2. Show that the roots of (a - x) (b - x) - h* = are real and are separated by a and by 6.
[Denoting the left-hand side by/(#), we have
By grouping terms, find an upper and a lower limit to the roots of the equations in Exx. 3-9.
3. x* - 3z 3 - 2* 2 + Ix + 3 -0. 4. x 5 - 10** - 5a? 3 + 6* 2 - 11* - 350 =0. [For an upper limit in Ex. 4, group thus : x*(x* - IQx - 5) + (6z 2 - llx - 350) .
For a lower limit, write x = -y and group thus :
5. x* + 4x 3 - llx* -9z- 50=0. 6. x*- 7. 2x*~llx*-I()x-l=Q. [Group thus: x(2x*~ llx-ll) + (x-l).] 8. z54-z 4 -6* 3 -8o,' 2 --15a;-10:-0. [Group thus : x(x* - 6z 2 - 15) + (z 4 - Sx 2 - 10).] 9. a; 5 - 3z 4 - 24z 3 + 95z 2 - 46x - 101 = 0.
10, For the equation f(x) =x n +p 1 x n - l +p 2 x n ~ 2 + ... +p n =0, if the n umerically
greatest negative coefficient is equal to - p 9 then p + 1 is an upper limit to the
roots. x n 1
[Let sol, then/(z)>0 if x n >p(x n ~ l + x n -* + ... + I), i.e. if x n >p .
.
x n x I
This holds if x n > p . ---- - , i.e. if x - 1 > p.]
11. For the equation f(x) ~x n +p l x n - l +piX n - 2 + ... +p n = Q, if the nu merically
greatest negative coefficient is equal to -p and the first negative coefficient is
p r , then Zjp 4- 1 is an upper limit to the roots.
This holds if x n >^ - r , i.e. if (x-l)x r ~ l >p, which holds if (*- I) r >p, i.e.
12. If the rule in Ex. 1 1 is applied to the equations in Exx. 3-7, show that the limits are (1) 4, - 4; (2) 351, - 5 ; (3) 9, - 51 ; (4) 6, - 1 ; (5) 7, - 6.
13. By using Ex. 11, find an upper limit to the roots of x 1 - 100# 2 - 237=0. Find the rational roots of the equations in Exx. 14-19.
14. * 3 -9z* + 22o;-24:=0. 15. a? 3 -5* 2 - 18*4-72=0.
16. 3* 3 - 2* 2 - 6* 4- 4=0. 17. 4# 4 i
18. 6s 4 -25^4-26^ + 4s-8=0. 19. 6a;H53^-95a; 2 -25ar+42=0.
ELEMENTARY SYMMETRIC FUNCTIONS 95
15. Symmetric Functions of the Roots of an Equation. Let a, /J, y, 8, c, ... be the roots of
x n +p 1 x n ^ 1 +p 2 x n - 2 + ... -f p w = 0, ... (A) then by Art. 2, 27a= -p l9 27aj9=]9 29 27a/?y = - j> 3 , etc ... (B) It will be shown that these equations can be used to express any
symmetric function of the roots in terms of the coefficients.
Functions of the type ZoPfPy* ... , where a, 6, c ... are positive integers, will be called elementary functions. These can be calculated in order by a process of multiplication, and any symmetric function of the roots can be expressed in terms of them.
Ex. 1. For equation (A) find the values of Za 3 yS and S**ffy.
(i) The product Sat, . 27aj9yS consists of terms of the types a 2 /?y8, a/?yf . The first
of these occurs once, in the product. The second occurs five times, namely as th e
product of any one of a, j5, y, 8, c and a term from 27a/?y8. Therefore Hoc . a88 =27a 2 8 +527<xSc, and
(ii) Consider the product 27aj8 . ZajSy. This consists of terms of the types a 2 2 y,
a s )8y8, a/tySe. The first of these occurs once. The term a 2 /?y8 occurs three times,
namely, as each of the products (a/?)(ay8), (ay)(a/?8), (a8)(ay). The term ajSyS c
occurs ten times, for we can select two out of the five, a, j, y, 8, c, in 10 wa y?.
Therefore ^aft . ^y - 27a 2 j8y + 3Za a ]8y8 + lOZ^ySe ... (C)
Using the last result and equations (B), we find that Z^Fr^PiPi-ptPa-Sps ... (D)
NOTE. These results can be tested by putting a=]8 = y ...=1. In this way (C) becomes CJ . C? = 3(7? + 12(72 + 10(7?.
