1042微微微甲甲甲01-04班班班期期期中中中考考考解解解答答答和和和評評評分分分標標標準準準 1. (8%) Determine whether the series
∞
n=1∑(−1)nsin ( 1
√n)ln (1 + 1
√n) is divergent, conditionally convergent or abso- lutely convergent.
Solution:
Let an=sin( 1
√n)ln (1 + 1
√n) Part1:
(1) lim
n→∞an= lim
n→∞sin( 1
√n)ln (1 + 1
√n) =0 (1pt) (2) an is decreasing (1pt)
Thus the series
∞
∑
n=1
(−1)nan is convergent by the Alternation Series Test. (2pt) Part2:
Consider the series
∞ n=1∑
∣(−1)nan∣ =
∞ n=1∑
sin( 1
√n)ln (1 + 1
√n) We use the Limit Comparison Test with
an=sin( 1
√n)ln (1 + 1
√n), bn= 1 n and obtain
n→∞lim an
bn
= lim
n→∞
sin(√1
n)ln (1 +√1 n)
1 n
= lim
n→∞
sin (√1 n)
√1 n
ln (1 +√1 n)
√1 n
=1 (2pt)
Since
∞ n=1∑
1
n is divergent, the series
∞
n=1∑∣(−1)nan∣diverges by the Limit Comparison Test.
Hence the series
∞
n=1∑(−1)nsin( 1
√n)ln (1 + 1
√n)is conditionally convergent. (2pt)
2. (8%) Find the sum of the series
∞
∑
n=0
x4n 2n + 1.
Solution:
Define
f (x) =
∞ n=0∑
x4n
2n + 1, g(x) =
∞ n=0∑
x2n+1 2n + 1 then f (0) = 1 and f (x) = 1
x2g(x2)for x ≠ 0.
g′(x) =
∞
∑
n=0
x2n= 1
1 − x2 for ∣x2∣ <1 ⇒ ∣x∣ < 1 g(x) =∫
1
1 − x2dx = 1 2∫ [
1 1 − x+
1
1 + x]dx = 1
2ln(1 + x 1 − x) +C By g(0) = 0 we know that C = 0 such that g(x) =1
2ln(1 + x 1 − x).
Therefore,
f (x) =
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩ 1
2x2ln(1 + x2
1 − x2) 0 < ∣x∣ < 1
1 x = 0
Note that f (x) diverges when ∣x∣ ≥ 1 by the Ratio Test and the Limit Comparison Test with ∑1
n at the end points.
(Another possible answer: since∫ 1
1 − x2dx = tanh−1(x2) +C, we also have f (x) = 1
x2tanh−1(x2)for 0 < ∣x∣ < 1.)
●Grading policy: 5 points for converting the sum into a function, 3 points for integration.
Page 2 of 12
3. (12%)
(a) Use a Riemann sum approximation of∫
n
1
ln tdt to show that ln(n!) ≥ n ln n − n + 1.
(b) Find the interval of convergence of the power series
∞
∑
n=1
(2n)!
n2n xn.
Solution:
(5 points for (a), 7 points for (b)) (a) f (t) = ln t is an increasing function:
From the figure, in [1, n] the upper sum (always taking the value on the right) is larger than the integral. Thus we have ln 2 + ln 3 + ⋯ + ln n ≥∫
n
1
ln t dt (2 points).
Since ln 1 = 0,
ln(n!) = ln 1 + ln 2 + ⋯ + ln n = ln 2 + ⋯ + ln n ≥∫
n
1 ln t dt = t ln t∣n1− ∫
n
1 1 dt = n ln n − n + 1 (3 points)
(b) Define an= (2n)!
n2n xn. Apply the Ratio Test:
nlim→∞∣ an+1
an
∣ = lim
n→∞
(2n+2)!
