Appendix Xi.

ON HORNER’S METHOD OF SOLVING EQUATIONS.

The rule given in this chapter is inserted on account of its excellence as an exercise in computation. The examples chosen will require but little use of algebraical signs, that they may be understood by those who know no more of algebra than is contained in the present work.

To solve an equation such as

2x⁴ + x² - 3x = 416793,

or, as it is usually written,

2x⁴ + x² - 3x - 416793 = 0,

we must first ascertain by trial not only the first figure of the root, but also the denomination of it: if it be a 2, for instance, we must know whether it be 2, or 20, or 200, &c., or ·2, or ·02, or ·002, &c. This must be found by trial; and the shortest way of making the trial is as follows: Write the expression in its complete form. In the preceding case the form is not complete, and the complete form is

2x⁴ + 0x³ + 1x² - 3x - 416793.

To find what this is when x is any number, for instance, 3000, the best way is to take the first multiplier (2), multiply it by 3000, and take in the next multiplier (0), multiply the result by 3000, and take in the next multiplier (1), and so on to the end, as follows:

2 × 3000 + 0 = 6000; 6000 × 3000 + 1 = 18000001

18000001 × 3000 - 3 = 54000002997

54000002997 × 3000 - 416793 = 162000008574207

Now try the value of the above when x = 30. We have then, for the steps, 60 (2 × 30 + 0), 1801, 54027, and lastly,

1620810-416793,

or x = 30 makes the first terms greater than 416793. Now try x = 20 which gives 40, 801, 16017, and lastly,

320340-416793,

or x = 20 makes the first terms less than 416793. Between 20 and 30, then, must be a value of x which makes 2x⁴ + x²-3x equal to 416793. And this is the preliminary step of the process.

Having got thus far, write down the coefficients +2, 0, +1,-3, and -416793, each with its proper algebraical sign, except the last, in which let the sign be changed. This is the most convenient way when the last sign is-. But if the last sign be +, it may be more convenient to let it stand, and change all which come before. Thus, in solving x³-12x + 1 = 0, we might write

-1 0 +12 1

whereas in the instance before us, we write

+2 0 +1 -3 416793

Having done this, take the highest figure of the root, properly named, which is 2 tens, or 20. Begin with the first column, multiply by 20, and join it to the number in the next column; multiply that by 20, and join it to the number in the next column; and so on. But when you come to the last column, subtract the product which comes out of the preceding column, or join it to the last column after changing its sign. When this has been done, repeat the process with the numbers which now stand in the columns, omitting the last, that is, the subtracting step; then repeat it again, going only as far as the last column but two, and so on, until the columns present a set of rows of the following appearance:

a b c d e f g h i k l m n o p

to the formation of which the following is the key:

f = 20a + b, g = 20f + c, h = 20g + d, i = e - 20h, k = 20a + f, l = 20k + g, m = 20l + h, n = 20a + k, o = 20n + l, p = 20a + n.

We call this Horner’s Process, from the name of its inventor. The result is as follows:

2 0 1 -3 416793 (20 40 801 16017 96453 80 2401 64037 120 4801 160

We have now before us the row

2 160 4801 64037 96453

which furnishes our means of guessing at the next, or units’ figure of the root.

Call the last column the dividend, the last but one the divisor, and all that come before antecedents. See how often the dividend contains the divisor; this gives the guess at the next figure. The guess is a true one, if, on applying Horner’s process, the divisor result, augmented as it is by the antecedent processes, still go as many times in the dividend. For example, in the case before us, 96453 contains 64037 once; let 1 be put on its trial. Horner’s process is found to succeed, and we have for the second process,

2 160 4801 64037 96453 162 4963 69000 27453 164 5127 74127 166 5293 168

As soon as we come to the fractional portion of the root, the process assumes a more methodical form.

