted within
the two branches of the parabola. But since y + 2x — 4
has now changed sign, all the admissible points, while
still keeping between the two branches of the parabola,
will cross the line y v The result will be that the only
admissible points will now be restricted to the blank
portion of the parabola cut off by the line y v instead
of being, as before, restricted to the shaded portion
within the two branches and extending indefinitely
in the positive direction towards positive infinity. A
glance at the diagram of § 142 will show that the
required result now is

1J-2'. 3'%. ' V\'. 3^1'. 2>

with, of course, the same table of limits.



CHAPTER XVII

A

144. The symbol — , when the numerator and denomi-
nator denote statements, expresses the chance that A is true
on the assumption that B is true; B being some state-
ment compatible with the data of our problem, but not
necessarily implied by the data.

A

145. The symbol denotes the chance that A is true

e

when nothing is assumed but the data of our 'problem. This
is what is usually meant when we simply speak of the
" chance of A."



§ 146, 147] CALCULUS OF LIMITS



129



146. The symbol^—, or its synonym S(A, B), denotes
B

A A

— — — ; and this is called the dependence* of the statement

A upon the statement B. It indicates the increase, or
(when negative) the decrease, undergone by the absolute

chance — when the supposition B is added to our data.



The symbol <5° D , or its synonym S°(A, B), asserts that the
B

dependence of A upon B is zero. In this case the state-
E E E




Fig. 1.



Fig. 2.



Fig. 3.



ment A is said to be independent oj the statement B ;
which implies, as will be seen further on (see S 149), that
B is independent of A.

147. The symbols a, b, c, &c. (small italics) respectively

ABC

represent the chances—, -, — , &c. (see S 145); and the

€ € €

symbols a! ', I/, c ', &c, respectively denote the chances

— , — , — , &c, so that we get
e e e

1 = n + a' = b + b' = c + c' = &c.



* Obscure ideas about ' dependence ' and ' independence ' in pro-
bability have led some writers (including Boole) into serious errors. The
definitions here proposed are, I believe, original.



130 SYMBOLIC LOGIC [§148

148. The diagrams on p. 129 will illustrate the pre-
ceding conventions and definitions.

Let the symbols A, B assert respectively as propositions
that a point P, taken at random out of the total number
of points in the circle E, will be in the circle A, that it
will be in the circle B. Then AB will assert that P will
be in both circles A and B ; AB' will assert that P will
be in the circle A, but not in the circle B ; and similarly
for the statements A'B and A'B'.



In Fig. 1 we have






A_ _ 3 . A'_ ,_10

7 _ft ~T3' 7~~ a ~1S


AB_ 1

e "l3 :


AB' 2
' ~T~13


In Fig. 2 we have






A_ _ 3 . A'_ ,_ <J .
7~ ~12' T~ ~12'


AB 1

e "li 1


AB'_ 2

~€ 12'


In Fig. 3 we have






A_ _ 3 . A'_ ,_ 8 .
e " ~fl ' T" ~ii '


AB 1

t ~ n ;


AB'_ 2

~€ li'


It is evident also that







111



™. - 1/A „ x A A AB A 1 3 1

Fig. 1, d(A, B)=- — - = -_ = _ — _= +_ ;

b K J B e B e 4 13 52'



in .big. 2, d(A, B)= — — - = — _ = =0;

° ' y ' ' B e B e 4 12

F „ }/ . B A A AB A 1 3 1

in Fig. 3. d(A, B) = - — _= — — = - = -

S B e B e 4 11 44

Similarly, we get

in Fig. 1, J(B,A)=+1;

in Fig. 2, 5(B, A)=0;

in Fig. 3, $(B,A)=-±



§§ 149, 150] CALCULUS OF LIMITS 133

149. The following formulae are easily verified :—

<•>£-*-?■£(•> *-}&

The second of the above eight formulae shows that if
any statement A is independent of another statement B,
then B is independent of A ; for, by Formula (2), it is
clear that <S°(A, B) implies S°(B, A). To the preceding
eight formulae may be added the following : —

AB = A B = B A AB_A B _B A

e " e *A e"B ; (10) Q^~Q'AQ~QBQ ;

(11)^± B = A + B _ AB . (12) A + B = A + B _^?

