Roots:zn=[r(cosθ+isinθ)} e+k.360 k=0,1,2,,n-1 12. Permutations ement(sequence) he number of per utations of n objects at a time Is P(n,r)=n(n-1)(n-2).(m-r+1) a permutation of positive integers is even or odd if the total number of inversions is an even integer or an odd integer, respectively. Inversions are counted relative to each integer j in the permutation by count- ing the number of integers that follow j and are less than j. These are summed to give the total f inversions. For example, the permutation four inversions three relative to 4 and one to 3. This permutation is therefore even. A combination is a selection of one or more objects from among a set of objects regardless of order. The number of combinations of n different objects taken Is C(n, r)= P(n, r) r!r!(m-r)!
7 Roots : z r i r k n n 1 1 36 / / (cos sin ) cos . = + = + [ ] 1/n θ θ θ 0 360 n i k n n + + = − sin θ . , k 0,1,2,. . ., 1 12. Permutations A permutation is an ordered arrangement (sequence) of all or part of a set of objects. The number of permutations of n objects taken r at a time is p n r n n n n r n n r ( ) ( )( ) ) ( ) , . . .( ! ! = − − − + = − 1 2 1 A permutation of positive integers is even or odd if the total number of inversions is an even integer or an odd integer, respectively. Inversions are counted relative to each integer j in the permutation by counting the number of integers that follow j and are less than j. These are summed to give the total number of inversions. For example, the permutation 4132 has four inversions: three relative to 4 and one relative to 3. This permutation is therefore even. 13. Combinations A combination is a selection of one or more objects from among a set of objects regardless of order. The number of combinations of n different objects taken r at a time is C n r P n r r n r n r ( , , ! ! ! ( ! ) ( ) ) = = −
Igebraic equations If ax+bx +c=0. and a+0. then roots are -b±√b2-4ac Cubic solve x'+bx+cx+d=0. let x=y-b/3. Then the reduced cubic is obtained y32+py+q=0 where p=c-(1/3)b2 and q=d-(1/3)bc +(2/27)b Solutions of the original cubic are then in terms of the reduced cubic roots y1. 2.3 x=y1-(1/3)bx2=y2-(1/3)b x3=y3-(1/3)b The three roots of the reduced cubic are y1=(A)+(B) y2=W(A)"+W2(B)3 y3=W2(A)"+W(B)"3 A=-q+(1 2y+19
8 14. Algebraic Equations Quadratic If ax bx c 2 + + = 0, and a ≠ 0, then roots are x b b ac a = − ± −2 4 2 Cubic To solve x bx cx d 3 2 + + + = 0, let x y b = − / 3. Then the reduced cubic is obtained: y py q 3 + + = 0 where p = c – (1/3)b2 and q = d – (1/3)bc + (2/27)b3. Solutions of the original cubic are then in terms of the reduced cubic roots y1,y2,y3: x y b x y b x y b 1 1 2 2 3 3 1 3 1 3 1 3 = − = − = − ( / ) ( / ) ( / ) The three roots of the reduced cubic are y A B y W A W B y W A 1 1 3 1 3 2 1 3 2 1 3 3 2 = + = + = ( ) ( ) ( ) ( ) ( ) / / / / 1 3/ 1 3/ + W B( ) where A q = − + p q + 1 2 1 27 1 4 3 2 ( / ) , • •
B=--q-1(1/27)p3+q2 W When(1/27)p+(1/4)p" is negative. A is complex in this case, A should be expressed in trigonometric form: A=r(cose+i sine), where 0 is a first or second roots of the reduced cubic are y1=2)cos(/3) +120° y=2(r)cos;+240° Figures 1. 2 to 1. 12 are a collection of common geo- metric figures. Area(A), volume(V), and other mea- For any right triangle with perpendicular sides a and b, the hypotenuse c is related by the formula c2=a2+b2 Thi tons Section 4.2
9 B q p q W i W i = − − + = − + = − − 1 2 1 27 1 4 1 3 2 1 3 2 3 2 2 ( / ) , , . When ( )1 27 1 4 3 2 / ( / ) p p + is negative, A is complex; in this case, A should be expressed in trigonometric form: A = r (cosθ + i sinθ), where θ is a first or second quadrant angle, as q is negative or positive. The three roots of the reduced cubic are y r y r 1 1 3 2 1 3 2 3 2 3 120 = = + ° ( ) ( / ) ( ) / / cos cos θ θ = + ° y r 2 240 1 3 ( ) / cos 3 θ 15. Geometry Figures 1.2 to 1.12 are a collection of common geometric figures. Area (A), volume (V), and other measurable features are indicated. 16. Pythagorean Theorem For any right triangle with perpendicular sides a and b, the hypotenuse c is related by the formula c2 = a2 + b2 This famous result is central to many geometric relations, e.g., see Section 4.2
FIGURE 1.2 Rectangle. A= bh. b FIGURE 1.3 Parallelogram.A= bh. FIGURE 1.4 Triangle. A=-bh
10 h b Figure 1.4 Triangle. A bh = 1 2 . b h Figure 1.3 Parallelogram. A = bh. b h Figure 1.2 Rectangle. A = bh
FIGURE 1.5 Trapezoid. A=-(a+b)h FIGURE 1.6 Circle. A=R: circumference= 2TR; arc length s= Re (0 in radians) FIGURE 1.7 Sector of circle. A R R(e-sine)
11 b a h Figure 1.5 Trapezoid. A a b h = + 1 2 ( ) . S R θ Figure 1.6 Circle. A π R = 2 ; circumference = 2πR; arc length S R = θ (θ in radians). θ R Figure 1.7 Sector of circle. A R sector = ; 1 2 2 θ Asegment A R = segment = − ( sin ). 1 2 2 θ θ