For reference, we give the following results, indicating briefly the process of reckoning.
// a, j3, y , . . . are ZAe roots of x n + fto; 71 -" 1 + p 2 ;rn "" 2 + . . . + y n == 0,
- (27 . aj8) 2 -
Here the elementary functions of the second, third and fourth degrees in a, j8, y, ... are calculated in order, and obviously the process can be continued so as to find any function of this kind. It is important to notice that the degree n of the equation does not occur in the reckoning.
96 DIFFERENCES OF ROOT3
16. Symmetric Functions involving only the Differences of the Roots of f(x)=0.
These are unaltered if the roots are diminished by any number h. Hence they may be calculated by using the equation f(y + h) = 0, where h is chosen so as to remove the second term.
A useful check is provided by the theorem in Ex. 1 below. The functions in Exx. 2, 3 are important.
Ex. 1. Let v be any symmetric function of the differences of the roots of (o c ,a 1 ,fl a ,...a n $a:,l) n =0 f
then if v is expressed in terms of a , a l9 a 2 ... , the sum of the numerical c oefficients is
zero.
The sum of the numerical coefficients is obtained by putting a^ a^ a 2 1 * n v - But in this case the given equation becomes (x + l) n , every root of which is - 1, and
since t; involves only the differences of the roots, in this case, v=0.
Ex. 2. // a, jS, y are the roots of ax 3 + 3bx* + 3cx + d =0, find the values of
(i) We have a (x - a) (x - jS) (x - y ) = ax 3 + 3bx- + 3cx + d. Substituting 4- 1 and - 1 for x in this identity, we find that
therefore a 2 (a 2 + 1 ) (j8 2 + 1 ) (y 2 + 1 ) - (a - 3c) 2 + (36 - d) 2 . (ii)27( j 8-y)(y-a)-^y-Za 2 -3
Ex. 3. // a, /3, y, o are the roots of
0, ... (A) find the values of
<!).( -/3) (ii)r(a-j8)VS f . + a-/3-8)(a+/3-y-8).
a) 2 - 8Za|8 -^ (6 2 - ac).
(ii) The equation whose roots are - , -i , - , - fi - is obtained by interchangi ng o and e,
6 and d, thus ^ y 8
(iii) Denote the function by v. Since v is a function of the differences of a, j 3, y, 8,
its value may be found by using the equation , 4# K ,
+ y+i-O, ... (B)
obtained by the substitution xy-b/a.
SYMMETRIC FUNCTIONS OF ROOTS 97
Let a', ', y', 8' be the roots of (B). Then since Za'=0,
a-8=jB' + /-a'-8' = -2(8' -fa 7 ), and -0/8 = (8'ta')(S'+jB')(8' + y')
.'. = - 827a'j8'y' = 320/a*.
#*. 4. 7/a + jS + y + 8 = 0, prove tftaf a 5 + )3 5 + y 5 + 8 5 - - SjScejB .
Let a, j3, y, 8 be the roots of & +p 2 %* +P& +jp 4 = 0, and let s r ~ZoL r > th en by Art. 15,
!=<), 5 2 -2^p 2 , 5 3 = -3p 3 . Also a 5 +p 2 <x?+p3<x. 2 +pi<xQ, with three si milar equa-
tions. Hence by addition,
17. Equations whose Roots are Symmetric Functions of a, j8, y. The following are typical examples.
^J&f'l. If a, /?, y are JAe roote of x? + qx + r=Q, find the equation whose root s are
(j8-y) 2 , (y-a) 2 , (a-jS) 2 .
Let 2~(j8 -y) 2 , then since 2?a = and a/?y = -r, z = (jg + y) 2 - 4j9y =a 2 f 4r/ a .
Hence the required equation can be found by eliminating a from a 3 +ga + r=0, a 3 -za-f4r=0.
By subtraction, (z -f q)oc = 3r.
Substituting for a in the first equation and simplifying, we have
which on expansion becomes -f 4g 3 + 27r 2 =0.
The artifice employed in the next example is often useful.
- Ex. 2. (i) Find the condition that the sum of two roots a, ft of xt+p^+p^+paX+p^Q ... (A)
may be zero.
