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Homework #13 with Solutions - First-Year Interest Group Seminar | N 1, Assignments of Health sciences

Material Type: Assignment; Class: FIRST-YEAR INTEREST GROUP SMNR; Subject: Nursing; University: University of Texas - Austin; Term: Unknown 1989;

Typology: Assignments

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This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page โ€“ find all choices before answering.

CalC12h01a 001 10.0 points Find the interval of convergence of the power series โˆ‘โˆž

n = 1

4 xn โˆš n

  1. interval of cgce = [โˆ’ 4 , , 4]
  2. interval of cgce = [โˆ’ 1 , 1) correct
  3. interval of cgce = (โˆ’ 4 , 4]
  4. interval of cgce = (โˆ’ 1 , 1)
  5. interval of cgce = (โˆ’ 1 , 1]
  6. interval of cgce = (โˆ’ 4 , 4)
  7. interval of cgce = [โˆ’ 1 , 1]
  8. interval of cgce = [โˆ’ 4 , 4)

Explanation: When an =

4 xn โˆš n

then โˆฃ โˆฃ โˆฃ โˆฃ

an+ an

xn+ โˆš n + 1

n xn

|x|

n โˆš n + 1

= |x|

n n + 1

Thus

lim n โ†’ โˆž

an+ an

โˆฃ =^ |x|^.

By the Ratio Test, therefore, the given series converges when |x| < 1, and diverges when |x| > 1.

We have still to check for convergence at x = ยฑ1. But when x = 1, the series reduces to (^) โˆž โˆ‘

n = 1

n

which diverges by the p-series test with p = 1 2 โ‰ค^ 1. On the other hand, when^ x^ =^ โˆ’1, the series reduces to โˆ‘^ โˆž

n = 1

(โˆ’1)n^

n

which converges by the Alternating Series Test. Thus the

interval of convergence = [โˆ’ 1 , 1).

keywords:

CalC12h04a 002 (part 1 of 2) 10.0 points For the series โˆ‘^ โˆž

n = 1

(โˆ’1)n n + 8

xn^ ,

(i) determine its radius of convergence, R.

  1. R = 1 correct
  2. R = 0

3. R =

4. R = (โˆ’โˆž, โˆž)

5. R = 8

Explanation: The given series has the form

โˆ‘^ โˆž

n = 1

an

with an = (โˆ’1)n^

xn n + 8

Now for this series,

(i) R = 0 if it converges only at x = 0, (ii) R = โˆž if it converges for all x,

while 0 < R < โˆž,

(iii) if it converges when |x| < R, and (iv) diverges when |x| > R.

But

lim n โ†’ โˆž

an+ an

โˆฃ = lim n โ†’ โˆž |x|

n + 8 n + 9

= |x|.

By the Ratio Test, therefore, the given series converges when |x| < 1 and diverges when |x| > 1. Consequently,

R = 1.

003 (part 2 of 2) 10.0 points (ii) Determine the interval of convergence of the series.

  1. interval convergence = (โˆ’ 8 , 8]
  2. converges only at x = 0
  3. interval convergence = [โˆ’ 8 , 8)
  4. interval convergence = [โˆ’ 1 , 1)
  5. interval convergence = (โˆ’ 8 , 8)
  6. interval convergence = (โˆ’ 1 , 1] correct
  7. interval convergence = (โˆ’ 1 , 1)

Explanation: Since R = 1, the given series

(i) converges when |x| < 1, and (ii) diverges when |x| > 1.

On the other hand, at the point x = 1 and x = โˆ’1, the series reduces to

โˆ‘^ โˆž

n = 1

(โˆ’1)n n + 8

โˆ‘^ โˆž

n = 1

n + 8

respectively. But by the Alternating Series Test, the first of these series converges. On the other hand, if we set

an =

n + 8

, bn =

n

then

lim n โ†’ โˆž

an bn

= lim n โ†’ โˆž

n n + 8

By the p-series Test with p = 1, however, the series

n bn^ diverges. Thus by the Limit Comparison test, the series

n an^ also di- verges. Consequently, the given series has

interval convergence = (โˆ’ 1 , 1].

keywords:

CalC12h07a 004 (part 1 of 2) 10.0 points For the series โˆ‘^ โˆž

n =

xn (n + 3)!

(i) determine its radius of convergence, R.

1. R =

2. R = 3

3. R = 0

  1. R = โˆž correct
  2. R = 1 Explanation: The given series has the form

โˆ‘^ โˆž

n = 1

anxn

with an =

(n + 3)!