150. Let A be any statement, and let x be any positive
proper fraction; then A x is short for the statement
A



— =%), which asserts that the chance of A is x.

\ € /

AB \
Similarly, (AB)* means — = x); and so on. This

convention gives us the following formulae, in which

A B

a and b (as before) are short for — and -.

e e

(1) A^:^=^- A V (2) A-B^AB/^A + B)-*-;

(3) (AB) x (A + By>:(x + y = a + b); (4) S°(A, B) = (AB) f '»;
(5) (AB)" = (A + B) a+& ;



132 SYMBOLIC LOGIC [§§150,151

< 6 >(s4)=(s=f)=*( A - B >;

„ /A B\ /A \
(7) [B = A) : \B = ! + {a = b):(AB)V + (a = h) -



It is easy to prove all these formulae, of which the last
may be proved as follows :

A_B\ /A_Z> A\ /K_b A\° (A/ 6\)°
B~Ay' ; \B~a'B/ : \B a'B/ : \ B\ X ~ a/ J

\ A V /A \

: jjj(a-&)| :( B = 0J + («-^)°:(ABr+(a = &).

The following chapter requires some knowledge of the
integral calculus.



CHAPTER XVIII

151. In applying the Calculus of Limits to multiple
integrals, it will be convenient to use the following
notation, which I employed for the first time rather
more than twenty years ago in a paper on the " Limits of
Multiple Integrals " in the Proc. of the Math. Society.

The symbols ^>{x)x m!n and x m - n (p(x), which differ in
the relative positions of <p(x) and x m >. n , differ also in
meaning. The symbol <J>(%)% m >. n is short for the integra-
tion (p(x)dx, taken between the superior limit x m and
the inferior limit x n \ an integration which would be

ex
commonly expressed either in the form ' m dx(p(x) or

fX

' ™<p(x)dx. The symbol x m . n <p(x), with the symbol

' v m\n to the left, is short for (j>(x m )— <p(% n )-

For example, suppose we have j <p(x)dx = ^(x). Then,

by substitution of notation, we get \ l m< p{ , ') ( ^ l ' = ( p{x)x m , n

J x n

= #m.»' v K#) = ^GO - ^G''») I so that we can thus entirely



§§ 151-153] CALCULUS OF LIMITS 133

dispense with the symbol of integration, /, as in the
following concrete example.

Let it be required to evaluate the integral

C z C'¥ C' 1 ' Table op Limits.

I "'-I dy I dx,
J«a JVi J-'o



z a = c



Vl = X «! = «

V2 = h h'o =0



the limits being as in the

given table. The full process is as follows, the order of

variation being z, y, x.

Integral z r . 2 y v . &. . = (z 1 - z 2 )y v . 2 x r . = (// - c)y v . 9 x v ,

= ?/r . 2 (k 2 - ^/>% . o = { (hA -cy-d- (hvl ~ cy 2 ) } x v ,

= { (I.* 2 - «b) - (W - cb) } x v . = (h^ 2 - ex - \tf + bc)x r .
= #i\ o(^^ 3 — ikr 2 — iH> 2 # + &«') = £a 3 — lea 2 — \b % ci + bca.

