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 (m—1)22 zm–+&c. therefore m (m-1) x 2m-2_&c. and (2+2)=1 (2-x)"=-mx 2" + 2 (z+r)"=>" + mx 2.1 + m (m-1) x? 21–2 +&c. 2 x2 +&c.... ; and 2 (2+) --1-m-+ +&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, + =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 +&c. xxl a 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 (1+xzji= + + &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, 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 + 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 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=: 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. 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 X" * 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 {1+ +&c.} (m+1) r", X" = (m+1)* .x", &c. Therefore "(1 (n-1) to (m+1) (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 น become a fluxion which has not yet been integrated. For this lu' 1 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 |