Now for this series,

by the Divergence test, therefore the series

โˆ‘^ โˆž

n = 1

an

diverges. On the other hand,

bn =

( 2 n + 7 7 n

)n( โˆ’

)n

= (โˆ’1)n

( 2 n + 7 2 n

)n = (โˆ’1)nan.

But, as we have seen,

lim n โ†’ โˆž an = e^7 /^2.

Thus bn oscillates between e^7 /^2 and โˆ’e^7 /^2 as n โ†’ โˆž; in particular,

lim n โ†’ โˆž bn

does not exist. Again by the Divergence Test, therefore, the series

โˆ‘^ โˆž

n = 1

bn

diverges. Consequently, the given power se- ries does not converge at x = ยฑ 7 /2 and so its

interval of convergence =

keywords:

CalC12h10a 007 10.0 points Find the interval of convergence of the power series โˆ‘โˆž

n = 0

5 n n!

xn^.

  1. interval =
  1. interval =
  1. interval = (โˆ’โˆž, โˆž) correct
  2. interval = (โˆ’ 5 , 5)
  3. interval =

[

  1. interval = [โˆ’ 5 , 5]
  2. interval =

[

]

Explanation: We apply the ratio test to the infinite series

โˆ‘^ โˆž

n = 0

5 n n!

|x|n^.

For this series,

an+ an

5 n+1n! 5 n(n + 1)!

|x| =

5 |x| n + 1

as n โ†’ โˆž. Thus the given power series converges for all x and so

interval = (โˆ’โˆž, โˆž).

CalC12h11s 008 10.0 points Find the radius of convergence and interval of

convergence of the series

โˆ‘^ โˆž

n=

(โˆ’8)n^ xn โˆš (^3) n + 5

1. R =

, I =

[

  1. diverges everywhere

3. R =

, I =

4. R = 8, I = (โˆ’ 8 , 8]

5. R =

, I =

]

correct

Explanation:

an =

(โˆ’8)n^ xn โˆš (^3) n + 5 , so

lim nโ†’โˆž

an+ an

โˆฃ = lim nโ†’โˆž

8 n+1|x|n+ โˆš (^3) n + 6 ยท

โˆš (^3) n + 5

8 n|x|n

= lim nโ†’โˆž 8 |x| 3

n + 5 n + 6 = 8|x|, so by the Ratio Test, the series converges

when 8|x| < 1. Then |x| <

, so R =

When x = โˆ’

, we get the divergent p-series

โˆ‘^ โˆž

n=

โˆš (^3) n + 5. When x =

, we get the series

โˆ‘^ โˆž

n=

(โˆ’1)n โˆš (^3) n + 5 , which converges by the Alternat-

ing Series Test. Thus, I =

]

CalC12h15a 009 10.0 points

Find the radius of convergence, R, of the power series

โˆ‘^ โˆž

n = 1

โˆš (^4) n(x + 6)n (^).

1. R = โˆž

  1. R = 1 correct

3. R = 0

4. R =

5. R = 6

Explanation: The given series has the form

โˆ‘^ โˆž

n = 1

an(x + 6)n

where an = 4

n.

Now for this series,

(i) R = 0 if it converges only at x = โˆ’6, (ii) R = โˆž if it converges for all x,

while 0 < R < โˆž

(iii) if it converges for |x + 6| < R, and (iv) diverges for |x + 6| > R.

But โˆฃ โˆฃ โˆฃ

an+ an

โˆš (^4) n + 1 โˆš (^4) n = 4

n + 1 n

so lim n โ†’ โˆž

an+ an

โˆฃ = 1^.

By the Ratio test, therefore, the given series

(a) converges for all |x + 6| < 1, and (b) diverges for all |x + 6| > 1.

Consequently,

R = 1.

CalC12h18a 010 10.0 points

Determine the radius of convergence, R, of the power series

โˆ‘^ โˆž

n = 1

(โˆ’5)n โˆš n

(x โˆ’ 2)n^.

1. R = 2

2. R = 0

3. R =

4. R = โˆž

5. R =

correct

6. R = 5

1. 2 R 1 = R 2 = 5

2. 2 R 1 = R 2 =

3. R 1 = 2R 2 = 5

4. R 1 = R 2 =

5. R 1 = 2R 2 =

  1. R 1 = R 2 = 5 correct

Explanation: When

lim n โ†’ โˆž

|cn|^1 /n^ =

the Root Test ensures that the series

โˆ‘^ โˆž

n = 0

cntn

is

(i) convergent when |t| < 5, and (ii) divergent when |t| > 5.

