Sidebilder
PDF
ePub

2

75. Let now y=m(a + c)-* *, of which the fluent is y=(a+2)"; we shall have

by the above theorem X = m (a +5)*-', -=m (m-1) (a +x)–,

=m(m-1)(m—2) (2+1)*-*, &c. Therefore y, or (a +x) =C+mr (a + c)m_1_

-1_m(m-1) r*(a+sjone + m (m-1) (m-2) (a+1)-5-&c.

Let x=0, and we shall 2.3 have Cra", and (a +1)" =" t.mx (a+*)*-

m (m-1)

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

m (m1)22 zm–+&c. therefore

m (m-1) x 2m-2_&c. and

[ocr errors]
[ocr errors]

(2+2)=1

[ocr errors]

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

2 (z+r)"=>" + mx 2.1 +

m (m-1)

x? 21–2 +&c.

2
r

x2
=1+m* +
m (m—1)

+&c.... ; and

2 (2+)

--1-m-+
m (m+1)x_m (m+1)(m+2) 5"

+&c 2

2.3 Now if z+x-6, we shall have

x?

m (m+1) (6—)"=" (1-met

&c.); and 2 (6-2)

mix (6+x)"—" (1+

m (m+1) x2

+&c.) b+

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

Å (art. 18. page 444) y=adla. Therefore X=ala, sa t'a,

+

[ocr errors]
[ocr errors]

=auf a, &c, which gives y, or Q=C+& la ** P Q R ***--&c. Let x=0, and we shall have C=1, and

2

2.3

[ocr errors]
[ocr errors]

+&c.

xxl a

[ocr errors]

a=1+* la q*_*r? a

at + &c.; dividing by as we shall obtain 2

wx la l-a-+xla

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 + xla+ + +&c., as we already know.

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

1 y = ; but as oy making X=

we should obtain a very 1 + x2 complicated series for the value of the are y, we must somewhat modify the preceding method. First, it is evident that y=

flux

+ 1 + xx

Itxx

1+XX

[ocr errors]

(1+xzji=

+

[ocr errors]
[ocr errors]
[ocr errors]

+ &c.

Again it is also evident (art. 62.) that s1+ 2.4. x42

2.4.. +

2.4. is Similarly S (1 + xx) 3 xx) 3°

3 (1+xx)3 3.5 (1+xx) 3 2.4.6. &c. Therefore arc tan x=

2 x3 +

q+ 3.5 (1+rx +

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

2.4.6. x? +

Therefore in general (since 3.5 (1 + xx)3 ' 3.5.7(1+xx)*

tan A sin A cos A = ) we shall have y=cos y (sin y + i 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
sin 2y (1+4 sin oy +

2.4

2.4.6

sin Oy +&c.) 3.5

3.5.7 if y=45°, we shall have

1 1.2 1.2.3 s, or the semi-circumference – 2 (1+ t: + + &c.)

1.3 1.3.5' 1.3.5.7

sin' y +

sin 4y +

[merged small][ocr errors][merged small][merged small][ocr errors][merged small]
[graphic]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

79. Supposing X to be always a function of x, and that it were required to integrate Å 1" x; we must write this expression under the form SX1".<=XI"c_n

ng Xi

r, by the formula for integra

SXl

tion by parts; and then supposing S4=X', we shall have by the

[ocr errors]
[ocr errors]

same formula s*= -=X'*-==(n=15$'31 If we make sy==x",

=x", we shall have sx"p=+ ==X" (p*==(n=2)539.3.3-) 3, &c. Therefore

SÅt' -=X75-n X'71 +n (n-1) X" /* 1*(1-1) (%) X"?-3 x+&c.; an expression which depends upon the integration of algebraic quantities only, and which will have a finite number of terms when n is a positive whole number.

For example. Let X=xma, and we shall have X=:

[ocr errors]
[ocr errors]
[merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]
[ocr errors]
[ocr errors]
[ocr errors]

Consequently Smilr=_

n (nl) m+1

m+1 (m+1)? Mr_n (n+1) (n+2)

fa-3

*+&c. .(m+1)' The only case which does not come within the general formula is that of m=-1, and then we have

pa +1

n+1 80. The foregoing formula applies equally to the case in which » is negative. But as then we obtain for the fluent an infinite series, we shall explain another mode of integration.

[ocr errors]

X.c

Suppose the quantity were which being written under the

( (hr) (lo which it is evidently equal, because flux lx=*,) (2x)

Xc gives

Xx

+ in1)

flux (Xx). Now call

form Xx. flux lt

to ,

2 급

1

[ocr errors]
[ocr errors]

X" *

[ocr errors]

ing flux (Xr) = X':, flur (X' x)=X", flux (X" x)=X",; we shall have by continuing the process X -XX

X' x S

(lx) (n-1) 7-x (11-1) (1-2)r (n-1)(n-2)(11–3)-** -&c., to a term of the form

(n-1) (n=2) (n=3)...2.1 DE the integration of which, if possible, will give that of the formula proposed.

For example, let X=r", we shall have X'=1" (m+1), X"=

1

[ocr errors]
[ocr errors]
[ocr errors]

{1+ +&c.}

(m+1) r", X" = (m+1)* .x", &c. Therefore

"(1

(n-1) to
m+1
(m + 1)

(m+1)
12 2 (16--2).(n-3) (n-2) (1-3) (144)

(m+1)-?
+

-S

therefore the proposed integral (11-1) (1-2)...lo 20

r

r

is reduced to that of

If we make 3M+1 = u, this quantity will

become a fluxion which has not yet been integrated. For this

lu'

[ocr errors][merged small][merged small][merged small]

1

[ocr errors]

1

a series, except in the case of m=-1; for then we find by the preceding series, or indeed without its aid, s

Ir. 81. Let it now be required to find the fluent of the exponential formula a' Xč. I observe first that a' i la = flux (a"); therefore Sa+ i= a*; and since, integrating by parts, Sa* Xi = Xfa* :SX fai,

we have fa* Xi=mX_1 X. Let Å=X' :, we shal

la

la

[merged small][merged small][merged small][ocr errors][ocr errors][merged small]
« ForrigeFortsett »