The series in (a) is (convergent/divergent

(1)

1. (18%) Determine whether the given series converges or diverges. Give reasons for your answers.

(a)

∞

X

n=1

sin θⁿ n!

, where θ 6= 0 is a fixed nonzero constant,

(b)

∞

X

n=1

1 + (−1)ⁿ2n² 3n²

ⁿ₃ ,

(c)

∞

X

n=1

n!

(2n + 1)(2n + 2) · · · (2n + n)

27 4

n

.

Ans. The series in (a) is (convergent/divergent) ;

the series in (b) is (convergent/divergent) ;

the series in (c) is (convergent/divergent) .

Solution:

(a) Let a_n = ^θ_n!ⁿ.

n→∞lim

|a_n+1|

|an| = lim

n→∞

˛

θn+1 (n+1)!

˛

|^θn_n!| = lim

n→∞

_n+1^θ

= 0 < 1, By ratio test,P∞

n=1

^θ

n

n!

converges ⇒ P∞ n=1

θⁿ

n! converges.

Let b_n= sin ^θ_n!ⁿ

n→∞lim

|bn|

|an| = lim

n→∞

|^sin(^θnn!)|

|^θnn!| = 1 and P |a_n| converges, By limit comparison test,P∞

n=1

sin ^θ_n!ⁿ

converges ⇒P∞

n=1sin ^θ_n!ⁿ converges.

(b) Let a_n = (¹⁺⁽⁻¹⁾_3n2ⁿ²ⁿ²)ⁿ³.

n→∞lim

p|an _n| = lim

n→∞

1+(−1)ⁿ2n² 3n²

1

3 = lim

n→∞

_3n¹₂ + ²₃(−1)ⁿ

1 3 < 1, By root test, P∞

n=1

(¹⁺⁽⁻¹⁾_3n2ⁿ²ⁿ²)ⁿ³

converges ⇒P∞

n=1(¹⁺⁽⁻¹⁾_3n2ⁿ²ⁿ²)ⁿ³ converges (c) Let a_n = (2n+1)(2n+2)···(2n+n)^n!

27 4

n

= ^n!(2n)!_(3n)! ²⁷₄n

, a_n> 0 ,∀n.

n→∞lim

an+1

an = lim

n→∞

(n+1)!(2n+2)!

(3n+3)! (²⁷₄)ⁿ⁺¹

n!(2n)!

(3n)! (²⁷₄)ⁿ = lim

n→∞

(n+1)(2n+1)(2n+2)

(3n+1)(3n+2)(3n+3) ·²⁷₄ = 1 We cannot decide from the Ratio Test whether the series converges.

Notice that ^aⁿ⁺¹_a

n = (n+1)(2n+1)(2n+2)

(3n+1)(3n+2)(3n+3)· ²⁷₄ = 2(n+1)(2n+1)

3(3n+1)(3n+2)· ²⁷₄ = ⁿ²⁺

3 2n+¹₂ n²+n+²₉ > 1, then a_n+1 is always greater than a_n,

Therefore all terms are greater than a₁ = ⁹₄, and a_n 9 0. The series diverges.

1

(2)

2. (10%) Find the radius and interval of convergence of the power series p(x) =

∞

X

k=2

3^k

k ln k(2x−1)^k. For what values of x does the series converge absolutely/conditionally?

Ans. (a) The radius of convergence is ;

(b) the interval of convergence is ;

(c) {x | p(x) converges absolutely} = ;

(d) {x | p(x) converges conditionally} = .

Solution:

Let ak= _{k ln k}³^k (2x − 1)^k

k→∞lim

ak+1

ak

= lim

k→∞

3k+1

(k+1) ln(k+1)(2x−1)^k+1

3k

k ln k(2x−1)^k

= |3(2x − 1)|

If |3(2x − 1)| < 1 ,p(x) is convergence absolutely ⇒ |x − ¹₂| < ¹₆ ⇒ R = ¹₆, ¹₃ < x < ²₃

∗ If x = ¹₃, then p(¹₃) = P∞ k=2

3^k

k ln k(−¹₃)^k =P∞ k=2

(−1)^k

k ln k is an alternating series.

P∞ k=2

(−1)^k

k ln k converges, since _{k ln k}¹ > (k+1) ln(k+1)¹ > 0 ∀k ≥ 2, and lim

k→∞

1 k ln k = 0.

But P∞ k=2

(−1)^k k ln k

=P∞ k=2

1

k ln k diverges by integral test (∵R∞ 2

dx

x ln x = ∞).

Therefore p(¹₃) converges conditionally.

∗ If x = ²₃, then p(²₃) = P∞ k=2

3^k

k ln k(¹₃)^k =P∞ k=2

k

k ln k diverges by integral test.

