Determine whether the sequence is convergent or divergent with an argument

(1)

1052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. (10%) Let p ∈ (0, 1). A sequence {xn}^∞n=1 is given by

x1=√p and xn+1=√

p + xn, for n ≥ 1.

Determine whether the sequence is convergent or divergent with an argument. If it is convergent, find the limit.

Solution:

• Claim 1: {xn} is bounded (4 points) Prove that 0 < xn< 2 ∀n ∈ N:

Base case: 0 < x1=√p <√ 1 < 2.

Assume that 0 < xk < 2 for k ≥ 1, we have 0 < x^k+1 =√

p + xk <√

1 + 2 < 2, thus the claim is proved by mathematical induction.

• Claim 2: {xn} is increasing (4 points) Base case: x2=√

p + x1=q

p +√p >√p = x1. Assume that xk > xk−1 for k ≥ 2, we have x^k+1 = √

p + xk > pp + xk−1 = xk, thus the claim is proved by mathematical induction.

By Claim 1 and 2, {xn} is monotonic (increasing) and bounded (above), thus it converges by the Monotonic Sequence Theorem.

• Find the limit: (2 points)

Since the sequence converges, assume that lim

n→∞xn= L, then

n→∞lim xn+1= lim

n→∞

√p + xn

⇒ L =pp + L

⇒ L²− L − p = 0

⇒ L = 1 ±√ 1 + 4p

2 , take L = 1 +√ 1 + 4p

2 since 0 < L < 2.

Page 1 of 12

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2. (12%)

(a) (5%) Find the values of p for which the series

∞

X

n=1

n

(1 + n³)^p is convergent.

(b) (7%) Determine whether the series

∞

X

n=1

(−1)ⁿtan⁻¹n

n is absolutely convergent, conditionally convergent, or divergent.

Solution:

(a) (5%)

(method 1) Let bn= n

n^3p = 1

n^3p−1. an, bn> 0 for all n > 0. For all p ∈ R, we have

n→∞lim an

bn

= lim

n→∞

n (1+n³)^p

n n^3p

= lim

n→∞( n³

1 + n³)^p= lim

n→∞( 1

1

n³ + 1)^p = 1 (3%) so both series

∞

X

n=1

an and

∞

X

n=1

bn are convergent or divergent (1%) by the Limit Comparison Test. Since

∞

X

n=1

1

n^3p−1 is convergent if and only if 3p − 1 > 1, which implies p > 2

3, we know that

∞

X

n=1

an is convergent if and only if p >2

3. (1%) (method 2)

(1). p ≤ 0: lim

n→∞

n

(1 + n³)^p = ∞, so it is divergent by the Limit Divergence Test. (1%) (2). 0 < p ≤ 2

3: 0 < n

(n³+ n³)²³ ≤ n

(n³+ n³)^p ≤ n

(1 + n³)^p for all n > 0. By Comparison Test,

∞

X

n=1

n (n³+ n³)²³ =

∞

X

n=1

2⁻²³1

n is divergent, so

∞

X

n=1

n

(1 + n³)^p is also divergent when 0 < p ≤ 2 3. (2%) (3). p > 2

3: 0 < n

(1 + n³)^p ≤ 1

n^3p−1 for all n > 0. By Comparison Test,

∞

X

n=1

1

n^3p−1 is convergent when p > 2

3,

∞

X

n=1

n

(1 + n³)^p is also convergent when p > 2 3. (2%) By (1)(2)(3),

∞

X

n=1

n

(1 + n³)^p is convergent if and only if when p > 2 3. (b) (7%)

First, we show that

∞

X

n=1

(−1)ⁿtan⁻¹n

n is NOT absolutelt convergent.

Let an= tan⁻¹n

n , and bn= 1

n. Then lim

n→∞

an

bn

= lim

n→∞tan⁻¹n =π 2 > 0.

By limit comparison test, we know

∞

X

n=1

(−1)ⁿtan⁻¹n n

=

∞

X

n=1

tan⁻¹n

n diverges since

∞

X

n=1

1

n diverges. (3%) Next, we show that

∞

X

n=1

(−1)ⁿtan⁻¹n

n is convergent.

