Examples of the Processes of the Differential and Integral Calculus |
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Page 21
... equal to the second side of ( 1 ) , and as this last consists of a finite number of terms having positive indices , the terms in the product of ( 2 ) and ( 3 ) which contain negative indices must disappear of themselves . Hence taking ...
... equal to the second side of ( 1 ) , and as this last consists of a finite number of terms having positive indices , the terms in the product of ( 2 ) and ( 3 ) which contain negative indices must disappear of themselves . Hence taking ...
Page 71
... equal to 1 . ¿ sin x = 1 + x + ( 3 ) Let x02 1.2 = - Hence we find 3x 1.2.3.4 = esin - 1x . u = ε 1 ( 1 − x2 ) 3 --- · esin - 1 8x05 1.2.3.4.5 3x6 1.2.3.4.5.6 + & c . du Then dx 1.3 1.3.5 and ( 1 − x2 ) − § x2 ) - $ = 1 + 1/1/2 x2 + ...
... equal to 1 . ¿ sin x = 1 + x + ( 3 ) Let x02 1.2 = - Hence we find 3x 1.2.3.4 = esin - 1x . u = ε 1 ( 1 − x2 ) 3 --- · esin - 1 8x05 1.2.3.4.5 3x6 1.2.3.4.5.6 + & c . du Then dx 1.3 1.3.5 and ( 1 − x2 ) − § x2 ) - $ = 1 + 1/1/2 x2 + ...
Page 73
... equal to the coefficient of ( cos x ) " in the original series multiplied by n2 : equating these we have the condition Ap + 2 ( n2 - p2 ) ( p + 1 ) ( p + 2 ) aps by means of which any coefficient is given in terms of that two places ...
... equal to the coefficient of ( cos x ) " in the original series multiplied by n2 : equating these we have the condition Ap + 2 ( n2 - p2 ) ( p + 1 ) ( p + 2 ) aps by means of which any coefficient is given in terms of that two places ...
Page 77
... multiplied by 22 " ( - ) " . By what has preceded it appears , therefore , to be equal to g ” ( 23 – 1 ) - 1.2 ... ( 2n ) Hence , giving n successive values , we find tan B2n - 1 DEVELOPMENT OF FUNCTIONS . 77 PART II.
... multiplied by 22 " ( - ) " . By what has preceded it appears , therefore , to be equal to g ” ( 23 – 1 ) - 1.2 ... ( 2n ) Hence , giving n successive values , we find tan B2n - 1 DEVELOPMENT OF FUNCTIONS . 77 PART II.
Page 78
... equal to Hence we have multiplied by ( - ) " 2 " : € 1 ( − ) " 93 " Bon- \ 1.2.3 ... ( 2n ) cot = 2 % 21 Ᏼ Ꮎ . B03 - & c . Ꮎ I 2 1.2.3.4 CHAPTER VI . EVALUATION OF FUNCTIONS WHICH FOR CERTAIN VALUES 78 DEVELOPMENT OF FUNCTIONS ,
... equal to Hence we have multiplied by ( - ) " 2 " : € 1 ( − ) " 93 " Bon- \ 1.2.3 ... ( 2n ) cot = 2 % 21 Ᏼ Ꮎ . B03 - & c . Ꮎ I 2 1.2.3.4 CHAPTER VI . EVALUATION OF FUNCTIONS WHICH FOR CERTAIN VALUES 78 DEVELOPMENT OF FUNCTIONS ,
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a² b2 a²x² angle arbitrary constant assume asymptote becomes branches C₁ Cambridge circle co-ordinates condition Crelle's Journal curvature curve cycloid determine differential coefficients differential equation dx dx dx dy dx dx² dy dx dy dy dy dy dz dz dz eliminate ellipse equal Euler factor formula fraction function Geometry gives Hence hypocycloid infinite intersection John Bernoulli Let the equation lines of curvature locus logarithmic logarithmic spiral Multiply negative origin parabola perpendicular radius SECT singular points singular solution spiral Substituting subtangent surface tangent plane theorem triangle University of Cambridge vanish whence x²)³