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BC Fall Review

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BC Fall Review. Hint: You don’t need to generate each term. Check your notes/book. Give the first 4 non-zero terms of the MacClaurin Series for the function:. Use P ( x ) to find the first 4 non-zero terms for arctan( x ). Since we already know that arctan(0) = 0, we know that C = 0. - PowerPoint PPT Presentation

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BC Fall Review

Give the first 4 non-zero terms of the MacClaurin Series for the

function:

21

1)(

xxf

Hint: You don’t need to generate each

6421)( xxxxP

Use P(x) to find the first 4 non-zero terms for arctan(x)

21)arctan(

x

dxx

Since we already know that arctan(0) = 0, we know that C = 0

dxxxx 6421

Cxxx

xxR 753

)(753

753)(

753 xxxxxR

Use this result to approximate

4

Find the maximum error involved in this approximation.

Use this result to approximate

4

Find the maximum error involved in this approximation.

)1arctan(4

7

1

5

1

3

11)1( R 72381.0

1.09

1error

Because this is an alternating series, we use the next term to find the error.

Use this result to approximate

4

Find the maximum error involved in

this approximation.

)1arctan(4

7

1

5

1

3

11)1( R 72381.0

1.09

1error

Because this is an alternating series, we use the next term to find the error.

Generate a 2nd degree polynomial to approximate ln(0.9) and ln(1.1). Find the error bound involved in each approximation.

0)1( f

xxf ln)(

1)1( f

1)1( f

2)1( f

2

)1()1(ln

2x

xx

2

)19.0()19.0(9.0ln

2

2

)1.0(1.0

2

105.0

2

)11.1()11.1(1.1ln

2

2

)1.0(1.0

2

095.0

Generate a 2nd degree polynomial to approximate ln(0.9) and ln(1.1). Find the error bound involved in each approximation.

2

)1()1(ln

2x

xx

2

)19.0()19.0(9.0ln

2

2

)1.0(1.0

2 105.0

2

)11.1()11.1(1.1ln

2

2

)1.0(1.0

2

095.0

Because this is an alternating series, we can just use the next term…

3 3( 1) (1.1 1).0003

3 3

xerror

Because this is no longer an alternating series, we need the Lagrange error bound for the 3rd term…

Generate a 2nd degree polynomial to approximate ln(0.9) and ln(1.1). Find the error bound involved in each approximation.

2

)19.0()19.0(9.0ln

2

2

)1.0(1.0

2 105.0

Because this is no longer an alternating series, we need the Lagrange error bound for the 3rd term…

3

2( )f c

c And the number between 0.9 and 1 that maximizes

this value is 0.9

3 3

3

(0.9 1) 2(0.9 1)(0.9) .000457

3! 6(0.9)error f

Find the radius and interval of convergence:

1

1

5

)2()1(

n

nn

n

x

n

n

n x

n

n

x

)2(

5

15

)2(lim

1

1)2(lim

xn

1)2(1 x 31 x

1

1

5

)12()1(

n

nn

n

1

1

5

)32()1(

n

nn

n

1

1

5

1)1(

n

n

n

1

1

5

)1()1(

n

nn

n

31 x

Use the integral test to determine if the given series converges:

12 1

1

n n

b

b x

dx1 2 1

limb

bx

1

1tanlim

1tantanlim 11

b

b

42

4

…and therefore it converges

22 1

1

n nUse the integral test to show that the given series converges

1)

Find the sum to which it converges.2)

2 2 1x

dxdx

x

B

x

Ab

b 11lim

2

1

1

11 2

xx

B

x

A 1 BBxAAx

0 BA

1 BA

2

1A

2

1B

dxxx

b

b 1

1

1

1lim

2

12

12

12ln

1

1lnlim

2

1

b

bb

1)

3ln2

1

22 1

1

n nUse the integral test to show that the given series converges

1)

Find the sum to which it converges.2)

2 2 1x

dxdx

x

B

x

Ab

b 11lim

2

12

12ln

1

1lnlim

2

1

b

bb

1)

3

1ln1ln

2

1

11

1lim

b

bb

…or more importantly, since we have a finite limit, the series converges.

22 1

1

n nUse the integral test to show that the given series converges

1)

Find the sum to which it converges.2)

2)

22 1

1

n n 1

1

1

1

2

1

2

nnn

...

6

1

4

1

5

1

3

1

4

1

2

1

3

1

1

1

2

1

4

3

dxx

e x

4/

0 2

tan

cos

Evaluate the integral: dxxe x

4/

0

2tan sec

xu tan

dxxdu 2sec

dueu1

0

0tanu 0u

)4/tan(u 1u

1

0

ue 1e

422 xx

dxEvaluate the integral:

2

x

24 xtan2x

ddx 2sec2

4tan4tan4

sec222

2

d

sec2tan4

sec22

2 d

dcos

1

sin

cos

4

12

2

d

2sin

cos

4

1

422 xx

dxEvaluate the integral:

2

x

24 xtan2x

ddx 2sec2

d 2sin

cos

4

1 sinu

ddu cos

24

1

u

du C

u4

1 C

sin4

1C csc

4

1

Cx

x

4

4 2

xxx

lnlim0

0 Indeterminate

x

xx 1

lnlim

0

L’Hopital’s Rule

2

0 1

1

lim

x

xx

x

x

xx

2

0lim 0

yxdx

dy 2 20 y

dxxy

dy 2 dxxy

dy 2C

xy

3ln

3

Cx

ey

3

3

3

3x

Aey 3

03

2 Ae

2A

3

3

2x

ey