微甲

992微甲^01-05班期中考解答和評分標準

1. (10%) Find the values of real number p for which the series is divergent, conditionally convergent, or absolutely convergent.

(a)

∑∞ n=2

(−1)ⁿ 1 n(ln n)^p. (b)

∑∞ n=1

(−1)ⁿsin ( 1

n^p

)· ln n.

Sol:

(a) Since d dx

( 1

x(ln x)^p )

=−(ln x)^p+ x· p(ln x)^{p−1 1}_x

x²(ln x)^2p =−(ln x)^p−1(ln x + p)

x²(ln x)^2p < 0 for x > e^−p. Hence ∀p ∈ R, we have 1

n(ln n)^p is decreasing for n large enough. (n > e^−p) If p > 0, then lim

n→∞

1

n(ln n)^p = 0. And if p < 0, then by L’hospital Law, we have

nlim→∞

1

n(ln n)^p = lim

n→∞

(ln n)^−p

n = lim

n→∞

−p(ln n)^{−p−1 1}_n

1 = lim

n→∞

−p(ln n)^−p−1 n

=· · · = lim

n→∞

(−1)^[−p]+1(p)(p + 1)(p + 2)· · · (p + [−p])(ln n)−p−([−p]+1)

n = 0.

Hence by alternating series test ,

∑∞ n=2

(−1)ⁿ 1

n(ln n)^p is convergent for all p∈ R.

Now consider a_n ≡ ¯¯

¯¯(−1)ⁿ 1 n(ln n)^p

¯¯¯¯ = 1

n(ln n)^p. Suppose p 6= 1, since we have proven an is decreasing for n large enough, thus ”integral test” is available here

∫ _∞

2

1

x(ln x)^pdx =

∫ _∞

ln 2

u^−pdu¯¯

¯¯u=ln x

= lim

b→∞

(u^1−p 1− p

)b ln 2

= lim

b→∞

b^1−p− (ln 2)^1−p

1− p =







∞ for p < 1 (ln 2)^1−p

p− 1 for p > 1

⇒











∑∞ n=2

1

n(ln n)^p , is convergnet for p > 1

∑∞ n=2

1

n(ln n)^p , is divergnet for p >< 1 .

Suppose p = 1, we also use integral test, since

∫ _∞

2

1

x(ln x)dx =

∫ _∞

ln 2

1 udu¯¯

¯¯ = lim

b→∞(ln b− ln(ln 2)) = ∞.

So we get the answer is

∑∞ n=2

(−1)ⁿ 1

n(ln n)^p is







absolutely convergent for p > 1 (2.5 pts) conditionally convergent for p≤ 1 (2.5 pts).

(b) For p≤ 0, since lim

n→∞sin( 1

n^p) ln n6= 0, thus

∑∞ n=1

(−1)ⁿsin( 1

n^p) ln n is divergent.

Now suppose p > 0, by limit comparison test, since

nlim→∞

sin(_n¹p) ln n

1

n^pln n = 1 ∀p ∈ R⁺= (0,∞) Thus we can examine

∑∞ n=1

1

n^p ln n to get the answer.

For 0 < p≤ 1 ,by comparison , since ln n n^p ≥ 1

n^p for n > 3, and

∑∞ n=1

1

n^p is divergent for 0 < p≤ 1, so is ∑^∞

n=1

ln n n^p . For p > 1 then

ln n n^p = 1

n^p+12 · ln n n^p−12 . Since lim

n→∞

ln n

n^p−12 = 0, thus ln n

n^p−12 < 1 for n large enough, that is ln n

n^p < n^−p+12 for n large enough, since p + 1

2 > 1, thus by compare with

∑∞ n=1

n^−p+12 , we get

∑∞ n=1

ln n n^p converges for p > 1.

So we get the answer is

∑∞ n=1

(−1)ⁿsin( 1

n^p) ln n is













divergent for p≤ 0 (1 pt)

conditionally convergent for 0 < p≤ 1 (2 pts) absolutely convergent for p > 1 (2 pts)

評分標準^:

1. 照詳解上的配分方式給分^.

2. alternating series test 前提條件不算分^, 有寫有加分^. 3. 寫答案未寫理由者不予給分^.

