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Material Type: Assignment; Class: FIRST-YEAR INTEREST GROUP SMNR; Subject: Nursing; University: University of Texas - Austin; Term: Unknown 1989;
Typology: Assignments
1 / 11
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 โโ
n = 1
4 xn โ n
Explanation: When an =
4 xn โ n
then โฃ โฃ โฃ โฃ
an+ an
xn+ โ n + 1
n xn
|x|
n โ n + 1
= |x|
n n + 1
Thus
lim n โ โ
an+ an
โฃ =^ |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 (^) โ โ
n = 1
n
which diverges by the p-series test with p = 1 2 โค^ 1. On the other hand, when^ x^ =^ โ1, the series reduces to โ^ โ
n = 1
(โ1)n^
n
which converges by the Alternating Series Test. Thus the
interval of convergence = [โ 1 , 1).
keywords:
CalC12h04a 002 (part 1 of 2) 10.0 points For the series โ^ โ
n = 1
(โ1)n n + 8
xn^ ,
(i) determine its radius of convergence, R.
Explanation: The given series has the form
โ^ โ
n = 1
an
with an = (โ1)n^
xn n + 8
Now for this series,
(i) R = 0 if it converges only at x = 0, (ii) R = โ if it converges for all x,
while 0 < R < โ,
(iii) if it converges when |x| < R, and (iv) diverges when |x| > R.
But
lim n โ โ
an+ an
โฃ = lim n โ โ |x|
n + 8 n + 9
= |x|.
By the Ratio Test, therefore, the given series converges when |x| < 1 and diverges when |x| > 1. Consequently,
003 (part 2 of 2) 10.0 points (ii) Determine the interval of convergence of the series.
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 = โ1, the series reduces to
โ^ โ
n = 1
(โ1)n n + 8
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 โ
an bn
= lim n โ โ
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 = (โ 1 , 1].
keywords:
CalC12h07a 004 (part 1 of 2) 10.0 points For the series โ^ โ
n =
xn (n + 3)!
(i) determine its radius of convergence, R.
โ^ โ
n = 1
anxn
with an =
(n + 3)!
Now for this series,
by the Divergence test, therefore the series
โ^ โ
n = 1
an
diverges. On the other hand,
bn =
( 2 n + 7 7 n
)n( โ
)n
= (โ1)n
( 2 n + 7 2 n
)n = (โ1)nan.
But, as we have seen,
lim n โ โ an = e^7 /^2.
Thus bn oscillates between e^7 /^2 and โe^7 /^2 as n โ โ; in particular,
lim n โ โ bn
does not exist. Again by the Divergence Test, therefore, the series
โ^ โ
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 โโ
n = 0
5 n n!
xn^.
Explanation: We apply the ratio test to the infinite series
โ^ โ
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 โ. Thus the given power series converges for all x and so
interval = (โโ, โ).
CalC12h11s 008 10.0 points Find the radius of convergence and interval of
convergence of the series
n=
(โ8)n^ xn โ (^3) n + 5
correct
Explanation:
an =
(โ8)n^ xn โ (^3) n + 5 , so
lim nโโ
an+ an
โฃ = lim nโโ
8 n+1|x|n+ โ (^3) n + 6 ยท
โ (^3) n + 5
8 n|x|n
= lim nโโ 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 = โ
, we get the divergent p-series
โ^ โ
n=
โ (^3) n + 5. When x =
, we get the series
โ^ โ
n=
(โ1)n โ (^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
โ^ โ
n = 1
โ (^4) n(x + 6)n (^).
Explanation: The given series has the form
โ^ โ
n = 1
an(x + 6)n
where an = 4
n.
Now for this series,
(i) R = 0 if it converges only at x = โ6, (ii) R = โ if it converges for all x,
while 0 < R < โ
(iii) if it converges for |x + 6| < R, and (iv) diverges for |x + 6| > R.
But โฃ โฃ โฃ
an+ an
โ (^4) n + 1 โ (^4) n = 4
n + 1 n
so lim n โ โ
an+ an
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
โ^ โ
n = 1
(โ5)n โ n
(x โ 2)n^.
correct
Explanation: When
lim n โ โ
|cn|^1 /n^ =
the Root Test ensures that the series
โ^ โ
n = 0
cntn
is
(i) convergent when |t| < 5, and (ii) divergent when |t| > 5.
On the other hand, since
lim n โ โ |n cn|^1 /n^ = lim n โ โ |cn| 1 /n ,
the Root Test ensures also that the series
โ^ โ
= 1
n cn tnโ^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 โ 6
n = 0
6 n+^
tn
n = 0
(โ1)nโ^16 n+1^ tn
n = 0
6 n^ tn
n = 0
6 n+^
tn^ correct
n = 0
(โ1)n 6 n^ tn
Explanation: We know that
1 โ x
= 1 + x + x^2 +... =
n = 0
xn^.
