1624 sequence
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12. 1 A sequence is…
(a) an ordered list of objects.
(b) A function whose domain is a set of integers.
Domain: 1, 2, 3, 4, …,n…
Range a1, a2, a3, a4, … an…
1 1 1 11, , , , ...2 4 8 16
{(1, 1), (2, ½), (3, ¼), (4, 1/8) ….}
Finding patterns
Describe a pattern for each sequence. Write a formula for the nth term
1 1 1 11, , , , ...2 4 8 16
1 1 1 11, , , , ...2 6 24 120
1 4 9 16 25, , , , ...4 9 16 25 36
11
2n
1!n
2
2( 1)nn
Write the first 5 terms for
1 2 3 4 10, , , , ... ...2 3 4 5
nn
The terms in this sequence get closer and closer to 1. The sequence CONVERGES to 1.
On a number line As a function
1nnan
Write the first 5 terms
1 2 3 4 10, , , , . ... ...2 3 4 5
nn
The terms in this sequence do not get close toAny single value. The sequence Diverges
1( 1) 1n
nn
an
Write the terms for an = 3
The terms are 3, 3, 3, …3
The sequence converges to 3.
y= L is a horizontal asymptote when sequence converges to L.
A sequence that diverges
1( 1) 1n
nn
an
Sequences
Write the first 5 terms of the sequence. Does the sequence converge? If so, find the value.
1( 1)2 1
n
na n
1 1 1 11, , , ,3 5 7 9
1( 1)lim 0
2 1
n
n n
1 1( 1) 1nna n
1 2 3 40, , , ,2 3 4 5
1lim ( 1) 1n
n does not existn
The sequence converges to 0.
The sequence diverges.
12.2 Infinite Series
1
1 1 1 1 1 ...2 4 8 162nn
11 1 1 1 1 ...
n
The series diverges because sn = n. Note that the Sequence {1} converges.
Represents the sum of the terms in a sequence. We want to know if the series converges to a single value i.e. there is a finite sum.
1
1( 1)n n n
1
1 1 1 1 1 1 ...( 1) 2 6 12 20 30n n n
Partial sums of
11
1122
s
223
1 11 2 2 3
s
31 1 1
1 2 2 443
3 3s
1 1 1 1...1 2 2 3 3 4 ( 1 1)n
nsn nn
If the sequence of partial sums converges, the series converges
and
1 2 3 4 5, , , , ... ...2 3 4 5 6 1
nn Converges to 1 so series converges.
1
1 1 1 1 1 1 ...( 1) 2 6 12 20 30n n n
Finding sums1
1( 1)n n n
1
1
1( 1)
1 1 1 1 1 1 1 1 1(1 ) ( ) ( ) ... ...1 2 2 3 3 4 1
n
n
n n
n n n n
Can use partial fractions to rewrite
1
1 1lim (1 ) 1( 1) 1n
n n n n
.The partial sums of the series 1
1( 1)n n n
Limit
Geometric Series
1
1 1 1 1 1 ...2 4 8 162nn
Each term is obtained from the preceding number by multiplying by the same number r.
1 1 1 1 ...5 25 125 625
2 4 8 16 ...3 3 3 3
Find r (the common ratio)
Is a Geometric Series
1 1 1 1 ...2 4 8 16
3 12 48 192 ...5 25 125 625
2 4 8 16 ...3 3 3 3
1
1
n
nar
Where a = first term and r=common ratio
Write using series notation1
1
1 12 2
n
n
1
1
3 45 5
n
n
1
1
2 23
n
n
The sum of a geometric series
2 3 ... nnrs ar ar ar ar
2 3 1... nns a ar ar ar ar
nn ns rs a ar
( 11
1 ,)1
n
n
na ars a rr
rr
1 .| 0| , nr as nif r | | 1
1,n
as rr
Geometric series converges to
If r>1 the geometric series diverges.