So

(a) The radius of convergence is ¹₆. (b) the interval of convergence is [¹₃,²₃)

(c) {x | p(x) converges absolutely} = (¹₃,²₃) (d) {x | p(x) converges conditionally} = {¹₃}

2

(3)

3. (12%) Represent sin²x as the sum of its Taylor series centered at π 3.

Ans. sin²x = .

Solution:

Method1:

f (x) = sin²x, f⁰(x) = 2 sin x cos x = sin 2x, f⁰⁰(x) = 2 cos 2x, . . . , f⁽²ⁿ⁾(x) = (−1)ⁿ⁺¹2²ⁿ⁻¹cos 2x, f⁽²ⁿ⁺¹⁾(x) = (−1)ⁿ2²ⁿsin 2x . . . f (^π₃) = ³₄, f⁰(^π₃) =

√ 3

2 , f⁰⁰(^π₃) = −1, . . . , f⁽²ⁿ⁾(^π₃) = (−1)ⁿ·2²ⁿ⁻², f⁽²ⁿ⁺¹⁾(^π₃) = (−1)ⁿ2²ⁿ⁻¹√ 3, . . .

⇒ f (x) =P∞ n=0

f⁽ⁿ⁾(^π₃)

n! (x −^π₃)ⁿ= ³₄ +P∞ n=1

2^k−1cos(²₃π+^k₂π)

n! (x − ^π₃)ⁿ Method2:

sin²x = 1 − cos 2x 2

= 1 2 −1

2cos[2(x − π 3) + 2

3π]

= 1 2 −1

2[cos2

3π cos 2(x −π

3) − sin2

3π sin 2(x − π 3)]

= 1 2 +1

4

∞

X

n=0

(−1)ⁿ2²ⁿ

(2n)! (x −π 3)²ⁿ+

√3 4

∞

X

n=0

(−1)ⁿ2²ⁿ⁺¹

(2n + 1)! (x − π 3)²ⁿ⁺¹

= 3 4 +

∞

X

n=1

2^k−1cos(²₃π + ^k₂π)

n! (x −π

3)ⁿ

3

(4)

4. (10%) Evaluate the sum of the infinite series 1² 0!+2²

1!+3² 2!+4²

3!+· · ·+(n + 1)²

n! +· · · = S .

Ans. S = .

Solution:

e^x =P∞ n=0

xⁿ n!

x · e^x =P∞ n=0

xⁿ⁺¹ n!

d

dx(x · e^x) = (1 + x)e^x =P∞ n=0

(n+1)xⁿ n!

(x + x²)e^x =P∞ n=0

(n+1)xⁿ⁺¹ n!

d

dx[(x + x²)e^x] = (1 + 3x + x²)e^x =P∞ n=0

(n+1)²xⁿ n!

Let x = 1 ⇒ S =P∞ n=0

(n+1)² n! = 5e.

4

(5)

5. (10%) Assume that all the first and the second partial derivatives of u(x, y) are con- tinuous and that u(x, y) satisfies _∂x∂y^∂²^u ≡ 0. Let a, b be two constants and f (x, y) = u(x, y)e^ax+by. Find the values of a, b such that _∂x∂y^∂²^f − 2^∂f_∂x − 3^∂f_∂y + abf ≡ 0.

Ans. a = , b = .

Solution:

∂f

∂x = ^∂u_∂xe^ax+by+ u(x, y) · ae^ax+by

∂f

∂y = ^∂u_∂ye^ax+by+ u(x, y) · be^ax+by

∂

∂y

∂f

∂x = _∂y∂x^∂²û eâx+by+ ^∂u_∂x· beâx+by+ ^∂u_∂y · aeâx+by+ u(x, y) · abeâx+by

∂

∂x

∂f

∂y

= _∂x∂y^∂²û eâx+by +^∂u_∂y · aeâx+by +^∂u_∂x · beâx+by + u(x, y) · abeâx+by

∵ _∂x∂y^∂²^f − 2^∂f_∂x− 3^∂f_∂y + abf ≡ 0

∴ _∂y∂x^∂²û eâx+by+^∂u_∂x· beâx+by+^∂u_∂y· aeâx+by+ u(x, y) · abeâx+by −2 · [^∂u_∂xeâx+by+ u(x, y) · aeâx+by]

−3 · [^∂u_∂yeâx+by + u(x, y) · beâx+by] +ab · u(x, y)eâx+by

=^∂u_∂xeâx+by(b − 2) + ^∂u_∂yeâx+by(a − 3) + u(x, y)eâx+by(2ab − 2a − 3b) ≡ 0.

⇒ b = 2, a = 3.

5

(6)

6. (15%) Let F (x, y, z) = x²+ 2z +Rz y

pt3 ²+ 1 − y²dt.

(a) Find the plane tangent to the surface {(x, y, z) ∈ R³| F (x, y, z) = 2} at (2, −1, −1).

(b) Let z = z(x, y) be the function implicitly defined by F (x, y, z) = 2 around (2, −1, −1).

