1032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. (18%) Test the series for absolute convergence, conditional convergence or divergence.
(a)
∞
∑
n=2 (−1)n
n(ln n)2. (b)
∞
∑
n=1
(−1)ntann1. (c)
∞
∑
n=1
(−1)n
1 12+1
22+⋯+ 1
n2
Solution:
(a) (Total: 6 points)
Step (1): Apply Integral Test to
∞
∑
n=2
∣ (−1)n n (ln n)2∣ =
∞
∑
n=2
1
n (ln n)2 (3 points).
Step (2): Correctly calculate the integral ∫
∞ 2
1
x (ln x)2 dx = −1 ln x∣
∞
2
= 1
ln 2 (3 points).
Step (3): Thus by Integral Test,
∞
∑
n=2
(−1)n
n (ln n)2 is absolutely convergent.
Grading Policies:
(1) As long as you applied Integral Test, you are granted 3 points regardless of the correctness of your calculation of the integral∫
∞ 2
1
x (ln x)2 dx.
(2) If your calculation of the integral ∫
∞ 2
1
x (ln x)2 dx is wrong, 1 or 2 points is granted depending on how many errors you make in that calculation.
(3) If you correctly proved that “
∞
∑
n=2
(−1)n
n (ln n)2 is convergent” by Alternating Series Test, you are also granted 3 points. However, these 3 points do not stack with points granted from Step (1) or Step (2).
(b) (Total: 6 points)
Step (1): Apply Limit Comparison Test to
∞
∑
n=1
∣(−1)ntan1 n∣ =
∞
∑
n=1
tan1
n to compare it with
∞
∑
n=1
1
n (1 points).
Step (2): Correctly derive the limit: lim
n→∞
tann1
1 n
=1 (1 points).
Step (3): Correctly state that
∞
∑
n=1
1
n is divergent. (1 points).
Step (4): Thus by Limit Comparison Test,
∞
∑
n=1
tan1
n is divergent.
Step (5): Apply Alternating Series Test to
∞
∑
n=1
(−1)ntan1
n. (1 points).
Step (6): Correctly state that: lim
n→∞tan1
n=0 (1 points).
Step (7): Correctly state that: tan1n is decreasing. (1 points).
Step (8): Thus by Alternating Series Test,
∞
∑
n=1
(−1)ntan1
n is convergent.
Step (9): Therefore,
∞
∑
n=1
(−1)ntan1
n is conditionally convergent.
Grading Policies:
Step (1) to Step (4) can be replaced by the following:
Step (1’): Apply Comparison Test to
∞
∑
n=1
tan1
n to compare it with
∞
∑
n=1
1
n (1 points).
Step (2’): Correctly state that: tan1n> n1, ∀n (1 points).
Step (3’): Correctly state that
∞
∑
n=1
1
n is divergent. (1 points).
Step (4’): Thus by Comparison Test,
∞
∑
n=1
tan1
n is divergent.
(c) (Total: 6 points)
Step (1): Correctly state that: lim
n→∞
(−1)n
1
12 +212+ ⋯ +n12
≠0. (6 points). (For example, by p-series, p = 2 > 1, we have lim
n→∞( 1 12+
1 22 + ⋯ +
1 n2) =
∞
∑
k=1
1
k2 is finite. So the above limit follows.) Step (2): Thus by Test for Divergence,
∞
∑
n=1
(−1)n
1
12 +212 + ⋯ +n12
is divergent.
Grading Policies:
(1) If you only correctly proved the divergence of
∞
∑
n=1
1
1
12+212 + ⋯ +n12
, you are granted 3 points.
(2) If you applied Alternating Series Test and claimed that “because lim
n→∞
1
1
12+212 + ⋯ +n12
≠ 0, i.e., the conditions for Alternating Series Test is not satisfied, therefore
∞
∑
n=1
(−1)n
1
12+212 + ⋯ +n12
is divergent ”, you will be deducted 3 points because the logic is incorrect.
2. (10%)
(a) Find the Maclaurin series for cos−1x. (Write down the general term explicitly.) (b) What is the radius of convergence of the series in (a).
(c) Let f (x) = cos−1(x2). Find f(10)(0).
Solution:
(a) Observe that d cosdx−1x= −√1
1−x2, so it suffices to find the Maclaurin series for √1
1−x2. 1
√
1 − x2 = (1 − x2)
−1/2
=
∞
∑
n=0
(
−1 2
n) (−x2)
n
=1 +(−12)
1! (−x2) +
(−12) ⋅ (−32) 2! (−x2)
2
+ ⋯+
(−12) ⋅ (−32) ⋅ ⋯ ⋅ (−12 −n + 1)
n! (−x2)
n
+ ⋯
=1 + 1 2 ⋅ 1!x2+
1 ⋅ 3
22⋅2!x4+ ⋯ +
1 ⋅ 3 ⋅ ⋯ ⋅ (2n − 1)
2n⋅n! x2n+ ⋯
=1 +
∞
∑
n=1
1 ⋅ 3 ⋅ ⋯ ⋅ (2n − 1) 2n⋅n! x2n
=
∞
∑
n=0
(2n)!
