Radius of Convergence - Calculus Two - Solved Exam, Exams for Calculus. Biju Patnaik University of Technology, Rourkela

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This is the Solved Exam of Calculus Two which includes Interval of Convergence, Radius of Convergence, Reasoning Clearly, Converges Conditionally, Power Series etc. Key important points are: Radius of Convergence, Inter...
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Solution: APPM 1360 Exam 3 Spring 2012

1. Determine whether the following series are conditionally convergent, absolutely convergent, or divergent.

(a) (10 pts)

∞∑ n=10

(−1)n lnn n

(b) (10 pts)

∞∑ n=6

(−1)n(n + 1)5n

n32n+1

Solution:

(a) Checking for absolute convergence, we have

∞∑ n=10

∣∣∣∣(−1)n lnnn ∣∣∣∣ = ∞∑

n=10

lnn

n >

∞∑ n=10

1

n , which is a diver-

gent p-series (p = 1), so ∞∑

n=10

lnn

n diverges by the Direct Comparison Test (or could also use the Integral

Test.)

Now, since the series is alternating, so we use the Alternating Series Test with bn = lnn

n . For f(x) =

lnx

x ,

we have f ′(x) = x(1/x)− lnx(1)

x2 =

1− lnx x2

< 0 for x ≥ e, so bn is decreasing for n ≥ 10. We also need to

show that bn → 0 as n→∞. Note, lim n→∞

ln(n)

n

L′H = lim

n→∞

1

n = 0, so the series converges by the Alternating

Series Test. Therefore, ∞∑

n=10

(−1)n lnn n

is conditionally convergent .

(b) Using the Ratio Test we have

lim n→∞

∣∣∣∣an+1an ∣∣∣∣ = limn→∞ (n + 2)5n+1(n + 1)32n+3 · n32n+1(n + 1)5n = limn→∞ 5(n2 + 2n)32(n2 + 2n + 1) = 532 = 59 < 1

or, using the Root Test, we have

lim n→∞

n √ |an| = lim

n→∞ n

√ (n + 1)5n

n32n+1 = lim

n→∞

n √ n + 1 · 5

n √ n · (32n)1/n · (3)1/n

= 5

32 =

5

9 < 1

where we have used the fact that lim n→∞

n √ n = lim

n→∞ n √ n + 1 = lim

n→∞ 31/n = 1, (see limits given on the

formula sheet) so the series is absolutely convergent by the either the Ratio Test or the Root Test.

2. Consider the function f(x) =

∞∑ n=2

(x− 1)3n

8n ln(n)

(a) (10 pts) What is the interval of convergence of f(x)? (Be sure to check the endpoints.) (b) (5 pts) What is the radius of convergence of f(x)? (c) (5 pts) If x = −2 is f(−2) conditionally convergent, absolutely convergent or divergent? Why? Solution: (a) Using the Ratio Test, we get

lim n→∞

∣∣∣∣an+1an ∣∣∣∣ = limn→∞

∣∣∣∣∣ (x− 1)3(n+1)8n+1 ln(n + 1) 8n ln(n)(x− 1)3n ∣∣∣∣∣ = limn→∞ |(x− 1)3|8 · ln(n)ln(n + 1) = |x− 1|38 limn→∞ ln(n)ln(n + 1) L′H= |x− 1|38

Now setting the limit less than 1 yields,

|x− 1|3

8 < 1⇒ |x− 1|3 < 8⇒ |x− 1| < 2⇒ −2 < x− 1 < 2⇒ −1 < x < 3

Now, check the endpoints, when x = 3 we have,

∞∑ n=2

(2)3n

8n ln(n) = ∞∑ n=2

8n

8n ln(n) = ∞∑ n=2

1

ln(n)

and ∞∑ n=2

1

ln(n) > ∞∑ n=2

1

n , so the power series is divergent when x = 3 by the Direct Comparison Test (or

could also use the Limit Comparison Test), and when x = −1 we have,

∞∑ n=2

(−2)3n

8n ln(n) = ∞∑ n=2

(−1)n8n

8n ln(n) = ∞∑ n=2

(−1)n

ln(n)

Note that bn = 1

ln(n) >

1

ln(n + 1) = bn+1 and lim

n→∞ bn = lim

n→∞

1

ln(n) = 0 and so the series is conditionally

convergent by the Alternating Series Test and the work done above. Thus, the interval of convergence is

−1 ≤ x < 3

(b) The radius of convergence is R = 3− (−1)

2 = 2.