Multiply each term by r
Sum of n terms
subtract
Find the sum of a Geometric Series
1
1, | | 1
1n
n
aar rr
Where a = first term and r=common ratio
1
1
1 12 2
n
n
1
1
3 45 5
n
n
1
1
2 23
n
n
The series diverges.
12 111
2
3
3 154 9 315
Repeating decimals-Geometric Series
2 4 6 88 8 8 80.080808 ...
10 10 10 10
21
12
8810
11 99110
n
n
aarr
The repeating decimal is equivalent to 8/99.
2 28 1
10 10a and r
11
2 21 1
8 110 10
nn
n nar
Series known to converge or diverge
1. A geometric series with | r | <1 converges
2. A repeating decimal converges
3. Telescoping series convergeA necessary condition for convergence:Limit as n goes to infinity for nth term in sequence is 0.
nth term test for divergence:If the limit as n goes to infinity for the nth term is not 0, the series DIVERGES!
Convergence or Divergence?
1
1010 1n
nn
1
1 12n n n
1
1.075 n
n
1
42nn
A series of non-negative terms convergesIf its partial sums are bounded from above.
A sequence in which each term is less than or equal to the one before it is called a monotonic non-increasing sequence. If each term is greater than or equal to the one before it, it is called monotonic non-decreasing.
A monotonic sequence that is boundedIs convergent.
12. 3 The Integral Test
Let {an} be a sequence of positive terms. Suppose that an = f(n) where f is a continuous positive, decreasing function of x for all xN. Then the series and the corresponding integral shown both converge of both diverge.
nn N
a
( )
N
f x dx
f(n) f(x)
The series and the integral both converge or both diverge
Exact area under curve is between
If area under curve is finite, so is area in rectangles
If area under curve is infinite, so is area in rectangles
Area in rectangle corresponds to term in sequence
Using the Integral test
21 1n
nn
22 2 1
1 12
1 2lim lim ln( 1)21 1
lim (ln( 1) ln 2)
b bb b
b
x xdx dx xx x
b
Thus the series diverges
2( )1n
na f nn
2(1
) xfx
x
The improper integral diverges
Using the Integral test
21
11n n
2 2 11 1
1 1lim lim arctan1 1
lim (arctan arctan1)2 4 4
bb
b b
b
dx dx xx x
b
Thus the series converges
2( ) 11na f n
n
21( )
1f x
x
The improper integral converges
Harmonic series and p-series
1
1p
n n
Is called a p-series
A p-series converges if p > 1 and divergesIf p < 1 or p = 1.
1
1 1 1 1 1 11 .... ....2 3 4 5n n n
Is called the harmonic series and it diverges since p =1.
11 3
1
n n
Identify which series converge and which diverge.
1
1
n n
1 3
1
n n
1
1
n n
21
100
n n
1
1
3 45 5
n
n
12. 4Direct Comparison testLet
1n
na
be a series with no negative terms
1n
na
Converges if there is a series
Where the terms of an are less than or equal to the terms of cn for all n>N.
1n
nc
1n
na
Diverges if there is a series
Where the terms of an are greater than or equal to the terms of dn for all n>N.
1n
nd
Limit Comparison test
both converge or both diverge:
Limit Comparison test
lim , 0nx
n
a c cb
Then the following series1n
na
1n
nb
and
lim 0nx
n
ab Amd
1n
nb
Converges then
1n
na
Converges.
lim nx
n
ab
1n
nb
Diverges then
1n
na
Diverges.
and
and
Convergence or divergence?
1
12 3nn
21 1n
nn
1
13 2n n
Alternating Series
A series in which terms alternate in sign
1( 1)n n
na
1
1( 1)n n
na
or
1
1
1 1 1 1 1( 1) ...2 4 8 162
nn
n
1
1 1 1 1( 1) 1 ...2 3 4
n
n n
Alternating Series Test1
1 2 3 41( 1) ...n
nn
a a a a a
Converges if:
an is always positive
an an+1 for all n N for some integer N.
an0If any one of the conditions is not met, theSeries diverges.