(ii) Use the result to find the equation whose roots are the six values of |~(a- f /?), where
a, j8 are any two roots of
a^ + 46a? + 6cz 2 + 4da; + e=0 ... (B) (i) Since a and -a are roots of (A),
a 4 + P&* + p 2 a 2 + P 3 a + Pi - 0, a 4 -2>i<x a + # 2 2 ~^3 +^4 =0 ;
.*. a 4 -fj 2 a 2 +^4=0 and a^a 2 ^ p a )=0.
Now a 9^0 unless p 4 =0, therefore 2>1 2 P4~2>1#2JP3+2?3 2=: 0> ... ... (C)
which is the required condition.
98 ROOTS OF THE CUBIC EQUATION
(ii) Let z -\ (a + j8), then we have (a - z) + (j8 - z) =0.
If then we diminish the roots of (B) by z, writing x~y + z, the sum of two roots of
the resulting equation in y will be zero. This equation is
ay* + 4By* + GCy* + 4Dy + E=Q, ... (D) where B=az + b, 0,2* + 2bz + c,
D-(a, b, c,<%, I) 3 , E = (a, b, c, d,e$z, I) 4 . For equation (D), the condition (C) gives
an equation of the sixth degree in 2, whose roots are the values of J(a EXERCISE XII
SYMMETRIC FUNCTIONS OF ROOTS
If a, j3, y are the roots of ax* + 36# 2 4- 3cx + d=0 and H = ac-b 2 , G = a*d - 3a6c + 2b\
prove that :
1. a^a 3 ^ -3(a' 2 d-9a&c-f9& 3 ). 2.* a^V 3 = 3 (^ 2 a - 9^6 -h 9c 3 ).
3. a*(a^ 2 -f ^y 2 + ya*)(ay 3 + j8a 2 -h yj8 2 ) = a*(3a 2 ]8V 8 = 9 (aW - 6a6aZ + 3ac 3 -f 36 8 rf).
4. a 2 {(j8-y) 2 -f(y-a) 2 4-(a-j8) 2 }-18(6 2 -a^).
5.* a f {a a (j3-y) 8 + j8 8 (y-) 8 + y 2 (a-j8)} = 18(c 8 -c). *Explain how to derive (2) from (1) and (5) from (4).
6. a 2 {a(j3~ 7. a 4 {(j8-y 8. a 3 (^4-y
9. a 3 (2a-~y)(2j3-y-a)(2y-a--/3)^ -27(9.
10. (i) The condition that a, 0, y are in A.P. is ^=^0. (ii) The condition that they are in O.P. is oc 3 =6 3 d.
(iii) The condition that they are in H.P. is rf 2 a - 3dcb + 2c 3 =0. [(i) If 2a = /M-y, then a= -b/a. (ii) Eliminate x between u = and ax 3 + d=0. (iii) Put x~l/y and use condition (i).]
11. The condition that a, /?, y may be connected by the equation
is aC 3 =5 3 D, where B=
12. The equation whose roots are jgy-a 2 yq-j8 2 )8 - y 2
2j8' a-f j8-2y
is a(aa: 2 -h 26a; + c) 3 = (aa^ + 36x 2 4- 3ca; + d) (ox -f Show that this reduces to a cubic equation. [Use Ex. 11.]
SYMMETRIC FUNCTIONS OF ROOTS 99
13. If a, 0, y are the roots of &+qx + r =0, prove that the equation whose rootsare - + -, - + -, ~ + - is
y ft a y a
[Let 2 = - 4- - . Prove that a 8 -f 2ga - rz = 0. Eliminate a between this equat ion
14. If a, ]8 are any two roots of x* + qx + r=Q, the equation whose roots are th e
six values of a/0 is
[Let z = - ; /. a = /fe, hence the required equation is got by eliminating j8 fr om
^ 3 +g0-fr=0 and z 3 8 + qzp + r = 0.] 15. If a + & + c=0 show that
(i) a* + b b + c 6 =5abc(bc + ca + ab) ; (ii) a 1 + b 7 + c 7 ~labc(bc + ca + ab )*.
[Let a, 6, c be the roots of a: 3 -f qx + r =0.] 16. Use the last example to show that
(x -f a) 5 - x 5 - a 5 5ax(# -f a) (# 2 -f a.r -f a 2 ), (# -fa) 7 - x* - a 1 = lax(x-{-a)(x z -f aa; + a 2 ) 2 . 17. Solve(x-fa-f-6) 5 -x 6 -a 5 -6 5 -:0.