2. (8%) Determine whether the series

∑∞ n=1

(−1)ⁿln (

1 + tan⁻¹ 1 n

)

is absolutely convergent, conditionally convergent, or divergent.

Sol:

Let f (x) = ln(1 + tan⁻¹ 1

x), then f⁰(x) = 1

1 + (tan⁻¹(¹_x))²

−1

x²+ 1 < 0 , ∀x > 0.

That is , a≡ ln(1 + tan⁻¹(1

n)) is decreasing , and

nlim→∞ln(1 + tan⁻¹ 1 n) = 0.

Hance by alternating series test , we have

∑∞ n=1

(−1)ⁿln(1 + tan⁻¹ 1

n) converves.

(8 pts) Now consider b_n= ln(1 + tan⁻¹ 1

n) > 0, since by Taylor expansion ln(1 + tan⁻¹ 1

n) = (1

n − 1 3(1

n)²+· · · )

− 1 2

(1 n − 1

3(1

n)²+· · · )2

+· · ·

Hance compare with

∑∞ n=1

1

n, we have

nlim→∞

ln(1 + tan^{−1 1}_n)

1 n

= lim

t→0⁺

ln(1 + tan⁻¹t)

t = lim

t→0⁺ 1 1+tan⁻¹t

1 1+t²

1 = 1

So we get

∑∞ n=1

ln(1 + tan⁻¹ 1

n) diverges , that is

∑∞ n=1

(−1)ⁿln(1 + tan⁻¹ 1

n) is conditionally convergent 評分標準^:

1. 照詳解上的配分方式給分^.

2. alternating series test 前提條件不算分^, 有寫有加分^. 3. 寫答案未寫理由者不予給分^.

3. (15%) The Fibonacci sequence {fn} is defined as f1 = f₂ = 1, f_n= f_n−1+ f_n−2 ∀n ≥ 3.

Let a_n= fn+1

f_n .

(a) Show that an= 1 + 1

a_n−1 and 1≤ an ≤ 2 for all n.

(b) Show that{a2n+1} is a monotonic sequence and is convergent.

(c) Find lim

n→∞a_2n+1.

(d) Find the radius of convergence of the power series

∑∞ n=1

fnsin (nπ

2 )

xⁿ.

Sol:

(a) a_n = f_n+1 fn

= f_n+ f_n−1 fn

= 1 + 1

fn

f_n−1

= 1 + 1

an−1 (2 pts) 1≤ a1 = 1

1 = 1≤ 2, assume that 1≤ ak≤ 2, for n = k + 1, 1

2 ≤ 1

a_k ≤ 1 ⇒ 1 ≤ 1 +1

2 ≤ 1 + 1

a_k ≤ 1 + 1 = 2 ⇒ 1 ≤ ak+1 ≤ 2 Hence, 1≤ an≤ 2 for all n by induction. (2 pts)

(b) a1 = 1 ≤ 3 2 = a3,

assume that a_2k−1 ≤ a2k+1 ⇒ 1 + 1

a_2k−1 ≥ 1 + 1

a_2k+1 ⇒ a2k ≥ a2k+2

⇒ 1 + 1

a_2k ≤ 1 + 1

a_2k+2 ⇒ a2k+1≤ a2k+3 (3 pts)

Hence, {a2n+1} is increasing by inducion and bounded by (a) ⇒ convergent. (1 pt) (c) Assume lim

n→∞a_2n+1 = α a_2n+1 = 1 + 1

a_2n = 1 + 1 1 + _a ¹

2n−1

= 2a_2n−1+ 1

a_2n−1+ 1 (2 pts)

nlim→∞a_2n+1 = lim

n→∞

2a_2n−1+ 1 a_2n−1+ 1

⇒ α = 2α + 1

α + 1 ⇒ α²− α − 1 = 0 ⇒ α = 1 +√ 5

2 (1 pt) (d)

∑∞ n=1

f_nsin (nπ 2 )xⁿ=

∑∞ n=0

(−1)ⁿf_2n+1x²ⁿ⁺¹ (1 pt) a_2n+2 = 1 + 1

a2n+1 ⇒ lim

n→∞a_2n+2 = 1 + 1 limn→∞a2n+1

= 1 + 1

α = 1 +√ 5

2 (1 pt) By ratio test, lim

n→∞

¯¯¯¯f_2n+3 f_2n+1x²¯¯

¯¯ < 1 ⇔¯¯

¯¯ limn→∞

f_2n+3 f_2n+2 lim

n→∞

f_2n+2 f_2n+1x²¯¯

¯¯ < 1 (1 pt)

