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But as we have in column I three times 5 with the 2 annexed, or 152, we need only multiply the last figure, 2, by 3, and annex the third figure of the root, 6, which gives 1566, the first factor of the correction sought, or the second term in column I. Multiplying this number by the 6, we obtain 9396, the correction sought; adding the correction to the trial divisor, we have 820596, the complete divisor; multiplying the complete divisor by the 6, subtracting the product from the dividend, and bringing down the next period, we have 249150447 for a new dividend We may now regard the first three figures of the root, 526, as tens of the next inferior order, and proceed as before till the entire root, 5263, is extracted.

661. From these principles and illustrations we deduce the following

RULE. I. Point off the given number into periods of three figures each, counting from units place toward the left and right.

II. Find the greatest cube that does not exceed the left hand period, and write its root for the first figure in the required root; subtract the cube from the left hand period, and to the remainder bring down the next period for a dividend.

III. At the left of the dividend write three times the square of the first figure of the root, and annex two ciphers, for a trial divisor; divide the dividend by the trial divisor, and write the quotient for a trial figure in the root.

IV. Annex the trial figure to three times the former figure, and write the result in a column marked I, one line below the trial divisor, multiply this term by the trial figure, and write the product on the same line in a column marked II; add this term as a correction to the trial divisor, and the result will be the complete divisor.

V. Multiply the complete divisor by the trial figure; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.

VI. Add the square of the last figure of the root, the last term in column II, and the complete divisor together, and annex two ciphers, for a new trial divisor; with which obtain another trial figure in the root.

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VII. Multiply the unit figure of the last term in column I by 3, and annex the trial figure of the root for the next term of column I; multiply this result by the trial figure of the root for the next term of column II; add-this term to the trial divisor for a complete divisor, with which proceed as before.

Notes.-1. If at any time the product be greater than the dividend, diminish the trial figure of the root, and correct the erroneous work.

2. If a cipher occur in the root, annex two more ciphers to the trial divisor, and another period to the dividend; then proceed as before with column I, annexing both cipher and trial figure.

EXAMPLES FOR PRACTICE

1. What is the cube root of 389017?

Ans. 73. 2. What is the cube root of 44361864 ?

Ans. 354. 3. What is the cube root of 10460353203 ?

Ans. 2187. 4. What is the cube root of 98867482624 ? Ans. 4624 5. What is the cube root of 30.625 ? Ans. 3.12866 +. 6. What is the cube root of 111/? Ans 4.8076 + 7. What is the cube root of .000148877 ? Ans .053. Find the values of the following expressions. 8. 122615327232 ?

Ans. 4968. 9. V 134217728 ?

Ans. 8. 10. *393042?

Ans. 1156. 11. 3648

Ans. 33 12. How much does the sun of the cube roots of 50 and 31 exceed the cube root of their sum?

Ans. 2.4986 +.

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3119

CONTRACTED METHOD. 662. In applying contracted decimal division to the extraction of the cube root of numbers, we observe,

1st. For each new figure in the root; the terms in the operation extend to the right 3 places in the column of dividends, 2 places in the column of divisors, and 1 place in column I. Hence,

2d. If at any point in the operation we omit to bring down new periods in the dividend, we must shorten each sycceeding divisor 1 place, and each succeeding term in column I, 2 places.

1. What is the cube root of 189, correct to 8 decimal places ?

OPERATION.