(n+1)2n+2 (2n)!
n2n
∣x∣ = lim
n→∞
(2n + 2)(2n + 1) (n + 1)(n + 1) (
n
n + 1)2n∣x∣ = 4 e2∣x∣
in which (by using l’Hospital’s Rule)
n→∞lim( n
n + 1)2n=exp[ lim
n→∞2n ln(1 − 1
n + 1)] =exp[2 lim
n→∞
(n+1)2−1
1−n+11
−n12
] =e−2.
Thus the radius of convergence is e2
4. (4 points) At x =e2
4 , with ln(n!)n ≥ ln n − n + 1 ⇒ n! ≥nne
en ⇒ (2n)! ≥ (2n)2ne e2n , an=
(2n)!
n2n e2n
22n = (2n)! e2n (2n)2n ≥
(2n)2ne e2n
e2n
(2n)2n =e ≠ 0 Since lim
n→∞an≠0, ∑ an diverges by the Test for Divergence.
At x = −e2 4, lim
n→∞an does not exist (alternating with absolute values larger than e), thus the series also diverges.
In conclusion, the interval of convergence is (−e2 4,e2
4 ). (3 points)
4. (8%) Find the Maclaurin series expansion of the function ln(1 + 3x + 2x2). Write out the general terms. What is the radius of convergence?
Solution:
Recall that, ln (1 + x) = x −x2 2 +
x3 3 −
x4
4 +... for ∣x∣ < 1 ln (1 + 3x + 2x2) =ln (1 + x) + ln (1 + 2x) =
∞
n=1∑(−1)n−1xn n +
∞
n=1∑(−1)n−1(2x)n
n =
∞
n=1∑(−1)n−12n+1 n xn (6 points)
Because the radiu of convergence of
∞
n=1∑(−1)n−1(2x)n n is 1
2, and the radiu of convergence of
∞
n=1∑(−1)n−1(x)n n is 1, the radiu of convergence of
∞
n=1∑(−1)n−12n+1 n xn is 1
2 (2 points)
Page 4 of 12
5. (12%)
(a) Find the Maclaurin series for sinh−1x.
(b) Find lim
x→0
sinh−1(x2) −x2
x6 .
Solution:
(sinh−1(x))′= (1 + x2)
−1 2
By binomial expansion, (1 + x2)
−1 2 =
∞ n=0∑(
−1 2
n)(x2)n (
−1 2
n) =
(−12)(−32)...(−12 −n + 1)
n! =
(−12)(−32 )...(−2n−12 )
n! = (−1)n (2n)!
22n(n!)2 We can find the Maclaurin series of
sinh−1(x) = C +∫
∞
n=0∑(−1)n (2n)!
22n(n!)2x2ndx = C +
∞
n=0∑(−1)n (2n)!
22n(n!)2(2n + 1)x2n+1
Because sinh−1(0) = 0 ⇒ C = 0 ⇒ the Maclaurin series of sinh−1(x) is
∞
∑
n=0
(−1)n (2n)!
22n(n!)2(2n + 1)x2n+1 (8 points)
limx→0
sinh−1(x2) −x2
x6 =lim
x→0
∑∞n=0(−1)n22n(n!)(2n)!2(2n+1)x4n+2−x2 x6
=lim
x→0
x2+−16x6+ ∑∞n=2(−1)n22n(n!)(2n)!2(2n+1)x4n+2−x2
x6 =
−1
6 (4 points)
6. (12%) Consider the curve C ∶ x = t3, y = 3t, z = t4. (a) Find the curvature of C at the point (−1, −3, 1).
(b) Is there a point on the curve C where the osculating plane is parallel to the plane x + y + z = 1?
