Short Review of Basic Mathematics
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1 Short Review of Basic Mathematics Tomas Co January 1, 1
2 c 8 Tomas Co, all rights reserve
3 Contents 1 Review of Functions 5 11 Mathematical Ientities 5 1 Drills 7 1 Answers to Drills 8 Review of ODE Solutions 11 1 First Orer Differential Equations 11 High Orer Linear ODEs with Constant Coefficients 1 Drills 14 4 Answers to Drills 15 Review of Matrix Operations 17 1 Basic Operations 17 Determinants 18 Matrix Inverse 19 4 Drills 5 Answer to Drills 1
4 4 c 8 Tomas Co, all rights reserve c 7 Tomas Co, all rights reserve 1 1 AA T B ACA T B 1 4 ( 1 si C 5 BC 5 Answer to Drills AA T AA T B ACA T /14 4/14 1/14 B 1 6/14 1/14 /14 4/14 /14 5/14 4 (si C 1 1 s 4s + 1 ( s 1 s 5 Not applicable! Sizes of B an C are not conformable for the prouct BC
5 c 8 Tomas Co, all rights reserve (c The inverse of A is given by A 1 1 A BT 1 Example: Let A then 1 b 11 ( ( ; b 1 1 ; b 1 ( b 1 ; b 1 b 1 ( 1 ( ( 1 ; b 1 1 ( ( ; b 1 ; b 4 6 B Expaning along row, ( A + 1 ( Thus, the inverse is given by A Checking: 4 Drills A 1 A 4/17 7/17 6/17 6/17 /17 9/17 /17 1/17 4/17 T ( 1 ( /17 7/17 6/17 6/17 /17 9/17 /17 1/17 4/ ( ( ( Let A 4, B 1 1, C an I an let s be a scalar variable Evaluate the following: Chapter 1 Review of Functions 11 Mathematical Ientities 1 e b 1 e b e a+b e a e b e ln(x x 4 ln (e x x 5 ln (ab ln(a + ln(b ( a 6 ln ln(a ln(b b 7 ln (x a a ln(x 8 log (x ln(x ln(1 ln(x 9 cos (a + b cos(a cos(b sin(a sin(b 1 sin (a + b cos(a sin(b + sin(a cos(b 5
6 6 c 8 Tomas Co, all rights reserve c 7 Tomas Co, all rights reserve cos (x + sin (x 1 1 cos (x sin (x cos (x 1 sin(x cos(x sin (x 14 ax + bx + c x b ± b 4ac a 15 (a + b (a b a b 16 (a + b a + ab + b ( 17 (a + b n a n + na n 1 b + + n! (n k!k! a n k b k + + b n where n! (n(n 1 (((1 an! 1 Define the imaginary number: i 1 18 (a + ib (a ib a + b 19 e it cos(t + i sin(t e it cos(t i sin(t 1 cos(t eit + e it sin(t eit e it i Metho : column expansion (a Given matrix A, choose any column, say column j (b Let A ij be matrix A with row i an column j remove (c The eterminant can be recursively etermine by A n ( 1 i+j Aij i1 Example: Expan along column, 1 1 ( ( 1 1 ( ( ( ( Matrix Inverse 1 Matrices: ( a11 a 1 a 1 a 1 1 ( a a 1 A a 1 a Example: ( 1 ( ( 1 1 1/ 1/ 1 1 /4 1/4 Checking: A 1 A n n Matrices: ( ( 1 ( 1/ 1/ 1 1 /4 1/ t ta a t a 1 t at a t ln(a t eat ae at (a Given matrix A, let A ij be matrix A with row i an column j remove (b Buil a new matrix B, where b ij ( 1 i+j A ij b 11 b 1n B b n1 b nn
7 18 c 8 Tomas Co, all rights reserve c 7 Tomas Co, all rights reserve 7 4 Transpose: Example: ( 1 ( ( ( ( ( ( Example: Determinants Notation: et(a A T a 11 a 1n a 11 a 1m a m1 a mn a n1 a nm T 1 ( Matrices: ( a11 a 1 a a 1 a 11a a 1a 1 ( Example: n n Matrices: Metho 1: row expansion (a Given matrix A, choose any row, say row i (b Let A ij be matrix A with row i an column j remove (c The eterminant can be etermine recursively by n A ( 1 i+j Aij j1 Example: Expan along row, 1 1 ( ( ( ( ( 1 + ( t ln (at 1 t sin(t cos(t t cos(t sin (t t 9 Derivative of Sums: Derivative of Proucts: 1 Derivative of Fractions: Chain Rule: 4 b a b a ( f(t + g(t f t ( f(tg(t t t ( t f u(t f u u t ( p f(t + q g(t t p f(tt + c 5 Integration by parts: 1 Drills 1 Fin Fin b f(tt c a v(b v(a ( e 1+t t + 1 t (ln (at e t+it t Fin the roots of : x + x 4 t + g t ( f t ( f(t 1 g(t f (t b a f(tt u v f(tt + q 4 Fin the solution of : log (x + log (x Expan (x + y 4 into a polynomial in x an y g(t + f(t ( f(t g t g(t b a g(tt ( u(bv(b u(av(a ( g t ( f t u(b u(a v u
