Math 262 Equation Sheet
Math 262 Equation Sheet INFINITE SEQUENCES
ak .
f (x) =
Review Let
{a n } be a sequence. If nlim an → +∞
=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 → +∞
lim(an +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
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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 ⎜⎜ ⎟ u v ⎝ ⎠
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:
www.prep101.com f ( a + h, b ) − f ( a )
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 dt
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
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
∂ ∂ ∇f ( x, y ) = $i ( x, y ) + $j ( 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.”
ak .
f (x) =
Review Let
{a n } be a sequence. If nlim an → +∞
=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 → +∞
lim(an +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 ⎜⎜ ⎟ u v ⎝ ⎠
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:
www.prep101.com f ( a + h, b ) − f ( a )
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 dt
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
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
∂ ∂ ∇f ( x, y ) = $i ( x, y ) + $j ( 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.”
