q Here putting n = 0, 1, 2...q-1, we get a different values, since 2πi is the smallest value of x for which the expansional value of e" is 1. After this the values recur. Hence If x= Ρ q ; p and q being real integers prime to one another, a has q different values; which form a geometrical progression, whose common ratio is a qth root of unity, viz. e e (2) Let x be wholly imaginary = yi. ψλια are those of e -2ny. This gives a different value for every value of n. Hence, if x is wholly imaginary, a has an infinite number of values, which form a geometrical progression whose common ratio is the expansional value of e (3) Let x be complex = y + zi. 2απί of values in geometric progression: and the number of values of 2nymi is equal to the denominator of y. 644. To trace the changes in the expansional value of ei, where x is real. Writing f(xi) the expansional value of exi, we have ƒ (xi) ׃ (yi)=ƒ {(x+y) i}. Let x(x) and it (x) be the real and imaginary parts of f(xi). Then {x (x) + i↓ (x)} {x (y) + i¥ (y)} = x (x + y) + i↓ (x + y). Hence equating the real and imaginary terms Now since x(x) contains only even and (x) only odd powers x (y) −x (x) = 24 {1 (x + y)} . ↓ {† (x − y)}....... .(7), 4(x) − 4 (y) = 2x {† (x + y)} • ↓ {} (x − y)}....... ..(8). Putting in (5) y=x, we have (9). 1 = {x (x)} + {(x)}2. Hence x(x) and 4 (x) are arithmetically not greater than 1. Now f(2) = 1, .. f (xi + 2mi) =f(xi); i.e. the values of f(xi) recur whenever x is increased by 2π. Again, since ƒ (2πi) = 1, .'. ƒ (πi) = √1 = + 1 or -1. But f(i) cannot be +1, because, by definition, 2′′ is the smallest value of x for which f (xi) = 1. (-)=0. i.e. x(x)=-x(x) and (x + π) = − ¥ (x). X Again, since so that ¥ ƒ (πi) = − 1, ·· ƒ (}πi) = √( − 1) = + i or − i; = We must show that, from x = 0 to x = π, x (x) and ↓ (x) are both positive. If x is a small positive quantity, x(x) and ↓ (x) are clearly positive. Hence, being continuous, they cannot become negative without passing through the value zero. If possible, let x (y) or ↓ (y) = 0 for some value of x between O and . Then, since {x (y)}2 + {v (y)}2 = 1 and ƒ (yi) = x (y) + i¥ (y), Hence f(2yi) would be f(yi) would equal ± 1 or ±i. 1 and f(4yi) would be 1. But f(xi) is not 1, until x = 2π, .. neither (y) nor x (y) can vanish if y is between 0 and 1π. .. (x) and x (x) are always positive from x = :. ƒ({πi) =+i; :: x(−1)=0 and 4 ( − 1π) = − 1. .. f(xi+i)= if (xi); i.e. x(x+1)=-(x) and (x + 1π) = × (x), Finally, by (7) and (8), as long as 1⁄2 (x + y) and 1⁄2 (x − y) lie between 0 and 1⁄2π, † (x) must increase and x(x) must decrease as x increases. Thus, between 0 and T, (x) continually increases from 0 to 1; and x(x) continually decreases from 1 to 0. Hence the values of (x) and x (x) cannot be repeated as x passes from 0 to π. 645. To find a value for λe (1+ y), i.e. a root of the equation The value of λ (1+ y) has to satisfy the condition e Writing f(x) for the series 1 + x + 1⁄2 x2 + 222x2 + f(x) × ƒ (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 ƒ (x) = (1 + y) (1 + 2) = 1 + y + (1 + y) z. That is, we may write λe (1 + y) + λe (1 + ≈) = λ, {1 + y + (1 + y) z} ; .. λe (1 + y) + ≈ + c222 + ... = {y + (1 + y) z} + ... + c, {y + (1+y) z} ̃ + ... z Equating the coefficients of z, we have 1 = 1 + y + ... + rc, yo−1 (1 + y) + (r + 1) Cr+1 y′′ (1+ y) + ... Equating to zero the coefficient of y", we have Putting r successively equal to 1, 2, 3... we have 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. Then λ (1 + zi) = i (≈ − }1⁄23 + 125 − ... ) + 1⁄2 (≈2 − 1≈4 + 3~6 — .....) 1 + zi .. 2i (≈ – †ï3 + ↓↓o — ...) = 2λ ̧ (1 + zi) − d ̧ (1 + 22) = d ̧ 1+22 1 + zi Thus e-zi series = 1 is wholly imaginary. Now when ≈ equals 0, the 0; hence as z varies from 0 to 1 or from 0 to 1, the series changes continuously. Now between - 1⁄2” and + 1⁄2π there is only one value of x for which the expansion of ei has any given value. Hence the series 2i (≈ – 1~3 + 1≈5 — ...) = the value of No e 1 + zi 1-zi (3) Let y be complex = y + zi say. Then λe (1 + y + zi) = 1⁄2λ, (1 + 2y + y2 + 22) + 1λe which reduces to cases (2) and (3). 1 + zi/(1+ y) 646. Hence we can find an expansion for π. 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. e Thus given that ab, we have The general values of a are the expansional values of (b+2n), for the several integral values of n. And the general values of log, a are λό + 2ηπί 1+2mπi' |