Math review. Math review

Transcription

1 Math review 1

2 Math review 3 1 series approximations 3 Taylor s Theorem 3 Binomial approximation 3 sin(x), for x in radians and x close to zero 4 cos(x), for x in radians and x close to zero 5 2 some geometry 6 right triangles 6 triangles in general 7 circles 8 3 exponentials 8 4 differential equations 9 5 Vectors 11 Scalar product 12 Cross product 12 Differentiation of vectors 13 Integration of vectors 13 Line integrals 14 6 Vector transformation properties 15 Under rotations 15 Matrix multiplication and rotation matrices 16 How unit vectors transform under rotations 18 Successive rotations in three dimensions 19 Kronecker delta symbol 20 Levi-Civita symbol 21 Tensors 22 7 Non-cartesian coordinate systems 24 cylindrical coordinates 25 spherical coordinates 25 8 partial derivatives 26 using partial derivatives to approximate changes in functions28 gradient operator 29 divergence operator 31 curl operator 32 2

3 9 quadratic formula 32 Math review 1 series approximations Taylor s Theorem 1 n f ( x) = f ( x0 ) + df ( x x d f 0) ( x x dx x= x + + 0) 0 n! dxn x= x + 0 This can be handy (Recall: n! = [n-1] n) n Binomial approximation Work out an approximation for ( 1 x ) Use x = 0 0 d n n 1 n x ( 1 x dx + ) = ( 1+ ) n + when x is small: 2 ( 1 n + x) = d n 1 n ( 1+ x ) = ( 1)( 1+ ) 2 d dx dx n 2 nn x 3

4 so n n n 1 ( 1 x) ( 1 x) n( 1 x) ( x 0) + = + x= x= n 2 2 n( n 1)( 1+ x) x= 0 ( x 0 ) + 2! 1 1 nx nn + + x 2 = ( ) 2 when 1 + x << we end up with ( ) n 1+ x 1+ nx This works when n is a fraction, too sin(x), for x in radians and x close to zero x = 0 0 dsin( x) d sin( x) = cos( x) = sin( x) 2 dx dx dcos( x) d cos( x) = sin( x) = cos( x) 2 dx dx 2 2 4

5 2 2 dsin( x) x d sin( x) sin( x) = sin(0) + x + dx 2! dx 3 3 x d sin( x) 3 x= 0 + 3! dx x= 0 2 x= x x = sin(0) + xcos(0) sin(0) cos(0) + 2! 3! x = 0 + x+ 0 + x for x1 3! cos(x), for x in radians and x close to zero again, x = dcos( x) x d cos( x) cos( x) = cos(0) + x + dx 2! dx 3 3 x d cos( x) 3 x= 0 + 3! dx x= 0 2 x= x x = for x 1 2! 2! 5

6 2 some geometry right triangles θ c a b a2+ b2= c 2 Integers n 1, n 2, n 3 for which 2 2 Pythagorean triples n + n = n are called Examples: 3, 4, 5; 5, 12, 13; 7, 24, 25; 8, 15, 17; 9, 40, 41; Trigonometric functions for right triangles: sin θ = b/ c cos θ= a/ c tan θ= b/ a csc θ = cb / sec θ= ca / cot θ=a / b Useful trig identities: sin2θ cos2θ 1 1 cot2θ csc2θ 1 tan 2θ=sec 2 = + = + + θ 6

7 ( ) ( ) Math review sin α + β = sinαcos β + cosαsin β cos α + β = cosαcos β sinαsin β ( ) ( ) ( ) ( ) ( ) ( ) sinα sin β = cos α β cos α + β cosα cos β = cos α β + cos α + β sinα cos β = sin α + β + sin α β triangles in general θ b c a θ c θ a b a2 + b2 2 ab cos θ = c2 (law of cosines) c ( θ ) ( θ ) ( θ ) sin sin sin a = b = c (law of sines) a b c 7

8 circles θ r s r tangent line 1 tangent and radius are perpendicular s= rθ θ in radians 2 ( ) 3 exponentials e x takes off like a rocket for large x de dx x d e x = e so = e n dx x n x Taylor s theorem: e x 2 x x = 1 + x ! n! n 2 1 This will be handy: for i 1 (so that i = ), 8