The equation being of the fourth degree, annex four ciphers to the dividend, three to the divisor, two to the antecedent, and one to the previous antecedent, leaving the first column as it is; then find the new figure by the dividend and divisor, as before, and apply Horner’s process. Annex ciphers to the results, as before, and proceed in the same way. The annexing of the ciphers prevents our having any thing to do with decimal points, and enables us to use the quotient-figures without paying any attention to their local values. The following exhibits the whole process from the beginning, carried as far as it is here intended to go before beginning the contraction, which will give more figures, as in the rule for the square root. The following, then, is the process as far as one decimal place:

Various exceptions may arise when an equation has two nearly equal roots. But I do not here introduce algebraical difficulties; and a student might give himself a hundred examples, taken at hazard, without much chance of lighting upon one which gives any difficulty.

This form might be also applied to the integer portions; but it is hardly needed in such instances as usually occur. See the article Involution and Evolution in the Supplement to the Penny Cyclopædia.

After the second step, the trial will rarely fail to give the true figure.

2 0 1 -3 416793(213 40 801 16017 96453 80 2401 64037 ----- 120 4801 ----- 274530000 160 ---- 69000 47339778 --- 4963 74127000 --------- 162 5127 -------- 164 529300 75730074 166 ------ 77348376 1680 534358 ---- 539434 1686 544528 1692 1698 1704 ----

If we now begin the contraction, it is good to know beforehand on what number of additional root-figures we may reckon. We may be pretty certain of having nearly as many as there are figures in the divisor when we begin to contract--one less, or at least two less. Thus, there being now eight figures in the divisor, we may conclude that the contraction will give us at least six more figures. To begin the contraction, let the dividend stand, cut off one figure from the divisor, two from the column before that, three from the one before that, and so on. Thus, our contraction begins with

| | | | |0002 1|704 5445|28 7734837|6 47339778 | | | |

The first column is rendered quite useless here. Conduct the process as before, using only the figures which are not cut off. But it will be better to go as far as the first figure cut off, carrying from the second figure cut off. We shall then have as follows:

| | | 1|704 5445|28 7734837|6 47339778(6 | 5455|5 7767570|6 734354 5465|7 7800364|8 5475|9 | |

At the next contraction the column 1|704 becomes |001704, and is quite useless. The next step, separately written (which is not, however, necessary in working), is

| | 54|759 780036|48 734354(0 | |

Here the dividend 734354 does not contain the divisor 780036, and we, therefore, write 0 as a root figure and make another contraction, or begin with

| | |54759 78003|648 734354(9 | 78008|5 32277 78013|4 |

At the next contraction the first column becomes |0054759, and is quite useless, so that the remainder of the process is the contracted division.

| 7801|34)32277(4137 | 1072 292 58 3

and the root required is 21·36094137.

I now write down the complete process for another equation, one root of which lies between 3 and 4: it is

x³ - 10x + 1 = 0

1 0 -10 -1(3·1110390520730990796 3 -1 2000 6 1700 209000 9 0 1791 19769000 9 1 188300 743369000000 9 2 189231 172311710273000 9 30 19016300 991247447681 9 31 19025631 39462875420 9 32 1903496300 0 0 1391491559 9 33 0 1903524299 0 9 58993123 9 33 1 1903552298 2 7 0 0 1886047 9 33 2 1903560698 0 5|9|1 172835 9 33 30 0 1903569097 8 5|6|3 1515 9 33 30 3 1903569144 5 2|2| 183 9 33 30 6 1903569191 1|8|8 12 9 33 30 90 1903569193 0|6| 1 9 33|30|99 1903569194|9|3| 9 33|31|08 | | | |09|33|31|17 | | | |

The student need not repeat the rows of figures so far as they come under one another: thus, it is not necessary to repeat 190356. But he must use his own discretion as to how much it would be safe for him to omit. I have set down the whole process here as a guide.

The following examples will serve for exercise:

1. 2x³ - 100x - 7 = 0 x = 7·10581133.

