THEOREM 2 Sum of a Geometric Series Let c 0. If |r| < 1, then

If |r| ≥ 1, then the geometric series diverges.

2 3

0 1n

n

ccr c cr cr cr

r

1

2 31

1

N

NN

c rS c cr cr cr cr

r

1 2 3

1

Mn M M M M

n M

crcr cr cr cr cr

r

N

Sum of an Infinite Geometric Series (80)

0 0

1 1 11 1 & 5

5 5 1

5

/ 44 5

nn

n n

cc r

r

Evaluate

3

3

37

3 47

4 1 7 6/ 4

27

1

n M

n

cr

r

2 3

0 1n

n

ccr c cr cr cr

r

1 2 3

1

Mn M M M M

n M

crcr cr cr cr cr

r

3 4 5

3

3 3 3 37 7 7 7 .

4 4 4 4

n

n

Sum of an Infinite Geometric Series (80)

0

Evaluate 5 .n

n

0

2 3Evaluate .

5

n

nn

S

Write S as a sum of two geometric series. This is valid by Theorem 1 because both geometric series converge:

0 0 0

1 3 1 32 2

5 5 5 5

2 1 5 5

4 /5

5 2 / 5 2 2

n n n n

n n n

S

Sum of an Infinite Geometric Series (80)

Does the given series converge?

THEOREM 1 Ratio Test Assume that the following limit exists:

(i) If ρ < 1, then na converges absolutely.

(ii) If ρ > 1, then na diverges.

(iii) If ρ = 1, the test is inconclusive (the series may converge or diverge).

20

!

n

nn an

2 3 42 2 2 2

11! 2! 3! 4!

S

1lim n

nn

a

a

Used for interval of convergence questions :-)

Ratio Test (Convergence) (81)

1

1 12 ! 2 lim 0

1 ! 2 1

nn n

n nn n

a an

a n n a

Prove that 1

2

!

n

n n

converges.

THEOREM 1 Ratio Test Assume that the following limit exists:

(i) If ρ < 1, then na converges absolutely.

(ii) If ρ > 1, then na diverges.

(iii) If ρ = 1, the test is inconclusive (the series may converge or diverge).

Compute the ratio and its limit with Note that2

.!

n

nan

1 ! 1 !n n n

the series converges,

by the ratio test.

1lim n

nn

a

a

R

(81)

Does2

1 2nn

n

converge?

2 21

1 2 2

1 2 1 2 1

2 2

nn

nn

na n n

a n n

1 1

lim2

n

nn

a

a

the series converges, by the ratio test. R

Ratio Test (Convergence) (81)

Does 1

!1

1000n

nn

n

converge?

the series diverges, by the ratio test.

11

1 ! 1000 1lim lim lim

1000 ! 1000

nn

nn n nn

na n

a n

R DNE

Ratio Test (Convergence) (81)

11

1

2 1 1lim lim lim

2 2 2

n nn

n nn n nn

a xx x

a x

Using the Ratio Test Where does converge?

2

n

n n

xa

Step 1. Find the radius of convergence.

Let and compute the ratio ρ of the Ratio Test:

We find that1

1 if 1 that is, if 22

x x

0 2

n

nn

xF x

Ratio Test

Where's our ?n

The radius of convergence 2.2 is 2 Rx

2 or 2 diverges x x F x

now has a variable hang on to the na

Ratio Test (Interval of Convergence) (81)

Using the Ratio Test Where does converge? 0 2

n

nn

xF x

Step 2. Check the endpoints.

The Ratio Test is inconclusive for x = ±2, so we must check these cases directly:

Both series diverge. We conclude that F(x) converges only for |x| < 2 (Figure 2).

0

22 1 1 1 1 1 1

2

n

nn

F

0 0

22 1 1 1 1 1 1 1

2

nn

nn n

F

The radius of convergence 2.2 is 2 Rx

Ratio Test (Interval of Convergence) (81)

Maximum Error in the Approximationof an Alternating Series

(82)

Alternating Harmonic Series Show that

converges conditionally. Then:

(a) Show that

(b) Find an N such that SN approximates S with an error less than 10−3.

1

1

1n

n

Sn

6

1

7S S

The terms are positive and decreasing, and

Therefore, S converges by the Leibniz Test. The harmonic series

diverges, so S converges conditionally but not absolutely.

1na

n lim 0.nn

a

1

1

n n

1

1 1

11n

n nn n

Maximum Error in theApproximationof an Alternating Series

(82)

3 3110 1 10 999

1N N

N

Alternating Harmonic Series Show that

converges conditionally. Then:

(a) Show that

(b) Find an N such that SN approximates S with an error less than 10−3.

Now, applying Eq. (2), we have

We can make the error less than 10−3 by choosing N so that

1

1

1n

n

Sn

6

1

7S S

1

1

1N NS S aN

6 7

1.

7S S a

"The distance between the sum and any partial sum is less than the next term in the sequence."

1na

n is +, , & converges to 0.na

by the Leibniz Test

the harmonic series diverges

Maximum Error in the Approximation of an Alternating Series(82)

Partial Sums & Taylor Series (83)7.

