161. Exponential Series. By the binomial theorem we may expand (1+4) (1) This is true for all values of x and n, provided n > 1. If n is not greater than 1 the series is not convergent; that is, the sum approaches no definite limit. The further discussion of convergency belongs to the domain of algebra. When x = 1 we have 1 2 to... () 2! 3! (1-7(1-2) n + X2 that is, et = 1 + x + + 2! 3! In particular, if x = 1 we have 1 1 2! 3! We cherefore see that we can compute the value of e 11.000000 by simply adding 1, 1, 1 of 1, $ of 1 of 1, and so on, 2 1.000000 3 0.500000 indefinitely, and that to compute the value to only a few 4 0.166667 decimal places is a very simple matter. We have merely 50.041667 to proceed as here shown. 6 0.008333 Here we take 1, 1, 1 of 1, $ of 4 of 1, 4 of } of f of 1, 7 0.001388 8 0.000198 and so on, and add them. The result given is correct 90.000025 to five decimal places. The result to ten decimal places 0.000003 is 2.7182818284. e = 2.71828. i sin x = ܕ6: COS X COS X = 162. Expansion of sin x, cos x, and tan x. Denote one radian by 1, and let cos 1 + i sin 1=k. Then cos x + i sin x =(cos 1 + i sin 1)* = kt, and, putting a for x, = k-x. That is, cos x + i sin x = and i sin x = k-3. By taking the sum and difference of these two equations, and dividing the sum by 2 and the difference by 2 i, we have 1 (K* + k--), 1 and sin x = (k* — k-r). ** (log k)? _ ** (log k) + 3! x2 (log k) 28 (log k) and e-clogk = 1- x log k + + 2! 3! 1 ** (log k), ** (log k) .. COS X = (k® + k-x)=1+ + + 2 2! 1 moo (log k) 8 206 (log k) and x log k + + + 3! 5! Dividing the last equation by x, we have 1 x2 (log k) ** (log k)' log k + + + 3! 5! 2 i 4! sin x = { ...}. sin x X log k. But remembering that a represents radians, it is evident that the smaller x is, the nearer sin x comes to equaling x; that is, the more nearly the sine equals the arc. Therefore the smaller a becomes, the wearer comes to 1, and the nearer the second member of the equation comes to log k. We therefore say that, as x approaches the limit 0, the limits of these two members are equal, and 1 = log k; whence log k = i, and kref. From the last two series we obtain, by division, 28 2 26 177 tan x = = x + + + 3 15 315 sin x COS X By the aid of these series, which rapidly converge, the trigonometric functions of any angle are readily calculated. In the computation it must be remembered that x is the circular measure of the given angle. Thus to compute cos 1, that is, the cosine of 1 radian or cos 57.29578°, or approximately cos 57.3o, we have 1 1 1 1 + 163. Euler's Formula. An important formula discovered in the eighteenth century by the Swiss mathematician Euler will now be considered. We have, as in $ 162, ix3 xt By multiplying by i in the formula for sin x, we have ix ixó ix? i sin x = ix + + 3! 5! 7! Adding, ixó cos x + i sin x = 1+ ix + + 2! 3! 4! 5! By substituting ix for in the formula for et, we see that 22x2 24x4 5. eix = 1+ ix + + + + + 2! 3! 4! 5! 22 ixs XC4 =1+ ix + + 2! 3! 4! 5! In other words, ex = cos x + i sin x. 2828 ixo or e 164. Deductions from Euler's Formula. Euler's formula is one of the most important formulas in all mathematics. From it several important deductions will now be made. Sin = cos x + i sin x, in which a may have any values, we may let x = 7. We then have ett = COS TT + i sin =- =-1+0, ein =-1. In this formula we have combined four of the most interesting numbers of mathematics, e(the natural base), i (the imaginary unit, V-1), # (the ratio of the circumference to the diameter), and – 1 (the negative unit). Furthermore, we see that a real number (e) may be affected by an imaginary exponent (in) and yet have the power real (-1). Taking the square root of each side of the equation ein =-1, we have TE V-1=i. Taking the logarithm of each side of the equation ein =-1, we have in = log (-1). Hence we see that - 1 has a logarithm, but that it is an imaginary number and is, therefore, not suitable for purposes of calculation. Since cos $ + i sin 0 = cos (2 km + $)+ i sin (2 km + o), we see that ex, which is equal to cos $ + i sin ø, may be written e(2 kn + ), or we nay write = cos $ + i sin • = cos (2 km + $)+ i sin (2 km + ) Hence (2 km + ) i = log[cos (2 km + )+ i sin (2 km + )]. If = 0, 2 Επί = log 1. If k = 0, this reduces to 0 = log 1. If k=1 we have 2 mi log 1; if k = 2, we have 4 ni = log 1, and so on. In other words, log 1 is multiple-valued, but only one of these values is real. If $ =, (2 km + T)i=(2k + 1) i = log (-1). Hence the logarithms of negative numbers are always imaginary; for if k = 0 we have ni = log (-1); if k=1 we have 3 ni = log (-1); and so on. If we wish to consider the logarithm of some number N, we have Ne2 kmi = N(cos 2 km + i sin 2 kt). log N + log 1 = log N. That is, log N = log N + 2 kni. Hence the logarithm of a number is the logarithm given by the tables, + 2 kmi. If k = 0 we have the usual logarithm, but for other values of k we have imaginaries. epi = e(2 km + Øli Exercise 83. Properties of Logarithms Prove the following properties of logarithms as given in $ 159, using b as the base : 1. Properties 1 and 2. 3. Property 4. 5. Property 6. 2. Property 3. 4. Property 5. 6. Property 7 Find the value of each of the followiny : 7. 5! 8. 7! 9. 6! 10. 8! 11. 10! Simplify the following : 10! 10! 7! 15! 20! 12. 13. 14. 15. 16. 17! 1 1 + + 2! 3! 1 1 1 18. Find to five decimal places the value of (2 + + + 2! 3! By the use of the series for co8 x find the following: 19. cost. 20. cos } 21. cos 2. 22. cos 0. 2 By the use of the series for sin x find the following: 26. sin 0. By the use of the series for tan x find the following: 27. tan 0. 28. tan 1. 29. tant. 30. tan 2. Prove the following statements : 31. eant = 1. 32. e = . 33. e* = V-1. 34. e= V1. TT 2 Given log 2 = 0.6931, find two logarithms (to the base e) of: 35. 2. 36. 4. 37. V2. 38. - 2. Given log 5 = 1.609, find three logarithms (to the base e) of: 39. 5. 40. 25. 41. 125. 42. 5. 46. V10. Given log 10 = 2.302585, find two logarithms (to the base e) of : 43. 100. 44. 10. 45. 1000. 47. From the series of 162 show that sin(-6)=- sin 0. 48. Prove that the ratio of the circumference of a circle to the diameter equals – 2 log ()=-2 i log i. |