On the other hand, since

lim n โ†’ โˆž |n cn|^1 /n^ = lim n โ†’ โˆž |cn| 1 /n ,

the Root Test ensures also that the series

โˆ‘^ โˆž

= 1

n cn tnโˆ’^1

is

(i) convergent when |t| < 5, and (ii) divergent when |t| > 5.

Consequently,

R = R 2 = 5.

CalC12i03c 013 10.0 points

Find a power series representation for the function f (t) =

t โˆ’ 6

  1. f (t) =

โˆ‘^ โˆž

n = 0

6 n+^

tn

  1. f (t) =

โˆ‘^ โˆž

n = 0

(โˆ’1)nโˆ’^16 n+1^ tn

  1. f (t) = โˆ’

โˆ‘^ โˆž

n = 0

6 n^ tn

  1. f (t) = โˆ’

โˆ‘^ โˆž

n = 0

6 n+^

tn^ correct

  1. f (t) =

โˆ‘^ โˆž

n = 0

(โˆ’1)n 6 n^ tn

Explanation: We know that

1 โˆ’ x

= 1 + x + x^2 +... =

โˆ‘^ โˆž

n = 0

xn^.

On the other hand,

1 t โˆ’ 6

1 โˆ’ (t/6)

Thus

f (t) = โˆ’

โˆ‘^ โˆž

n = 0

t 6

)n = โˆ’

โˆ‘^ โˆž

n = 0

6 n^

tn^.

Consequently,

f (t) = โˆ’

โˆ‘^ โˆž

n = 0

6 n+^

tn

with |t| < 6.

CalC12i05s 014 10.0 points

Find a power series representation for the function

f (x) =

2 โˆ’ x^3

  1. f (x) =

โˆ‘^ โˆž

n = 0

x^3 n 2 n+^

correct

  1. f (x) = โˆ’

โˆ‘^ โˆž

n = 0

2 nx^3 n

  1. f (x) =

โˆ‘^ โˆž

n = 0

x^3 n 23 n

  1. f (x) = โˆ’

โˆ‘^ โˆž

n = 0

x^3 n 23 n

  1. f (x) =

โˆ‘^ โˆž

n = 0

2 nx^3 n

  1. f (x) = โˆ’

โˆ‘^ โˆž

n = 0

xn 2 n+

Explanation: After simplification,

f (x) =

2 โˆ’ x^3

1 โˆ’ (x^3 /2)

On the other hand, we know that

1 1 โˆ’ t

โˆ‘^ โˆž

n = 0

tn^.

Replacing t with x^3 /2, we thus obtain

f (x) =

โˆ‘^ โˆž

n=

x^3 n 2 n^

โˆ‘^ โˆž

n=

x^3 n 2 n+^

keywords:

CalC12i15s 015 10.0 points

Find a power series representation for the function f (z) = ln(3 โˆ’ z).

  1. f (z) = ln 3 +

โˆ‘^ โˆž

n = 1

zn 3 n

  1. f (z) = โˆ’

โˆ‘^ โˆž

n = 1

zn n 3 n

  1. f (z) = ln 3 โˆ’

โˆ‘^ โˆž

n = 0

zn 3 n

  1. f (z) =

โˆ‘^ โˆž

n = 0

zn n 3 n

  1. f (z) = ln 3 โˆ’

โˆ‘^ โˆž

n = 1

zn n 3 n^

correct

  1. f (z) = ln 3 +

โˆ‘^ โˆž

n = 0

zn n 3 n

Explanation: We can either use the known power series representation

ln(1 โˆ’ x) = โˆ’

โˆ‘^ โˆž

n = 1

xn n

or the fact that

ln(1 โˆ’ x) = โˆ’

โˆซ (^) x

0

1 โˆ’ s

ds

โˆซ (^) x

0

n = 0

sn^

ds

โˆ‘^ โˆž

n = 0

โˆซ (^) x

0

sn^ ds = โˆ’

โˆ‘^ โˆž

n = 1

xn n

For then by properties of logs,

f (z) = ln 3

z

= ln 3 + ln

z

so that

f (z) = ln 3 โˆ’

