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Hence the series (5) and (6) are equal for every value of y between 0 and any positive finite quantity.

Hence we may equate the coefficients of the corresponding powers

of y in the two series. Equating the coefficient of y,

202

za
loge (1 + x) = -

+(-1)"-1.
3
2

+... .....(7), for all values of , numerically not greater than 1.

, Example. Expand log. sec in powers of tan 6, if tan 0<l. log, seco={ logsec? A=1 log (1+tan? A)={tana 0 – 1tan4 6+tane 8 - ...

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474. The student should note and compare the two series (4) and (7) of the last article ; viz.,

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e

The power of e is expressed by (1) factorial denominators and (2) positive signs.

The logarithm to base e is expressed by (1) simple numeral denominators and (2) alternate signs.

It is important to note that the series on the right is to be equated to the one positive value of eil in the first case; and to the one positive or negative value of loge (1+x) in the second

case.

475. The series for loge (1 + x) gives directly the logarithms only of (positive) numbers up to 2 inclusive.

Now any number greater than 1 is the reciprocal of some number less than 1: and log n=- - log - Hence the series gives almost directly the logarithms of any number whatever.

.

n

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476. The series for loge (1 + 2) does not converge rapidly unless x is very small. Hence it is not immediately useful for the arithmetic calculation of logarithms. But we have

loge (1 + x) = + x - * ** + 5 x –...

loge (1 – a) = – X – 1 22 – } c3 – ... :. by subtraction, (if x is not >1 numerically)

1

- 1

1 + 2C log

= 2(x+ 2 + 2 + ...)...............(8). 1 XC

e

m - n

If now m and n are positive and m > n, then

is positive

m+n

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Hence we can find in succession, from loge 1=0, the logarithms of any other numbers by a rapidly converging series. Thus put n=1 and m = 2. Thus

loge 2 = 2 {} + } ()+ } (3)3 + ...}.

1

Put n= 2 and m = 3. Thus

loge 3 – loge 2 = 2 {} + } (3)3 + } (x)" + ...}

+

and so on.

477. In order to calculate logarithms to base 10, we have to note that

log 10 n = loge n = loge 10. Hence every log to base e must be multiplied by the constant modulus 1 = loge 10 to give the corresponding log to base 10.

As an example in the calculation of logarithms, we will find the value of 1 = loge 10 to 8 decimal places.

5

3

478. To find the modulus 1 - loge 10. We have

loge 10 = log 5 + loge 8 = loge 1 + 3 loge : In Art. 476, (9), put m=

5 and n=

4; also put m= 2 and n=1. Thus

loge 1 = 2 {} + } ()* +3 () +}()' + ...},

{73 * ++
3 loge 2 =2 {1+} (3) +} (3) + (+ ...},
= 2 {1+ }+} () + ()* + ...}.

2
Thus we have expressed loge 10 in powers of .

By successively dividing by 9 we may place in one column the several powers of , and dividing these by the required numbers, we may place the results in a second column and add. Thus

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This result has to be multiplied by 2.
Thus

loge 10 = 2:30258509.
And, by division,

1 modulus

logio e = •43429448.

=

loge 10

479. Logarithms to the base 10 are called Common Logarithms: those to the base e are called Napierian Logarithms.

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m

+

+

n

+

+

+

+

+

+

+

n+1

Thus

1 1 1 1+1+

12 13 Multiply both sides by n; then m (n-1=an integer

1
1

1
n+1' (n + 1)(n+ 2)" (n + 1) (n + 2)(n+3)
But the series
1
1

1
(n + 1) (n + 2) (n + 1) (n + 2)(n+3)
is less than the geometrical series

1
1

1
+

(n + 1) (n + +1)3
1
1

1

1 1-1/(n+1)

n+1-1 This series is therefore a fraction less than 1.

Thus the difference of two integers is less than 1 but greater than 0 : which is impossible.

Hence e cannot be expressed as a fraction with finite integral numerator and denominator; i.e. it is irrational or incommensurable (with unity).

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EXAMPLES XVIII.

[Here the series (n)' is used as an abbreviation for the series whose nth term is (n)':]

1. The series nyx", where y is finite and a positive, is convergent if x < 1, and divergent if x > 1.

2. Prove the theorem of Art. 454, in the converse form ; viz. "The series + (n) is convergent or divergent according as the series 6" (6) is convergent or divergent, where (n) is positive

» and decreases as n increases, and b > 1.”*

3. Hence show that the series logn 2, where x is positive, is divergent.

4. From the known theorem that the series a" is convergent or divergent according as x is or is not less than 1, show that the series un is convergent or divergent according as the limit, when n is infinite, of Un is or is not less than 1.

5. Hence the series (a, +an+a,/n+ ...)" is convergent or divergent according as a, is or is not less than 1.

6. From the known theorem that the series no is convergent or divergent according as x is or is not less than – 1, show that the series Un is convergent or divergent according as the limit, when n is infinite, of logn Un is or is not less than – 1.

7. Hence the series no. +&q{n+a, in*+... is convergent or divergent according as a, is or is not less than – 1.

8. The series
1 1.3

5
1.3.5. 7

is convergent and = 1. 4 4. 6

8 4.6.8.10

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0

+

+

1.3. 4.6.

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* This is generally the form in which the theorem should be used, when the series, whose convergency has to be determined, is given.

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