9 ix e ( ix) ( ix) ( ix) ( ix) = ! 3! 4! x x x i + 2! 3! 4! = 1 + ix x x i x x 2! 4! 3! = = cos x + i sin x 4 differential equations It is common in physics to describe how quantities change in response to external circumstances Because of this, calculus is the natural language for describing the physical world Many of our statements about how things work are phrased as differential equations An example: consider a box full of radioactive atoms The more atoms there are in the box, the more decays there should be per unit time In addition, the greater the decay rate (the shorter the half-life) for this species of atom, the more decays there ll be per unit time N(t) number of surviving atoms inside the box at time t Γ 1 τ is the decay rate where τ is the mean life of an atom 9

10 The number of decays per unit time is equal in magnitude, opposite in sign, to the change per unit time of the surviving population in the box Therefore, the statement The number of decays per unit time equals the number of surviving atoms in the box times the decay rate for that species of atom is equivalent to the differential equation ( ) dn t = N() t Γ or dt dn() t = Γ Nt () dt How can we solve this differential equation for N(t)? Note that dn() t dt is proportional to N(t) since Γ is a constant From a few pages back, after replacing x with t: de dt t at t de = e so = ae dt at Note that at de dt is proportional to e at since a is a constant Compare: dn() t dt = Γ Nt () and de dt at at = ae 10

11 It looks like Math review at Nt () e and a Γ works: Nt () = e Γt is a solution to the differential equation But solution! Nt () = 2e Γt is also a What to do? We have a first order differential equation, so we can determine the unique solution with the help of one initial condition (A second order differential equation would require two initial conditions) If we know N(0) we can use that: say there are N 0 atoms in the box at t = 0 The only version of N(t) that solves the differential equation and satisfies the initial condition is Nt ( ) = Ne Γ t 0 5 Vectors Vectors have a length and a direction b θ a Unit vectors in cartesian coordinates: xˆ, y, ˆ z ˆ or iˆ, ˆj, k ˆ Addition, subtraction, you know about 11

12 Scalar product a b = a b cosθ Component of a along b is a cosθ = ( ab ) b Also, if a = ax xˆ + ay yˆ + az zˆ b = b x xˆ + b y yˆ + b z zˆ then a b = a b + a b + a b x x y y z z Cross product a b = a b sinθ Direction is given by right hand rule For the vectors as drawn above: a b b (out of the paper) a (into the paper) a b = a b a b z + a b a b x + a b a b yˆ ( x y y x) ˆ ( y z z y) ˆ ( z x x z) 12

13 Differentiation of vectors Go back to the basic definition of a derivative: d at ( ) dt lim Δ t 0 at t at Δt ( +Δ ) ( ) y u t () u t t u t u t ( +Δ ) ( ) ( +Δ t) particle trajectory x d vt ( ) = xt ( ) dt is an example where this is useful Integration of vectors Same idea: an integral is just a big sum ti = t2 () =lim ( i ) { Δ t = t } t2 a t dt a t t 0 i 1 1 Δ t t t2 ti = t2 For example, v () t dt =lim { ( i ) t = t v t Δ t } t 1 Δ t 0 i 1 13

14 Recall that (for constant velocity) velocity time = distance When Δ t is small, v ( t i ) t i +Δ t () Δt displacement between t i and t2 Therefore, v t dt is the net displacement between t and t t Line integrals A S B Travel from point A to point B along the path labeled S Say there s an applied force acting, which might be a function of F s position along the path S: ( ) Recall that the work done by the force through a small interval is F i as long as F is almost constant over the interval Δs Net work done is F( s ) all intervals i i Δsi In the limit that the intervals become infinitesimal, Δ s Δ s 14

15 B Net work = F ds A 6 Vector transformation properties Under rotations rotate vector by θ a y a θ a y a ϕ a x a x let a a ax = a cos ϕ ay = a sinϕ a = a since lengths of vectors are invariant under rotations ( θ ϕ) ( θ ϕ ) a = a cos + a = a sin + x useful trig identities: ( ) ( ) cos θ + ϕ = cosθcos ϕ sinθsinϕ sin θ + ϕ = sinθcos ϕ + cosθsinϕ y 15

16 use this to rewrite a = acosϕ cos θ asinϕsinθ x = a cos θ a sinθ x a = asinϕ cos θ + acosϕsinθ y = a y cos θ + a y x sinθ Matrix multiplication and rotation matrices ( au + bw) ( av + bx) a b u v = c d w x ( cu dw) ( cv dx + + ) also: ( ax+ by) a b x = c d y cx dy ( + ) 16

