(10%) Find the values of ρ for the convergence of the series below (a

1022微微微甲甲甲07-11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. (10%) Find the values of ρ for the convergence of the series below

(a)

∞

X

n=0

e^{n(ρ2−ρ−2)}, (b)

∞

X

n=1

eⁿ¹ − 1 n^ρ .

Solution:

(a) This geometric series converges if and only if the ratio

e^ρ2−ρ−2< 1, or ρ²− ρ − 2 < 0 (3 points) (ρ − 2)(ρ + 1) < 0 ∴ −1 < ρ < 2 (2 points)

(b) Since e^x=

∞

X

n=0

xⁿ n!, x ∈ R

eⁿ¹ − 1 ≈ 1

n (3 points)

Apply the limit comparison test toXeⁿ¹ − 1

n^ρ andX 1 n^ρ+1

lim

n→∞

e¹ⁿ− 1 n^ρ

1 n^ρ+1

= lim

n→∞

eⁿ¹ − 1 1 n

= lim

1 n=t→0⁺

e^t− 1

t = 1 ∈ (0, ∞)

Thus two series both converges or diverges. (1 point)

X 1

n^ρ+1 is a p-series, which converges if and only if 1 + ρ = p > 1

∴ Xeⁿ¹ − 1

n^ρ converges if and only if ρ > 0 (1 point)

2. (10%) (a) Prove ln(n + 1) < 1 + 1

2 + · · · + 1

n < 1 + ln n.

(b) Test for convergence of

∞

X

n=1

a_n, where a_n= 1

1 +¹₂+ · · · + ¹_n.

Solution:

(a) f (x) = 1

x, g(x) = 1 1 + x ˆ n

0 g(x)dx 6 1 +1

2 + · · · + 1

n 6 f (1) + ˆ n

1

f (x)dx you can graph the picture.

(b) by (a) an> 1

1 + lnn > 1 1 + n By comparison testX

an div.

3. (15%) Let f (x, y) = sin(x − y)e^−x2−y2, P = (√ 2,√

2).

(a) Find the maximum rate of change of f at P

(b) Find the direction in which the maximum rate of change occurs.

(c) Find the directional derivative Du(P ), where u =

1 2,

√3 2

 .

Solution:

(a) 9 points

hf (x, y) =< cos(x − y)e^−x2−y2− 2x sin(x − y)e^−x2−y2, − cos(x − y)e^−x2−y2− 2y sin(x − y)e^−x2−y2 >

(fx 2 points,fy 2 points) hf (√

2,√

2) =< e⁻⁴, −e⁻⁴>

(e⁻⁴ 1 point,−e⁻⁴ 1 point)

the maximum rate of change is |h f (√

2,√ 2)| =

√2 e⁴ (formula 2 points,answer 1 point)

(b) 3 points

the direction is the same ash f (√

2,√

2),which is the direction of < e⁻⁴, −e⁻⁴>

(any vector that is the same direction as < 1, −1 > will be valid and get 3 points.But if |h f (√

2,√ 2)| is wrong in (a),one can get only 0 point! If one just claim ”the direction is parallel to h

f (√ 2,√

2)” but no write down ”< e⁻⁴, −e⁻⁴ >” explicitly,he can get only 2 points because opposite directions are included and they cause minimum not maximum. )

(c) 3 points

Duf (P ) =h

f (P ) · u = 1 −√ 3 2e⁴ (formula 2 points,answer 1 point)

4. (15%) Let r(t) = t⁴ 2, t,4

5t⁵²

for t ≥ 0.

(a) Find the length of the arc 0 ≤ t ≤ 2 of r(t).

(b) Find the curvature κ(t).

(c) Find T(1), N(1) and B(1), the principal unit normal vector and the binormal unit vector when t = 1 respectively.

Solution:

−→

r(t) = (t⁴ 2, t,4

5t⁵²) (a)

−→

r⁰(t) = (2t³, 1, 2t³²) . Then |

−→

r⁰(t) | = 1 + 2t³. Therefore, L = ˆ 2

0

(1 + 2t³) = 10 [3pts]

(b)

−→

r⁰⁰(t) = (6t², 0, 3t¹²). Thenr^−→0(t) ×r^−→00(t) = (3t¹², 6t⁷², −6t²) and |r^−→0(t) ×r^−→00(t) | = 3t¹²(2t³+ 1) Therefore, the curvature k(t =|r^−→0(t) ×r^−→00(t) |

|r^−→0(t) |³

= 3√

t

(2t³+ 1)²[3pts]

(c)

−→

T (t) =

−→

r⁰(t)

|r^−→0(t) |

= 1

1 + 2t³(2t³, 1, 2t³²) ⇒T (1) = (^−→ 2 3,1

3,2 3) [3pts]

−→

T⁰(t) = 1

(1 + 2t³)²(6t², −6t², −6t⁷² + 3t¹²) ⇒

−→

T⁰(1) ) = (2 3, −2

3, −1 3)

Therefore,N (1) =^−→

−→

T⁰(1)

|T^−→0(1) |

= (2 3, −2

3, −1 3) [3pts]

−→

B(1) =

−→

T (1) ×

−→

N (1) = (1 3,2

3, −2 3) [3pts]

5. (15%) Find the extreme values of f (x, y) = x²y − xy + xy²on x²+ xy + y²− x − y = 1.

Solution:

Define g(x, y) = x²+ xy + y²− x − y Use the method of Lagrange multipliers:

 Of = λOg

g = 1

⇒







∂/∂x : y(2x + y − 1) = λ(2x + y − 1) · · · (1)

∂/∂y : x(x + 2y − 1) = λ(x + 2y − 1) · · · (2) x²+ xy + y²− x − y = 1 · · · (3) (6% up to this point.)

