Imágenes de páginas



log. 9=1—(.04357,50878,58+.00218,24027,01)=


To find log. 101 and log. 99, we have from the preceding calculation, simply by changing the position of the decimal point,

[blocks in formation]


[blocks in formation]

log. 1001 = 3.00043,40775 log. 999 2.99956,54882

log. 101=2+(.00434,30895,93-.00002,17158,10)=



log. 99-2-(.00434,30895,93+.00002,17158,10)=

We employ twelve places of decimals in the computation, but retain only ten in the logarithms, which may therefore be relied upon as correct in the last place.

In the same manner we should find

[blocks in formation]

116. These logarithms prepared, we may proceed as follows. We have 1024-210, and log. 1024-10 log. 2; therefore in (53) d let n=1000, d=24, and =.024; then



10 log. 2= log. 1000+M[.024-(.024)3+ (.024)3 — &c.]

whence, by taking six terms of the series,

log. 2(3+.01029,99566,4)=0.30102,99956,64,

log. 3= log. ✓9= log. 9, log. 5= log. 10- log. 2.

To find log. 7, in (64) let p-9800-79.2.10°, p+1=9801=

119.34; then we have

2 log. 11+4 log. 3




2 M

-2 log. 7— log. 2— 2 log. 10=

[ocr errors]

whence we have

log. 7= log. 11+2 log. 3— log. 2—1— M(

Let p+1-8464

Here the second term of the series is less than



; the first term will therefore suffice in finding the logarithm to eleven or twelve places. The same degree of accuracy is obtained with the use of only one term of the series in the following examples.

In (64) let p+1=6656—2o.13, p=6655—5.113; then we have log. (p+1)=9 log. 2+ log. 13, log. p= log. 5+3 log. 11, and the formula gives

log. 13= log. 5+3 log. 11-9 log. 2+ Let p+1

14400-24.3.10%, p=14399-119.7.17; then

2M 28799

log. 17= log. 14400 — log. 847—( + &c.) Let p+1-5776-199.24, p=5775-; then log. 19=(log. 5775 — log. 16)+; -+ &c.


Let p+1-8281-7.13, p=8280-; then
log. 23
log. 8281 — log. 360 —(


+ &c.)


Let p+1=13225=5o.23o, p=13224-; then

log. 29 = log. 13225— log. 456—(2



[merged small][merged small][ocr errors]





2M 16927

+ &c.)

= 232.2*, p=84633.7.13.31; then

+ &c.)

+ &c.

+ &c.

Let p+1-5625=5*.3o, p=5424-23.19.37; then


log. 5625 — log. 152 — (11249 + &c.)

log. 37= log. 5625

Let p+1=6561=3o, p=6560=25.5.41; then


log. 41 — log. 6561 — log. 160—(13121


117. To continue the computation for primes above 41, we may employ Borda's formula, (65), without any transformations such as the above; and the first term will suffice in finding the logarithms to twelve places. Thus, to find log. 43 we make p=41, and find

log. 43=2(log. 42-log. 40)+log. 39+

In like manner

log. 47=2(log. 46-log. 44)+log. 43+

+ &c.) *

2M 34399

log. 53=2(log. 52 — log. 50)+log. 49+

2M 45495

2M 66249

+ &c.

+ &c.

M 21217

+ &c.

and with this formula we may compute the logarithms of all the prime numbers under 100.

Above 100 it will be more convenient to use (67), which requires only two logarithms to be known, and gives the logarithm correctly to twelve places with the use of only one term of the series. For example, when q=103, we have

log. 103=(log. 102+log. 104)+

Log. 104 is known before log. 103, since 104 is a composite number8.13. In general, q being a prime number, q-1, and q+1 must be composite, and their logarithms are found before that of q.

+ &c.

* These transformations may be obtained, by inspection, from a table of composite numbers.

118. Above 500 the formulæ of Art. 110 may be employed, and the series omitted entirely except in computing very extended tables. Thus, in (70), if n=500 it is evident that is less than


24 ; the n5 1012 series therefore will not affect the eleventh or twelfth place. Omitting the series, the formula may be put in the following form:

log. (n+3)= log. (n − 2) + 5 [log. (n + 2) — log. (n−1)]— 10[ log. (n+1)-log.n].

(74) For example, the common logarithm of 509 may be found from the five logarithms immediately preceding it in the table, by a computation like the following, where we put

D= log. (n+2)— log. (n—1), D'= log. (n+1)— log. n,

and the characteristics are omitted.

[merged small][ocr errors][ocr errors][merged small][ocr errors]


506 70415,05168,40 n+1 507 70500,79593,33 n+2 508 70586,37122,84 | log. (n+3)=70671,77823,40

10D' =


71529,22072,70 857,44249,30

When n=10000 in (68), we may omit the series, which will not affect the eleventh place of decimals; and we shall have

log. (n+2)= log. (n−1) — 3[ log. (n+1)— log. n.] (75) Or we may omit the series in (65), and we shall have

log. (p+2)= log. (p-2)+2[log. (p+1)—log. (p-1.)] (76) With these formulæ we may continue the computation as far as log. 100,000, which is the usual limit of the tables.

119. Above 100,000 we may use (67) without the series, if we require only ten places; we shall then compute by the formula

log. q = [ log. (q− 1) + log. (q+1)],,


from which it appears that the logarithm of any number above 100,000 is an arithmetical mean between the log. immediately preceding and that immediately succeeding it at least as far as the tenth place of decimals. With this formula, then, we may extend the table at pleasure.

120. By retaining the series in the formulæ, and using one, two, or more terms, we may compute logarithms to twenty or more places, as has been done by Briggs and others.

The variety of the formulæ here given will enable the computist to test the correctness of his work by occasionally computing the same logarithm by two or more methods; he will thus detect the amount of the error occasioned by the neglected terms of the series, and may operate in such a manner that this error shall not affect the last figure retained.

121. Formula (41) of Art. 91, enables us to compute the number corresponding to any given logarithm. We may thus obtain the approximate numbers corresponding to exact logarithms. These numbers have been called anti-logarithms, and tables of them are found convenient in some astronomical computations. In these tables the usual arrangement is reversed; the exact logarithms are placed first, increasing regularly by 1 from 1 to 10000, (or, as in Dodson's Anti-Logarithmic Canon, to 100,000,) and the corresponding nearest numbers in the columns opposite, with their differences and proportional parts.

[blocks in formation]


1+ M

[ocr errors][merged small]
[ocr errors]
[merged small][merged small][merged small][ocr errors]

In applying this to naperian logarithms we make M=1; we then have

m=n(1+D+2+23+ &c.)


[ocr errors]
« AnteriorContinuar »