A TABLE OF SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS 8 1.4142136 1.259921 0.5 27 1.7320508 1 442250 0.3333333 125 2.2360680 1.709976 0.2 343 2.6457513 1.912931 0 1428571 1331 3.3166248 2 223980 0.0909091 2197 3.6055513 2351335 0.0769230 Cube, Square Root. Cube Root. 1 17 289 19 361 23 529 29 841 31 961 37 4913 41231056 2.571282 0.0588235 6859 4.3588989 2.668402 0.0526316 12167 4.7958315 2843867 0.0434783 24389 5.3851648 3072317 0.0344828 29791 5.5677644 3.141381 0.0322581 1369 50653 60827625 3.332222 0.0270270 68921 6.4031242 3.448217 0.0243902 41 1681 43 1849 79507 6.5574385 3.503398 0.0232558 47 2209 103823 6.8556546 3.608826 0.0212766 53 2809 148877 7.2801099 3756286 0.0188679 59 3481 205379 7.6811457 3.892996 0.0169492 61- 3721 226981 7.8102497 3.936497 0.0163934 67 4489 300763 8.1853528 4.061548 0.0149254 71 5041 357911 8.4261498 4 140818 0.0140845 73 5329 389017 8.5440037 4 179339 0.0136986 79 6241 493039 8.8881944 4 290840 0.0126582 83 6889 571787 9-1104336 4·362071 0.0120482 89 7921 704969 9.4339811 4 464745 00112360 97 9409 912673 9.8488578 4·594701 0.0103093 101 10201 1030301 10 0498756 4 657009 0.0099010 103 10609 1092727 10 1488916 4 687548 0.0097087 107 11449 1225043 10-3440804 4747459 0.0093458 109 11881 1295029 10-4403065 4 776856 0.0091743 113 12769 1442897 10-6301458 4.834588 0.0088496 By the aid of the foregoing table we may by multiplication determine the square root, cube root, and reciprocal of any number which does not contain a prime factor greater than 113. Suppose we wish the square root of 365, we decompose 365 into the prime factors 5 and 73. Now, it is obvious that the root of the product of any number of factors is equal to the product of their roots, hence, the square root of 365 is equal to the square root of 5 × 73, which is equal to the square root of 5 multiplied by the square root of 73. By our table we find √365=19.1049731 nearly. For the cube root of the same number, we have the following work : 73 =4179339 5 =1.709976 4.179339 2925537 37614 3761 292 25 365=7146568 nearly. For the reciprocal of 365, we have as follows: reciprocal of 73=0.0136986 66 5=0.2 0.00273972=reciprocal of 365. This last work for finding the reciprocal of a composite number, reposes upon the principle, that the reciprocal of the product of any number of factors is equal to the product of their reciprocals. The roots of composites are very readily obtained when the number can be separated into two factors, one of which can have its root accurately found: thus, the square root of 5537=49×113, is equal to 7 times the square root of 113=10-6301458×7=74·4110206 nearly. The cube root of 664= the cube root of 8 times 83= 2 times the cube root 83 = 2 times 4·362071=9-724142 nearly. LXXXIII. This rule for finding the sum of all the terms of an infinite decreasing geometrical progression, has a beautiful application in the following problem, which is sometimes so stated as to appear paradoxical. At what time next after 1 o'clock will the hour-hand of a clock overtake the minute-hand? The distance which the hour-hand is forward of the minute-hand, is at first of the circumference of the dial plate. When the minute-hand shall have moved over this, the hour-hand will have advanced of it, or of of the circumference, which is the distance which now separates them. Again, while the minutehand moves over this last distance of of of the cucumference, the hour-hand moves over of it, that is, 1 1 1 1 of of 12 12 12123' which is the distance which now separates them. Again while the minute-hand moves over this last distance, the hour-hand advances of it, or over a distance = of of of of the circumference, which is the distance which now separates them. Hence, as the hour-hand always moves as far as the distance last moved by the minute-hand, it follows that the minute-hand cannot overtake the hour-hand. But we know that it does overtake it, inasmuch as it completes the circuit of the dial plate, and therefore must of necessity pass it. How then is this seeming contradiction to be explained? Although we have assumed an infinite number of partial movements made by the hands, each succeeding move being of the preceding one, still the time required to make this infinite number of moves is finite. These successive moves made by the minute-hand, considering the circumference of the dial plate as the unit, form this progression, This being a decreasing geometrical progression, whose ratio as well as first term is, its sum is found to be TT. Now, as the minute-hand can move through the whole circumference of the dial plate in one hour, it will require of one hour, which equals 5 minutes and TT of a minute to move over of the circumference of the dial plate, which is the exact time required for the minute-hand to overtake the hour-hand. From which it appears that the foregoing fallacy consists in inferring that, because there was an infinite number of successive operations, it must require an infinite length of time to perform them, which, as we have seen, is not a legitimate conclusion. Question 67, page 282, affords a second example of an infinite number of successive operations being performed in a finite time. For the number of bounds made by this elastic ball, provided it move in accordance with the law mentioned in the question, is infinite, since every time it falls it must bound, and every time it bounds it must of necessity fall. We have already given 300 feet as the whole distance through which the ball moved. We will now show that the time required is finite. The successive distances, expressed in feet, through which the ball fell, form the series 100, 50, 25, 12, &c., to an infinite number of terms. By the law of falling bodies, the distances fallen through are to each other as the squares of the times employed in falling; and 16 feet is the distance which a body falls in the first second of its descent; consequently, the times, expressed in seconds, which were required to fall through the respective distances which are given in the above series, form the √√2. Therefore its sum, as found by rule under 93 Article 83, is 10√금을 1-√2 Again, the distances passed through in bounding form |