Formulary_HS15.pdf - Chair of Mathematics University of...

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Chair of Mathematics, University of St.Gallen Mathematics A: Formulary Autumn Semester 2015 1 Quadratic equation The solutions to the quadratic equation a x 2 + b x + c = 0 , where a, b, c 2 R , a 6 = 0, are given by x 1 , 2 = - b ± p b 2 - 4 a c 2 a . 2 Geometric sequences and series The partial sums s n of a geometric sequence with q 6 = 1 satisfy s n = n X k =1 a k = n X k =1 a 1 q k - 1 = a 1 1 - q n 1 - q . Moreover, lim n !1 s n = 1 X k =1 a 1 q k - 1 = a 1 1 - q if | q | < 1. 1
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Mathematics A: Formulary, Autumn Semester 2015 2 3 Euler number The following holds: (i) lim n !1 ( 1 + p n ) n = e p , for p 2 R . (ii) lim n !1 P n k =0 p k k ! = P 1 k =0 p k k ! = e p , for p 2 R . 4 Trigonometric functions For all x 2 R we have: (i) sin 2 ( x ) + cos 2 ( x ) = 1 (Pythagorean theorem); (ii) sin( - x ) = - sin( x ) and cos( - x ) = cos( x ); (iii) sin( k ) = 0 and cos ( 2 + k ) = 0 for all k 2 Z ; (iv) sin ( x + 2 ) = cos( x ) and cos ( x + 2 ) = - sin( x ). 5 Contour lines (conic sections) (i) Circle with radius r > 0 and centre P = ( u, v ) 2 R 2 : ( x - u ) 2 + ( y - v ) 2 = r 2 . (ii) Ellipse with semi-axes a, b 2 R ++ and centre P = ( u, v ): ( x - u ) 2 a 2 + ( y - v ) 2 b 2 = 1 . (iii) Let a, b 2 R , a 6 = 0, b 6 = 0. Hyperbola open left and right: x 2 a 2 - y 2 b 2 = 1 . Hyperbola open upwards and downwards: y 2 b 2 - x 2 a 2
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