⇔¯¯¯ lim

n→∞a_2n+2 lim

n→∞a_2n+1x²¯¯¯ < 1 ⇔¯¯

¯¯¯(1 +√ 5 2 )²x²

¯¯¯¯

¯< 1 ⇔ |x| < 2 1 +√

5 =

√5− 1 2 Hence, R =

√5− 1

2 (1 pt)

4. (8%) Find the Maclaurin series for ln (x +√

1 + x²). (You must write down the n th term.)

Sol:

f (x) = ln(x +√

1 + x²)

⇒ f⁰(x) = (1 + √ ^x 1+x²) (x +√

1 + x²) = 1

√1 + x² (2 pts)

⇒ f⁰(x) =

∑∞ k=0

(−¹₂ k

)

x^2k (2 pts) ∀ |x| < 1 (1 pt)

⇒ f(x) = C +

∫

f⁰(x)dx = C +

∑∞ k=0

(−¹₂ k

) x^2k+1

2k + 1 ∀ |x| < 1 (2 pts) f (0) = ln 1 = 0 ⇒ C = 0 (1 pt)

⇒ f(x) =

∑∞ k=0

(−¹₂ k

) x^2k+1 2k + 1

(

=

∑∞ k=0

(−1)^k(2k)! x^2k+1 2^2k(k!)²(2k + 1)

)

∀ |x| < 1

5. (15%) Let r(t) =ht,√

2 ln cos t, tan t− ti, −π

2 < t < π

2, and P be the point r (π

4 )

. (a) Find the length of the arc r(t), 0≤ t ≤ π

4.

(b) Find the unit tangent vector T, the principal unit normal vector N and the binormal vector B at P .

(c) Find the curvature κ at P .

(d) Find the center of the osculating circle (the circle of curvature) at P . Sol: r⁰(t) = (1,−√

2 tan t, tan²t) ⇒ |r⁰(t)| = sec²t (a)

∫ ^π

4

0

sec²t dt (1 pt) = tan t¯¯¯

π 4

0

= 1 (2 pts) (b) T(t) = r⁰(t)

|r⁰(t)| (1 pt) T_P = (1

2,−

√2 2 ,1

2) (1 pt) N(t) = T⁰(t)

|T(t)| T⁰(t) = (− sin 2t, −√

2 cos 2t, sin 2t) (1 pt) N_P = (−

√2 2 , 0 ,

√2

2 ) (1 pt) B(t) = T(t)× N(t) (1 pt) B_P = (−1

2,−

√2 2 ,−1

2) (1 pt) (c) κ(t) = |T⁰(t)|

|r⁰(t)| (2 pts) κP =

√2

2 (1 pt) (d) The radius r = 1

κ. The center C_P = P + rN (2 pts) C_P = (π

4 − 1, −√ 2 ln√

2, 2− π

4) (1 pt) 6. (12%) Let f (x, y) = x¹³y²³.

(a) Find f_x(0, 0) and f_y(0, 0).

(b) Let L(x, y) be the linear approximation of f at (0, 0). Does lim

(x,y)→(0,0)

|f(x, y) − L(x, y)|√ x²+ y² exist?

(c) Find the directional derivative of f at (0, 0) in the direction h1, mi.

(d) Is f (x, y) differentiable at (0, 0)?

Sol:

(a) By definition of partial derivative, f_x(0, 0) = lim

h→0

f (h, 0)− f(0, 0)

h = 0.

Similarly f_y(0, 0) = 0.

(b) By definition, L(x, y) = f (0, 0) + f_x(0, 0)x + f_y(0, 0)y.

Since f_x(0, 0) = f_y(0, 0) = 0, we have L(x, y) = 0. Therefore

lim

(x,y)→(0,0)

|f(x, y) − L(x, y)|√

x²+ y² = lim

(x,y)→(0,0)

|x¹³y²³|

√x²+ y²

Along x = y,

lim

(x,y)→(0,0)

|x¹³y²³|

√x²+ y² = lim

x→

√|x|

2|x| = 1

√2. Along x = 0,

lim

(x,y)→(0,0)

|x¹³y²³|

√x²+ y² = 0.