(2n⋅n!)2x2n (another expression) Thus
cos−1x =∫ −
∞
∑
n=0
(2n)!
(2n⋅n!)2x2ndx = −
∞
∑
n=0
(2n)!
(2n⋅n!)2(2n + 1)x2n+1+C Now since cos−10 =π2, we have C =π2. Therefore
cos−1x = π 2 −
∞
∑
n=0
(2n)!
(2n⋅n!)2(2n + 1)x2n+1 (b) R = 1
(c) By (a), we have cos−1(x2) = π2 −
∞
∑
n=0
(2n)!
(2n⋅n!)2(2n+1)(x2)
2n+1
= π2 −
∞
∑
n=0
(2n)!
(2n⋅n!)2(2n+1)x4n+2. Thus f(10)(0) is obtained when n = 2, and then f(10)(0) = −403 ⋅10!.
評分標準
(1) 三題配分分別是6分、2分和2分。
(2) 組合數沒展開扣1分;忘了常數項π2扣1分。
(3) a小題微分微錯最多得4分。
(4) c小題須答案達一定程度才給分。
3. (8%) Find the interval of convergence of the series
∞
∑
n=0
(n + 3)xn, and compute the sum.
Solution:
We can use the ratio test to find the radius of convergence
(1 point), since
n→∞lim n + 4 n + 3=1,
the radius is 1
(1 point). To get the inteval of convergence , we need to check the end points, 1 and −1. Since
n→∞lim (n + 3) and
n→∞lim (−1)n(n + 3)
are both nonzero, the series is not convergent at −1 and 1. So the inteval of convergence is [−1, 1]
(2 points).
To get the sum, we can find what Σ∞n=03xn and Σ∞n=0nxn are. Since Σ∞n=0xn= 1−x1
(1 point),
Σ∞n=03xn= 3 1 − x. And if we put S = Σ∞n=0nxn,
(1 − x)S = x + x2+x3+ ⋯ = x 1 − x
S = x
(1 − x)2
(2 points).
So the sum of the series is
3 1 − x+
x
(1 − x)2
(1 point).
4. (8%) Compute the sum of the series S = (1)(1
2) + (1 −1 3)(
1
2)3+ (1 −1 3+
1 5)(
1
2)5+ (1 −1 3 +
1 5 −
1 7)(
1
2)7+ (1 −1 3+
1 5−
1 7+
1 9)(
1 2)9+ ⋯.
(Hint: imitate the method of deriving the sum of a geometric series.)
Solution:
Let the sum be S, then
(1 − 1 4)S =3
4S =1 2 −
1 3(
1 2)
3
+ 1 5(
1 2)
5
− 1 7(
1 2)
7
+ 1 9(
1 2)
9
− ⋯
(2 points).
Let
f (x) = x −1 3x3+
1 5x5−
1 7x7+
1 9x9− ⋯
,
f′(x) = 1 − x2+x4−x6+x8− ⋯ = 1
1 + x2
(2 points).
Then
f (x) =∫ f
′(x) dx = arctan x + C where the constant C is clearly 0
(2 points). So
S = 4
3arctan1
2
(2 points).
5. (8%) A curve consists of two pieces of curves:
C1∶ r(t) = (t2+3t)i + (t3−4t + 1)j, t ≤ 0,
C2∶ y = p(x), x > 0, where p(x) is a polynomial of degree 2.
Find the polynomial p(x) so that this curve is continuous and has continuous slope and continuous curvature.
Solution:
Let p(x) = ax2+bx + c for some parameters a, b, and c to be specified.
Consider x = 0 when t = 0, this curve is continuous.
r(t = 0) =< 02+3 × 0, 03−4 × 0 + 1 >=< 0, 1 >, (1) and
<x = 0, y = p(x = 0) >=< 0, a02+b0 + c >=< 0, c > . (2) Let (1) equals (2), we get c = 1.
Get 1 points with (1) or the conclusion c = 1.
Consider x = 0 when t = 0, this curve has continuous slope.
dy dx∣
t=0=
dy dt dx dt
∣t=0= 3t2−4
2t + 3 ∣
t=0=
−4
3 (3)
and
dy
dx∣x=0=2ax2+b∣x=0=b (4)
Let (3) equals (4), we get b = −43.
