Math 262 study sheet

Math 262 study sheet - Math 262 Equation Sheet INFINITE...

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Unformatted text preview: Math 262 Equation Sheet INFINITE SEQUENCES ak . f ( x) = Review Let {a n } be a sequence. If nlim a n → +∞ =L lim a n [ then forms: use Divergence Test lim a n = L n → +∞ Name Root Test lim c = c lim ca n = c lim a n = cL n → +∞ n→ +∞ n→ +∞ n→ +∞ n→+∞ n→+∞ n→+∞ n→+∞ n→+∞ lim a n n→+∞ a lim n = n→+∞ = n → +∞ bn lim bn M M ≠0 Squeeze Theorem For Sequences Let that {a n }, {bn }, and {cn } be sequences such {a n } ≤ {cn } ≤ {bn } . If lim a n = L and ∑ 1 ∑1 bk ∞ a k ≤ bk for all k . converges, then ∑ ∞ 1 ∑ ∞ 1 ak ak arctan x = x − ex = 1+ ∞ 1 ak and Name Alternating Series Test x3 x5 x7 + − +. 3 5 7 x x2 x3 x 4 + + + + ... 1! 2! 3! 4! cos x = 1 − diverges. ∑ x2 x3 x4 + − + .... 2 3 4 x3 x5 x7 x9 + − + .... sin x = x − 3! 5! 7! 9! diverges, then also Suppose that = 1 + 2 x + 3 x 2 + 4 x 3 + ..... bk ∑ ∞ 1 −1 ≤ x ≤ 1 ∞ = ∑ xn ∞ = ∑ nx n −1 −1 ≤ x ≤ 1 x 2 x 4 x6 x8 + − + − .... 2! 4! 6! 8! n ∞ = ∑ (− 1) 1 n ∞ = ∑ (− 1) 1 ∞ =∑ 1 ∞ ∞ (2n + 1)! n −1 ≤ x ≤ 1 −1 ≤ x ≤ 1 (− ∞, ∞ ) x 2 n +1 n = ∑ (− 1) 1 x 2 n +1 2n + 1 xn n! = ∑ (− 1) 1 xn n x 2n (2n)! (− ∞, ∞ ) (− ∞, ∞ ) bk are two series with positive terms with ak bk If p > 0 ⇒ either both series n → +∞ TESTS TO DETERMINE THE CONVERGENCE/DIVERGENCE OF A SERIES ak ∞ VECTORS AND COORDINATE GEOMETRY k → +∞ lim bn = L ⇒ lim c n = L n → +∞ ∞ Radius of Convergence 1 p = lim n → +∞ and ∑ (1 − x )2 ln ( x + 1) = x − are two series with positive terms ∑1 bk Limit Comparison Test n → +∞ p = 1 ⇒ no conclusion. ∞ L provided that Sigma Notation 1 = 1 + x + x 2 + x 3 + x 4 . + ... 1− x 1 1 If Form 1 converges. lim(anbn ) = liman • limbn = LM Quotient POWER OF SERIES AND RADIUS OF CONVERGENCE p = lim k a k Suppose that If lim(an − bn ) = lim an − lim bn = L − M Product ak +1 ak is a series of positive terms such that n → +∞ ( liman +bn ) = liman + limbn = L+M Difference k a k +1 ak is a series of positive terms k If p > 1 ⇒ the series diverges. If p < 1 ⇒ the series converges. Comparison Test p = lim If p > 1 ⇒ the series diverges. k → +∞ p = lim ∑a If is a series of nonzero terms If p < 1 ⇒ the series converges. If p = 1 ⇒ no conclusion. k → +∞ n → +∞ Constant Multiple Additive =±∞ lim a k ≠ 0 diverges if k p = 1 ⇒ no conclusion. with and Property Constant k k → +∞ k → +∞ a ∑a If then the following results are true n → +∞ Convergence If p > 1 ⇒ the series diverges. If p < 1 ⇒ the series converges. Properties of Sequences lim bn = M ∫ f ( x)dx k →+∞ f ( x) f ' ( x) = lim lim x→a g ( x) x →a g ' ( x) sequences with ∑a with {a n } and {bn } be two convergent ∑a Ratio Test Absolute with +∞ lim a k = 0 decreasing, and 2. a Ratio Test L’Hoptial’s Rule is defined. The series f ( x)dx diverges if L’Hopital’s Rule to evaluate the limit. Suppose that ) +∞ ∫ 0∞ 0 0∞ , ,0 , ∞ ,1 , ∞ − ∞ 0∞ The series converges if and only if 1. a k +1 ≤ a k ; the series is if results in either one of the following n → +∞ If the function is decreasing and continuous on a,+∞ , then the series converges the sequence converges. If − a1 + a 2 − a3 + a 4 − .... and converge, or both series diverge. What Does It Say Suppose that ak > 0 for k = 1,2,3,... then the series has the Vectors & Vector Dot Products Let u be a vector having n components, then the length or norm of is u defined to be form of either: Name Integral Test ∑a k What Does It Say is a series of positive terms a1 − a 2 + a3 − a 4 +…. or u= (n1 )2 + (n2 )2 + (n3 )2 + ... + (n n )2 We’ve helped over 50,000 students get better grades since 1999! Need help for exams? Check out our classroom prep sessions - customized to your exact course - at www.prep101.com u t 2. s = s(t ) = ∫ r ' (t ) dt v Let and be two vectors, then their dot product is defined to be: lim ( x , y ) → ( a ,b ) f ( x , y ) = f ( a , b) t0 Partial Derivatives u ⋅ v = u v cos θ Formulas for Curvature, Torsion, Radius of Curvature Properties Name Symmetry u ⋅v = v⋅u k Linearity Suppose that u is a vector having n components, then its unit vector is found be dividing each component of the vector by its norm. That is, 1 u= u (n1 ) Similarly the partial with respect to y, at the point (a, b ) , denoted as f , is defined to be 1 y κ v×a v×a f y = lim k →0 (v × a ) da f ( a , b + k ) − f (b ) k In practice we use the following f x = f1 = ∂f ∂x ⇒ f y = f2 = ∂f ∂y ⇒ dv 2 T + v κN dt dt v×a h 2 differentiate the function f with respect to x while holding all other variables as constants. differentiate the function f with respect to y while holding all other variables as constants. Frenet – Serret Formulas + (n2 ) + (n3 ) + ... + (n n ) 2 a= τ= 1 2 3 dT N = B × T = dt dT dt Tangential and Normal Components Torsion Unit Vectors = B= h→0 v×a ρ= Unit Normal Vector ⎛u ⋅v⎞ ⎟ θ = cos −1 ⎜ ⎜ ⎟ uv⎠ ⎝ unit κ= Binormal Vector ⇔ f x (a, b) = lim v Radius of Curvature A. Show orthoganality: u ⋅ v = 0 u⊥v B. Finding the angle between two vectors: w ww.prep101.com f (a) f ( a + h, b ) − v v Uses of Dot Product u T= Curvature R x , denoted as f x , at the point (a, b) is defined to be Formula Unit Tangent Vector (k u )⋅ v = k (u ⋅ v) ∈ u (v + w) = u ⋅ v + u ⋅ w Multiplication with a scalar Given a function f ( x, y ) , the partial with respect to 2 2 u dT = κN ds dB = −τ N ds dN = −κ T + τ B ds The idea can easily be extended to functions having more than 2 variables. Higher Derivatives VECTOR VALUED FUNCTIONS Tangent Directions If to the points on the curve is a position vector (x(t ), y(t ), z(t )) then we can find the following quantities v avg = r (t + ∆t ) − r (t ) ∆t = , ∆t change in time Velocity v (t ) = r ' (t ) = dr dx dy [ y(t )] j + dz [z (t )]k = [x(t )]i + dt dt dt d t lim ( x , y ) →( a ,b ) f ( x, y ) = L Implies that A. Every neighborhood of (a, b) contains points of the domain of f ( x, y ) different from (a, b). B. For every ε (positive) there exists a positive δ such that f ( x, y ) − L < ε whenever (x, y) is in the domain of inequality: Speed 0< f and satisfies the Arc – length a(t) =r''(t) = ∂ 2 f = ∂ ⎛ ∂f ⎞ ⎜⎟ ∂x 2 ∂x ⎝ ∂x ⎠ ∂ 2 f ∂ ⎛ ∂f ⎞ = 2= ⎜ ⎟ ∂y ∂y ⎜ ∂y ⎟ ⎝⎠ f yy = f 22 Second derivative of f with respect to x Second derivative of f with respect to y f xy = f 12 = ∂2 f ∂ ∂f =⎛⎞ ⎜⎟ ∂y∂x ∂y ⎝ ∂x ⎠ Mixed partial f yx = f 21 = ∂2 f = ∂ ⎛ ∂f ⎞ ⎜⎟ ∂x∂y ∂x ⎜ ∂y ⎟ ⎝⎠ Mixed partial (x − a )2 + ( y − b )2 < δ v(t ) Acceleration Second derivatives for functions of 2 variables f xx = f 11 = Definition of Limit on a plane ˆ ˆ r (t ) = x(t )i + y (t ) ˆ + z (t )k j Average Velocity PARTIAL