Examples of the Processes of the Differential and Integral CalculusJ. and J.J. Deighton, 1846 - 529 стор. |
З цієї книги
Результати 6-10 із 88
Сторінка 72
... assume u = A + A1x + А2x2 + & c . + A „ x ” + & c . 0 By taking the logarithmic differentials we find ( n + 1 ) An + 1 = a , 4 , + 2α2 A „ -1 + & c . + ( n + 1 ) an + 1 Ap⋅ Also since u = 6 % when x = 0 , A = € % , so that we have 0 2 ...
... assume u = A + A1x + А2x2 + & c . + A „ x ” + & c . 0 By taking the logarithmic differentials we find ( n + 1 ) An + 1 = a , 4 , + 2α2 A „ -1 + & c . + ( n + 1 ) an + 1 Ap⋅ Also since u = 6 % when x = 0 , A = € % , so that we have 0 2 ...
Сторінка 73
D. F. Gregory. Assume cos n x = a + a ̧ cosx + & c ... + a , ( cos x ) 2 + Differentiating , ... + ap + 2 ( COS x ) P + 2 + & c . -1 n sin na = { a1 + 2 a2 cos x + ... .. + pa2 ( cos x ) 3 − 1 + Differentiating again , n2 x ... + ( p + ...
D. F. Gregory. Assume cos n x = a + a ̧ cosx + & c ... + a , ( cos x ) 2 + Differentiating , ... + ap + 2 ( COS x ) P + 2 + & c . -1 n sin na = { a1 + 2 a2 cos x + ... .. + pa2 ( cos x ) 3 − 1 + Differentiating again , n2 x ... + ( p + ...
Сторінка 76
... assume we have x € x 1 = a + a ̧ x + a2x2 + az ‹ x3 + & c . , = αo а1 x + а2x2 — az ‹ x3 + & c .; - and subtracting the latter from the former , x ( 1 - e ) € * ---- 1 = − x = 2 { a1x + Azx2 + & c . } , and comparing the coefficients ...
... assume we have x € x 1 = a + a ̧ x + a2x2 + az ‹ x3 + & c . , = αo а1 x + а2x2 — az ‹ x3 + & c .; - and subtracting the latter from the former , x ( 1 - e ) € * ---- 1 = − x = 2 { a1x + Azx2 + & c . } , and comparing the coefficients ...
Сторінка 85
... assume the form a + h for substitute in both numerator and denominator , and develop both according to powers of h ; reduce the new fraction to its simplest form , and then make h = 0 ; the result will be the true value of the fraction ...
... assume the form a + h for substitute in both numerator and denominator , and develop both according to powers of h ; reduce the new fraction to its simplest form , and then make h = 0 ; the result will be the true value of the fraction ...
Сторінка 86
... assume x = a W = h ; ( 2a ) the result is 1 + 31⁄2 a ' when h - 0 , which is the real value of the fraction . tan 7 Ꮡ - π X ( 28 ) u = when a 0 . = " 2x tan a Let x = 0 + h , or h ; then , expanding the circular function by the formula ...
... assume x = a W = h ; ( 2a ) the result is 1 + 31⁄2 a ' when h - 0 , which is the real value of the fraction . tan 7 Ꮡ - π X ( 28 ) u = when a 0 . = " 2x tan a Let x = 0 + h , or h ; then , expanding the circular function by the formula ...
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