(16 points) Prove that the series ∞ X n=2 (−1)n n(n − 1) 1 2n converges absolutely, and find its sum

1002微微微甲甲甲01-05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. (16 points) Prove that the series

∞

X

n=2

(−1)ⁿ n(n − 1)

1

2ⁿ converges absolutely, and find its sum.

Solution:

Part I (6pts) Prove the series converges absolutely

let an=

∞

X

n=2

(−1)ⁿ n(n − 1)

1 2ⁿ

then the series convergent absolutely if

∞

X

n=2

|an| converges

use ratio test:

n→∞lim |an+1

a_n | = ( 1

(n + 1)n2ⁿ⁺¹/ 1

(n)(n)2ⁿ) =1 2 < 1 use comparison test:

compared an to bn = 1

n² or bn = 1 2ⁿ

show that an< bn everywhere and prove that

∞

X

n=2

|bn| converges

Part II (10pts)find its sum

f (x) = 1

1 + x = 1 − x + x²− x³+ x⁴.... =

∞

X

n=0

(−1)ⁿxⁿ

g(x) =

Z 1

1 + xdx = ln(1 + x) = x −1 2x²+1

3x³.... + c1 =

∞

X

n=1

(−1)ⁿ⁻¹1

nxⁿ+ c2 For x = 0 → c1 = c2 = 0

h(x) = Z

ln(1 + x)dx = 1

1 × 2x²− 1

2 × 3x³+ 1

3 × 4x⁴+ .... + c3 =

∞

X

n=2

(−1)ⁿ 1

n(n − 1)xⁿ+ c4

(1 + x) ln(1 + x) − (1 + x) =

∞

X

n=2

(−1)ⁿ 1

n(n − 1)xⁿ+ c4 For x = 0 → c4 = −1

let x =1

2 ∞

X

n=2

(−1)ⁿ 1

n(n − 1)xⁿ= 3 2ln3

2 −1 2

2. (a) (8 points) The Binomial Theorem implies that

(1 − x)⁻¹² = 1 +

∞

X

n=1

(2n)!

kⁿ(n!)²xⁿ

for some constant k. Find k, and find the interval of convergence of the power series.

(b) (8 points) Estimate the error if one uses x = −1

4, and the first five non-zero terms in (a) to approximate 1

√ 5.

Solution:

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(a) (1 − x)⁻¹² =

∞

X

n=0

(2n)!

4ⁿn!²xⁿ. So k = 4.(2 points) Let an = (2n)!

4ⁿn!². R = lim an

a_n+1 = 1. (2 points)

At x = −1, since an is decreasing and converge to 0,X

an(−1)ⁿ converges. (2 points) At x = 1, sinceX

a_n≥X

a_nxⁿ= (1 − x)⁻¹² for all x ∈ (−1, 1) and lim

x→1⁻

(1 − x)⁻¹² = ∞.

Xan diverges. (2 points)

Hence, the interval of convergence is [−1, 1).

(b) 1

√5 = 1

2((1 − (−1

4))⁻¹² = 1 2(

∞

X

n=0

(2n)!

4ⁿn!²(−1

4)ⁿ) (1 point) Since it is an alternating series (2 points),

| 1

√5 −1 2(

∞

X

n=0

(2n)!

4ⁿn!²(−1

4)ⁿ)| ≤ 10!

2 ∗ 4¹⁰(5!)² = 126

4¹⁰(5 points)

3. (16 points) Consider the curve

r(t) = t² i + (sin t − t cos t) j + (cos t + t sin t) k, t ≥ 0 Find T(t), N(t), B(t), the curvature κ(t) and the torsion τ (t).

Solution:

r(t) = (t², sint − tcost, cost + tsint)

r⁰(t) = v(t) = (2t, cost + tsint − cost, −sint + tcost + sint) = (2t, tsint, tcost) (1% ) r⁰⁰(t) = a(t) = (2, sint + tcost, cost − tsint)

r⁰⁰⁰(t) = a⁰(t) = (0, 2cost − tsint, −2sint − tcost)

|r⁰(t)| =p

4t²+ t²sin²t + t²cos²t =√

5t ( 1% ) T (t) = r⁰(t)

|r⁰(t)| = 1

√5(2, sint, cost) =

√5

5 (2, sint, cost) ( 2% ) To get full points (3%) please answer both question correctly.

T⁰(t) = 1

√5(0, cost, −sint) , |T⁰(t)| = 1

√5

pcos²t + sin²t = 1

√5 N (t) = T⁰(t)

|T⁰(t)| = B(t) · N (t) = (0, cost, −sint) ( 3% ) To get full points (3%) please answer both question correctly.