17 We can write a as a x a, a similarly to get y a x cos θ sinθ = a y sin θ cosθ ax a y Note the utility of this: successive rotations can be represented as products of matrices a x cos β sin β ax = a sin β cosβ y ay = cos β sin β sin β cosβ cos α sinα ax sin α cosα a y = cos β sin β cos α sinα sin β cos β sin α cosα a x a y since matrix multiplication is associative This works fine in 3 dimensions: 17

18 ax ax ay = R ay where R is 3 3 rotation matrix az az How unit vectors transform under rotations x ˆ = ( 1 0 0) In the rotated frame it becomes R R R 1 R R R R 0 R = 21 R R R 0 R (Note that 1 st index is row, 2 nd is column) So: the 1 st column of R is the same as the representation of the new frame after rotation ˆx in Similarly, the 2 nd column of R is the same as the representation of ŷ and the 3 rd is ẑ xˆ yˆ= 0, yˆ zˆ = 0, zˆ xˆ= 0, regardless of which coordinate system we use (scalar product is invariant under rotations) so each column of R is perpendicular to every other column 18

19 xˆ yˆ = 0, yˆ zˆ = 0, zˆ xˆ = 0 so xˆ yˆ = R R + R R + R R = yˆ zˆ = R R + R R + R R = z ˆ xˆ = R R + R R + R R = Transformations of this sort are called orthogonal transformations: quantities that are orthogonal before remain orthogonal after Successive rotations in three dimensions Successive rotations (where the two rotation axes might be different) can be described like so: a = R Ra 2 1 In general, rotations do not commute: 19

20 Kronecker delta symbol δ ij 1 i= j = 0 i j The Kronecker delta symbol δ ij gives us another way of writing dot products: ab = ab ij i j δ ij 20

21 Also, since xˆ xˆ = 1, yˆ yˆ = 1, zˆ zˆ = 1 xˆ yˆ = 0, yˆ zˆ = 0, zˆ xˆ = 0 we have 3 R R = δ and k= 1 ki kj ij 3 R R = δ k= 1 ik jk ij Levi-Civita symbol ε ijk = 1 0 ( ) ( ) + 1 if ijk is an even permutation of , 312, or 231 if ijk is an odd permutation of , 132, or 321 if any two (or more) of ijk,, are equal It s not hard to see that i= 3 j= 3 k= 3 ˆ also written as ˆ a b = εijk ab i j k εijk ab i j k i= 1 j= 1 k= 1 i, j, k When working with these symbols, you can always write things out explicitly, replacing the indices i, j, and k with numbers and writing out each term in the sums It s messy, but clear 21

22 One can even show that be so moved! k ε ijk ε lmk = δ il δ jm δ im δ should you jl Tensors Let s say we constructed a 3 3 object out of vectors like so: Start with a, b ; define T so that ab ab ab z xx xy x T = a yx b a yy b a y b z ab z x ab z y ab z z In terms of components, T ij = ab i j (This is an outer product ) If we switched to a (rotated) coordinate system, determined the coordinates of a, b in this system, we d calculate T using the rule T ij = a i b j How is T related to T? We know: a = Ra and b = Rb In terms of components, a i = R a and b R k ik k j = b l jl l This comes from the definition of matrix multiplication 22

23 As a result, T ij = a b i j = R a R b k ik k l jl l For a particular choice of i, j, k, l, each of R, R, a, b ik il k l just numbers so we can change the order of our sums: are T a b ij = i j = R a R b R a k ik k = l jl l k = R a R b = R R a b ik k jl l l k k l ik jl k l R b ik k jl l l If we want, we can group things: T ij = R T R l k ik kl jl This is the th il component of the product R T The transpose of R is defined this way: RT R ij = ji (Useful: the transpose of a rotation matrix is also its inverse) This lets us write ( ) il ( ) T ij = RT R T l lj 23

24 From the definition of matrix multiplication, the sum is simply the ij th component of the product of the matrices RT and RT As a result, { RT} = T = T R R T R T I find I prefer to write it in terms of components: T ij = R R T kl ik jl kl Something that transforms this way is called a tensor (of rank 2) We ll see later how tensors can be useful 7 Non-cartesian coordinate systems Sometimes these will be convenient: don t be put off by their unfamiliarity! 24

