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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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