κ(t) = |T⁰(t)|

|r⁰(t)| = |v(t) × a(t)|

[v(t)]³ = 1

√5

√1 5t= 1

5t ( 3% ) To get full points (3%) please answer both question correctly.

v(t) × a(t) = (−t², 2t²sint, 2t²cost) (1% )

|v(t) × a(t)| =p

(−t²)²+ (2t²sint)²+ (2t²cost)²=p

t⁴+ 4t⁴=√ 5t² If your calculation is wrong, you will receive some partial credit.

B(t) = v(t) × a(t)

|v(t) × a(t)| = 1

√5(−1, 2sint, 2cost) =

√5

5 (−1, 2sint, 2cost) ( 2% ) To get full points (3%) please answer both question correctly.

τ (t) = [v(t) × a(t)] · a⁰(t)

|v(t) × a(t)|² =2t²sint(2cost − tsint) + 2t²cost(−2sint − tcost) 5t⁴

=(4t²sint · cost − 2t³sin²t) + (−4t²sint · cost − 2t²cost²t) 5t⁴

=−2t³ 5t⁴ = −2

5t ( 3% )

I will also grant partial credit for partial solutions and solutions with minor flaws. I will give no credit for wildly incorrect answers which are obviously only there in the hopes of getting partial credit.

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4. (16 points) Let F (x, y, z) = x²+ 2z + Z z

y

p3

(t²+ 7)y²dt. Find the tangent plane of the surface F (x, y, z) = 2 at the point (2, −1, −1).

Solution:

F 1 =∂F

∂x = 2x (1 points) F 2 =∂F

∂y =2 3y⁻¹³

Z z y

p3

t²+ 7dt − y²³p³

y²+ 7 (5 points) F 3 =∂F

∂z = 2 + y²³p³

z²+ 7 (4 points) F 1(2, −1, −1) = 4 (1 points)

F 2(2, −1, −1) = −2 (2 points) F 3(2, −1, −1) = 4 (1 points)

4(x − 2) − 2(y + 1) + 4(z + 1) = 0 (2 points)

5. Suppose that z = f (x, y) has continuous second order partial derivatives, and x = s²− t², y = 2st. Define F (s, t) = f (s²− t², 2st)

(a) (6 points) Express F_s, F_tin terms of f_x, f_y, s and t.

(b) (10 points) Show that Fss+ Ftt= h(s, t)(fxx+ fyy) for some function h(s, t). Find h(s, t) explicitly.

Solution:

5.(a)

F_s(s, t) = 2sf_x(s²− t², 2st) + 2tf_y(s²− t², 2st)...(3pts) F_t(s, t) = −2tf_x(s²− t², 2st) + 2sf_y(s²− t², 2st)...(3pts) (b)

F_ss(s, t) = 2f_x+ 2s[2sf_xx + 2tf_xy] + 2t[2sf_xy+ 2tf_yy]

= 2f_x+ 4s²f_xx+ 8stf_xy + 4t²f_yy...(3pts) F_tt(s, t) = −2f_x+ 4t²f_xx− 8stf_xy+ 4s²f_yy...(3pts)

F_ss(s, t) + F_tt(s, t) = 4(s² + t²)f_xx+ 4(s²+ t²)f_yy...(2pts)

= 4(s² + t²)(f_xx+ f_yy) ⇒ h(s, t) = 4(s²+ t²)...(2pts)

6. Let f (x, y, z) = yx²+ xz²− y.

(a) (10 points) Find all critical points of f (x, y, z) and classify them.

(b) (10 points) Find the maximum and minimum of f on the region x²+ y²+ z²≤ 1.

Solution:

(a)

f1= 2xy + z² (1 point) f2= x²− 1 (1 point) f3= 2xz (1 point)

⇒ critical points at f1= f2= f3= 0

⇒ (±1, 0, 0) (2 points)

Hessian





2y 2x 2z

2x 0 0

2z 0 2x



 (1 point) ⇒





0 ±2 0

±2 0 0

0 0 ±2



 (2 points)

def H 6= 0 ⇒ neither positive definite nor negative definite

⇒ both saddle points (2 points)

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(b)

x²+ y²+ z²− 1 = g(x, y, z) ≤ 0

Solve the system

 ∇f = λ∇g (1 point) g = 0











2xz + z²= 2λx (1 point) x²− 1 = 2λy (1 point) 2xz = 2λz (1 point) x²+ y²+ z²= 1

case 1: z = 0, then 2xy = 2λx. If x = 0, y = ±1 ⇒ (0, ±1, 0) ⇒ f (0, ±1, 0) = ∓1 If x 6= 0, y = λ ⇒ x²= 2y²+ 1 = 1 − y² ⇒ (±1, 0, 0) ⇒ f (±1, 0, 0) = 0 (1 point)

case 2: z 6= 0, x = λ ⇒ x²= 2xy + 1 = (2x²− z²) + 1 ⇒ (0, 0, ±1) ⇒ f (0, 0, ±1) = 0 (1 point) max = 1 at (0, −1, 0) (2 points), min = −1 at (0, 1, 0) (2 points)

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