Solution:
(a) Let r(t) = t3i + 3tj + t4k
r′(t) = 3t2i + 3j + 4t3k ⇒ r′(−1) = 3i + 3j − 4k (1pt) r′′(t) = 6ti + 0j + 12t2k ⇒ r′′(−1) = −6i + 0j + 12k (1pt) r′(−1) × r′′(−1) = 36i − 12j + 18k (1pt)
∣r′(−1)∣ =
√
32+32+ (−4)2=
√
34 (1pt)
∣r′(−1) × r′′(−1)∣ =
√
362+ (−12)2+182=42 (1pt) Hence κ(−1) =∣r′(−1) × r′′(−1)∣
∣r′(−1)∣3 = 21 17√
34= 21√
34 578 (1pt) (b)
N(t) =r′′(t)∣r′(t)∣2−r′(t)(r′′(t) ⋅ r′(t))
∣r′(t)∣3
=r′′(t)( 1
∣r′(t)∣) −r′(t)(r′′(t) ⋅ r′(t)
∣r′(t)∣3 )(2pt) Since < 1, 1, 1 >⊥ T(t) = r′(t)
∣r′(t)∣ and < 1, 1, 1 >⊥ N(t)
⇒<1, 1, 1 >⊥ r′(t) and < 1, 1, 1 >⊥ r′′(t)
⇒ { <1, 1, 1 > ⋅r′(t) = 0 ⇒3t2+3 + 4t3=0 . . . (1)
<1, 1, 1 > ⋅r′′(t) = 0 ⇒6t + 12t2=0 . . . (2) (1pt) by (2) we have t = 0 or −1
2 and take it into (1)
⇒
⎧⎪
⎪
⎨
⎪⎪
⎩
3 ⋅ 0 + 3 + 4 ⋅ 0 ≠ 0 3 ⋅ (−1
2)2+3 + 4 ⋅ (−1
2)3≠0 (1pt)
Hence there is no point on the curve C such that the osculating plane is parallel to the plane x + y + z = 1. (2pt)
Page 6 of 12
7. (12%) Let f (x, y) =
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
(x2+y2)sin 1
x2+y2 if (x, y) ≠ (0, 0),
0 if (x, y) = (0, 0).
(a) Is fxcontinuous at (0, 0)?
(b) Write down the linear approximation L(x, y) of f at (0, 0).
(c) Find the limit lim
(x,y)→(0,0)
f (x, y) − L(x, y)
√
x2+y2 .
Solution:
(a)
For (x, y) ≠ (0, 0), fx(x, y) = 2x sin( 1 x2+y2) −
2x
x2+y2cos( 1
x2+y2) (2)
∵lim
t→0fx(t2, 0) = lim
t→02t2sin(1 t4) −
2t2 t4 cos(1
t4) =0 − lim
t→0
2 t2cos(1
t4)
= −2 lim
u→0+
1 ucos( 1
u2) = −2 lim
v→∞v cos(v2)
vlim→∞v cos(v2)does not exists.
∴fxis not continuous at (0, 0). (2) (b)
L(x, y) = f (0, 0) + fx(0, 0)∆x + fy(0, 0)∆y
∵fx(0, 0) = lim
t→0
f (t, 0) − f (0, 0)
t =lim
t→0
t2sin(1/t2) −0
t =lim
t→0t sin(1 t2)
=0 (2) fy(0, 0) = lim
t→0
f (0, t) − f (0, 0)
t =lim
t→0
t2sin(1/t2) −0
t =lim
t→0t sin(1 t2)
=0 (2)
∴L(x, y) = 0 + 0∆x + 0∆y = 0 (2) (c)
(x,y)→(0,0)lim
f (x, y) − L(x, y)
√ x2+y2
= lim
(x,y)→(0,0)
√
x2+y2sin( 1 x2+y2) Let x = r cos θ and y = r sin θ, then we have:
(x,y)→(0,0)lim
√
x2+y2sin( 1
x2+y2) = lim
r→0+
√r sin(1 r)
∵ −
√r ≤√ r sin(1
r) ≤
√r and lim
r→0+
√r = 0
∴lim
r→0+
√r sin(1
r) =0 by the squeeze theorem Hence lim
(x,y)→(0,0)
f (x, y) − L(x, y)
√
x2+y2 =0 (2)
8. (12%) Let f (x, y) =
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
sin(x3) −sin(y3)
x2+y2 if (x, y) ≠ (0, 0),
0 if (x, y) = (0, 0).