Absolute and Conditional Convergence
• A series is absolutely convergent if the corresponding series of absolute values converges.
• A series that converges but does not converge absolutely, converges conditionally.
• Every absolutely convergent series converges. (Converse is false!!!)
nn N
a
nn N
a
Is the given series convergent or divergent? If it is convergent, its it absolutely convergent or conditionally convergent?
( 1) / 2
1
( 1)3
n n
nn
1
( 1)ln( 1)
n
n n
1
( 1)n
n n
1
1
( 1) ( 1)n
n
nn
a) Is the given series convergent or divergent? If it is convergent, its it absolutely convergent or conditionally convergent?
( 1) / 2
1
( 1) 1 1 1 1 ...3 9 27 813
n n
nn
This is not an alternating series, but since ( 1) / 2
1 1
( 1) 13 3
n n
n nn n
Is a convergent geometric series, then the givenSeries is absolutely convergent.
b) Is the given series convergent or divergent? If it is convergent, its it absolutely convergent or conditionally convergent?
1
( 1) 1 1 1 ......ln( 1) ln 2 ln3 ln 4
n
n n
Diverges with direct comparison with the harmonicSeries. The given series is conditionally convergent.
1
( 1) 1 1 1 ......ln( 1) ln 2 ln3 ln 4
n
n n
Converges by the Alternating series test.
c) Is the given series convergent or divergent? If it is convergent, its it absolutely convergent or conditionally convergent?
1
1
( 1) ( 1) 2 3 4 51 2 3 4
n
n
nn
By the nth term test for divergence, the seriesDiverges.
d) Is the given series convergent or divergent? If it is convergent, its it absolutely convergent or conditionally convergent?
1
( 1) 1 1 1 11 2 3 4
n
n n
Converges by the alternating series test.
1
( 1) 1 1 1 11 2 3 4
n
n n
Diverges since it is a p-series with p <1. TheGiven series is conditionally convergent.
The Ratio Test
Let be a series with positive terms and nn N
a
1lim nn
n
aa
Then• The series converges if ρ < 1• The series diverges if ρ > 1• The test is inconclusive if ρ = 1.
The Root Test
Let be a series with non-zero terms and nn N
a
lim | |nn na L
Then• The series converges if L< 1• The series diverges if L > 1 or is infinite• The test is inconclusive if L= 1.
1
2!
n
n n
2 11
32
n
nn n
Convergence or divergence?
2
1
n
nn
en
. Procedure for determining Convergence
Power Series (infinite polynomial in x)
20 1 2
0..... ...n n
n nnc x c c x c x c x
20 1 2
0( ) ( ) ( ) ..... ( ) ...n nn n
nc x a c c x a c x a c x a
Is a power series centered at x = 0.
Is a power series centered at x = a.
and
Examples of Power Series 2 3
01 ...
! 2 3!
n
n
x x xxn
2
0
( 1) 1 1 1( 1) 1 ( 1) ( 1) ..... ( 1) ...3 93 3
nn n
n nn
x x x x
Is a power series centered at x = 0.
Is a power series centered at x = -1.
and
Geometric Power Series
2 3 4
01 ...n n
nx x x x x x
1 , 11 1
1aS xr x
a and r x
12
22 3
3
1
1
1
P x
P x x
P x x x
. The graph of f(x) = 1/(1-x) and four of its polynomial approximations
Convergence of a Power Series
There are three possibilities 1)There is a positive number R such that the
series diverges for |x-a|>R but converges for |x-a|<R. The series may or may not converge at the endpoints, x = a - R and x = a + R.
2)The series converges for every x. (R = .)
3)The series converges at x = a and diverges elsewhere. (R = 0)
What is the interval of convergence?
0( 1)n
n
Since r = x, the series converges |x| <1, or -1 < x < 1. In interval notation (-1,1).Test endpoints of –1 and 1.
