5 Steps to a 5 AP Calculus BC 2019

362 STEP 4. Review the Knowledge You Need to Score High

Step 2: For three decimal place accuracy, we want to find the value of nfor which

the remainder is less than or equal to 0.0005. Choose x 0 =1 in the interval

(0, 1). All derivatives ofex are equal toex, and therefore|f(n+1)(x)|≤e,so

M=e·

∣

∣∣

∣Rn

(

1

2

)∣∣

∣∣≤ e

(n+1)!

∣

∣∣

∣

1

2

− 1

∣

∣∣

∣

n+ 1

and

e

(n+1)!

∣

∣∣

∣

1

2

− 1

∣

∣∣

∣

n+ 1

=

e

2 n+^1 ·(n+1)!

≤

0 .0005 whenn≥4.

Step 3:

√

e=

∑^4

n= 0

1

2 n·n!

= 1 +

1

2

+

1

22 ·2!

+

1

23 ·3!

+

1

24 ·4!

= 1. 6484

Example 2

Estimate sin 4◦accurate to five decimal places.

Step 1: 4 ◦=

π

45

radians. Substitute

π

45

forxin the MacLaurin series that represents

sinx·sin

π

45

=

π

45

−

(π/45)^3

3!

+

(π/45)^5

5!

−

(π/45)^7

7!

+···

Step 2: For five decimal place accuracy, we must find the value ofnfor which the absolute

error is less than or equal to 5∣ × 10 −^6. For allx,|f(n+1)(x)|≤1. Choosex 0 =0.

∣

∣∣Rn

(

π

45

)∣∣

∣∣≤^1

(n+1)!

∣∣

∣∣π

45

− 0

∣∣

∣∣

n+ 1

=

(π/45)n+^1

(n+1)!

. The absolute error is less than or
equal to 5× 10 −^6 forn≥3.

Step 3: sin 4◦=sin

π

45

=

π

45

−

(π/45)

3!

3

+

(π/45)^5

5!

−

(π/45)^7

7!

= 0. 069756

14.8 Rapid Review

1. Find the sum of the series 81+ 27 + 9 + 3 + 1 +

1

3

+···.

Answer:This is a geometric series with first term 81 and a ratio of

1

3

,

soS=

81

1 − 1 / 3

=

243

2

.

2. Determine whether the series

∑∞

n= 1

5 n

n!

converges or diverges.

Answer:nlim→∞

5 n+^1

(n+1)!

·

n!

5 n

=nlim→∞

5

n+ 1

=0so

∑∞

n= 1

5 n

n!

converges by ratio test.

3. Determine whether the series

∑∞

n= 1

lnn

√

n

converges or diverges.

Answer:

lnn

√

n

>

1

√

n

forn>e.

∑∞

n= 1

1

√

n

is ap-series withp<1, so it diverges.

∑∞

n= 1

lnn

√

n

=

ln 2

√

2

+

∑∞

n= 3

lnn

√

n

and

∑∞

n= 3

lnn

√

n

diverges by comparison to

∑∞

n= 3

1

√

n

, so the

series diverges.