Therefore, the limit does not exist.

(c) First normalize the vector < 1, m > to unit vector u = 1

√1 + m² < 1, m >. By definition of the directional derivative, we have

D_uf (0, 0) = lim

t→0

f (0 + √ ^t

1+m², 0 + √^tm

1+m²)− f(0, 0) t

= lim

t→0

m²³

√1 + m² = m²³

√1 + m².

(d) By (c), D_uf (0, 0) = m²³

√1 + m². By (a), 5f(0, 0) · u = 0.

Therefore,5f(0, 0)·u 6= Duf (0, 0) in general. Hence f is not differentiable at (0, 0).

Grading Policy :

Question (a) worth 2 pts. Only correct answer would get 2 pts.

Question (b) worth 5 pts. You would get 1 pt if you write down the linear approximation L(x, y) of f at (0, 0). Another 4 pts depends on your answer to the limit

lim

(x,y)→(0,0)

|f(x, y) − L(x, y)|√ x²+ y² .

Question (c) worth 3 pts. You would receive 2 pts if you did not normalized the vector

< 1, m > and the answer you give is m²³.

Question (d) worth 2 pts. If you only answer NO, you would get 1 pt. The other 1pt depends on your explnation .

7. (12%) Let f (x, y) = xye^−xy2.

(a) Find the gradient of f .

(b) Find the directional derivative of f at the point (1, 1) in the direction h 2

√5, 1

√5i.

(c) Find the tangent plane of z = f (x, y) at the point (1, 1,1 e).

(d) Let z = f (x, y) and x = u²+ 3v, y = uv− 3v. Find ∂z

∂v

¯¯¯

(u,v)=(2,−1). Sol:

(a) 5f(x, y) = e^−xy2(y− xy³, x− 2x²y²).

(b) Since f is differentiable on R², we have

D_<√²

2,√¹

5>f (1, 1) =5f(1, 1)· < 2

√2, 1

√5 >= √−1 5e. (c) The tangent plane of z = f (x, y) at the point (1, 1,1

e) is (z− 1

e) = f_x(1, 1)(x− 1) + fy(1, 1)(y− 1) = −(y − 1).

(d) By chain rule,

∂z

∂v = ∂z

∂x

∂x

∂v +∂z

∂y

∂y

∂v = (y− xy³)e^−xy2 · 3 + (x − 2x²y²)e^−xy2 · (u − 3).

When (u, v) = (2,−1), (x, y) = (1, 1). Therefore,

∂z

∂v

¯¯¯

(u,v)=(2,−1)= 0· 3 + −1

e · (−1) = 1 e.

Grading Policy :

Question (a) worth 3 pts. You would get 2 pts if you only give correct formula to either f_x or f_y.

Question (b) worth 3 pts. You would get 1 pt if you know

D_<√²

2,√¹

5>f (1, 1) =5f(1, 1)· < 2

√2, 1

√5 > .

Another 2 pts depends on your calculation.

Question (c) worth 3 pts. You would get 1 pt if you know

(z− 1

e) = f_x(1, 1)(x− 1) + fy(1, 1)(y− 1) = −(y − 1).

Another 2 pts depends on your calculation.

Question (d) worth 3 pts. You would get 1 pt if you know ∂z

∂v = ∂z

∂x

∂x

∂v +∂z

∂y

∂y

∂v. Another 2 pts depends on your calculation.

8. (10%) Find the local maximum and minimum values and the saddle points, if exist, of

f (x, y) = x³+ x²+3

2x²y + 2xy + 2y²+ 1 2y.

