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the hyperbolic logarithm of N, by article 8. This series, however, if x be a whole number, does not converge.

1 Let M be a whole number, and M= x is less than 1. For, multiplying both sides of the equation by 1-X, we have M-Mx=1, and therefore 1

1 1-M=x. Now, let M=i =1+x\P. Then we have

1 1+as

P=14x
=

1(

(by putting r= ) 1-X = 1—- rx +rx x x2 —4 x x x x3 + &c. But for the same reasons as above, r must be indefinitely small, and therefore may be rejected from the factors 7--1, r--2, r--3, &c. Consequently, taking i from each side of the above equation, a=-rx

2 rx4

&c. But --r= }, and therefore, dividing 4 5 the left hand side of the equation by 5, and the other

x2

X5 by-r, we have ap=x+ + + + + &c. = the

2 3

5 hyperbolic logarithm of M. 11. As, by the last article, the hyperbolic logarithm

22

x4 x5 xo x7 of N or 1+x is x +

+

&c. 2 3

5 6 7

rx?

r23

3

rx5

23

X4

4

x 3

+

4

1

is x +

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and as the hyperbolic logarithm of M or x2

23 25 xo x7 2

+
+
+ +

+ &c. the hyperbolic log3 4 5 6 7

1 + x arithm of NX M, or is equal to the sum of these

2x3 2.35 2x7 two series, that is, equal to 2x + + + + &c.

3 5 This series converges faster than either of the preceding, and its value may be expressed thus: 2 x (x +

x7
+
-3
5+

7
+ &c.).

n+1
12. The logarithm of = 2 x logarithm of

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n

2

2

2

2

2

2

2

2
Х

2

2

n? +n

n

212-42

2n + 1 + the logarithm of

For, as the 2n +1

2n+1)

-1° addition of logarithms answers to the multiplication of the numbers to which they belong, the logarithm of the square of any number, is the logarithm of the number multiplied by 2. Hence the 2n + 2

2n+2. logarithm of is 2 x logarithm of But 2n +1

2n + 1 2n+2) 2n +1 2n+21 4n? + en + 4

2n+1) 2n +11 -1 =

4no +4n n? + 2n +1 n+1xnx1

n+1.

al X n +1 From the preceding articles, hyperbolic logarithms may be calculated, as in the following examples. Example 1. Required the hyperbolic logarithm of n+1

2n+2 4 2. Put = 2, and then n=1,

3' 2n+19

9 8

In order to proceed by the series in 2n + 1 1+x 4

1 article 11, let

and then x=
1-X
'

x=3 Consequently, X =0.14285714286

=0.00097181730.
3
205

=0.00001189980
5

n

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=0.00000017347 7

=0.00000000275
9
x11

=0.00000000004
11

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1 + x

2n+2 4 Log. of

or

or 0.28768207244 1 -X 2n + 1 The double of which is 0.57536414488, and answers to the first part of the expression in article 12.

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2

9 or -1

8

2

2n +11 Log. of

- 0.11778303566 2n +11 which answers to the second part of the expression in article 12. Consequently the hyperbolic logarithm of the number 2 is 0.57536414488 +0.11778303566= 0.69314718054.

The hyperbolic logarithm of 2 being thus found, that of 4, 8, 16, and all the other powers of 2 may be obtained by multiplying the logarithm of 2, by 2, 3, 4, &c. respectively, as is evident from the properties of logarithms stated in article 6. Thus by multiplication, the hyperbolic logarithm of 4=1.38629436108

of 8=2.07944154162

&c. From the above, the logarithm of 3 may easily be obtained. 4

3 = 4 x 3; and therefore as the logarithm 4 of was determined above, and also the logarithm of 4, 3 From the logarithm of 4, viz. - - 1.38629436108,

4 Subtract the logarithm of 3

viz. -- 0.28768207244,

For 4 •

of

And the logarithm of 3.is ---- 1.09861228864. Having found the logarithms of 2 and 3, we can find, by addition only, the logarithms of all the powers of 2

and 3, and also the logarithms of all the numbers which can be produced by multiplication from 2 and 3. Thus,

To the logarithm of 3, viz. 1.09861228864
Add the logarithm of 2, viz. --- 0.693 14718054

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ac

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And the sum is the logarithm of 6 - 1.79175946918 To this last found add the logarithm of 2, and the sum 2.48490664972 is the logarithm of 12.

The hyperbolic logarithms of otber prime numbers may be more readily calculated by attending to the following article.

13. Let a, b, c, be three numbers in arithmetical progression, whose common difference is 1. Let 6 be the prime number, whose logarithm is sought, and a and c even numbers whose logarithms are known, or easily obtained from others already computed. Then, a being the least of the three, and the common difference being 1, a =6–1, and c=b+i. Consequently a xc=b-1x6+1 =b-1, and ac+1=b; and therefore

This is a general expression for the fraction which it will be

1 + x proper to put =

that the series expressing the hyperbolic logarithm may converge quickly. For as 1+x ac+1

, ac + acx=ac+1 - acx - X, and therefore 1-X

1 2acx +x=,

and x =

2act 1 Example 2. Required the hyperbolic logarithm of 5.

1 Here a=4,c=6, and x=

Consequently, 2ac +1

49
x=0.0204081632
2013

=0.0000028332
2015

=0.0000000007 5

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ac

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1

3

Sum of the above terms

0.0204109971

2

Log. of

1+x 25

or
1-X 24

-- 0.0408219942

25 But x 8 x 3=25, and the addition of logarithms

24 answers to the multiplication of the natural numbers to which they belong. Consequently,

25 To the log. of

0.0408219942 Add the log. of 8

2.0794415422 And also the log. of 3 1.0986122890

24

97,

And the sum is the log. of 25-3.2188758254 The half of this, viz. 1.6094379127, is the hyperbolic logarithm of 5; for 5 x 5=25. Example 3. Required the hyperbolic logarithm of 7.

1 1 Here a=6, c=8, and x=

ac +1

1+1 and" 2ac + 1

ac 1-7 49

Consequently, 48

x =0.01030927835
2009

=0.00000036522
3
x5

=0.00000000002
5

Sum of the terms

0.01030964359

2

49

Log. of

- 0.02061928718

48

To which add log. of 8 --- 2.07944154162
And also log. of 6

1.79175946918

The sum is the log. of 49 - 3.89182029798

49 For

x6x8=49. Consequently the half of this, viz.

48 1.94591014899, is the hyperbolic logarithm of 7; for 7x7=49.

If the reader perfectly understand the investigations and examples already given, he will find no difficulty in calculating the hyperbolic logarithms of higher prime numbers. It will only be necessary for him, in order to guard against any embarrassment, to compute them as they advance in succession above those already men

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