Get 1 points with (3) or r′(0) =< 3, −4 >.
Get 1 points with the conclusion b = −43.
Consider x = 0 when t = 0, this curve has continuous curvature.
κ = ∣r′×r′′∣
∣r′∣3
=
∣ <2t + 3, 3t2−4, 0 > × < 2, 6t, 0 > ∣
∣ <2t + 3, 3t2−4, 0 > ∣3 (5)
=
∣12t2+18t − 6t2+8∣
(4t2+12t + 9 + 9t4−24t2+16)32 (6) Then,
κ∣t=0= 8 2532 =
8
125 (7)
and
Get 2 points with (7).
κ∣x=0=
∣p′′(x)∣
(1 + p′(x)2)
3 2
∣x=0=
∣2a∣
(1 + (2ax + b)2)
3 2
∣x=0=
∣2a∣
(1 + b2)
3 2
(8)
Let (7) equals (8), we get ∣a∣ = (1258 )(12527)(12) = 274. Get 2 points with (8).
Two polynomials of degree 2 are our solutions.
p1(x) = 4 27x2−
4
3x + 1 (9)
p2(x) = −4 27x2−
4
3x + 1 (10)
Get 1 points with two solutions.
6. (12%) Find the limit, if it exists, or show that the limit does not exist.
(a) lim
(x,y)→(0,0) x5+x2y3
x4+y6 . (b) lim
(x,y,z)→(0,0,0) exyz−1 x2+y2+z2.
Solution:
(a) The limit does not exist, because the limit approaches different values along with x = 0 and x2=y3. Along with x = 0,
y→0lim
05+02y3 04+y6 =lim
y→0
0
y6 =0, (11)
for all y ≠ 0.
Get 3 points with one of limit value like equation (11).
Along with x2=y3,
f (x, y) = f (y3/2, y) = y15/2+y6 y6+y6 =
y3/2+1
2 (12)
y→0lim
y3/2+1
2 =
0 + 1 2 =
1
2 (13)
Get another 3 points with another limit value like equation (13).
(b) The limit approaches to zero.
Case 1: xyz = 0. Because (x, y, z) → (0, 0, 0) means (x, y, z) ≠ (0, 0, 0) and x2+y2+z2≠0. Then, f (x, y, z) = e0−1
x2+y2+z2 =0. (14)
Case 2: xyz ≠ 0.
Transfer Cartesian coordinates to spherical coordinate.
x = r sin θ cos φ, y = r sin θ sin φ,
z = r cos θ, (15)
where r > 0 because of xyz ≠ 0. Then,
f (x, y, z) =exp{r3δ} − 1
r2 , (16)
where δ = sin2θ cos φ sin φ cos θ. By L’Hospital’s Rule, lim
r→0
exp{r3δ} − 1 r2 =lim
r→0
exp{r3δ}3r2δ
2r ==lim
r→0r(3δ exp{r3δ}
2 ) =0, (17)
for all δ is finite.
Get 1 point if you conclude the limit 0 only through a specific direction like x = 0, y = 0, x = z = 0, or x = y = z, ...
Get 5 points if you proof the limit 0 by transforming spherical coordinate but with only wrong spherical expression.
Get 0 point if you try to proof the limit 0 with wrong logical derivations.
7. (10%) Let f (x, y) =
⎧⎪
⎪⎪
⎪
⎨
⎪⎪
⎪⎪
⎩ x3−y3
x2+y2, for (x, y) ≠ (0, 0) 0, for (x, y) = (0, 0).
(a) Find fx(x, y) and fy(x, y).
(b) Are the functions fx and fy continuous at (0,0)?
Solution:
(a) (3pts) For (x, y) ≠ (0, 0),
fx(x, y) = 3x2 x2+y2 −
2x(x3−y3) (x2+y2)2
=
x4+3x2y2+2xy3 (x2+y2)2
,
fy(x, y) = −3y2 x2+y2 −
2y(x3−y3) (x2+y2)2
=
−y4−3x2y2−2yx3 (x2+y2)2
.
(3pts) For (x, y) = (0, 0),
fx(0, 0) = lim
h→0
f (h, 0) − f (0, 0)
h =lim
h→0
h3 h2⋅
1 h=1, fy(0, 0) = lim
k→0
f (0, k) − f (0, 0)
k =lim
k→0− k3 k2⋅
1 k = −1.
