Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small]

1.2.3.4.5 +....

Reverting the series, we obtain

so c=st

1.2
+ +

+
1.2.3 1.2.3.4 1.2.3.4.5
Restoring the value of s,

1
1
1

1
c=1+ + +
1:2 1.2.3 1.2.3.4

1 + By taking 12 terms of this series, we find the approximate value of c to be 1.7182818. But the base is 1+c; hence, adding 1 to the result, and representing the sum by e, the usual symbol for the Naperian base, we have

e= 2.7182818, which is the base of the Naperian system.

411. In the general formula, (A), the quantity M, which depends upon the base, is called the modulus of the system. Thus, the modulus of the Naperian system is unity.

Let us here designate Naperian logarithms by nap. log., and logarithms in any other system by log., simply. Then,

po pop*
log. (1+p)=

po
+
+

(1) 2

5

[ocr errors]
[ocr errors]

M =

p

po
+

(2) 2 3

5 Dividing (1) by (2), we obtain log. (1+p)

(3) nap. log. (1+p) or,

{nap. log. (1+P) XM = log. (1+p), , (4) where M is the modulus of the system in which the logarithm of the second member is taken. Hence,

The modulus of any particular system is the constant multiplier which will convert Naperian logarithms into the logarithms of that system.

412. Formula (A) can be employed for the computation of logarithms, only when p is less than unity; for if p be greater than unity, the series will be diverging. The series, however, may be transformed into another which will be always converging. Let us resume the logarithmic series,

p
log:(1+p) = M

P-
+

(1) 2 3 4 5 If in this equation we substitute -P

for

P, we shall have

p p* po M

(2) 2 3 4 5 If we subtract equation (2) from equation (1), observing that

[ocr errors]

P 1

+ -....).

[ocr errors]

...).

log:(1+p)-log:(1-P) = log.

we shall have

log. (1+0),

...). (8)

+

1+p

pe

po = 2M ( pt

+

+

3 5 7 1

1+p Assume P =

; whence we obtain

22+1 These values substituted in equation (3), give

z+1

1-P

[ocr errors]
[ocr errors]
[ocr errors]

+..).

1
1
1

1
2M
+

+

(4) 2z+1 3(2x+1) 5(2x+1) 7(2x+1)' The first member of this equation is equivalent to log. (2+1), log. 2. Hence, finally, we have og (z+1)-log.z= 1 1 1

1 2M + + +

+.

(B) 3(2x+1) 5(2z+1) 7(2x+1)' This series is rapidly converging, and may be employed with facility for the computation of logarithms, in the Naperian, or in the common system. To commence the construction of a table, first make z=

-1; then log. z = 0, and the formula will give the value of log. (2+1), or log. 2.

Next make z = 2; then the formula will give the value of log. (2+1), or log. 3 ; and so on.

(22 +1

+

..).

nap. log. 2

[ocr errors]
[ocr errors]

It is necessary to compute directly the logarithms of prime num bers only, in any system; for, according to (404, 3), the logarithm of any composite number may be obtained, by adding the logarithms of its several factors.

413. We will now illustrate the use of formula (B), by com. puting the Naperian logarithms of 2, 4, 5, and 10.

Make z=1; then nap. log. 2 = 0, and nap. log. (z+1)= nap log. 2; and since M=1, we have

1 1 1 2

+ 3.30

+ +

5:36 7.37 We first form a column of numbers, by dividing by 3', or 9, continually; then dividing the first of these members by 1, the second by 3, the third by 5, and so on, we obtain the several terms of the series.

32
90.66666666 ; 1= .66666666
9 7407407 ; 3 2469136
9 823045 ; 5

161609
9
91449 ; 7

13064
9
10161 ; 9

1129
9
1129 ; 11 =

103
9
125 : 13

10
14 - 15

1 .69314718

= nap. log. 2.

2 Whence, by (404, 5),

1.38629436 = nap. log. 4. Next make z=4 ; then z+1=5; and 2z+1=9; and we have

1 1

1

1 nap. log. 5=2

+
3.98

+
5.96

+ +...+nap. log. 4

7.97
912
9% = 81 0.22222222 : 1= .22222222
81 274318 ; 3=

91449
81 3387 ; 5

677
42 - 7 =

6
.22314354, sum of series

.

1.9

M м

M=

Το

.22314354 Add

nap. log. 4 = 1.38629436

1.60943790 = nap. log. 5. Add

nap. log. 2 = .69314718 Whence, by (404, 3),

2.30258508 = nap. log. 10. 414. In order to compute common logarithms, we must first determine the modulus of the common system.

From (411), equation (3), we have

log.(1+p)

nap. log:(1+p) In this equation, make 1+p = 10, the base of the common system. Then we have

1
= .43429448,

(1) 2.30258508 the value of the modulus sought. Substituting this value in formula (B), we obtain the formula for common logarithms, as follows:

log. (2+1)-log.z =
1
1
1

1
.8685
6858896

+
+

(0)

1) -1)' To apply this formula, assume z =

10; then
log. z=, and 2x+1= 21.

21 | .86858896
21 = 441.04136138 ; 1= .04136138
441 9379 ; 3=

3126
5

4

.04139268, sum of series. Add

log. z = 1.0

log. (2+1) = 1.04139268 = log. 11. If we make z=99, then z+1= 100, and 22+1 = 199. In this case, the formula will give the logarithm of 99; for, log. (z+1)-log. 2= log. 100—log. 99 = 2— log. 99.

199 .86858896 1999 = 39601 436477 ; 1= .00436477

11 ; 3 =

4 .00436481, sum of serios.

21

log. 99.

Therefore, we have

2-log. 99 = .00436181, whence,

1.99563519 = Subtract

log. 11 = 1.04139268

.95424251 = log. 9 And by (404, 6), log. 9 = 47712126 = log. 3.

Thus we may compute logarithms with great facility, using the formula for prime numbers only.

USE OF TABLES.

415. The following contracted tables will illustrate the princi. ples of logarithms, and the methods of using the larger tables. The logarithms are taken in the common system.

[blocks in formation]
« ΠροηγούμενηΣυνέχεια »