18. If a, , y, 8 are the roots of x*+p&*+pjc+pt=0, prove that (1)^ = 2(^-2^), (ii)^ 7 - ~lp*(p**-pt).
If a, j8, y, 8 are the roots of ax* + 4bx*-\- 6cx 2 + 4dx + e=Q and //~ac~6 2 , Ga*d- 3abc -f 26 3 , prove that :
19. a 2 a 2 =4(4& 2 -3ac). 20.* a 2 21. a*i;a z p = I2(ad~2bc). 22.* a 2
*Note that (20) follows from (19) and (22) from (21). 23. a*Z(* - J 8) 2 (y 2 + y8 + S 2 ) = 144(c 2 - bd). 24. a 3 2;(a - ]8)(a - y)(a - 8) = - 32(7.
25. If a, 0, y, ... are the roots of x n +p l x n - 1 +p 2 x n - 2 + ...+p n = 0 , prove that
Ev* j3 2 = - ptf
26. Show that the squares of the roots of
a<F n - a^- 1 4- a^- 2 - . . . -i- ( - l) n a n =0 are the roots of
V n - M*"" 1 + M n ~ 2 - ... 4- ( - 1 ) n 6 w = 0, where
&o = oS ft i = i 2 - 2a 2 ^2 == a a - %<*>ia* + 2a a 4 , . . .
b r =a r * - 2a r _ 1 a r+ ! -f 2a r ^ 2 a r+2 - 2a r ^30 r+8 + . . . . 27. If the equation whose roots are the squares of the roots of the cubic
is identical with this cubic, prove that either a =6=0 or a =6 =3, or a, b are the roots of 3 a
CHAPTER VII
PARTIAL FRACTIONS
1 . Rational Fractions. An expression of the form P/Q, where P
and Q are polynomials in x, is called a rational fraction. If P is of lower degree than Q, P/Q is called a proper fraction. If P is not of lower degree than Q y P/Q is called an improper fraction. By means of the division
transformation, an improper fraction can be expressed as the sum of an integral function and a proper fraction. Thus
** + ! , 2
Theorem. If A+^ A' + -^-, , where the letters denote polynomials in x ix #
and X/Y, X'/Y f are proper fractions, then A^A' and X/YsX'/Y'. X' X YX'-XY'
r or A A ^7 -y. yy *
Also X and X' are respectively of lower degrees than Y and Y', and therefore YX'-XY' is of lower degree than YY'.
Hence A - A' =0; for otherwise, the polynomial A - A' would be identi- cally equal to a proper fraction, which is impossible. Thus Az=A' y and consequently X/Y^X'/Y'.
2. Partial Fractions. To resolve a given fraction into partial
fractions is to express it as the sum of two or more simpler fractions. Fundamental Theorem. // C/AB is a proper fraction and the factors
A, B are prime to each other, proper fractions X/A and Y/B can be found such that
CX Y
For since A is prime to B } polynomials X', Y r can be found so that BX'+AT = 1 9 (Ch. Ill, 18.)
j^ * C CX> CT
and therefore - = j- + -~- .
FORMS OF PARTIAL FRACTIONS 101
If CX.' I A and CY'/B are not proper fractions, by division, we can find polynomials Q, Q', X, Y 9 such that
CX' X , CT n Y and ~ = Q +>
where X/A and Y/B are proper fractions, and then C _ X Y _ _,
Now X and 7 are respectively of lower degrees than A and B, therefore BX + AY is of lower degree than AB. Thus (BX + A Y)/A B is a proper fraction : so also is C/AB, therefore
3. To resolve a Proper Fraction P/Q into its simplest set of Partial Fractions. We shall prove that
(i) To a non-repeated factor x-a of Q there corresponds a fraction of the A
form
x-a
(ii) To a factor (x - b) n of Q there corresponds a group of the form
-i -o
1 J -- ?_ J -- 2 -- L -L. / T\O'/ 1\*> '***',
x-b (x-b) 2 (x-b)* (x-b) n *
(lii) To a non-repeated quadratic factor x 2 +px + q of Q there corresponds a , 4 . - . .
fraction of the form
(iv) To a factor (x 2 +px + q) n of Q there corresponds a group of the form +-
'" (x 2 +px + q) n '
Here A, B I} B 2 , ... , C v (7 2 , ... are independent of x.
Proof (i). Let $ = (#-a) . S, then since x-a is a non-repeated factor, B is prime to x - a, and so
P/Q = */(*- a) +7/B,
where the fractions on the right are proper fractions. Hence X is a constant.
