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arithms are so very useful in calculations of the highest importance.

Le 8. If the indices of the powers of l+a, be multiplied by a, the products are called the hyperbolic logarithms of the numbers equal to the powers of 1+a. Thus, if the number N be equal to 1+al", then na is the hyperbolic logarithm of N; and if the number M be equal to 1+am, then ma is the hyperbolic logarithm of M. Hyperbolic logarithms are not those in common use, but they can be calculated with less labour than any other kind, and common logarithms are obtained from them.

9. If successive powers of a very small fraction be raised, they will successively be less and less in value. This truth appears most evident by putting the value in the form of a vulgar fraction. Thus

2

1

1000001

3

1

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n

m

-1

2

1 10000000000 100000 1000000000000000

&c. 10. Let it be required to determine the hyperbolic logarithm L, of any number N. Using the same notation as in the preceding articles, 1+a" = N, and by extracting the nth root of each side of the equation, 1+a=N. Put m=1, and 1+x=N, and then N7

N1+x)=(by the binomial theorem) 1 tmx +mex

m-1

m-2 X x2 +mx X x x + &c. = 1+a. Now, as

3 a is indefinitely small, the power of 1+a, which is equal to the number N, must be indefinitely high; or, which is the same thing, n must be indefinitely great. Consequently m must be indefinitely small, and therefore may be rejected from the expressions m—1, m-2, m-3, &c. Hence 1 being taken from each side of the

mx3 mx4 above equation, we have a = mx

2

mx?

+

+

2

3

mx 5

-&c. Each side of this equation being divided by

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24

25 + -&c. But m=",

= 3 4 5

x2 x3 X5 anx +

+

&c. =L, 2 3

5

x4

a and therefore

10 m

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1

1

M=x. Now, let M=

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Х

2

3

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3

4

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

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

1

1+0p. Then we have

1-X 1. 1+a=

-X. p=12* (by putting r=-) - XP -rx + rxx 32

Т Х

x 23+ &c. But for the same reasons as above, r must be indefinitely small, and therefore may be rejected from the factors r--1, -2, -3, &c. Consequently, taking 1 from

rr2 each side of the above equation, a=--12–

2 rx4 r25 &c. But

and therefore, dividing the left hand side of the equation by, and the other

22:23

15 by -r, we have ap=x+

+
+

+ &c. = the 3

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

x2 xt x xo x7 of N or 1+x is x +

&c. 5 6tig

1 and as the hyperbolic logarithm of M or ..22 23. 24 x5 x6 x7 + 3

+

+4 + 2

+ &c. the hyberbolic log5 6 7

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

1

2.3 2.75 2.27 two series, that is, equal to 2x + + + + &c.

3 5 7 This series converges faster than either of the preceding, and its value may be expressed thus: 2 x (x + x3 x5

X7

+ &c.). 3

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

24

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+

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2

3

4

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2

2

2

2n+21

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Х.

2

-1

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2n + 2

2n + 1 + the logarithm of

For, as the 2n +1

-1°

2n+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+21"

4n? t 8n + 4 2n+1)

2n +1)
* 2n +1

4n2 + 4n na + 2n +1_n+1 xnx1 _n+1.

nxn+1 From the preceding articles, hyperbolic logarithms may be calculated, as in the following examples. Example 1. Required the hyperbolic logarithm of nt1

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

and 2n + 1

3 2n + 1)" 9

In order to proceed by the series in 212 to 1

1+x 4 article 11, let

and then x=

7
2=0.14285714286
23

=0.00097181730
3

N2 +n

n

n

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8

co

[blocks in formation]

1 +*

2n + 2 4 Log. of

or

or 0.28768207244 1

2n +1 3 The double of which is 0.57536414488, and answers to the first part of the expression in article 12.

1

1+9 Secondly, let and then 8+ 8x9-94, and

8

bogubose 7739 x now is equal to 17

Consequently,
30 =0.0588235294 I

=0.00006784721
3
5

=0.00000014086
5
27

=0.00000000035
7

3

Sum of the aboye terms

0.05889151783

2

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9 or

8

2n +11 Log. of

0.11778303566 2n +12 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 3

4

For4

of

4

was determined above, and also the logarithm of 4,

From the logarithm of 4, viz. -- 1.38629436108,

Subtract the logarithm of, viz. -- 0.28768207244,

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,

Pok
To the logarithm of 3, viz. --- 1.09861228864
Add the logarithm of 2, viz. --- 0.69314718054

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 other 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 a 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+1. Consequently a xo=6–1*6+1 = 12-1, and ac+1=b'; and therefore

62_ac + 1

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

ас

ac

1-X

= 1-X

ac

1.

1 2acx +x=1, and x=

2ac + 1 Example 2. Required the hyperbolic logarithm of 5.

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

Consequently, Zac +149

x=0.0204081632 ve

2013

-=0.0000028332
3
=0.0000000007

5
Base Sum of the above terms ---0.0204109971
Dias
1+x 25

0.0408219942 1-X 24

xo

obotLog. of

or

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