25 cylindrical coordinates z r a x = r cosθ y = r sinθ z y θ x r, θ, z spherical coordinates z z = r cos θ θ r a x = r sin θ cos ϕ y = r sin θ sin ϕ y x ϕ r, θ, ϕ 25

26 The unit vectors changes which way they point as x moves around in space if we are working in a non-cartesian coordinate system This complicates the taking of derivatives Note the unfortunate change in the meaning of the angle θ when we switch from cylindrical to spherical coordinates 8 partial derivatives We ll work with functions of several variables Sometimes we ll want to know how the function changes if we change one variable while keeping all the others fixed For example, imagine we define a function h(x,y) that represents the height above sea level in Champaign as a function of latitude (x) and longitude (y) 2 2 Here s a graph of the function ( 03 ) 2 2 x y 5 3 h= x y e 26

27 The definition of a partial derivative with respect to y is this: (, ) hxy (, + Δy) hxy (, ) hxy y Note that x is held constant lim Δ y 0 Δy This amounts to measuring the slope as (in this case) you move parallel to the y axis It s easy to take partials: you treat all the variables except the selected one as if they were constants For example, if f 2 3 ( x, y, z) = x + 6 xy z + xz2 then 27

28 ( x2 ) f = + 6xy z + xz y y y y = xy z If you look in a small region around a particular x,y point any smooth function will look like a plane: using partial derivatives to approximate changes in functions How much does h change if we go from x,y to x+ Δx, y+ Δ y? ( ) ( ) ( ) ( ) Δ h = h x+δ x, y+δy h x, y { h x+δ x y+δy h x y+δ y } + { h( x y+δy) h( x y) } =,,,, 28

29 I ve just subtracted, then added the same thing For small Δx, Δy we have (, ) h x y+δy hx ( +Δ xy, +Δy) h( x, y+δy) Δx x from the definition of a partial derivative Also, as long as the partial derivative doesn t change violently with position, (, ) (, ) hxy+δy hxy x x As a result, Δh h Δ x + h Δy x y gradient operator The gradient operator (in two dimensions) is defined this way: xˆ + x yˆ y We use it like so: 29

30 hxy (, ) = xˆ h ( xy, ) + yˆ ( xy, ) x y If we move a small distance away from the point( x, y ) along the direction δ = xˆδ x + yˆδ y we find Δ h = h δ h x x + δ y y With our definition for, we can rewrite this as Δ h = δ ( h) In three (Cartesian) dimensions f x + y + z f x y z (,, x y z) ˆ ˆ ˆ ( x, y, z) Useful fact: the direction of changes most rapidly h is the direction in which hxy (, ) In cylindrical coordinates r, θ, z we have ˆ θ f r z r + + z f r z r r θ z (, θ, ) ˆ ˆ (, θ, ) 30

31 In spherical coordinates r, θ, ϕ we have ˆ θ ˆ ϕ f r r + + f r r r θ rsinθ ϕ (, θ, ϕ) ˆ (, θ, ) ϕ divergence operator The divergence operator acts on vector fields and is defined as specified below Cartesian coordinates: A A A y A z ( xyz,, ) ( xyz,, ) xˆ + ( xyz,, ) ˆ + ( xyz,, ) ˆ x y z A ( xyz) (,, ) ( ),, A xyz A ( xyz,, x y z ) A xyz,, + + x y Cylindrical coordinates: A r z A r z r A r z A r z (, θ, ) ( ) ˆ ( ) ˆ r, θ, + θ, θ, θ + z(, θ, ) 1 ( rar ) 1 Aθ Az Ar (, θ, z) + + r r r θ z Spherical coordinates: Arθϕ A rθϕ r A rθϕθ A rθϕϕˆ (,, ) (,, ) ˆ + (,, ) ˆ + (,, ) r θ 2 ( r Ar ) ( A sinθ ) A θ ϕ Ar (, θϕ, ) r r rsinθ θ rsinθ ϕ 31 z z zˆ

32 curl operator The curl operator acts on vector fields Its form in Cartesian coordinates is this: A ( xyz,, ) A y ( xyz,, ) A ( xyz,, ) x ( xyz,, ) A ( xyz,, ) A z + y ( xyz,, ) A ( xyz,, ) A x + z The forms in cylindrical and spherical coordinates are more complicated in appearance See a math reference for them x y z y z x zˆ xˆ yˆ 9 quadratic formula Very handy: if 2 ax bx c + + = 0 then x = ± 2 b b 4a 2a c 32