First we note that when Og =−→

0 , (x, y) = (1/3, 1/3) is not on g(x, y) = 1.

Case 1: 2x + y − 1 = 0

⇒ y = 1 − 2x, λ: any.

Substitute into (3) we have (x, y) = (−1 3 ,5

3) or (1, −1) Case 2: 2x + y − 1 6= 0 ⇒ λ = y

(I) x + 2y − 1 = 0 ⇒ x = 1 − 2y Substitute into (3) we have (x, y) = (5

3,−1

3 ) or (−1, 1) (II) x + 2y − 1 6= 0 ⇒ λ = x = y

Substitute into (3) we have (x, y) = (−1 3 ,−1

3 ) or (1, 1) (6% for the above discussions.)

Substitute these points into f we find that f (−1

3 ,5

3) = f (5

3,−1

3 ) = f (−1

3 ,−1 3 ) = −5

27 are the minima, while

f (1, −1) = f (−1, 1) = f (1, 1) = 1 are the maxima.

(3% for the conclusion.)

6. (10%) Find the local maximum, and local minimum values and saddle point(s) of f (x, y) = y³+ 3x²y − 3x²− 3y²+ 3.

Solution:

f_x= 6xy − 6x = 0, 6x(y − 1) = 0, x = 0 or y = 1

fy= 3y²+ 3x²− 6y = 0, 3x²+ (3y²− 6y + 3) = 3, x²+ (y − 1)²= 1 if x = 0, y = 0, 2

if y = 1, x = −1, 1

We have 4 critical points: (0, 0), (0, 2), (1, 1), (−1, 1) fxx= 6y − 6

fxy= fyx= 6x fyy = 6y − 6

D(x, y) = (6y − 6)²− (6x)²= 36[(y − 1)²− x²]

D(0, 0) = 36 > 0, f_xx(0, 0) = −6 < 0, local maximum. f (0, 0) = 3 D(0, 2) = 36 > 0, f_xx(0, 2) = 6 > 0, local minimum. f (0, 2) = −1 D(1, 1) = −36 < 0

D(−1, 1) = −36 < 0

Ans: local maximum f (0, 0) = 3 local minimum f (0, 2) = −1

Two saddle points at (1, 1) and (−1, 1) Grading Policy:

fxand fy: 2 points 4 critical points: 2 points D(x, y): 2 points

local maximum/minimum values: 2 points 2 saddle points: 2 points

7. (10%) (a) Find the radius of convergence and the interval of convergence of the power series

∞

X

n=0

(x − 1)ⁿ (−2)ⁿ√

n. (b) Let f (x) =

∞

X

n=0

(x − 1)ⁿ (−2)ⁿ√

n when the power series is convergent. Evaluate f⁽³⁾(1).

Solution:

(a) By the ratio test, the series converges when

n→∞lim |a_n+1 an

| = lim

n→∞|

√n

√n + 1||x − 1

−2 | = |x − 1

−2 | < 1 ⇒ |x − 1| < 2 = R (3%) Check the endpoints:

(1%)(1)x = 3,

∞

X

n=1

(−1)ⁿ

√n converges by alternating series test(1%).

(1%)(2)x = −1,

∞

X

n=1

√1

n diverges by p-series with p = 1 2 < 1.

By(1)&(2), the interval of convergence I = (−1, 3].

(b) Since f (x) =

∞

X

n=1

(x − 1)ⁿ (−2)ⁿ√

n =

∞

X

n=1

f⁽ⁿ⁾(1)

n! (x − 1)ⁿ (3%) Hence, f⁽³⁾(1) = 3!

−8√ 3 = −

√3 4 (1%)

If you compute f⁰(x), f⁰⁰(x), and f⁰⁰⁰(x). Then you get zero point if the answer is not correct.

8. (15%) (a) Write down the general terms the MacLaurin series of sin x and sin⁻¹x.

(b) Find their radii of convergence.

(c) Find lim

x→0

sin x · sin⁻¹x − x²

x⁶ .

Solution:

(a)(b)

sin x =

∞

X

n=0

(−1)ⁿ

(2n + 1)!x²ⁿ⁺¹...(4 pts) radius of convergence = ∞...(1 pt) sin⁻¹x =

ˆ x 0

√ dt

1 − t² = ˆ x

0

∞

X

n=0

−1 2

k



(−t²)^kdt

=

∞

X

n=0

−1 2

k

 (−1)^k

ˆ x 0

t^2kdt =

∞

X

n=0

−1 2

k

 (−1)^k

2k + 1x^2k+1...(5 pts) , where

−1 2

k



= (⁻¹₂ )(⁻¹₂ − 1) · · · (⁻¹₂ − k + 1) k!

radius of convergence = 1...(1 pt)

(c)

We first compute the power series of the product

sin x sin⁻¹x = [x − x³ 6 + 1

120x⁵+ · · · ] · [x + (−1 2 )(−1

3 )x³+(⁻¹₂ )(⁻¹₂ − 1) 2!

x⁵ 5! + · · · ]

= [x − x³ 6 + 1

120x⁵+ · · · ] · [x + x³ 6 + 3

40x⁵+ · · · ]

= (x²+ 1

18x⁶+ 1

30x⁸+ · · · )...(2 pts) So

x→0lim

sin x sin⁻¹x − x²

x⁶ = lim

x→0

(x²+₁₈¹x⁶+₃₀¹x⁸+ · · · ) − x² x⁶

= lim

x→0 1

18x⁶+₃₀¹x⁸+ · · ·

x⁶ = lim

x→0( 1 18+ 1

30x²+ · · · ) = 1

18...(2 pts)