(ii) Let Q = (x - b) n . B. It is assumed that x - b is not a factor of JB, so that JS is prime to (x - 6) n , and consequently
where the fractions on the right are proper fractions.
Hence X is of degree n - 1 at most, and it can be put in the form Bi(x-b) n ~ l + B 2 (x-b) n ~*+...+ B n , (Ch. Ill, 4.)
where 5 1 , 2 are constants, which proves the statement in question. (iii) The proof is similar to that of (i), but X is of the form Cx + D.
(iv) The proof is similar to that of (ii), but each of the set J9 1 , B 2 , ... is of
the form Cx + D.
Various ways of finding the constants are explained in the following examples.
X*
Ex. 1. Express j^- as the sum of an integral function of x and three proper fractions. ' ~ *
We have (x - a) (x - 6) (x - c) = x* - a: 2 . 2a + xZab - abc.
Hence it is easy to see that the quotient in the division of x 4 by (x - a)(x ~b )(x~ c)
is x + a -f b + c. We may therefore assume that re 4 4 5 C
(A)
(x~a)(x-b)(x-c) x-a x-b x-c*
To find A, multiply each side of (A) by a; -a, and put x=a.
=
' (a-b)(a-cY
The values of B and C can be found in the same way ; a 4
(x-a)(x-b)(x-c) (a-b)(a-c)' x-a
Ex. 2. Resolve into the simplest possible partial t fractions
6 c
a? 8 -5
m , . a; 2 -5 21 43
Therefore - - r^rr - - = -- ,- -f ; - =-^ + T - rr* + - ^ (#-l) s (a;-2) x-l (*~1) 2 (a;-!) 8 x -2
Second method. Find d as before. Next, multiply each side of (A) by (x - 1) 8 , and
then put a? = l ; l'-5 .
C r~ ^~ 4.
Next, multiply each side of (A) by a?, and let #->oo ; ; /. a= -2.
Finally, put a: =0 in (A), /. --= ~a-f 6 ~c~- , .'. 6 = 1. 2i JL
METHODS OF QUADRATIC DENOMINATORS 103
Particular attention is called to the last two steps. In this way two equations connecting the coefficients can always be written down. The steps are equivalent to the following : Multiply each side of (A) by
(x~-l) 3 (x-2) and equate the coefficients of sc 3 (the highest power of x) and the absolute terms on each side.
... T 33? - 2x 2 - 1 ax + b cx + d ... W
Multiplying by (x z + x -fl) (x* - x + 1), we find that 3x*~2x 2 -l=(a
and, equating coefficients,
J = -2, a-6 + c + d=0, 6 + d=-l ... (B) Whence a=2, 6 = 1, c = l, d- ~2.
Note that the first and last equations in (B) can be found by (i) multiplying (A )
by x and letting #->oo , and (ii) putting x=0. Two other equations may be quickl y
obtained by putting x~l and x- - 1.
The following method is sometimes useful. From (A), 3s 3 - 2x 2 - 1 _ ca^-f (c + d)
x 2 -x -fl ~~ aX x 2 - x 4- 1 whence by division,
The integral functions and also the fractions are identically equal, and so a + c = 3 9 b + 2c + d = l, 2c + 2d= -2, -2c= -2,
giving a = 2, 6 = 1, c = l, rf= -2. 1 __ a 6 c c? ea;+/ .
(m) - " + ^ + ^l> + (^^ + ^ *" ( }
We first find e and /. Multiply both sides of (A) by x 2 -f x + 1, and then let This gives
D = * B+/ ' when
We have & 2 = - a? ~ 1 and # 3 = 1, whence we find that
and
It follows that l=3(
/. 3(/-2c)*-3(e+/) = l.
This linear relation holds for two values of x (namely, the roots of x 2 4- x + 1=0).
It is therefore an identity, and so
/-2e=0 and l=-3(e+/), giving e=-l/9 and /= -2/9.