(c) f(−2) is divergent since x = −2 is outside the interval of convergence.

3. (a) (10 pts) Find a Maclaurin Series for f(x) = x2e−x 2 . Give your answer in

∑ -notation and combine

all like terms. (There are some useful formulas on the formula sheet!)

∫ 1/2 0

∑ -notation.

(c) (10 pts) Estimate the integral in part (b) using the first two non-zero terms of the series. Find a bound on the error of your estimate using the Alternating Series Error Estimation Theorem.

Solution:

(a) From the formula sheet, we know ex = ∞∑ n=0

xn

n! , so x2e−x

2 = x2

∞∑ n=0

(−x2)n

n! =

∞∑ n=0

(−1)nx2n+2

n!

(b) Using the series from part (a) yields∫ 1/2 0

x2e−x 2 dx =

∫ 1/2 0

∞∑ n=0

(−1)nx2n+2

n! dx

= ∞∑ n=0

∫ 1/2 0

(−1)nx2n+2

n! dx =

[ ∞∑ n=0

(−1)nx2n+3

(2n + 3)n!

]1/2 0

= ∞∑ n=0

(−1)n(12) 2n+3

(2n + 3)n!

=

∞∑ n=0

(−1)n

22n+3(2n + 3)n!

(Note that we can interchange the sum and the integral because the sum is absolutely convergent for all x)

(c) Since

∫ 1/2 0

x2e−x 2 dx =

∞∑ n=0

(−1)n

22n+3(2n + 3)n! from part (b), we have

∫ 0.5 0

x2e−x 2 dx ≈ 1

23 · 3 − 1

25 · 5 =

17

480 ≈ 0.0354

Since the series is alternating, the error bound is |E| < ∣∣∣∣ 17 · 272!

∣∣∣∣ = 11792 ≈ 0.000558. (This is the first unused term of the series)

4. (10 pts) Suppose we approximate y = ln(x) by a Taylor Polynomial of order 2, T2(x), centered at a = 4.8 for 3.6 ≤ x ≤ 5, use Taylor’s Formula to estimate the error of the approximation.

Solution: Here the corresponding Taylor Remainder term is R2(x) = f (3)(z)

3! (x− 4.8)3 where

f(x) = ln(x)⇒ f ′(x) = 1/x⇒ f ′′(x) = −1/x2 ⇒ f ′′′(x) = 2/x3

and note that 3.6 < z < 5 implies z3 > (3.6)3 and so 1/z3 < 1/(3.6)3 and moreover 3.6 ≤ x ≤ 5 implies −1.2 ≤ x− 4.8 ≤ 0.2, so |x− 4.8| ≤ 1.2 and so |x− 4.8|3 ≤ (1.2)3 and thus the error bound we get is

|R2(x)| = |f (3)(z)|

3! |x− 4.8|3 = 2

z3 · 3! |x− 4.8|3 < 2

(3.6)3 · 3! (1.2)3 =

1

3 · (

1.2

3.6

)3 =

1

34 =

1

81

so by Taylor’s Formula we have |error| = |R2(x)| < 1/81.

(a) Suppose we are told that |f ′′(x)| ≤ 6 for 0 ≤ x ≤ 1, then in order to guarantee that the Midpoint

Rule approximation for

∫ 1 0

f(x) dx is accurate to within 0.01 we would need n ≥ 4.

(b) The Trapezoidal Rule approximation of the integral

∫ 2 1

1

x dx with n = 5 is given by

∫ 2 1

1

x dx ≈ 0.1

[ 1

1 +

2

1.2 +

2

1.4 +

2

1.6 +

2

1.8 +

1

2

]

(c) 2 =

∞∑ n=0

1

2n

(d) 2106 =

106∑ n=0

( 106

n

) (e) Consider {fm}∞m=0, if limm→∞ fm 6= 0 then {fm}

∞ m=0 diverges.

Solution:

(a) F (b) AT (c) AT (d) AT (e) F

Comments: (a) If n = 4 then K(b− a)3

24n2 =

6

24 · 16 =

1

64 , so the desired accuracy is not guaranteed if

n = 4, we need n ≥ 5.

(c) This is a Geometric Series with a = 1 and r = 1/2 so a

1− r =

1

1− 1/2 = 2.

(d) Note that

( 106

n

) = 0 for n > 106, and so using the binomial series we get (1 + x)106 =

106∑ n=0

( 106

n

) xn,

now let x = 1 then 2106 =

106∑ n=0

( 106

n

) .

(e) This is not true in general since {fm}∞m=0 is a sequence.