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Here putting n= 0, 1, 2...q-1, we get a different values, since 2ni is the smallest value of x for which the expansional value of all is 1. After this the values recur. Hence

If x=%; p and q being real integers prime to one another,

=

9

=

P

;

9 am has

9
different values; which form a geometrical progression,

2μπί whose common ratio is a qth root of unity, viz. e

(2) Let x be wholly imaginary = yi.
Then the expansional values of e

(dea +2nni)yi

are those of e iynea é e-2nya.

This gives a different value for every value of n. Hence, if x is wholly imaginary, aw has an infinite number of values, which form a geometrical progression whose common ratio is the expansional value of e

(3) Let a be complex = y + zi.
Then the expansional values of e

(dea +2nni)(y+zi)

are those of eenca pizdea ,—2nzt. e2nymi. The factor e

has an infinite number of values in geometric progression ; and the number of values of e2nywi is equal to the denominator of y.

2επί

- 2n ZT

:

644. To trace the changes in the expansional value of eai, where x is real.

Writing f (xi) = the expansional value of exi, we have

f (xi) xf (yi) = f {(x+y) i}. Let x (a) and iy (a) be the real and imaginary parts of f(xi). Then

{x (2) + 14 (x)} {x(y) + iy (y)}= x (x + y) + 2y (oC + y). Hence equating the real and imaginary terms

x (x + y) = x (x). (y)-4 ().(y) (1), y (x + y) = x (x): x(y) + x(x). + (y) .........(2).

.

Now since x(Qc) contains only even and y (x) only odd powers

of X,

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:: x(-x) = x (x) and (-x)=-4 ()

.(3), also

and
4 (0) = 0...

.(4), and x (x - y)= x (x). (y) + y (x). (y)

(5), y (ac y) = x (x) . x(y) x(2). ¥ (y)

.. (6). From (1), (2), (5), and (6) we have

x(y)-x(x) = 24 {1 (x+y)}. {1 (x - y)}. .....(7),

V (x) - (y) = 2x {} (x + y)}.4 {} (x - y)}.........(8). Putting in (5) y=x, we have 1 = {x (x)} + {v (x)}.

·(9). Hence x (oc) and y (a) are arithmetically not greater than 1,

Now f (2+i) = 1, .. f (xi + 27i) = f (xi); i.e. the values of f (xi) recur whenever x is increased by 27. Again, since

f (2mi) = 1, .: S (T2) = J1= + 1 or - 1. But f (tri) cannot be + 1, because, by definition, 21 is the smallest value of x for which f (xi)=1. :: f(ni)=-1; i.e. X (7)=-1 and v (T) = 0,

x 1 ,
.: x(-7)=-1 and ( - )= 0.

– 7
.: f(xi + mi) = -f(xi);
i.e. x(x + 7) = -x (x) and y (x + 7) = -4 (x).

.
:: X(T. - x) =-x(x) and v (- x) = 4 (2).
Again, since
f(T) = -1, :: f(i) = (-1) = + i or-i;

x(37) = 0 and ¥ (17) = +i or - i. We must show that, from x = 0 to x = 37, x (x) and y (x) are

X both positive. If x is a small positive quantity, x(x) and y (x) are clearly positive. Hence, being continuous, they cannot become negative without passing through the value zero.

If possible, let x (y) or y(y) = 0 for some value of x between O and . Then, since

=

so that

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i.e.

{x (y)} + {y (y)} = 1 and f (yi) = x (y) + iy (y), f (yi) would equal +1 or #i. Hence f (2ye) would be = 1 and

+

( f (4yi) would be 1. But f (xi) is not 1, until x = 27, .. neither 4 (y) nor x(y) can vanish if y is between 0 and 17. :: y(x) and x (x) are always positive from x = 0 to x = }.

4
:. f(ini) = + i; i.e. x(!) = 0 and 4 (17)=1.

.: x(-) = 0 and 4 (-1) = -1.
.. f (aci + žti) = if (xi);
x (x + 3) =- \ (oC) and y (x + ) = x (2),
:: x({ – 3) = ¥ (20) and y (17 – 2) = x (x).

1 :: x(47) =ų (IT) since {x (2)} + {y (x)} = 1: and so on.

