Let & be the value of a, which renders A=1, then

ε

(e-1)-(-1)2+(e—1)3-} (e−1)' + ... =1

....

**=1+2+12+1·2·3+1-2-3-4+***

Now, since this equation is true for every value of x, let x=1; then

1

1

1

e=1+1+1-2+1-2-3+1·2·3·4+

=1+1+1 (1) + } ( 11 ) + ( 1.3.3) +....

=2.718281828459......

223. We add another method of calculating the logarithm of any given number.

Let N be any given number whose logarithm is x, in a system whose base is a; then

aN and a2=N2.

Hence, by the exponential theorem, we have from the last equation

x222

1+Axz+A2 1.2+...=1+A1~+A121-2+

and equating the coefficients of z, we get Ar=A1; hence

because A =(a —1) —¦(a −1)2+}(a —1)3—....... in the expansion of aTM, A1=(N-1)-(N − 1)2+} (N—1)3-... in the expansion of N1.

and

224. To find the logarithm of a number in a converging series. We have seen that if a N, then

=

(N-1)-(N-1)2+(N-1)3-(N-1)+...

x= (a1)-(a -1)+(a -1)-(a −1)'+...

Now the reciprocal of the denominator is the modulus of the system;*and, representing the modulus by M, we have

x= log. N=M{(N−1)—¦(N−1)2+}(N−1)3—¦(N−1)'+.....} Put N=1+n; then N-1-n, and we have

log. (1+n)=M(+ n − }n2 + }n3 — }n1+}n3—.......) · · ·[A]
log. (1-n) M(—n — } n2 — } n3 — }n* — }n3 + ...)
log. (1―n)=2M(n+}n3+}n3+}n2+...)

Similarly,
... log. (1+n)

* If, in the expression for a1, deduced in (Art. 222), we make x=

if e be the base of the system of logarithms expressed by log. Therefore

is,

A log. a

by a previous definition (Art. 221), the modulus for passing from the system whose base is ɛ to that whose base is a. If log. a refers to the base a, becomes equal to log. ɛ.

S

1

1

1

1

log. (P+1)— log. P=2M { 2P+ i+3(2P+1)3+5(2P+1)+

1

1

.'. log. (P+1)= log. P+2M {2P+1+3(2P+1)2+5(2P+1) + ... }

Hence, if log. P be known, the log. of the next greater number can be found by this rapidly converging series.

By substituting the series of natural numbers for N in this formula, the corresponding values of r will be their logarithms.

224. To find the Napierian logarithms of numbers.

In the preceding series, which we have deduced for log. (P+1), we find a number M, called the modulus of the system; and we must assign some value to this number before we can compute the value of the series. Now, as the value of M is arbitrary, we may follow the steps of the celebrated Lord Napier, the inventor of logarithms, and assign to M the simplest possible value. This value will therefore be unity, and we have

log. (P+1)=log.

1

1

1

P+2{2P+1+3(2P+1)3+5(2P+1)+

Expounding P successively by 1, 2, 3, 4, &c., we find

1

1

1

2 +

3.33

5.35+

;+ ...) =

+...).

....}

=6931472

7.57+...). =1.0986123

In this manner the Napierian logarithms of all numbers may be computed.

225. To find the common logarithms of numbers.

The base of the Napierian system is &=2·718281828..., and the base of the common system is b=10, the base of our common system of arithmetic; then we have b=10, and a=ɛ=2·718281828..., and consequently, if N denote any number, we shall have

*To find the value of the Napierian base, observe that, since com. log. N=43429448X Nap. log. N., if we make in this expression Ne, the Napierian base, we have

com. log. =43429448.

From a table of common logs., therefore, we find the number corresponding to the loga

and the modulus of the common system is, therefore,

Hence, to construct a table of common logarithms, we have

1

1

1

log. (P+1)= log. P+·86858896 2P+1+
2P+1+3(2P+1)3+5(2P+1)1⁄2' +

Expounding P successively by 1, 2, 3, &c., we get

...

P-1 1

3 1

-1 +3 P+1

5

P+

and thus we have a series for computing the logs. of all numbers, without knowing the log. of the previous number.

EXAMPLES.

(1) Given the log. of 2=0.3010300, to find the logs. of 25 and 0125.

100 102 Here 25= = 4 22

therefore log. 25-2 log. 10-2 log. 2=1.3979406.

... log. 0125= log. 1— log. 10-3 log. 2=-1-3 log. 2=2.0969100 (2) Calculate the common logarithm of 17.

Ans. 1.2304489.

(3) Given the logs. of 2 and 3 to find the logarithm of 22.5.

Ans. 1+2 log. 3-2 log. 2.

(4) Having given the logs. of 3 and 21, to find the logarithm of 83349. Ans. 6+2 log. 3+3 log. 21.

rithm 43429448, which is 2-7182818, the Napierian base. This also furnishes us with another definition of the modulus of the common (or any other) system of logarithms; it is the common (or, &c.) logarithm of the Napierian base. See further note at the end of Progressions.

PROGRESSIONS.

ARITHMETICAL PROGRESSION.

227. WHEN a series of quantities continually increase or decrease by the addition or subtraction of the same quantity, the quantities are said to be in Arithmetical Progression. A more appropriate name is Progression by Differences.

Thus the numbers 1, 3, 5, 7, which differ from each other by the addition of 2 to each successive term, form what is called an increasing arithmetical progression, or progression by differences, and the numbers 100, 97, 94, 91, which differ from each other by the subtraction of 3 from each successive term, form what is called a decreasing progression by differences. Generally, if a be the first term of an arithmetical progression, and 8 the common difference, the successive terms of the series will be

in which the positive or negative sign will be employed, according as the series is an increasing or decreasing progression.

Since the coefficient of 8 in the second term is 1, in the third term 2, in the fourth term 3, and so on, in the nth term it will be n-1, and the nth term of the series will be of the form

In what follows we shall consider the progression as an increasing one, since all the results which we obtain can be immediately applied to a decreasing series by changing the sign of d.

228. To find the sum of n terms of a series in arithmetical progression.

Let a

first term.

Then

S= sum of the series.

S=a+(a+)+(a+28) +...+l.

Write the same series in a reverse order, and we have

S=

2 + (1—8)+(1—28)+......+a

Adding, 28=(a+1)+(a+1)+(a+1) +......+(a+1)

=n(a+l), since the series consists of n terms.
n(a+1)

...S=

2

(2)

Hence, if any three of the five quantities a, l, d, n, S be given, the remaining two may be found by eliminating between equations (1) and (2).

It is manifest from the above process that

The sum of any two terms which are equally distant from the extreme terms is equal to the sum of the extreme terms, and if the number of terms in the series be uneven, the middle term will be equal to one half the sum of the extreme terms, or of any two terms equally distant from the extreme terms.

EXAMPLE I.

Required the sum of 60 terms of an arithmetical series, whose first term is 5 and common difference 10.

A body descends in vacuo through a space of 16 feet during the first second of its fall, but in each succeeding second 32 feet more than in the one immediately preceding. If a body fall during the space of 20 seconds, how many feet will it fall in the last second, and how many in the whole time?

To insert m arithmetical means between a and b.

Here we are required to form an arithmetical series of which the first and last terms, a and b, are given, and the number of terms =m+2; in order, then, to determine the series, we must find the common difference. Eliminating S by equations (1) and (2), we have

(4) Required the sum of the odd numbers 1, 3, 5, 7, 9, &c., continued to 101 terms?

Ans. 10201.

(5) How many strokes do the clocks of Venice, which go on to 24 o'clock, strike in the compass of a day?

Ans. 300.