In order to apply the alternating series test, we need to show that the sequence {an} is decreasing and

n→∞lim an= 0

(1){an} is decreasing (at least for n ≥ N): (3%)

x − tan⁻¹x

(3)

x→∞lim x

1 + x² = 0 and lim

x→∞tan⁻¹x = π 2,.

So there exsits N ∈ N such that n

1 + n²− tan⁻¹n < 0 for n ≥ N (2) lim

n→∞an= 0: (1%) (method 1):

n→∞lim

tan⁻¹n

n = lim

n→∞tan⁻¹n · lim

n→∞

1 n = π

2 · 0 = 0 (method 2):

Note that 0 ≤ tan⁻¹n

n ≤π/2

n for n ≥ 1. Since lim

n→∞

π/2

n = 0, we have lim

n→∞

tan⁻¹n

n = 0

So, by alternating series test, we know that

∞

X

n=1

(−1)ⁿtan⁻¹n

n is convergent.

And therefore,

∞

X

n=1

(−1)ⁿtan⁻¹n

n is conditionally convergent.

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3. (12%) A plane curve C is parameterized by r(t) = (cos t + t sin t, sin t − t cos t), t > 0, as Figure 1.

x y

Figure 1: The plane curve C.

(a) Compute the unit tangent vector T(t), the unit normal vector N(t), and the curvature κ(t).

(b) Show that all centers of osculating circles, r(t) + 1

κ(t)N(t), lie on a circle.

Solution:

(a) (10%) (Method 1)

r^′(t) = (− sin t + sin t + t cos t, cos t − cos t + t sin t) = (t cos t, t sin t) (1%), so T(t) = r^′(t)

|r^′(t)| (2%) = (cos t, sin t) (1%).

N(t) = T^′(t)

|T^′(t)| (2%) = (− sin t, cos t) (1%).

κ(t) = |T^′(t)|

|r^′(t)| (2%) = 1 t (1%).

(Method 2)

The plane curve can be view as r(t) = (cos t + t sin t)i + (sin t − t cos t)j + 0k. Thus, we have r^′(t) = t cos ti + t sin tj + 0k

r^′′(t) = (cos t − t sin t)i + (sin t + t cos t)j + 0k r^′× r^′′(t) = 0i + 0j + t²k (1%) T(t) = r^′(t)

|r^′(t)| (2%) = cos ti + sin tj + 0k (1%), so N(t) = T^′(t)

|T^′(t)| (2%) = − sin ti + cos tj + 0k (1%).

κ(t) = |r^′× r^′′(t)|

|r^′(t)|³ (2%) = 1 t (1%).

(b) (2%)

r(t) + 1

κ(t)N(t) = (cos t + t sin t, sin t − t cos t) + t(− sin t, cos t) = (cos t, sin t).

Thus, all centers of osculating circles lies on a circle. (2%)

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4. (10%) Consider the function

f (x, y) =





 x²y

x⁴+ 2y² if (x, y) 6= (0, 0) 0 if (x, y) = (0, 0).

(a) Find the limit lim

(x,y)→(0,0)f (x, y) or explain why the limit does not exist.

(b) Compute the directional derivative D^uf (0, 0), where u = (cos θ, sin θ) is any direction.

Solution:

(a) (5 points)

If we evaluate the limit along the curves y = mx²,

(x, y) → (0, 0)lim

y= mx2

f (x, y) = lim

x→0

x²· mx²

x⁴+ 2m²x⁴ = m 1 + 2m² which varies as the value of m varies. Thus the limit does not exist.

(b) (5 points in total) Note that since lim

(x,y)→(0,0)f does not exist, f is not continuous at (0, 0) and in turn f is not differentiable at (0, 0). Thus the relation D^uf = ∇f · u can NOT be used here.