โˆ‘^ โˆž

n = 1

zn n 3 n^

CalC12i18s 016 10.0 points

  1. f (z) =

โˆ‘^ โˆž

n = 0

z^4 n 4 n

  1. f (z) =

โˆ‘^ โˆž

n = 0

z^4 n+ 4 n + 2

correct

  1. f (z) =

โˆ‘^ โˆž

n = 0

(โˆ’1)nz^4 n+ 4 n + 2

  1. f (z) =

โˆ‘^ โˆž

n = 4

z^4 n 4 n + 2

  1. f (z) =

โˆ‘^ โˆž

n = 0

(โˆ’1)nz^4 n 4 n

Explanation: By the geometric series representation,

1 1 โˆ’ s

โˆ‘^ โˆž

n = 0

sn^ ,

and so s 1 โˆ’ s^4

โˆ‘^ โˆž

n = 0

s^4 n+^.

But then

f (z) =

โˆซ (^) z

0

n = 0

s^4 n+

ds

โˆ‘^ โˆž

n = 0

{โˆซ^ z

0

s^4 n+1^ ds

Consequently,

f (z) =

โˆ‘^ โˆž

n = 0

z^4 n+ 4 n + 2

CalC12i26a 018 10.0 points

Use the Taylor series representation cen-

tered at the origin for eโˆ’x

2 to evaluate the definite integral

I =

0

6 eโˆ’x

2 dx.

1. I =

โˆ‘^ โˆž

k = 0

(โˆ’1)k 2 k + 1

6 ยท 22 k+

2. I =

โˆ‘^ โˆž

k = 0

(โˆ’1)k k!(2k + 1)

6 ยท 22 k+1^ correct

3. I =

โˆ‘^ โˆž

k = 0

k!

6 ยท 22 k

4. I =

โˆ‘^ n

k = 0

k!(2k + 1)

6 ยท 22 k+

5. I =

โˆ‘^ โˆž

k = 0

(โˆ’1)k k!

6 ยท 22 k

Explanation: The Taylor series for ex^ is given by

ex^ = 1 + x +

x^2 +... +

n!

xn^ +...

and its interval of convergence is (โˆ’โˆž, โˆž). Thus we can substitute x โ†’ โˆ’x^2 for all val- ues of x, showing that

eโˆ’x

2

โˆ‘^ โˆž

k = 0

(โˆ’1)k k!

x^2 k

everywhere on (โˆ’โˆž, โˆž). Thus

I =

0

k = 0

(โˆ’1)k k!

x^2 k

dx.

But we can change the order of summation and integration on the interval of convergence, so

I = 6

โˆ‘^ โˆž

k = 0

( โˆซ^2

0

(โˆ’1)k k!

x^2 k

dx

โˆ‘^ โˆž

k = 0

[ (^) (โˆ’1)k k!(2k + 1)

x^2 k+^

] 2

0

Consequently,

I =

โˆ‘^ โˆž

k = 0

(โˆ’1)k k!(2k + 1)

6 ยท 22 k+^.

CalC12i38s 019 10.0 points

Determine the function f having power se- ries representation

f (x) =

โˆ‘^ โˆž

n = 2

n(n โˆ’ 1) xn+

on (โˆ’ 1 , 1).

  1. f (x) =

2 x^7 (1 โˆ’ x)^3

correct

  1. f (x) =

2 x^7 1 โˆ’ x

  1. f (x) =

2 x^5 (5 โˆ’ x)^3

  1. f (x) =

x^7 (1 โˆ’ x)^3

  1. f (x) =

x^5 5 โˆ’ x

  1. f (x) =

x^5 (5 โˆ’ x)^3

Explanation: The presence of the coefficients n(n โˆ’ 1) suggests that f should be related to the sec- ond derivative of some function since

d^2 dx^2

n = 0

anxn

โˆ‘^ โˆž

n = 2

n(n โˆ’ 1)anxnโˆ’^2

on the interval of convergence of the power series. Now โˆ‘^ โˆž

n = 2

n(n โˆ’ 1)xn+

= x^7

n = 2

n(n โˆ’ 1)xnโˆ’^2

On the other hand

1 1 โˆ’ x

โˆ‘^ โˆž

n = 0

xn^ ,

in which case,

1 (1 โˆ’ x)^2

d dx

1 โˆ’ x

โˆ‘^ โˆž

n = 1

nxnโˆ’^1 ,

and

2 (1 โˆ’ x)^3

d^2 dx^2

1 โˆ’ x

โˆ‘^ โˆž

n = 2

n(n โˆ’ 1)xnโˆ’^2 ,

Consequently,

f (x) =

2 x^7 (1 โˆ’ x)^3