152. The following formulae of integration are self-
evident : —

( 1 ) *W . n = - %n' . m ; (2) #«>*V . „ = ~ <£OX'. m J
(3) *W. „<£(■») = -X n '.rn<t>{z)\ (4) ^' m <. n + X n , mt = X ni . r J

( 5 ) #(«)(#»' . n + *»» . r) = 0(^>m' . r I

( 6 ) fe . n + #„' . r )<£<>) = a? m . . ,#*') ;

\ ' / //?»' . n' ' r' . s i/n' . rnfir' . s 2/m' . n^s' . >• — 2/«' . mP^s' . r 5

'. / '' m' . 71 ~r" "■ V . s "m' . s • " J r > . n '

( 9 ) (x m . . „ + x r , . s )(p(x) = (x m . . , + ^ , n )(p(x) ;

( 1 0) <p(x)(x m > . n + *V . s) = 0(#)(#m< . . + <?V. „)•

153. As already stated, the symbol , when A and B

are propositions, denotes the chance that A is true on the
assumption that B is true. Now, let x and y be any

numbers or ratios. The symbol - means - x — ; and

3/B y B

when either of these two numbers is missing, Ave may

suppose the number 1 understood.

r PU xA x A A 1 A

Ihus, - means - x — ; and — means - x — .

B IB xB x B



V34>



SYMBOLIC LOGIC [§§ 154, 155



x x =l


"l =1


z=A


•' 2 =1-2/




y + z — l


y 8 =l -a




! e = A:


= aj i'.o^i'.o 2! i'.


D





154. The symbol IntA(x, y, z) denotes the integral
Idxldyjdz, subject to the restrictions of the statement A, the
order of variation being x, y, z. The symbol hit A, or
sometimes simply A, may be used as an abbreviation for
Int A(x, y, z) when the context leaves no doubt as to the
meaning of the abbreviation.

155. Each of the

Table op Limits.

variables x, y, z is
taken at random be-
tween 1 and ; what
is the chance that the

. . z( 1 — x — y)

traction —

1-y-yz

will also be between

1 and ?

Let the symbol Q,

as a proposition, assert that the value of the fraction in

question will lie between 1 and ; and let A denote our

data .1',,?/,'^%. We have to find -, Avhich here =-

1 • (W 1 1 -0 A

(see § 145). Also, let N denote the numerator z(l — x — y),
and D the denominator 1 — y — yz of the fraction in ques-
tion ; while, this time, to avoid ambiguity, the letter n
will denote negative, and p positive (small italics instead of,
as before, capitals). We get

Q = N^D p (N - J)) n + N n D'\N - Bf.
Taking the order of variation x, y, z, as in the table, we
get, since z is given positive,

W={l-x-yY = {x-{l-y)Y=Xt
N" = ( 1 - x - yf ={.v-(l-y)Y = ,/- 2
E> p = (l -y-yzf= \y{\ +z)-l \-» = y 2 ,

D ,l = (l-?/-F) n =<?/(l+^)-l} P = ?/ 2
(N-T>r = (z-z,v + y-iy = (z,r- // -:+iy>






y + z- 1



= ■':,



(N - Vy = (z - zx + y- l) p = (z,c -y-z+l) n = x 3



§ 155] CALCULUS OF LIMITS 135

Substituting these results in our expression for Q, we
shall have

Multiplying by the given certainty x v -0 (see table), we get

X V. oH == lV i\ 1'. 3. 0^2' + -'V. r. 2.02/-2-

Applying Formulae (1) and (2) of § 135, we get (see
§ 137)

#3. = X l X Z - '<0 )" + *oK - ^ = ^3 + •%

t% _ r = x s (3C 3 - xj* + ar^ - x 3 ) n = X^e + x v n = x s ,

X 2. = ff«te f - ^ + • ?, o('' - V = X 2* + ^ = «*

Substituting these results in our expression for x v Q,,
we get

#r. oQ = %(%2/3 + A W3')y2' + •'3' . 2^/2

== X 2' . 31/2' . 3 "r ^2' . 0^3' . 2' "•" #3 '. 2^2'

We now apply Formula (3) of § 135 to the statements

"*2\ 3' ^2'.o' ''3'. 2' **nUS

'2' . 3 = ' V -l' . 3(^2 X i) = < r -2' . $%'
*2' .0 == ^2' . Q\ X 2 ^0) = ''V . e
"^3' . 2 = ■% . 2V^3 — ,?, 2 ) = X S . iVl-