12 Finally, by (7) and (8), as long as } (x + y) and } (x - y) lie between 0 and 3, 4 (x) must increase and x (2) must decrease as x increases.

Thus, between 0 and 17, V(x) continually increases from 0 to 1; and x () continually decreases from 1 to 0.

Hence the values of ų (x) and x (oc) cannot be repeated as x passes from 0 to 2.

To find a value for de (1+y), i.e. a root of the

645. equation

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е

+

+

=y;

The value of de (1 + y) has to satisfy the condition

{łe (1 + y)} {re (1 + y)}} 1. (1+y)+

2

B
3

Y
.. when de (1 + y)= 0, y = 0 and

1.

de (1 + y) If then de (1 + y) = co + C1y+ cuya + czy3 + ..., we must have co

:0, and

Ci 1; so that
1 (1 + y) = y + cya + czy + ... + cry" + ...

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...)

we have

=

a

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Writing f (c) for the series 1 + x + 3x + frites +

f (a) f (x) = f (x + x'); :: if a is a root of f(x) = 1 + y; and a' of f (x) = 1 +z; then a +a' is a root of f (c) = (1+y)(1 + 2) = 1 +y+(1+y) 2.

That is, we may write

1. (1+y)+1, (1 + x) = 1, {1+ y + (1 + y) }; :: 1 (1+y)+z+cqx2 + ... ={y +(1+y)2}+ ... +cr{y +(1+y) z}" +... + =);

; Equating the coefficients of , we have

1=1 + y + ... + rcr y^-*(1 + y) + (r + 1) Cr+1 Y" (1+y)+... Equating to zero the coefficient of y”, we have

e

+

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Putting r successively equal to 1, 2, 3... we have

(-1)" Cr+1

;

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:: 1 (1+y)=y- }y + y – 444 + ... which is a convergent series if mod. y is not greater than 1. [For, by Art. 596, mod. y" = (mod. y)". See also Arts. 616, 453, 445.]

We must here distinguish the cases according as y is wholly real, wholly imaginary, or complex.

(1) Let y be wholly real.
Then the series for 1 (1+y) is real;
:. y - 3y + 3y3 194 + the one real value of loge (1 + y).
(2) Let y be wholly imaginary = zi say.
Then de (1 + zi) - i (2 – 3:23 + 125 - ...) + 7 (zo – 12* + 328 - ...)
=i (2 – 3x3 + - ...) + } (1+z).

+11 (

1 + zi .:. 21 (3 – 4* + 3x3 - ...) = 21. (1 + zi) – de (1 + x^)=de ** 2i 2 )

1 1 + zi

is wholly imaginary. Now when z equals 0, the

1 series 0; hence as z varies from 0 to 1 or from 0 to - 1, the series changes continuously.

+

Thus de

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Now between – ja and + 37 there is only one value of a for which the expansion of eit has any given value.

1+zi Hence the series 2i (2 – 3x3 + 3x –...)= the value of de

1 which lies between - and + žt. :.de (1 + zi) = } the real value of loge (1+za)

1 + zi
+ the value of the which lies between - 1 and + .

1
(3) Let y be complex = y + zi say. Then

.

1 + zi/(1+y) A. (1 + y + 2) = x (1 + 2 + y + 2) + ,

1 – zi/(1+y)' which reduces to cases (2) and (3). 646. Hence we can find an expansion for 7.

1 + i

1 + i Thus f (i)=i=

..de 1 i

1-i

žti. Hence putting =1 in (2) of last article,

1

in = 1 - 3 +}-}+...
Other expansions for it may be found by observing that

1 + x + y + xy
λ, +λ.
1

1
y

1 - 2 y + xy Thus putting x = ji and y = ji, we have

1 + ži

1+ 3 +de

= mi. 1

-2

1

1 5

1 + x

1 + Y

=

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1 + i he

1-i

de

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1-i

1

647. The indeterminateness of the powers and logarithms of quantities, which arise from our introducing imaginary indices, gives rise to some difficulties in the management of equations.

Thus given that a=b, we have

The general values of aare the expansional values of (teb+2017), for the several integral values of n.

1b + 2nti And the general values of loge a are

1 + 2mai

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