0(1)n
n
Series diverges
Series diverges
2 3 4
01 ...n n
nx x x x x x
Geometric Power Series
1 3 31 3 ( 1) 41
11 ( 1)3
1 ( 1)3
a and r x
x
aSr x x
2
0
( 1) 1 1 1( 1) 1 ( 1) ( 1) ..... ( 1) ...3 93 3
nn n
n nn
x x x x
1 ( 13
1 1 1
)
( )3
r x
x
Find the function
Find the radius of convergence
2 4x
Geometric Power Series
0
( 1) ( 1)3
nn
nn
x
Find the radius of convergenceFind the interval of convergence
For x = -2,
0 0 0
( 1) ( 1) ( 1) 1( 2 1)3 3 3
n n nn
n n nn n n
Geometric series with r < 1, converges
0 0 0 0
( 1) ( 1) ( 3) 3( 4 1) 13 3 3
n n n nn
n n nn n n n
By nth term test, the series diverges.
For x = 4
2 4x Interval of convergence
Finding interval of convergence
1x
Use the ratio test: 1
1 1
n n
n nx xu and un n
1
1lim lim1
nn
n n nn
u x n xu n x
R=1For x = 1 For x = -1
0
n
n
xn
(-1, 1)
0
1
n n
0
( 1)n
n n
Harmonic series diverges
Alternating Harmonic series converges
11lim lim
1
nn
n n nn
u x n xu n x
[-1, 1)Interval of convergence
Differentiation and Integration of Power Series
20 1 2
0( ) ( ) ( ) ..... ( ) ...n nn n
nc x a c c x a c x a c x a
If the function is given by the series
Has a radius of convergence R>0, the on the interval (c-R, c+R) the function is continuous, Differentiable and integrable where:
1
0( ) ( )nn
nf x nc x a
0
( )( )1
n
nn
x af x dx C cn
The radius of convergence is the same but the interval of convergence may differ at the endpoints.
Constructing Power Series
If a power series exists has a radius of convergence = RIt can be differentiated
20 1 2( ) ( ) ( ) ..... ( ) ...n
nf x c c x a c x a c x a
2 11 2 3( ) 2 ( ) 3 ( ) ..... ( ) ...n
nf x c c x a c x a nc x a
22 3( ) 2 2*3 ( ) 3* 4( ) ....f x c c x a x a
23 4( ) 1* 2*3 2*3* 4 ( ) 3* 4*5( ) ...f x c c x a x a
( )( ) ! ( )nnf x n c terms with factor of x a
So the nth derivative is
( )( ) ! ( )nnf x n c terms with factor of x a
All derivatives for f(x) must equal the seriesDerivatives at x = a.
1
2
3
( )( ) 1* 2( ) 1* 2*3
f a cf a cf a c
( ) ( ) !nnf a n c
( )( )!
n
nf a cn
Finding the coefficients for a Power Series
( )2 3
0( )
( ) ( ) ( )( ) ( ) ( ) ( ) ...! 2! 3!
( ) ...!
k
kn
n
f a f a f af a f x a x a x ak
f a xn
If f has a series representation centered at x=a, the series must be
If f has a series representation centered at x=0, the series must be
( )2 3
0( )
(0) (0) (0)( ) (0) ( ) ( ) ...! 2! 3!
(0) ...!
k
kn
n
f f ff a f x a x ak
f xn
Form a Taylor Polynomial of order 3 for sin x at a =
n f(n)(x) f(n)(a) f(n)(a)/n!
0 sin x
1 cos x
2 -sin x
3 -cos x
22
22
22
22
22
22
22* 2!
22*3!
4
The graph of f(x) = ex and its Taylor polynomials
2 3 4
01 ...
! 2 3! 4!
n
n
x x x xxn
Find the derivative and the integral
2 1 3 5 7
0
1...
(2 1)! 3! 5! 7!
n n
n
x x x xxn
Taylor polynomials for f(x) = cos (x)
Converges only at x = 0
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