Examples of the Processes of the Differential and Integral Calculus |
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Page 21
... 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 the terms with ...
... 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 the terms with ...
Page 73
... Every term on the second side vanishes except the first , and there remains a = cos n ( 2r + 1 ) — · To find a1 , make x = ( 2r + 1 ) — in the second equation , when we obtain 2 π sin n ( 2 + 1 ) 2 π DEVELOPMENT OF FUNCTIONS . 73.
... Every term on the second side vanishes except the first , and there remains a = cos n ( 2r + 1 ) — · To find a1 , make x = ( 2r + 1 ) — in the second equation , when we obtain 2 π sin n ( 2 + 1 ) 2 π DEVELOPMENT OF FUNCTIONS . 73.
Page 105
... side of the hexagon , AP , the height of the prism , b , OSN = 0 . = = a , Then and QM = a cot 0 . ON = NM = a , and SN = a cosec 0 , The surface of ABPQ = ļa ( 2b - a cot 0 ) . 3a2 The surface of PQRS PR . SN = cosec 0 . 2 Whence the ...
... side of the hexagon , AP , the height of the prism , b , OSN = 0 . = = a , Then and QM = a cot 0 . ON = NM = a , and SN = a cosec 0 , The surface of ABPQ = ļa ( 2b - a cot 0 ) . 3a2 The surface of PQRS PR . SN = cosec 0 . 2 Whence the ...
Page 118
... sides , each intersecting pair makes equal angles with the side in which they meet ; consequently the triangle formed by these lines is the triangle of least perimeter which can be inscribed in the given triangle . See Cambridge ...
... sides , each intersecting pair makes equal angles with the side in which they meet ; consequently the triangle formed by these lines is the triangle of least perimeter which can be inscribed in the given triangle . See Cambridge ...
Page 125
... as 3 ; that its centre coincides with the centre of gravity of the triangle ; of contact bisect the sides of the triangle . and that the points Bérard , Ib . p . 284 . ( 20 ) To find a point within a triangle MAXIMA AND MINIMA . 125.
... as 3 ; that its centre coincides with the centre of gravity of the triangle ; of contact bisect the sides of the triangle . and that the points Bérard , Ib . p . 284 . ( 20 ) To find a point within a triangle MAXIMA AND MINIMA . 125.
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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