APPENDIX C.
CUBE ROOTS.
The mode of extracting the roots of exact cubes which I taught the boys, and which was probably that adopted by Zerah Colbourn, will be best shown by an example. Suppose the question to be, What is the cube root of 596,947,688? This looks like a formidable array of figures, and a schoolboy, resorting to the usual mode of extracting the root, would fill his slate with figures, and perhaps occupy an hour in the process. Zerah Colbourn or my class would have solved the question in a minute, and without making any figures at all. My class would have proceeded as follows: They would first fix in their memories the number of millions (596) and the last figure of the cube (8), disregarding all other figures. Then, knowing the cubes of all numbers from 1 to 12 inclusive, they would at once see that the first or left-hand figure of the root must be 8; and deducting the cube of 8 (512) from 596, they would obtain a remainder of 84. This they would compare with the difference between the cube of 8 (512) and the cube of 9 (729), that is to say, with 217; and seeing that it was nearly four-tenths of such difference, they would conclude that the second figure of the root was 4. The third or last figure of the root would require no calculation, the terminal figure of an exact cube always indicating the terminal figure of its root—thus 8 gives 2. The cube root, therefore, is 842. In this process there is some risk of error as regards the second figure of the root, especially when the third figure is large; but with practice an expert calculator is able to pay due regard to that and certain other qualifications which I could not explain without making this note unduly long. As already stated, Zerah Colbourn did occasionally blunder in the second figure; and this circumstance assisted me in discovering the above process, which I have little doubt is the one he followed. If, instead of an exact cube, another number of nine figures be taken, the determination of the third figure of the root, instead of being the easiest, becomes by far the most difficult part of the calculation.
[This part of the explanation was written by Sir Rowland Hill, as a note to the Prefatory Memoir, before the year 1871. What follows was added in 1875.]
Rule for extracting the roots of imperfect cubes divisible into three periods:—
1. Find first and second figures as described above.
2. Deduct cube of first figure from the first period (of the number whose root is to be extracted), modified, if necessary, as hereafter described.
3. Then multiply the number (expressed by both figures) by each figure in succession, and by 3.
4. Deduct the product (or the significant figures thereof—see example), from the remainder obtained as above. (See 2).
5. Divide the remainder now obtained by the square of the number expressed by both figures (see 3), multiplied by 3—dropping insignificant figures (see example),—and the quotient will be the last figure (or 3rd figure) of the root.
I can confidently affirm from experience that there is nothing in the above calculations too difficult for those who, possessing a natural aptitude, are thoroughly well practised in mental arithmetic. I doubt, however, whether the mode just described be exactly that which we followed; our actual mode, looking at the results as described above (which is in exact accordance with my Journal), must, I think, have been more facile; but as it is fully fifty years since I gave any thought to the subject, and as, in the eightieth year of my age, I find my brain unequal to further investigation, I must be contented with the result at which I have arrived.
It must be remarked, however, that cases will arise when some modification of the process will be necessary. As, for instance, when the first period of the cube is comparatively light, it may be necessary to include therein one or more figures of the second period treated as decimals; indeed, if the first period consist of a single figure, it will be better to incorporate it with the second period, and treat both together as one period,[377] relative magnitude in the first period dealt with being important as a means of securing accuracy in the last figure of the root. But expert calculators soon learn to adopt necessary modifications, and by the “give-and-take” process to bring out the correct result. Indeed, I find it recorded in my Journal that “small errors will sometimes arise which, under unfavourable circumstances, will occasionally amount to a unit.” These observations it must be understood to apply only to the extraction of the roots of imperfect cubes, which Zerah Colbourn invariably refused to attempt. When the cube is perfect, the last figure of the root, as shown in the text, requires no calculation at all.
Example.
What is the cube root of 596,947,687?
[Note.—This is the number treated above, except that in the unit’s place 7 is substituted for 8, in order to render the number an imperfect cube; so slight a change, however—though rendering it necessary to calculate the last figure of the root,—will still leave the root as before.]
Following the rule, we find the first and second figures of the root in the manner described above. They are 8 and 4.
We next calculate the third or last figure of the root.
As the first figure of the second period of the cube is so large, it will be unsafe to disregard it. Call the first period, therefore, 596·9; all other figures may be neglected.
| 596·9 | mill. | ||
| (2) | 8³ = | 512 | ” |
| 84·9 | ” | ||
| (3) deduct 84 x 8 x 4 x 3 = (roughly) | 80·6 | ” | |
| (5) divide by 84² x 3 = (88 x 80 x 3)[378] = 2·1 | 4·3 | ||
| 2 | |||
| Quotient—2, which is the third or last figure of the root. | |||
[Note.—I have not encumbered the above figures with the ciphers which should accompany them, as, to the expert calculator, this will be needless.]
The root, therefore, is 842.
It is stated in the text that my pupils could extract the cube roots of numbers ranging as high as 2,000,000,000. In the ordinary mode this number would be divided, as above, into four periods; but my pupils treated the 2,000 as one period, the approximate root of which is of course 12, the cube of 12 being 1,728.