(a) Calculate ∇f (0, 0).
(b) Use the definition of directional derivative to calculate Duf (0, 0), where u = 1
√2(i − j).
(c) Is f (x, y) differentiable at (0, 0)?
Solution:
(a)
fx(0, 0) = lim
t→0
f (t, 0) − f (0, 0)
t =lim
t→0
sin(t3)
t3 =1 (3) fy(0, 0) = lim
t→0
f (0, t) − f (0, 0)
t =lim
t→0− sin(t3)
t3 = −1 (3)
∴∇f (0, 0) = (1, −1) (b)
Du(0, 0) = lim
t→0
f (0 +√1
2t, 0 −√1
2t) − f (0, 0)
t (1)
=lim
t→0
sin( t3
2√
2) −sin(− t3
2√ 2)
t3 =
2 2√
2lim
t→0
sin( t3
2√ 2)
t3 2√ 2
= 1
√2 (2) (c)
If f (x, y) is differentiable at (0, 0), then Du(0, 0) = ∇f (0, 0) ⋅ u. (2) However,
∵Du(0, 0) = 1
√
2 and ∇f (0, 0) ⋅ u = (1, −1) ⋅ ( 1
√ 2, − 1
√ 2) =
√ 2
∴Du(0, 0) ≠ ∇f (0, 0) ⋅ u, i.e f (x, y) is not differentiable. (1)
Page 8 of 12
9. (12%) Suppose that x, y, z satisfy the relation x2+2y2+3z2+xy − z = 9. Find ∂2z
∂x2, ∂2z
∂x∂y and ∂2z
∂y2.
Solution:
Let F (x, y, z) = x2+2y2+z2+xy − z − 9 = 0
Then
∂z
∂x = − Fx
Fz
= − 2x + y
6z − 1 (3 points) And
∂z
∂y = − Fy
Fz
= − x + 4y
6z − 1 (3 points) Therefore
∂2z
∂x2 =
∂
∂x(−
2x + y 6z − 1) =
(−2)(6z − 1) + (2x + y)(6zx)
(6z − 1)2 =
−2
6z − 1−6(2x + y)2
(6z − 1)3 (2 points)
∂2z
∂x∂y =
∂
∂y(−
2x + y 6z − 1) =
(−1)(6z − 1) + (2x + y)(6zy)
(6z − 1)2 =
−1
6z − 1−6(2x + y)(x + 4y)
(6z − 1)3 (2 points)
∂2z
∂y2 = ∂
∂y(−x + 4y 6z − 1) =
(−4)(6z − 1) + (x + 4y)(6zy)
(6z − 1)2 = −4
6z − 1−6(x + 4y)2
(6z − 1)3 (2 points)
10. (12%) Find all critical points of the function f (x, y) = xye−x2−y2 and classify them.
Solution:
fx(x, y) = (1 − 2x2)ye−x2−y2 (1 points) fy(x, y) = (1 − 2y2)xe−x2−y2 (1 points)
Ô⇒ critical points are (0, 0), ( 1
√2, 1
√2), ( 1
√2, −1
√2), (−1
√2, 1
√2), (−1
√2, −1
√2)(2 points) fxx(x, y) = (−6xy − 4x3y)e−x2−y2
fyy(x, y) = (−6xy − 4xy3)e−x2−y2
fxy(x, y) = (1 − 2x2−2y2+4x2y2)e−x2−y2 (2 points)
∆(0, 0) = −1 < 0
∆( 1
√ 2, 1
√ 2) =
4
e2 >0 and fxx( 1
√ 2, 1
√ 2) =
−2 e <0
∆( 1
√ 2, −1
√ 2) =
4
e2 >0 and fxx( 1
√ 2, −1
√ 2) =
2 e >0
∆(−1
√ 2, 1
√ 2) =
4
e2 >0 and fxx(
−1
√ 2, 1
√ 2) =
2 e >0
∆(−1
√ 2, −1
√ 2) =
4
e2 >0 and fxx(
−1
√ 2, −1
√ 2) =
−2
e <0 (4 points) Hence
(0, 0) is a saddle point
( 1
√ 2, 1
√ 2), (−1
√ 2, −1
√
2)are local maximum points (
1
√ 2, −1
√ 2), (−1
√ 2, 1
√
2)are local minimum points (2 points)
Page 10 of 12
11. (12%) Among all planes that are tangent to the surface x2yz = 1, are there the ones that are nearest or farthest from the origin? Find such tangent planes if they exist.