104 PARTIAL FRACTIONS
Next, multiply (A) by x + 1, and then let x = - 1 ; 1 1
" a ~(-2) 3 .l~ 8*
Multiply (A) by x, and let -> oo; /. = a -f 6 -f e, giving 6 = ^ + ^ = ^-|. In (A), let xQ, .'. -l=a~& + e-d+/, giving c -i ;
1 ' (x-l)(x>- !)(**-!) 1 1 17 1 1 1 1 1 1 i _ !** _ 1 ^ /__ 1 \9 ' / 8 s + 1'72 *-l 4 (*-l) 2r 6 (x~l) 3 9 a; EXERCISE PARTIAL FRACTIONS
Resolve into partial fractions I. J 2 . 3( *~ 6) . 3. a: 3 H-a: 2 -f 1 " ' ' ' ' g --- a; 3 - I9x - 15 ' ' 10 11 12
' )(a? I -f2)- ' (a: 2 -fl) 2 (a; 2 -f-2)" *
15. Express - - : r v as the sum of a constant and two proper fractions. r (l-ax)(l~bx) r r
Hence show that if 1 -o&x 2 is divided by 1 -(a + b)x + dbx\ so as to obtain a quotient of (n 4- 1) terms, this quotient is
and find the corresponding remainder.
16. Express . - ~ - =~ - ; as the sum of a constant and three proper fractions <a-)(*-&)(*-c) ^ ^
17. Express ; - - r-- - - as the sum of a constant and three proper fractions (-a)(*-6)(a-c) p ^
18. Hence show that
_ ""
IDENTITIES FROM PARTIAL FRACTIONS 105 3.4
19. Express ~ rr-, . as the sum of an integral function of x and r (x-a)(x-b)(x-c) 6
three proper fractions. Hence show that
(a-b)(a-c)(a-d) (b-c)(b-d)(b-a) (c-d)(c-a)(c~b)
_ _ (d-a)(d-b)(d-c) is equal toa-ffc-fc-fd. 20. If ( 1 + x) n c + c v x + c 2 x 2 + ...+ c n x n , show that _c c t x x+ 1 [Assume that , ... " = ^ +
21. Hence show that
_ _ ___
v; 2 34 V ; 714-2
22. Use Ex. 21 and similar identities to prove that (\\ _^___fL_4__^__ -4_f_l\n
v; 1.2 2.3 3.4 " V ; ( , _ _. 3.4 4.5 1.2.3 2.3.4 v ; (n + l)( + 2)(n + 3) 2(n-f3) vlvy 2.3.4 370 T '"' 23. Prove that
24. Use Ex. 23 to prove that
[In Ex. 23, write n -f 1 for n, and put ! = !, a,=2, ... , a n =w, and a?=0.]
CHAPTER VIII
SUMMATION OF SERIES
1 . Meaning of Summation. Let u n be a function of the positive integral variable, and let
The function s n possesses this peculiarity it is the sum of n terms, and these terms cannot be added up, unless the value of n is specified.
Sometimes it is possible to express s n as the sum of p terms, where p is a number which does not depend on n. For example,
and the number of terms on the right (namely two) is independent of n.
To sum a series to n terms or to find the sum of the first n terms of a series is to express s n in the form just described ; but this is not always possible. Among the series capable of summation are arithmetic and geometric
series ; a harmonic series cannot be summed. The sum of the first n terms of the series
u i +
is often denoted by H r r ^iU r and sometimes by 2u n , or simply by s n . 2. Method of Differences. // we are able to express u n in the
form v n - v n _ 1 , where v n is some function of n, then we can sum the series to n terms.
For, by hypothesis,
whence by addition
METHOD OF DIFFERENCES 107
Ex. 1. Sum the series I . 2 . 3 + 2 . 3 . 4 -f 3 . 4 . 5 -f- ... to n terms. Here u n ~n(n + l)(n-t-2)
' u n= v n~ v n- where v w =-i(
and since v Q, we have by the preceding, J ) (n 4- 2) (n + 3).
3. Series in which u n is the Product of r successive terms of an A. P., beginning with the n-th.
The last example is an instance of this type, and the same method can be applied to the general case, which is as follows :
It is required to sum to n terms the series in which w n = (a + w&)(a + n + I . 6)...(a + w + r- 1 " . />), where a, b y r are constants.
We have __u n {(a + n + r .b)- _ __ _ u n (a -f n + r . b) (a -f n - 1 . b) u n n Therefore w n = v n - v n _^ where v n - -/ and consequently 5 W = v n - %
Here v is independent of n, hence the sum to n terms may be found by
the following rule : On the right of u n introduce as a factor the next term of the A. P. ; divide by the number of factors so increased and by the common difference of the A. p. and add a constant.
The constant is found by putting w = l, as in the next example ; or by substituting for n in v n .