2. x⁴ + x³ + x² + x = 6000 x = 8·531437726.

3. x³ + 3x² - 4x - 10 = 0 x = 1·895694916504.

4. x³ + 100x² - 5x - 2173 = 0 x = 4·582246071058464. _ 5. ∛2 = 1·259921049894873164767210607278.

6. x³ - 6x = 100 x = 5·071351748731.

7. x³ + 2x² + 3x = 300 x = 5·95525967122398.

8. x³ + x = 1000 x = 9·96666679.

9. 27000x³ + 27000x = 26999999 x = 9·9666666.....

10. x³ - 6x = 100 x = 5·0713517487.

11. x⁵ - 4x⁴ + 7x³ - 863 = 0 x = 4·5195507.

12. x³ - 20x + 8 = 0 x = 4·66003769300087278.

13. x³ + x² + x - 10 = 0 x = 1·737370233.

14. x³ - 46x² - 36x + 18 = 0 x = 46·7616301847, or x = ·3471623192.

15. x³ + 46x² - 36x - 18 = 0 x = 1·1087925037.

16. 8991x³ - 162838x² + 746271x - 81000 = 0 x = ·111222333444555....

17. 729x³ - 486x² + 99x - 6 = 0 x = ·1111..., or ·2222..., or ·3333....

18. 2x³ + 3x² - 4x = 500 x = 5·93481796231515279.

19. x³ + 2x² + x - 150 = 0 x = 4·6684090145541983253742991201705899.

20. x³ + x = x² + 500 x = 8·240963558144858526963.

21. x³ + 2x² + 3x - 10000 = 0 x = 20·852905526009.

22. x⁵ - 4x - 2000 = 0 x = 4·581400362.

23. 10x³ - 33x² - 11x - 100 = 0 x = 4·146797808584278785.

24. x⁴ + x³ + x² + x = 127694 x = 18·64482373095.

25. 10x³ + 11x² + 12x = 100000 x = 21·1655995554508805.

26. x³ + x = 13 x = 2·209753301208849.

27. x³ + x² - 4x - 1600 = 0 x = 11·482837157.

28. x³ - 2x = 5 x = 2·094551481542326591482386540579302963857306105628239.

29. x⁴ - 80x³ + 24x² - 6x - 80379639 = 0 x = 123.

30. x³ - 242x² - 6315x + 2577096 = 0 x = 123.

31. 2x⁴ - 3x³ + 6x - 8 = 0 x = 1·414213562373095048803.

32. x⁴ - 19x³ + 132x² - 302x + 200 = 0 x = 1·02804, or 4, or 6·57653, or 7·39543.

33. 7x⁴ - 11x³ + 6x² + 5x = 215 x = 2·70648049385791.

34. 7x⁵ + 6x⁴ + 5x³ + 4x² + 3x = 11 x = ·770768819622658522379296505.

35. 4x⁶ + 7x⁵ + 9x⁴ + 6x³ + 5x² + 3x = 792 x = 2·0520421768796053652140434012812019734602755995 45541724214.

36. 2187x⁴ - 2430x³ + 945x² - 150x + 8 = 0 x = ·1111...., or ·2222...., or ·3333...., or ·4444....

The solution of x³ + 0x² + 0x-2 = 0.

Taken from a paper on the subject, by Mr. Peter Gray, in the Mechanics’ Magazine.

Taken from a paper on the subject, by Mr. Peter Gray, in the Mechanics’ Magazine.

Taken from a paper on the subject, by Mr. Peter Gray, in the Mechanics’ Magazine.

Taken from the late Mr. Peter Nicholson’s Essay on Involution and Evolution.

Taken from the late Mr. Peter Nicholson’s Essay on Involution and Evolution.

Taken from the late Mr. Peter Nicholson’s Essay on Involution and Evolution.

Taken from the late Mr. Peter Nicholson’s Essay on Involution and Evolution.