ANALYSIS. We 15.73879355+, Ans. proceed by the usual 189.000000

method to extract I II 125

the cube root of the 7500 64 000

given number until 157 1099 8599 60 193

we have obtained 974700 3 807000

the three figures, 1713 5139 979839 2939517

5.73; the corres984987 867483*

ponding remainder 1719 1375 986362 789090

is 867483, and the

next trial divisor 98774 78393 17 12 98786 69150

with the ciphers

omitted is 984987. 9880

9243 8892

We now omit to

bring down a period 988

351
296

of ciphers, thus con

tracting the divid99

55 50

end 3 places; and

we contract the di10

5

visor an equal num5

ber of places by cmitting to annex the two ciphers, and regarding the right hand figure, 7, as a redundant figure. Then dividing, we obtain 8 for the next figure of the root. To complete the divisor, we obtain a correction, 1375, contracted 2 places by omitting to anrex the trial figure of the root, 8, to the first factor, 1719, and regarding the right hand figure, 9, as redundant in multiplying. Adding the contraction to the contracted divisor, we have the complete divisor, 986362, the right hand figure being redundant. Multiplying by 8 and subtracting the product from the dividend, we have 78393 for a new dividend. Then to form the new trial divisor, we disregard the square of the root figure, 8, because this square consists of the same orders of units as the two rejected places in the divisor; and we simply add the correction, 1375, and the complete divisor, 986362, and rejecting 1 figure, thus obtain 98774, of which the right hand figure, 4, is redundant. Dividing, we obtain 7 for the next root figure. Rejecting 2 places from the last term in column I, we have 17 for the next contracted term in this column. We then obtain, by the manner shown in the former step, the correction 12, the complete divisor, 98786, the product, 69150, and the new dividend, 9243. We then obtain the new trial

divisor, 9880; and as column I is terminated by rejecting the two places, 17, we continue the contracted division as in square root, and thus obtain the entire root, 5.73879355 £, which is correct to the last decimal place, and contains as many places as there are places in the periods used. Hence the following

RULE. I. If necessary, annex ciphers to the given number, and assume as many figures as there are places required in the root ; then proceed by the usual method until all the assumed figures have been employed

II. Form the next trial divisor as usual, but omit to annex the two ciphers, and reject one place in forming each subsequent trial divisor.

III In completing the contracted divisors, om it at first to annex the trial figure of the root to the term in column I, and reject 2 places in forming each succeeding term in this column.

IV. In multiplying, regard the right hand figure of each contracted term, in column I and in the column of divisors, as redundant.

Notes.-1. After the contraction commences, the square of the last root figure is disregarded in forming the new trial divisors.

2. Employ only full periods in the number.

EXAMPLES FOR PRACTICE.

1. Find the cube root of 24, correct to 7 decimal places.

Ans. 2.8844992 +. 2. Find the cube root of 12000.812161, correct to 9 decimal places.

Ans. 22.894801334 +. 3. Find the cube root of .171467, correct to 9 decimal places.

Ans. .555554730 +. 4. Find the cube root of 2.42999 correct to 5 decimal places.

Ans. 1.34442+. 5. Find the cube root of 19.44, correct to 4 decimal places.

Ans. 2.6888 +. 6. Find the value of $ to 6 places. Ans. .941035 £. 7. Find the value of V.571428 to 9 places.

Ans. .829826686 E.

8. Find the value of V 1.086743252 to 7 places.

Ans. 1.057023 £. 9. Find the value of 1.053 to 7 places.

Ans. 1.084715 +.

5

ROOTS OF ANY DEGREE.

663. Any root whatever may be extracted by means of the square and cube roots, as will be seen in the two cases which follow.

CASE I.

OPERATION.

664. When the index of the required root contains no othér factor than 2 or 3.

We have seen that if we raise any power of a given number to any required power, the result will be that power of the given number denoted by the product of the two indices, (643, III). Conversely, if we extract successively two or more roots of a given number, the result must be that root of the given number denoted by the product of the indices. 1. What is the 6th root of 2176782336 ?

ANALYSIS. The index of the 6 = 2 X 3 required root is 6

2 X 3; we * 2176782336 46657

therefore extract the square root V 46656

36, Ans. of the given number, and the

cube root of this result, and ob

tain 36, which must be the 6th 32176782336 1296

root required. Or, we first find 1296

36, Ans.

the cube root of the given num

ber, and then the square root of the result, as in the operation. Hence the following

RULE. Separate the index of the required root into its prime factors, and extract successively the roots indicated by the several factors obtained; the final result will be the required root.

Or,

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EXAMPLES FOR PRACTICE.

1. What is the 6th root of 6321363049 ? 2. What is the 4th root of 5636405776?

Ans. 43. Ans. 274.

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