This shows that the application of § 135, Form 3, intro-
duces no new statement in y ; so that we have finished
with the limits of x, and must now apply the formulas of
§135 to find the limits of y. Multiplying the expres-
sion found for tr ro Q by the datum y v Q , we get

#i'.o 7 /i'<r* ==, ''2 . s/Ar. r. 3.0 + '''2 . 0//3 . 2'.r. * x 3. 2?/i'.2. o-
By applying the formulae of § 135, or by simple inspec-
tion of the table, we get y% v = y.y ; y 3 . = y z ; y$ . 2 < . r = Vz' '■>
y. 2 = y 20 and substituting these results in the right-hand
side of the last equivalence, we get

<'V.2yi\oQ = ''V.3y2'.3 + <>2\oy3\o + <'V.2/'r.2



136 SYMBOLIC LOGIC [§§ 155, 156

The application of § 135, Form 3, to the y-statements
will introduce no fresh statements in z, nor destroy any
term by showing that it contains an impossible factor tj.
We have therefore found the nearest limits of y ; and it
only remains to find the limits of z. Multiplying the last
expression by the datum z ro we get

QA = Q,?v _ Q y v . oZ r . = Ov . 3y 2 . . 3 + dfe . o2/ 3 - . o + % . s#r . 2>r . o-

The application of § 135, Form 3, to the factor z r _ will
effect no change, since {z x — z ) p is a certainty. The pro-
cess of finding the limits is therefore over ; and it only
remains to evaluate the integrals. We get

A Int A ^

= Int(/e.r. $2 . 3 + x* . y s . o + %? . Hl\- . 2K' .

for Int A = Int x v ,& v . f fs l .. =l. The integrations are

easy, and the result is log 2 (Naperian base), which is



5
a little above -.

9

156. Given that a is positive, that n is a positive whole
number, and that the variables x and y are each taken
at random between a and — a, what is the chance that
{(x + y'T - a} is negative and {{x + y) n+1 - «} positive ?

Let A denote our data y Y _ 2 x r#2 (see Table); let Q de-
note the proposition {(x+y) n -a} s , and let R denote the
proposition {(.?• + y) n+1 - a} p , in which the exponent N
denotes negative, and the exponent P positive.

„ , , . QR , . , Int QRA

We have to find the chance — ^-, which = .

In this problem we have only to find the limits of
integration (or variation) for the numerator from the
compound statement QRA, the limits of integration for
the denominator being already known, since A = ?/ 1 .2^1.2.



156]



CALCULUS OF LIMITS



Vtf



Table of Limits



y 1 = a

y 2 = ~ a
i

y 3 = a n — X
1

y i = — a n — x

i
7i =a n + 1 —x

•J 5

1

y, = - « n+l - x


x x = a

,r 2 = -a

i
x 3 = a n — a
i
x. = a — a n

4

1

x =a + a n + 1

i
x e = a + a n

6

1

x 7 = a n+1 — a

i
x 8 = — a — a n + 1

i
a? g = « — a"+i


n+l
" 3 = (±) B


A = e = 3/r.2 a a'.2







We take the order of integration y, #. The limits
being registered in the table, one after another, as they
are found, the table grows as the process goes on. For
convenience of reference the table should be on a separate
slip of paper.

We will first suppose n to be even. Then



Q = j {pc + y) - a n \ \ (as + y) + a"



I

= y -



» — a?)| K | S



I



!/ + <a« +.!■)■, =y t ;.



= \t*



R= \(as+y) — «"•"*" x ;■



— ) p [ / 1
) t



//o-



Hence QR = 2/ 3U ?/ 5 = y 3 , 4i5 ; and multiplying by the
datum 7/V.2' we § e ^

Q%r . 2 = y 3 - . r . 2 . 4 . 5 = Os" ''3 + yvXgXy&A . 6 + y^e)
= (//3->'s + yv^3')(y^5 + y& x h) = y$. ^ + y^.^s . a + yr. s% ;

for by application of the formulae of § 135, ,r 4 5 = ,r 5



138 SYMBOLIC LOGIC [§156

#5 . 3 = «'■ s ; <'V. 5 = 1 ( an impossibili ty) ; and Xg . 5 - = .r 3 , For
when a> 1 we have ( 2a — a n ] , and when ct< 1 we have

( a"+i — a Tl ] ; so that x 6 — a; 3 is always positive.