Solution:
Preliminaries Let f (x, y, z) = x2yz, and let Trbe the tangent plane of a point r = ⟨a, b, c⟩ on the surface. The gradient of f on r is
∇f (r) = ⟨2abc, a2c, a2b⟩ = ⟨2 a,1
b,1 c⟩.
Note that here we use the condition a2bc = 1 since r is a point on the surface. Since ∇f (r) is also the normal vector of Tr, the equation of Tris
2 ax +1
by +1 cz = 4.
The distance between Tr and the origin is
d(Tr, 0) = ∣2a⋅0 +1b⋅0 +1c ⋅0 − 4∣
√ (2a)
2
+ (1b)
2
+ (1c)
2 =
4
√
4
a2+b12 +c12
.
Now, let
g(a, b, c) = 4 a2+
1 b2 +
1 c2
and find the maxima and minima subject to the constraint a2bc = 1. The maxima of g correspond to the nearest tangent planes, and the minima correspond the farthest. We will use several methods to solve this optimization problem.
Method 1 Applying the AM-GM inequality,
g(a, b, c) = 2 a2+
2 a2+
1 b2 +
1 c2 ≥4
√ 2 a2 ⋅
2 a2⋅
1 b2 ⋅
1 c2 =4
√ 4 a4b2c2 =8.
The maxima does not exist since g → ∞ when a → 0. And the minima of g occurs when 2/a2=1/b2=1/c2; that is, a2=2b2=2c2. By a2bc = 1, we have
a = ±4
√
2 and b = c = ± 1
√4
2. Method 2 Applying the Lagrange multiplier,
∇g + λ(f − 1) = 0;
that is,
− 8 a3 +λ2
a=0, − 2 b3 +λ1
b =0, and −
2 c3+λ1
c =0.
Therefore, the extrema occurs when λ = 4/a2=2/b2=2/c2; that is, a2=2b2=2c2. It follows that g(a, b, c) = 8.
Also, the extrema occurs when the derivative of g does not exist; that is, a = 0 or b = 0. Since g → ∞ when a → 0 or b → 0, these does are not exist, and we can guarantee that those extrema with g = 8 are global minima.
Method 3 Replacing c by 1/a2b,
g(a, b) = 4 a2+
1
b2 +a4b2. The first order partial derivatives are
8 3 2 2 4
Therefore, the extrema occurs when ga=0 and gb=0; that is, a = ±√4
2 and b = c = ± 1
√4
2. It follows that
D = gaagbb−g2ab=24 ⋅ 16 − (±8√
2) = 384 − 128 = 256 > 0;
that is, these extrema are local minima. Also, the extrema occurs when derivative of g does not exist; that is, a = 0 or b = 0. Since g → ∞ when a → 0 or b → 0, these does are not exist, and we can guarantee that those local minima are global minima.
Results After solving the optimization problem, we find the farthest tangent planes 23/4x + 21/4y + 21/4z = 1,
23/4x − 21/4y − 21/4z = 1,
−23/4x + 21/4y + 21/4z = 1,
−23/4x − 21/4y − 21/4z = 1.
The nearest tangent plane does not exist since g has no maxima.
Points
(2%) Find ∇f(r).
(2%) Find equation of Tr.
(2%) Find the distance between Tr and the origin.
(2%) Find the extrema of g.
(2%) Find farthest tangent planes.
(2%) Show that the nearest tangent plane does not exist.
Page 12 of 12