DIFFERENTIATION Chain Rule dy dz d2r dx = [x'(t)]i + [ y'(t)] j + [z'(t)]k dt dy dt2 dt Definition of Continuity on a Plane For continuity to happen, the following 2 conditions must hold: 1. lim ( x , y ) →( a ,b ) f ( x, y ) exists z = f ( x, y ) is a differentiable function of y with x = x(t ) and y = y (t ) then z ' (t ) = ∂z ∂x ∂z ∂y + ∂x ∂t ∂y ∂t Our Course Booklets - free at prep sessions - are the “Perfect Study Guides.” x and Need help for exams? Check out our classroom prep sessions - customized to your exact course - at www.prep101.com Example: z ( x, y ) = x sin y where x (t ) = e t and y (t ) = t Solution: z ' (t ) = ∂z ∂x ∂z ∂y + ∂x ∂t ∂y ∂t differentiable function of m variables, −∇f ( x, y ) = Direction of the steepest slope t1 , t 2 , t 3 ,..., t m then z is ultimately a function of t1 , t 2 , t 3 ,..., t m . So 2 downwards in the XY plane. ∇f ( x, y ) = Slope in steepest direction ∂z ∂z ∂x1 ∂z ∂x 2 ∂z ∂x n = + + ... + ∂t i ∂x1 ∂t i ∂x 2 ∂t i ∂x n ∂t m ∂x ∂y = et = 2t ∂t ∂t ∂z ∂z = sin y = x cos y ∂x ∂y z ' (t ) = e t sin y + 2tx cos y $ Du f = u ⋅∇f $ respect to u w ww.prep101.com $$ Tangent Planes & Gradients Suppose that z= f ( x, y ) has tangent planes everywhere, then the gradient is: z be a differentiable function of n variables, x1 , x 2 , x 3 ,..., x n and each x j is a Let u + kDu f = Tangent vector with respect to $ k − ∇f = Upward normal vector to the surface = Perpendicular to all tangent vectors = Normal vector to the tangent plane. ∇f ( x, y ) = Direction of the steepest slope upwards in the XY plane LOCAL EXTREMAS Suppose that the surface uv ⇒ ∇f D 2 u f ( x0 , y0 ) z = f ( x, y ) has tangent planes at every point. 2 u ( x0 , y0 , z0 ) on the surface where the tangent plane is horizontal = 0 , are called critical point (local extrema). p o s itiv e fo r a ll θ ⇒ ⇒ ⇒ D The points f n e g a tiv e f o r a ll θ ⇒ ⇒ ⇒ D u2 f c h a n g e s s l o p e ⇒ n e a rb y s lo p e s a r e u p su rfa c e b e n d s u p lo c a l m in im u m n e a r b y s lo p e s a r e d o w n su rfa c e b e n d s d o w n lo c a l m a x im u m n e a r b y s lo p e s is u p in s o m e d ir e c tio n s a n d d o w n in o th e r d ir e c tio n s ⇒ su rfa c e b e n d s u p a n d d o w n ⇒ s a d d le The test can be expressed in terms of the Jacobian matrix ⎡ ∂2 f ∂2 f ⎤ ⎢2 ⎥ ∂x∂y ⎥ ⎡a b ⎤ ∂x = D=⎢ 2 ∂ 2 f ⎥ ⎢c d ⎥ ⎢∂ f ⎣ ⎦ ⎢ ∂x∂y ∂y 2 ⎥ ⎣ ⎦ If d e t(D )< 0 ⇒ w e h a v e a s a d d le p o in t If d e t(D )< 0 and a+c> 0 ⇒ w e h a v e a lo c a l m in im u m If d e t(D )< 0 and a+c< 0 ⇒ w e h a v e a lo c a l m a x im u m Lagrange Multipliers To find the extreme values: 1. 2. First get it into a Lagrangian function: L ( x, y ) = f ( x, y ) + λ g ( x, y ) Solve for the critical point by finding out when 0= 3. $ u direction ∂ ∂ i j ∇f ( x, y ) = $ ( x, y ) + $ ( x, y ) ∂x ∂y The chain rule can be extended to include functions which have more than one variable. = Directional derivative of f with ∂L ∂L ∂L = f x ( x, y ) + λ g ( x , y ) , 0 = g ( x, y ) = f y ( x , y ) + λ g ( x, y ) , 0 = ∂x ∂λ ∂y Once the critical points have been found, plug it back into the function and test to see which point satisfies the given criteria. Our Course Booklets - free at prep sessions - are the “Perfect Study Guides.” ...
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