We must now apply § 135, Form 3, to the statements
in y. We get y s . 2 = y&. &#\ yz>. 5 z =yv.5 a i> Vv.^ — yv.^n-
Substituting these results, we get

Q%1'. 2 = VS. 2<?6\ 5 + y&. 5%. 3«1 + y V . 5%. 7-

Having found the limits of the variable y, we must
apply the three formulae of § 135 to the statements in x.
Multiplying by the datum x Vti , we get

Q%V. g»l'. 2 = 3fa. 2'' V. 6' . 5. 2 + 2/ 3 '. B®1'. 5'. 3 . 2«1 + ?/l'. 5'?V. 1'. 7 . 2
= 2/3'. 5^1'. 3 ft l + llv. b X Z'. V. 7 I

for x x i 5 = »/ ; ajj/ 5 - = ajji ; x Zm 2 = '''3 ! #7. 2 = ,: ^7*

We obtain these results immediately by simple in-
spection of the Table of Limits, without having recourse
to the formulae of § 13 5. Applying the formulae of
§ 135 to the statements in x which remain, we get

x Vm v = ./v",, + x r a 2 , ; # r 3 = x Vi 3 a. 2 ;
•%. 7 = iV 3'. n a \ I iV i' . 7 = '''v. 7 a 3-
Substituting these values, we get

Q% r . 2 Xy. 2 = QRA = // 3 - 5 X V . 3 «. ! . 2 + //r. 5 (#3'. 7«2 . 1 "1 #1'. ?"*. 3>
== . ? /3'.5' ?, l'.3 a l "I" VM.ffiv.'fl r \
— (Vb'.S^V. 3 + yv.sfls'. 7/ (6 l J

for ai.2 = «i = a 2.i3 an( i %.3 = > ? ( an impossibility).

This is the final step in the process of finding the
limits, and the result informs us that, when n is even,
QRA is only possible when a x ( which =1) is an inferior
limit of a. In other words, when n is even and a is not
greater than 1, the chance of QR is zero. To find the
chance when n is even and a is greater than 1, we have



§ L56] CALCULUS OF LIMITS 139

only to evaluate the integrals, employing the abbreviated
notation of § 151. Thus

Integral A = Int y v 2 ,?; r 2 = (y x — y 2 )^v. 2 — ( 2a)# r> 2

= x v , 2 {2ax) = 2cuc 1 — 2ax 2 = 4a 2
Integral QRA = y w 5 # r< 3 + y Vm 6 # 3 ,. 7

= (y-s - y^ v v. 3 + (Vi - y 5 )'%.7

= ( a Tl — a 1l + l W,_ 3 + ( a — ««+i+ ,i ] W 7



= ®V. 3 a?l ~ aH + 1 F + ' r 3'-7 *" ~ an+1 V + i^



= a n - a** 1 (^j - ^ 3 ) + ( a - a w +* )(a? 8 - x 7 ) + £(#?, - ^)



= '( a n — a n + l Y 2a - £ a " - | a ™+ 1



QR = Int QRA _ Int QRA
A 7w« A 4a 2

= — ( a" _ a ^+i Y 4a - a Tl - a"* 1

We have now to find the chance when % is odd. By
the same process as before we get

QR A = (y 3 , 5 ,/' r- 3 + y v . 5( %. 7 K + y & . &&. 2 «3-

Here we have £wo inferior limits of a, namely, a x and a 3 ,
so that the process is not yet over. To separate the
different possible cases, we must multiply the result
obtained by the certainty (a 1 +a r )(a 3 + a$), which here
reduces to a x +a v 3 + %, since a x is greater than a y

For shortness sake let M x denote the bracket co-
efficient (or co-factor) of a x in the result already obtained


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