4

7 4

1 1 20

3 15 24

11

6041

63

3 4 5

35 48 063

S

S S

THEOREM 1 Taylor Series Expansion If f (x) is represented by a power series centered at c in an interval |x − c| < R with R > 0, then that power series is the Taylor series

In the special case c = 0, T (x) is also called the Maclaurin series:

0 !

nn

n

f cT x x c

n

4

2 3 4

0

0 " 0 ''' 0 00 ' 0

! 2! 3! 4!

nn

n

f f f ff x x f f x x x x

n

We define the nth Taylor polynomial centered at x = a as follows:

2' "

1! 2! !

nn

n

f a f a f aT x f a x a x a x a

n

Taylor Series of f (x) centered at x = c:

2 3'' ''''

2! 3!

f c f cT x f c f c x c x c x c

(83)

4 5

3

' 3 , " 3 4 ,

and in general, 1 3 4 2nn n

f x x f x x

f x n x

Find the Taylor series for f (x) = x−3 centered at c = 1.

21

Note that 3 4 2 2 !.2

n n

1 3 1 4

2 3 2 5

3 3 3 6

i.e. ' 1 3 3

" 1 3 2 2 12

''' 1 3 4 3 2 60

f x x x

f x x x

f x x x

11 1 2 !

2nnf n

0 !

nn

n

f cT x x c

n

Patterns Factored FormExponent in terms of ???n

(83)

1 11

2

2 ! 1 2 1 !1 2 2! !

11

!

2

n n

n

n

n

n n n nfa

n n nn n

2 ! 2 1 !n n n n

we can write the coefficients of the Taylor series as:

The Taylor series for f (x) = x −3 centered at c = 1 is

0 !

nn

n

f cT x x c

n

2 3

0

1 3 1 6 1 10 1

2 11 1

2n n

n

T x x x x

n nx

11 1 2 !

2nnf n

Partial Sums & Taylor Series (83)

Lagrange Form of the Remainder (84)

2' "

1! 2! !

nn

n

f a f a f aT x f a x a x a x a

n

approximates centered at .f x a

1

1

1 !

nn

n

f zR x x a

n

,z a x

Lagrange Form of the Remainder (84)Approximate (near 1) with a 4th degree Taylor polynomial.

What is the remainder?

xy e x

2 3 4

4 1 1 1 10! 1! 2! 3! 4!

e e e e eP x x x x x

We are figuring out the remainder associated with our 4th

degree polynomial.

5

4

551 1

55 ! !

zf zR

exx x

Maclaurin S Converges terie o s r fof xf x

2 3 4 5

0

2 1 3 5 7

0

2 2 4 6

0

sin

cos

1 ! 2! 3! 4! 5!

1 2 1 ! 3! 5! 7

!

1 1 2 ! 2! 4

All

All

! 6

!

xn

n

nn

n

nn

n

x x x x xx

n

x x x xx

n

x x x x

n

e x

xx

x

2 3 4

0

1

1

All

1 1

n

n

x x x x xx

x

x

These 3 expansions should be memorized!

(85) (86) (87)

(88) sin cos ?m n

x x dx

2

is odd and positive make one factor of sine your and

convert everything else to cosine

is odd and positive make one factor of cosine your and

convert everything else

Otherw

to sine

s

ise use:

in

m du

n du

x

21 cos2 1 cos2 and cos

2 2

x xx

(88) sin cos ?m n

x x dx

2 2

2 4 2 4 4 2

5 3

5 3 4 2

2 2 3 2

5 31 1

c

sin 1 cos cos

sin cos cos

5 3Check...

1 1cos cos cos sin cos sin

5 3

sin cos c

cos cos

os 1

os si

sin co

n

s

5

3

x x x dx

x x x dx u u du u u du

u uC

dx x C x x x x

d

x x C

u

x

x x x x x

x du xdx

3 2sin cos ?x xdx

THEOREM 2 Slope of the Tangent Line Let c (t) = (x (t), y (t)), where x (t) and y (t) are differentiable. Assume that

' is continuous and ' 0. Thenx t x t

CAUTION Do not confuse dy/dx with the derivatives dx/dt and dy/dt, which are derivatives with respect to the parameter t. Only dy/dx is the slope of the tangent line.

'/

/ '

y tdy dy dt

dx dx dt x t

Derivative of a Parametric Curve (89)

Let c (t) = (t2 + 1, t3 − 4t). Find:(a) An equation of the tangent line at t = 3(b) The points where the tangent is horizontal.

