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+
2u?

a

a

(a+x). Now if we reduce into a series, either oy actual di

ats vision or by the method of indeterminate coefficients, art. 41, page 95,

e we shall find

+

&c. Therefore- orlog a+

a toto (a+x)= )

&c. +C. If we make x=0, the constant

3.x3 C=log a; consequently we shall have l (a +x)= la +

2a* 23 -&c.; and writing for , we obtain I (a) = la

- x? 3a?

x2

az' or x =

by 2a За?

+

a+2 substitution we shall find I (a-x) = 2la-1 (a+2) =la

atz z2

+ 2 (a+z) &c., a series converging with a degree of rapidity proportionate to the excess of a above z. For example 1 100=1 (99+1)=199+

1

100 +c(=

1

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(1 ++ +*+&c.)

. Suppose now that att

a

&c.; and therefore I (a+z) =la+aiz+ala+3)

(+

+

2(100)+&c. = 4.60517018; and I 11=/(10+1) =110+ 2.117+&c.=2.397, &c.*

,

t

1+I2

.

or arc tan ix

se If we had y = then y=arc tan x; but. being re

1+xx duced into a series, gives y=24xi+** 3°*+&c. · therefore y *==*++*+-+*+&c

. any arc, . its sine, or y=arc sin t, we shall have j=

x == (1-xx); this expresion expanded into a se v (1-4X

.X

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x? &c

3

5

7

Let now

y

be

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The logarithms found above are the hyperbolicor Naperean ; these are directly converted into the common logarithms, by multiylying them by the constant factor 0.434294482.

ries by the binomial theorem gives y==(1+++

*

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1.3 1.3.5

att

2.4 2.4.6

1 x3 1.3 aces 1.3.5 x? +&c.). Therefore y or arc sin x=x+. +

+
2 8 2.4
3

5 2.4.67 +&c., a fluent to which it is not necessary to add a correction. Letx=1,

1. and call the semicircumference ®, and we shall obtain

=

2=1+

+

+&c. If ra, the arc y becomes

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1

It

+

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XX

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2

fox *'. Let again * = X“ i and we shall have ** X X XV, &c.

1 1.3 1 1.3.5 1 +

. 3 2.4 5 2.4.6 7

6 1 1 1.3

1.3.5 1 +

+&c. 2 3.23' 2.4 5.25 2.4.67.27

' 74. These examples suffice to explain the preceding method. That which follows will merit attention.

The formula flux (xy)=ay+, gives xy=fxy+Syč; therefore, in general Sxy=xy-Syx, and if we denote by X any function of X, we shall in like manner have /Xi=Xx-fx8. Call Å=X' ;; then by the same principle fæ8, or SX'xx'=X' xx «

23
"s "_

2.3

2.3 Substituting these different values in the first expression, we shall obtain SX;=XzX'+

as
-X" X+ -X"-&c.

2.3 2.3.4 2.3.4.5 or, supposing * constant

** Š

+&c. 2.2 2.3.22 2.3.4.3

1 EXAMPLE. Let X=. and we shall have

ata

2

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75. Let now y=m(a +x)"-11, of which the fluent is y=(a + x)";

shall have by the above theorem X = m (a + 1)-!,

we

m(m—1) #*(0+2)

2

+

m (n-1) (a + x)",

=

= m (m-1)(m-2)(a +1)-,&c. Therefore y, or (a+*) =C+mr (a+2)-1_m(m-1)

x1– m (m-1) (m--2)

(a+r)^5 — &c. Let x=0, and we shall 2.3 have C=a", and (a +3)" - 21 t.mx (a +x)"}

m (m-1)

2 * (a +2)*-*+&c. Make a +r=%, and we shall obtain z"=(:-)"

m (m-1),o zm+&c. therefore + mazm

-122

2
(2-x)"=x-mx 2 +

m (m-1)

per 2-2_&c. and

2

m (m-1) r2 20–2+ &c.

(+5)*=1+m*+

r

m

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(z+)==-m+

نج

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(x+x)"=2" +mz za-1 +

2 zr

-1) g2 =+m +

+ &c....; and

2 (2+5) ; m (m+1) x? _m (m+1)(m+2) ***

t&c x

2

2.3 Now if z+x=b, we shall have (6—x)"=3* (1– 6 m (m+1)

&c.); and b

2 (6—x)?

mix (6+x)"=6" (1+

m (m+1).

z?

+ &c.)

2 (6+x) These latter series will be found useful in extracting the roots of numbers.

76. To find the value of y=a", take the fluxion, and we have (art. 18. page 444) y=a*šla. Therefore X=ala, a't'a,

+

b + x

aat

=a* Pa, &c. which gives y, or Q=C+ala_** lo a

2 Pa

*a'--&c. Let x=0, and we shall have C=1, and 2.3

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a

xxl a

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y = X

'+ P* a ď+&c.; dividing by a we shall obtain

a' * 2

x? ? a 1=rla +&c. Hence a-*=1-xla + -&c.; and 2

2 consequently if we suppose x positive, its odd powers must change their sign which renders the series all positive, and therefore a*-1 + xlat + + &c., as we already know.

2 2.3 77. Let now y be any arc, x its tangent, and we shall have .

1 ; but as oy making X= we should obtain a very 1 + xx

1 + xx complicated series for the value of the are y, we must somewhat modify the preceding method. = -fx

+ 1 + xx

+ xx

Itxx

2.r? Again it is also evident (art. 62.) that

) s (1 + 3x)2

(1+xx)? 2.4. xtc

2.4. .r+

2.4. rs +

Similarly s (1 + xx) 3(1+xx)

3 (1+xx) 3.5 (1 + xx)

3.5 (1+xx)3+ 2.4.6 zo è

2 x3 &c. Therefore arc tan r= +

+ 3.5 (1 +xx)+'

1 + xx 3 (1+xx) 2.4. rs 2.4.6. x7

Therefore in general (since 3.5 (1 + xx)' '3.5.7(1 + x.r)* sin A cos A

tan A
) we shall have y=cos y (sin y+j sino y +

12.4 1+tan’A

3.5 2.4.6

sin' y+&c.); or since sin A cos A= sin 2A, 3.5.7

First, it is evident that y=1 . -st flur Ga=

)

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r=+

+

+ &c.

sin' yt

sing 24 (1 + $ sin oy+

2.4

2.4.6 + +

sin y +

sin ”y + &c.) 2.

3.5

3.5.7 if y=45°, we shall have

1 1.2 1.2.3 s, or the semi-circumference – 2 (1+

+ + &c.) 1.3 1.3.5' 1.3.5.7

+

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OF THE INTEGRATION OF LOGARITHMIC AND EXPONENTIAL FLUXIONS.

78. To integrate the logarithmic fluxion Xi ls, supposing X to be any function of 2, let y=ht, and =Xi, we shall then have sXile=Sýs=yzsaj=kSXi se Therefore the integral of the quantity proposed is reduced to that of Xi, and of

op Si. Heace it may be found by the foregoing rules, if Xš does not contain any transcendental quantity.

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1

n+1

Therefore from lo =

Di an integral which is n+1 subject to no other exception than the case of n=. 1. But then we have so le=Se flus h=**. Again let X= and we shall have sX* =

and = 2-1

() +1(1–)= +(1-5)

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