Sol:

f_x = 3x²+ 2x + 3xy + 2y f_y = 3

2x² + 2x + 4y +1

2 (2 pts) critical points:







3x²+ 2x + 3xy + 2y = 0⇒ (3x + 2)(x + y) = 0 ⇒ x = −2

3 or x =−y 3

2x²+ 2x + 4y + 1 2 = 0 x =−2

3 ⇒ 3 2 4 9 − 22

3 + 4y + 1

2 = 0⇒ y = 1 24 x =−y ⇒ 3

2y²− 2y + 4y + 1

2 = 0 ⇒ 3y² + 4y + 1 = 0⇒ y = −1 or y = −1 3 So critical points is (−2

3, 1

24), (1,−1), (1 3,−1

3) (2 pts)

f_xx = 6x + 2 + 3y fyy = 4

f_xy = 3x + 2

D = f_xxf_yy − fxy² = 24x + 8 + 12y− (3x + 2)² (3 pts) D(−2

3, 1

24) =−16 + 8 +1

2 − 0 < 0 ⇒ (−2 3, 1

24) is saddle point D(1,−1) = 24 + 8 − 12 − 25 < 0 ⇒ (1, −1) is saddle point D(1

3,−1

3) = 8 + 8− 4 − 9 > 0, fxx = 2 + 2− 1 > 0

⇒ (1 3,−1

3) is local minimum with minimum value f (1

3,−1 3) =

(1 3

)3

+ (1

3 )2

−3 2

(1 3

)2

1 3− 21

3 1 3 + 2

(1 3

)2

+1 2

1

3 =− 2

27 (3 pts) 9. (10%) Let Γ be the ellipse with center at the origin that is the intersection of the plane

x + y + 2z = 0 and the surface x²+ 2y²+ 4z² = 35.

(a) Find the lengths of the major and the minor axes (長軸與短軸^{) of Γ.}

(b) Find the area of the region enclosed by Γ.

Sol:

(a) The length of the major axis: √

105. The length of the minor axis: 2√ 14 (b) The area of the region enclosed by Γ: 7

2

√30π

Let f (x, y, z) = x²+ y²+ z², g(x, y, z) = x + y + 2z, h(x, y, z) = x²+ 2y²+ 4z²− 35 Applying the method of Lagrange multipliers, we need to solve













∇f = λ∇g + µ∇h · · · (∗) g = 0

h = 0

or



























2x = λ + 2µx· · · (1) 2y = λ + 4µy· · · (2) 2z = 2λ + 8µz· · · (3) x + y + 2z = 0· · · (4)

x²+ 2y²+ 4z²− 35 = 0 · · · (5)

⇒













2x(1− µ) = λ · · · (1)⁰ 2y(1− 2µ) = λ · · · (2)⁰ 2z(1− 4µ) = 2λ · · · (3)⁰

Utilizing (4) and (5) and considering (1)·x+(2)·y +(3)·z we have x²+ y²+ z² = 35µ

Case 1: λ = 0

If µ = 1⇒ y = z = 0, ∴ x = 0 by (1), but contradicts (5). Similarly, µ = 1

2 ⇒ x = z = 0 and µ = 1

4 ⇒ x = y = 0 both lead to contradiction.

If µ6= 1,1 2,1

4 ⇒ x = y = z = 0, also a contradiction.

∴ λ 6= 0 Case 2: λ6= 0 Substitute x = λ/2

1− µ, y = λ/2

1− 2µ, z = λ

1− 4µ into (4):

λ{ 1

2(1− µ) + 1

2(1− 2µ) + 2

1− 4µ} = 0

Arranging the numerator, we have 20µ²− 23µ + 6 = 0

⇒ (5µ − 2)(4µ − 3) = 0

⇒ µ = 2 5 or 3

4. Therefore x²+ y²+ z² = 14 or 105 4 . The length of the major axis is 2×

√105 4 =√

105 and the length of the minor axis is 2√

14.

The area is π·

√105 2 ·√

14 = 7 2

√30π

• 2 points for (1) to (3) each (6 points in total).

If only (∗) is present (i.e., without (1) to (3) and without solving):

* If the scalar function f is reasonably defined ⇒ 3 points

* If f is a scalar function but not correctly defined ⇒ 2 points

* If f is not even a scalar function (i.e., in the form of constraint) ⇒ 1 point Note: Using g to eliminate one variable in f and h is acceptable, but then the two functions should only have two dimensions.

• If both lengths of the axes are correct, 3 points will be credited. Half the val- ues (i.e.,

√105

2 and √

14) are also regarded as correct answers.

* If only one of them is correct ⇒ 2 points

* If both are incorrect but the two values of µ have been solved ⇒ 1 point

• 1 point for the area.