Find the direction(s) at the point (x, y) = (2, −1) along which z(x, y) increases most rapidly. That is, find unit vector u ∈ R² such that (D_uz)(2, −1) attains its maximum.

(c) Evaluate _∂x∂y^∂²^z at (x, y, z) = (2, −1, −1) for the function z(x, y) given in (b).

Ans. (a) The equation of the tangent plane is ;

(b) the direction(s) is(are) ;

(c) At (x, y, z) = (2, −1, −1) ,_∂x∂y^∂²^z = .

Solution:

(a) ∇F (x, y, z) = (2x,Rz

y(¹₃)(t²+ 1 − y²)⁻²³)(−2y)dt −py³ ²+ 1 − y², 2 +pz³ ²+ 1 − y²)

∇F (2, −1, −1) = (4, −1, 2 + 1) = (4, −1, 3)

tangent plane: (4, −1, 3) · (x − 2, y + 1, z + 1) = 0 ⇒ 4x − y + 3z = 6.

(b) z(x, y) is differential at (2, −1) and D_uz = ∇z · u ≤ |∇z|

when u = _|∇z|^∇z , D_uz ’=’ |∇z|

F (x, y, z(x, y)) = 2

∂F

∂x + ^∂F_∂z · ^∂z_∂x = 0 ⇒ ^∂z_∂x = −

∂F

∂x

∂F

∂z

, ^∂F_∂y +^∂F_∂z · ^∂z_∂y = 0 ⇒ ^∂z_∂y = −

∂F

∂y

∂F

∂z

At (2, −1), ∇z = (−⁴₃, −⁻¹₃ ) = (−⁴₃,¹₃), |∇z| =

q16+1

9 =

√ 17

3 , u = (−^√⁴₁₇,^√¹₁₇).

(c) _∂x∂y^∂²^z = _∂y∂x^∂²^z = _∂y^∂

− ^2x

2+³

√

z²+1−y²

= 2x ·

1

3(x²+1−y²)^{− 2}³·{2z·^∂z_∂y−2y}

(2+³

√

z²+1−y²)²

At (2, −1, −1), _∂x∂y^∂²^z = 2 · 2 ·

1

3·1·{2(−1)·¹₃−2(−1)}

3² = ₂₇⁴ {2 −²₃} = ¹⁶₈₁.

6

(7)

7. (12%) Find all the local maxima, local minima, and saddle points of G(x, y) = x³ − x²+ 3x²y − 4xy − 3y³+¹³₂ y².

Ans. G has a local maximum = at (x, y) = ;

G has a local minimum = at (x, y) = ;

G has saddle point(s) at (x, y) = .

Solution:

G_x = 3x²− 2x + 6xy − 4y G_y = 3x²− 4x − 9y²+ 13y







G_xx = 6x − 2 + 6y G_xy = G_yx = 6x − 4 G_yy = −18y + 13

⇒ Hession of G =

6x − 2 + 6y 6x − 4 6x − 4 −18y + 13

解 G_x = 0

G_y = 0 ⇒ (x, y) = (0, 0), (14, −7), (²₃,¹₉), (²₃,⁴₃)

代入 Hession 判別: ⇒ (0, 0), (14, −7), (²₃, ⁴₃) 都是 saddle point

⇒ (²₃,¹₉) 是 minimum, G(²₃,¹₉) = −¹⁰⁹₄₈₆.

7

(8)

8. (13%) Find the extreme values of H(x, y, z) = xy + z³ on the intersection of the plane x − y = 0 and the ellipsoid x²+ 2y²+ 3z² = 9.

Ans. The maximum of H is at (x, y, z) = ;

the minimum of H is at (x, y, z) = .

Solution:

∇H = (y, x, 3z²)

∇(x − y) = (1, −1, 0)

∇(x²+ 2y² + 3z²) = (2x, 4y, 6z)

∃ λ, µ, s.t.











y = λ + µ · 2x · · · (1) x = −λ + µ · 4y · · · (2) 3z² = 0 + µ · 6z · · · (3) x − y = 0 · · · (4) x²+ 2y²+ 3z² = 9 · · · (5)

By (4): x = y, (1)+(2): 2x = 6xµ, x = y = 0 or µ = ¹₃ Case1:

If x = 0, y = 0, z = ±√

3, H = ±3√ 3 Case2:

If µ = ¹₃ plug in (3): 3z² = 2z ⇒ z = 0, z = ²₃ If z = 0, x = y plug in (5): 3x² = 9 ⇒ x = y = ±√

3 ,H = 3.

If z = ²₃, x = y plug in (5): 3x² = ²³₃ ⇒ x = y = ±

√23

3 , H = ⁷⁷₂₇ < 3.

∴ The maximum of H is 3√

3 at (0, 0,√

3), the minimum of H is −3√

3 at (0, 0, −√ 3)

8