'/

/ '

y tdy dy dt

dx dx dt x t

23 4

2

dy t

dx t

3

23

6t

dy

dx

2315 1015

63 10,c y x

0 ' 0 & ' 0dy

y t x tdx

2 2' 3 4 0

3y t t t

7 16,

3

2

33 3c

7 16,

2

3 3 3 3c

Derivative of a Parametric Curve (89)

2 2

3

2

2 3

2

2

3 42

12 6

( ) ( 1, 4 )2

6 2 3 4 2

4 8

82

d tdt t

c t t t tt

t t t

tt

t t

t

Let c (t) = (t2 + 1, t3 − 4t). Find the acceleration

'/

/ '

y tdy dy dt

dx dx dt x t

2

2.

d y

dx

2

2

d dyd y d dy dt dx

dxdx dx dxdt

Parametric Acceleration (90)

2

Definition of Arc Length

( ) is a curve on , The arc length of

between and is

1 '( )

( ) is

smooth

smootha curve on , The arc length of

betwe

b

a

y f x a b f

a b

s f x dx

x g y c d g

2

en and is

1 '( )d

c

c d

s g y dy

a b

c

d

Arc Length on a Function (91)

Arc Length from 0, / 4 ?

ln cosy x

1

2

2

Arc Length on a Function (91)

APPS; A+ Calc; Calc Pack 2; Arc Length...

5/6/13... I'm getting an error!

sintan

cos

dy xx

dx x

4

0

2/1 tan xs dx

/ 4

0sec xdx

/ 4

0ln sec tanx x

ln 2 1 ln1

0.882

21 '( )

b

as f x dx

'ln

d uu

dx u

sec ln sec tanudu u u C

ln cosy x

(91) Arc Length from 0, / 4 ?

Arc Length on a Parametric Curve (92)

2 2 2 2

2

2 2

' ' 4 1 cos 4sin

4 cos

1 c

1 2 cos

os8 1 cos 16

2

16si

in

n

s

2

t

x t y t t t

t

tt

t

t

Calculate the length s of one arch of the cycloid generated by a circle of radius R = 2.

sin , cosx Rt R t y R R t

2 2' '

b

a

s x t y t dt

2 sin , 2 1 cosx t t y t

' 2 1 cos , ' 2sinx t t y t t

2

0 0

0

0

4 sin 8 sin2

8 cos 8 co 1s

/ 2 / 2

6

ts dt udu

u u

u t du dt

2

& now, back to

our Trig Iden1

tic

tieos

s... n si2 2

t t

(92)

Calculate the length s of one arch of the cycloid generated by a circle of radius R = 2.

sin , cosx Rt R t y R R t (92)

TI-84Plus: PRGM; APCALC2; A; 2;

2 2sin ; 2 2cos ; 0;

16

2x t t y t a b

L

Since x = r cos θ and y = r sin θ, we use the chain rule.

sin cos/

/ cos sin

drrdy dy d d

drdx dx d rd

To find the slope of a polar curve r = f (θ), remember that the curve is in the x-y plane, and so the slope is .

dy

dx

Product Rule...

Slope of a Polar Curve (93)

Horizontal & Vertical Tangents of a Polar Curve (93) (94)The equation r = 4 sin θ defines a circle of radius 2 tangent to the x-axis at the origin. Find its horizontal and vertical lines,

0,2

0 .

Write the equation in parametric form.24sin cos , 4sinx y

2 2 34 cos sin 4cos2 0 , vertical tangents

4 4

8sin cos 4sin 2 0 0, horizontal tangents2

dx

ddy

d

2 2 2 24 2, 4 2

2 2 2 2x x

2,4

2,4

4 0 0, 4 1 4y x

0,0

4,2

x

RShell Method

c

d

0

Write and in terms of .

Note: 4 4

R L y

x x y

L

2d

c

V R Ldy

shellArea 2 RL

Shell Method (95)

y

a b

Shell Method

0

Write and in terms of .

Note: 4 4

R L x

y y x

2b

a

V R Ldx

shellArea 2 RL

R

L

Shell Method (95)

Shell Method (95)

1,0

3

Find the volume of the solid of revolution formed by

revolving the region bounded by

and the -axis 0 1 about the -axis.

y x x

x x y

Implies everything is in terms of x.

1 3

02 x x x dx

1 2 4

0

13 5

0

2

23 5

x x dx

x x

1 12

3 5

4

15

x

y3y x x

Shell Method (95)

1 2

0

1 3

0

14 2

0

2 ( 1)

2

24 2

x x dx

x x dx

x x

2

Find the volume of the solid formed by revolving the region

bounded by the graphs of 1, 0, and 1

about the -axis.

y x y x

y

1 12

4 2

3

2

2 1y x

1x

1

2

Simpson’s Rule (96) Given a table of a continuous function on , ,

estimate area under a curve using Simpson's Rule:

a b

0 1 2 3 14 2 4 43 n n

b af x f x f x f x f x f x

n

x –2 –1 0 1 2

y 6 3 2 3 6

46 4 3 2 2

404

33 6

12

2

APPS: A+ Calc; Calc Pack I; Simpson's Rule;

2; 2; 2; 4 Subintervalsf x x A B Nice!

Dot Product (97) (98)

1 2 1 2 1 2' 'd

r t r t r t r t r t r tdt

THEOREM 3 Product Rule for Dot Product Assume that r1(t) and r2(t) are differentiable. Then

1 2 1 2 1 1 2 2

Find the dot product...

, , which is a scalar.u u v v u v u v

2,3 2, 22 4 6