Ex. 1. Sum to n term* 1 . 3 . 5 + 3 . 5 . 7 -f 5 . 7 . 9 f . . . . Here u n ~ (2n- l)(2w-f l)(2w + 3), and applying the rule,
To find the constant C, put ?i = 1 in (A), noting that s l = I . 3 . 5 ;
-f 3)(2 + 5) + 15}.
* The expression in the bracket (} is formed by subtracting the term which prece des n from that
which follows (n-l-2) in the A.P. 1, 2, 3, ... . H B.C. A.
108 SUM OF INTEGRAL FUNCTIONS
4. Series in which u n is a Rational Integral Function of n.
In Ch. Ill, 4, (3), it has been shown that a rational .integral function of n of degree r can be expressed in the form
o + bn + cn(n + l)-f dn(n + l)(n + 2)-f ... to (r + 1) terms,
where a, 6, e, ... are constants whose values may be found by synthetic division. We can therefore use the rule of the last article to sum series of this type, as in the next example.
Ex. I. Sum the series
2.34-3.6 + 4. ll-K..-f(tt + l)(n 2 -f2). Here tt = (n + l)(n' + 2) 1 + 1 )* + 1 + 2 +2
+0
Dividing by n, n + 1, n + 2 in succession, as on the -2 right, the reckoning shows that 1 - 2
tt n = n( therefore
3n 8 + Wn* + 2ln + 38),
5. Series in which u w is the Reciprocal of the Product of r successive terms of an A. P., beginning with the n-th. A simple instance is the following :
Ex.l. Here 111 n 4 (2/i-l)(2w + l) 4 1 where t> n --
The general case is as follows : It is required to sum to n terms the series in which u n is the reciprocal of
(a-f n6)(a + w + 1 . 6)...(a-hn-t-r~l . 6), where a, 6, r are constants.
* The numerator is the difference between the first and last factors of the deno minator.
SUM OF RECIPROCALS 109
Proceeding as in the last example, we have _ 1 (a + n + r-1 . 6)-(a-f nb)
Therefore u n ^v n _ l -v n , and s n = v Q -v n , where 1 1
. b)(a + n + 2 . b) ... (a + n + r- 1 .6)
Now v is independent of n, hence the sum to n terms may be found by
the following rule : Remove the left-hand factor from the denominator of u n ; divide by the number of factors so diminished and by the common difference of the A.P. and subtract the result from a constant.
The constant is found either by putting n = 1, or by substituting for n in v n . Thus, in the last example, by the rule
Ex. 2. Sum the series 4- + ^-^ + . . . + l.Tc J* , O O.U W ( 7i "f 1
n n(n + 3) n(?i + l)(n + 2)(n 1
* *
To find the constant C, put w = l in (A), observing that s^ _L^ x 2 . c^ 11
" 1.4 3 3.2.3.4* " 18*
Alternative method, by Partial Fractions. We have l
n 4 5 u
. 8 = * , x 11 1 1 1
^ ""18
which is the same result as before, in a different form.
1 10 TWO IMPORTANT TYPES
The last example is an instance of a scries whose n-th term is
where P is a polynomial in n of degree m.
If m<r - 1, this can he summed by expressing P in the form a H- % (a + n&) + a 2 (a 4- w6) (a 4- n + 1/>) 4- . . . .
If w = r - 1, the series cannot be summed, for to do this we should have to sum a harmonical progression.
If w>r~l, the nth term consists of an integral and a fractional part which can be dealt with separately.
a a(a + \) a (a + 1) (a + 2) a(a+ 1) ... (a-f n - 1) 6 + b(b + 1 " then s n = - - K ( + *) - } For where r n = u n (a + n)/ (a - b + I ) (n > 1 ). AT (a-fl)-fc a Also w, = w t . v r / - T ~ = f r _- . a-b+1 a~fe+l a 1 Therefore s n = v n 7 r = = r {w n (a + n) - a}. a-6-hl a-6 + 1 ^ * '
7. The Series U 4-u 1 x-f-u 2 x 2 4-... + u n x n where u n is a Poly- nomial in n of Degree m.
If x = l, this series can be summed as in Art. 4. If x ^1, let
then s (1 - x) = w + v& -f t'^ 2 + . . . + v w a: n - u n x n+l , where ^i = %-^o> v 2 ^u z -u lt ... t? n = w n -w n . r
Now v n is a polynomial in n of degree m - 1 ; thus by repeatedly multiply- ing by 1 - x (m multiplications in all), we can reduce the problem of finding 8 to that of summing a geometric series.