8 8 c 8 Tomas Co, all rights reserve 6 Prove the following ientity: (a + ib e p+iq + (a ib e p iq e (a p cos(q b sin(q 7 Fin 1 ( + t e t t 8 Let r be a function of t Show that the following is true: ( πr + 5r (4πr + 5 r t t Chapter Review of Matrix Operations 9 Fin k an φ such that: cos (t + 4 sin (t k sin (t + φ 1 Fin 1 Answers to Drills 1 6e 1+t t t if t 1 tf(tt, where f(t t if t > 1 ( 1 + ln (at (1 + it e t+it t x x 1, x x 4 + 4x y + 6x y + 4xy + y 4 6 (a + ib e p+iq + (a ib e p iq e ((a p + ib (cos(q + i sin(q (a ib (cos(q i sin(q e (a p cos(q b sin(q 1 Basic Operations 1 Aition: a 11 a 1n b 11 b 1n + a m1 a mn b m1 b mn Example: Scalar Prouct: Example: (a 11 + b 11 (a 1n + b 1n (a m1 + b m1 (a mn + b mn ( ( ( a 11 a 1n γ a m1 a mn Matrix Prouct: a 11 a 1n b 11 b 1p a m1 a mn b n1 b np where, c ij γa 11 γa 1n γa m1 γa mn c 11 c 1p c m1 c mp n a ik b kj i 1,, m j 1,, p k1 17
9 16 c 8 Tomas Co, all rights reserve c 7 Tomas Co, all rights reserve e 8 Hint: use chain rule 9 cos(t + 4 sin(t k sin(t + φ Comparing coefficients on both sies (k sin(φ cos(t + (k cos(φ sin(t k cos(φ 4 k sin(φ k φ Atan ( t t + 1 ( ttt 1
10 1 c 8 Tomas Co, all rights reserve c 7 Tomas Co, all rights reserve x + (1 + t x sin(t subject to x( 1 t ( e t x e t subject to x( t x t + 4x t + 4x + e t sin (t x t e t 5 x t 6x subject to x( an x t ( x subject to x( an t ( 4 Answers to Drills 1 x(t exp (1 + ln( e t x(t 1 ( 1 + φ(t t φ(τ sin(ττ ( t + t where φ(t exp x(t e t e t 4 x(t 4 + ( t e t 1 e t cos(t 5 x(t e t + (t 1 e t
11 14 c 8 Tomas Co, all rights reserve Numerical Example: Drills x t + x t 6x e t sin (t subject to x( 1, Step 1: complementary solution x t ( s + s 6 r 1 ; r y 1 e t ; y e t Step : particular solution x c C 1 e t + C e t W (t 5e t u 1 (t 1 ( et cos(t sin(t u (t 1 ( 65 e t cos(t sin(t x p (t 1 5 e t ( cos(t 5 sin(t Step : etermine integration constant x x c + x p C 1 e t + C e t + 1 ( 5 e t cos(t 5 sin(t x( C 1 + C x t ( C 1 + C 11 5 C 1 7 C Solution: x(t 7 e t et + 1 ( 5 e t cos(t 5 sin(t Solve the following ifferential equations 1 x t e t (x + subject to x( 1 Chapter Review of ODE Solutions 1 First Orer Differential Equations 1 Separable Type Form : Solution: Example: x t f(tg(x subject to x( x o x then f(t t Linear First Orer Form : x o 1 g(x x t f(tt an g(x x, x t 5 x 1 + t x x 1 x x t x( ln (x ln (x 5 ( x ln 5 ln (1 + t x x x (1 + t t t ( ln (1 + t ln(1 x x (1 + t 5 x t + p(tx q(t subject to x( x o 11
12 1 c 8 Tomas Co, all rights reserve Solution: Example: ( Let φ exp then p(t t an q(t 4t, an p(tt be an integrating factor, then x 1 ( x + φ(t t q(τφ(ττ x t + tx 4t x( x φ(t e R tt e t x e t (x + t 4τe τ τ e t ( x + e t + (x e t High Orer Linear ODEs with Constant Coefficients Form : Solution: where n x a n t + + a x n 1 t + a x q(t subject to x( x o,,, x(t x c (t + x p (t x c is the complementary solution which solves n 1 x t n 1 ( x o,n 1 n x c a n t + + a x c n 1 t + a x c x p is the particular solution which solves n x p a n t + + a x p n 1 t + a x p q(t c 7 Tomas Co, all rights reserve 1 Example: Secon Orer Step 1: Fin x c : x a t + a x 1 t + a x q(t 1 Get the characteristic polynomial, a s + a 1 s + a Solve for the roots, also known as eigenvalues, r 1 a + a 4a 1 a r a a 4a 1 a a 1 a 1 Form the complementary solution to be where Step : Fin x p : x c C 1 y 1 (t + C y (t y 1 (t e r1t e rt if r 1 r y (t te r1t if r 1 r where C 1 an C are constants, evaluate later to fit the initial conitions Metho : Variation of Parameters y 1 Evaluate the Wronskian: W (t y 1 t y y 1 t Calculate u 1 an u : u 1 (t y (tq(t a W (t t u (t y1 (tq(t a W (t t Form the particular solution: x p (t u 1 (ty 1 (t + u (ty (t Step : Use initial conitions to etermine the values of C 1 an C
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