(b) (4pts) If fx, fy is continuous at (0, 0), then lim
(x,y)→(0,0)fx and lim
(x,y)→(0,0)fy exist and their values equal 1 and
−1 respectively. But along x = 0, lim
(x,y)→(0,0)
fx(x, y) = lim
a→0fx(0, a) = lim
a→0
0 a4 =0, which conflicts to fx(0, 0) computed in (a). Similarily along y = 0,
lim
(x,y)→(0,0)
fy(x, y) = lim
b→0fy(b, 0) = lim
b→0
0 b4 =0, which isn’t equal to −1(= fy(0, 0) = −1).
8. (8%) Find the tangent plane of the surface 4
πarctanz
2 =x2+ ∫
z xy
xy
√ 1 + t3dt at the point (1, 2, 2).
Solution:
Let F (x, y, z) = x2+xy∫
z xy
√
1 + t3dt −π4arctanz2. Note that ∇F (x0, y0, z0)is normal to the point (x0, y0, z0)on the surface F (x, y, z) = 0. Thus we compute ∇F at first,
Fx=2x + y∫
z xy
√
1 + t3dt + xy(
√
1 + (xy)3⋅ (−y)), (2pts)
Fy=x∫
z xy
√
1 + t3 dt + xy(
√
1 + (xy)3⋅ (−x)), (2pts)
Fz=xy
√ 1 + z3−
4 π⋅
1 2⋅
1 1 + (z2)2
. (2pts)
Hence, ∇F (1, 2, 2) = (2 + 0 − 4√
1 + 23, −2√
1 + 23, 2√
1 + 23−π2(1+11 )) = (−10, −6, 6 −π1) (1pt), and the tangent plane at (1, 2, 2) is −10(x − 1) − 6(y − 2) + (6 −1π)(z − 2) = 0. (1pt)
9. (8%) Find all points at which the direction of fastest change of the function f (x, y, z) = x2+y2+z2−2x − 4y − 6z is i + 2j + 3k.
Solution:
When ∇f = 2(x − 1, y − 2, z − 3) is parallel to (1, 2, 3)[ 1 pt ]
correct ans:
⎧⎪
⎪⎪
⎪
⎨
⎪⎪
⎪⎪
⎩
x−1
1 = y−22 = z−33 {(t, 2t, 3t) ∶ t ∈ R}
{(t, 2t, 3t) ∶ t ∈ R/{1}}
[ 8 pts ]
Note, “fastest” means increasing or decreasing most drastically.
Thus if you only consider the case:
⎧⎪
⎪
⎨
⎪⎪
⎩
{(t, 2t, 3t) ∶ t ∈ R, t ≥ 0}
{(t, 2t, 3t) ∶ t ∈ R/{1}, t ≥ 0} you’ll get [7 pts]
If you only write one or two points, such as (1, 2, 3), you’ll get [ 2 pts ]
10. (10%) Let g(x, y) = 4x3−13y3+6x2y + 3xy2−12x2−12xy − 30y2. Find the critical points of g(x, y), and classify them.
Solution:
∇f = 0 [ 1 pt ]
critical points: (0, 0), (2, 0), (23,−43), (83,−43) [ 1 pt for each]
If you mentioned D(x, y) [ 1 pt ]
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎩
(0, 0), D(0, 0) > 0, gxx<0, local maximum (2, 0), D(2, 0) < 0, saddle point
(23,−43), D(23,−43) <0, saddle point
(83,−43), D(83,−43) >0, gxx>0, local minimum
[ 1 pt for each]
11. (10%) Find the points on the intersection of the plane x + y + 2z = 2 and the paraboloid z = x2+y2 that are nearest to and farthest from the origin.
Solution:
Set
f (x, y, z) = x2+y2+z2 g(x, y, z) = x + y + 2z − 2 h(x, y, z) = x2+y2−z.
Assume that ∇f + λ1∇g + λ2∇h = 0. Then we obtain
2x + λ1+2λ2x = 0 2y + λ1+2λ2y = 0 2z + 2λ1−λ2=0.
We also have
x + y + 2z = 2 x2+y2−z = 0.
Hence we can yield critical points are (12,12,12) and (−1, −1, 2). Since f (12,12,12) = 34 and f (−1, −1, 2) = 6, we can conclude that (12,12,12)is the point on the intersection nearest to the origin and (−1, −1, 2) is the point on the intersection farthest to the origin.
評分標準：函數假設寫好以及 Largrange’s multiplier 的使用方式有做說明，這裡佔4%，方程式列出，解方程部份基本
上不看過程(但必須要寫)，不過中間若有明顯錯誤便會扣分(依錯的程度來斟酌)，最後要說明哪個點為最大值點，哪個點為最
小值點，此處必須解釋原因，只有寫下結論者扣4%，解釋不夠詳細會斟酌扣分，而筆墨分為2%，但空白者與不相干的過程皆
不給分。