On the other hand,
1 t โ 6
1 โ (t/6)
Thus
f (t) = โ
n = 0
t 6
)n = โ
n = 0
6 n^
tn^.
Consequently,
f (t) = โ
n = 0
6 n+^
tn
with |t| < 6.
CalC12i05s 014 10.0 points
Find a power series representation for the function
f (x) =
2 โ x^3
n = 0
x^3 n 2 n+^
correct
n = 0
2 nx^3 n
n = 0
x^3 n 23 n
n = 0
x^3 n 23 n
n = 0
2 nx^3 n
n = 0
xn 2 n+
Explanation: After simplification,
f (x) =
2 โ x^3
1 โ (x^3 /2)
On the other hand, we know that
1 1 โ t
n = 0
tn^.
Replacing t with x^3 /2, we thus obtain
f (x) =
n=
x^3 n 2 n^
n=
x^3 n 2 n+^
keywords:
CalC12i15s 015 10.0 points
Find a power series representation for the function f (z) = ln(3 โ z).
n = 1
zn 3 n
n = 1
zn n 3 n
n = 0
zn 3 n
n = 0
zn n 3 n
n = 1
zn n 3 n^
correct
n = 0
zn n 3 n
Explanation: We can either use the known power series representation
ln(1 โ x) = โ
n = 1
xn n
or the fact that
ln(1 โ x) = โ
โซ (^) x
0
1 โ s
ds
โซ (^) x
0
n = 0
sn^
ds
n = 0
โซ (^) x
0
sn^ ds = โ
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 โ
n = 1
zn n 3 n^
CalC12i18s 016 10.0 points
n = 0
z^4 n 4 n
n = 0
z^4 n+ 4 n + 2
correct
n = 0
(โ1)nz^4 n+ 4 n + 2
n = 4
z^4 n 4 n + 2
n = 0
(โ1)nz^4 n 4 n
Explanation: By the geometric series representation,
1 1 โ s
n = 0
sn^ ,
and so s 1 โ s^4
n = 0
s^4 n+^.
But then
f (z) =
โซ (^) z
0
n = 0
s^4 n+
ds
n = 0
{โซ^ z
0
s^4 n+1^ ds
Consequently,
f (z) =
n = 0
z^4 n+ 4 n + 2
CalC12i26a 018 10.0 points
Use the Taylor series representation cen-
tered at the origin for eโx
2 to evaluate the definite integral
0
6 eโx
2 dx.
k = 0
(โ1)k 2 k + 1
6 ยท 22 k+
k = 0
(โ1)k k!(2k + 1)
6 ยท 22 k+1^ correct
k = 0
k!
6 ยท 22 k
โ^ n
k = 0
k!(2k + 1)
6 ยท 22 k+
k = 0
(โ1)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 (โโ, โ). Thus we can substitute x โ โx^2 for all val- ues of x, showing that
eโx
k = 0
(โ1)k k!
x^2 k
everywhere on (โโ, โ). Thus
0
k = 0
(โ1)k k!
x^2 k
dx.
But we can change the order of summation and integration on the interval of convergence, so
k = 0
0
(โ1)k k!
x^2 k
dx
k = 0
[ (^) (โ1)k k!(2k + 1)
x^2 k+^
0
Consequently,
k = 0
(โ1)k k!(2k + 1)
6 ยท 22 k+^.
CalC12i38s 019 10.0 points
Determine the function f having power se- ries representation
f (x) =
n = 2
n(n โ 1) xn+
on (โ 1 , 1).
2 x^7 (1 โ x)^3
correct
2 x^7 1 โ x
2 x^5 (5 โ x)^3
x^7 (1 โ x)^3
x^5 5 โ x
x^5 (5 โ x)^3
Explanation: The presence of the coefficients n(n โ 1) suggests that f should be related to the sec- ond derivative of some function since
d^2 dx^2
n = 0
anxn
n = 2
n(n โ 1)anxnโ^2
on the interval of convergence of the power series. Now โ^ โ
n = 2
n(n โ 1)xn+
= x^7
n = 2
n(n โ 1)xnโ^2
On the other hand
1 1 โ x
n = 0
xn^ ,
in which case,
1 (1 โ x)^2
d dx
1 โ x
n = 1
nxnโ^1 ,
and
2 (1 โ x)^3
d^2 dx^2
1 โ x
n = 2
n(n โ 1)xnโ^2 ,
Consequently,
f (x) =
2 x^7 (1 โ x)^3