It follows that s = P/(l ~x) mfl where P is a polynomial in x. Ex. 1. Sum the series I 2 + 2 2 z + 3 2 z a + . . . + n*x n ~ l . Denoting the sum by s, we have
s ( I - x) = 1 + So: H- 5x* + . . , + (2 H - 1 ) x n - 1 - Let 5' =
SUM OF POWERS 111
.__
8. The Series 1 r + 2 r + 3 r -f ...-f n r . We shall write
and cr = n(n-f-l), a'
(1) Theorem. // r is a positive integer, S r can be expressed as a poly- nomial in n of which the highest term is n r + l /(r + 1 ).
For we can find a l9 a z , ... a r , independent of n, such that
n r ^a l n + a 2 n(n-l) + ... +a r n(n-l)(n~2) ... (n-r + 1), ... (A) where obviously a r = l. Writing n- 1, n-2, ... 1 in succession for n and adding, by Art. 3, we have
+ -a r (w + l)w(tt-l) ... (n-r + 1) ... (B)
The right-hand side, when expanded, is a polynomial in n of which the highest term is n r+1 . a r /(r + 1), that is, n r+l /(r -f- 1). #*. 1. Show that S^ln(n + l)(2n + l) 9 S 3 ^4n 2 (n + l) 2 . We have 7i 2 -n-f W(TI~ 1) ;
Also
(2) If /S r = 6 1 n + 6 2 n 2 -ffe 3 n 3 + ...-f6 r4 . 1 n r " f - 1 =/(n), the values of & r41 , 6 r ,
fe r _j, ... may be found as follows.
/(tt) = l r + 2 r + ...+n r , and /(n-l)-l r -f 2 r -f ... + (n-l) r , therefore w r =/(n) -/(n - 1 ) ,
so that n r = 6 r41 {w r+1 - (n - l) r+1 } + b r {n r - (n - l) r } + . . . + 6 r
Expanding and equating coefficients, it will be found that 7, J A 1 h r h n h r(r-l)(r-2)
6r+1 ""' r = ' ftr - 1 = ' 6r - 2= ' ftr - 8=
and so on. If r is large, b lf 6 2 , ... cannot be found conveniently in this wa y,
and we use other methods.
ODD AND EVEN POWERS
(3) The values of 8^ S 3 , S 5> ... , may be found in succession by consider- ing the function v n = n r (n + l) r -(n-l) r n r .
We have v 1 + v 2 + ...+v n =n r (n + l) r , and, expanding by the Binomial theorem,
Putting n - 1, n - 2, ... 1 in succession for n and adding,
_ 5 + ... , ... (A)
where a- n(n + 1). The last term is rS r+l or S r , according as r is even or odd.
When r = l, 2, 3, ... this formula gives
$3 =4
-87 =
Thus if r is odd and >1 and S r =f(n), then n 2 (n + 1) 2 is a factor of/(n), and the remaining factor is a rational integral function of n(n + l). Of
course, it does not follow that S r is arithmetically divisible by n z (n + 1) 2 .
(4) The values of S 2 , /S 4 , S 6 , ... , may be found as follows. Let w n = n r (n + l) r (2n + 1) - (n - l) r r (2n - 1) ;
then W 1 + w 2 + ... +w n =ri r (n + l) r (2n + l)=a r a, where a=n(n + l) and cr'
On expansion, we have
*2r-4 .
(B)
the last term being S r or (r + 2)S r+1 , according as r is even or odd. When r 1, 2, 3, ... this formula gives
r = l Jour' =3S 2 r = 2
r-3
r=5 |
' (3o* - lOa 8 + ITo 2 - 15a + 5).
BERNOULLI'S NUMBERS
113
Thus if r is even and $ r =/(n), then n(n + l)(2n + 1) is a factor of /(ft), and the remaining factor is a rational integral function of n(n + 1).
It will presently appear that the formula (B) is immediately deducible from (A) of the last section.
(5) Bernoulli's Numbers. We define B r as the coefficient of n in S r , when this is expressed as a polynomial in n.
Thus ^ = 2, and 7? 3 , B^ B ly ... are all zero; for S 1 =ln(n + l) and
$ 3 , $ 5 , 5 7 , ... all contain the factor n 2 (n-f I) 2 . The numbers B 2> -B 4 , B Q9
- # 8 , . . . are the famous numbers of Bernoulli, in terms of which a large number of functions can be expanded. For convenience we shall refer to