By definition, when u = hcos θ, sin θi (already unit length), Duf (0, 0) = lim

h→0

f (0 + h cos θ, 0 + h sin θ) − f(0, 0)

h (2 points)

= lim

h→0

h³cos²θsin θ h⁴cos⁴θ+2h²sin²θ− 0

h

= lim

h→0

cos²θ sin θ h²cos⁴θ + 2 sin²θ

=





 cos²θ

2 sin θ, sin θ 6= 0 (θ 6= nπ, n ∈ Z; 2 points) 0, sin θ = 0 (θ = nπ, n ∈ Z; 1 point)

(Note: if you tempted to use D^uf = ∇f · u and calculated fx(0, 0) = 0 and fy(0, 0) = 0 correctly by definition, 2 points will be credited.)

Page 5 of 12

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5. (10%) A differentiable function f (x, y) has the following properties:

• f(0, 0) = 1.

• D^uf (0, 0) = 2, where u = 3 5,4

5

.

• D^vf (0, 0) = 3

√2, where v =

1

√2, 1

√2

.

(a) What is the maximal rate of increase of f (x, y) at (0, 0)?

(b) Use the linearization of f (x, y) at (0, 0) to estimate f (0.07, −0.05).

Solution:

(a) Let ∇f(0, 0) = (a, b) then we have 3

5a +4

5b = 2 and 1

√2a + 1

√2b = 3

√2

⇒ ∇f(0, 0) = (2, 1) (3 pts)

Maximum rate of change = |(2, 1)| =√

5 (2 pts)

(b) L(x, y) = f (0, 0) + fx(0, 0) · (x − 0) + fy(0, 0) · (y − 0)

= 1 + 2x + y (4 pts)

L(0.07, −0.05) = 1 + 0.14 − 0.05 = 1.09 (1 pts)

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6. (10%) Find the local maximum and minimum values and saddle point(s) of the function f (x, y) = x³− y³+ 3x²+ 3y²− 9x.

Solution:

First, we find all critical points by calculating^⇀∇f(x, y) = 0.

i.e.,

(fx = 3x²+ 6x − 9 = 0(2%) fy = −3y²+ 6y = 0(2%)

Solving the equation, we get four points: P1= (1, 0), P2= (1, 2), P3= (−3, 0), and P⁴= (−3, 2).

Next, we compute the Hessian matrix of f : Hess(f ) =6x + 6 0 0 −6y + 6

(2%)

At P1, we have D(P1) = 72 > 0, and fxx(P1) = 12 > 0, so P1 is a local minimum with f (1, 0) = −5 (1%) At P2, we have D(P2) = −72 < 0, so P2is a saddle point (1%)

At P3, we have D(P3) = −72 < 0, so P2is a saddle point (1%)

At P4, we have D(P4) = 72 > 0, and fxx(P4) = −12 < 0, so P4is a local maximum with f (−3, 2) = 31 (1%)

Page 7 of 12

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7. (14%) Viviani’s curve, sometimes also called Viviani’s window, is the intersection of the cylinder (x − 1)²+ y²= 1 and the sphere x²+ y²+ z²= 4, as Figure 2.

Figure 2: Viviani’s curve.

(a) Find the tangent line equation of the Viviani’s curve at P (1, 1,√ 2).

(b) Find the points on the Viviani’s curve that are nearest to and farthest from Q(2, 0, 2).

Solution:

(a) Let F1(x, y, z) = (x − 1)²+ y²− 1 = 0 and F2(x, y, z) = x²+ y²+ z²− 4 = 0.

We compute

∇F1(x, y, z) = (2(x − 1), 2y, 0) ⇒ ∇F1(1, 1,√

2)//(0, 1, 0) (1pt)

∇F²(x, y, z) = (2x, 2y, 2z) ⇒ ∇F²(1, 1,√

2)//(1, 1,√

2) (1pt) The directional vector of the tangent line is

(0, 1, 0) × (1, 1,√

2) = (√

2, 0, −1). (2pts) So the tangent line equation is







x(t) = 1 +√ 2t y(t) = 1 z(t) =√

2 − t

, t ∈ R (1pt)

(b) Consider the Lagrange function L(x, y, z, λ, µ) = (x−2)²+y²+(z−2)²−λ((x−1)²+y²−1)−µ(x²+y²+z²−4).

Then we will find all critical points of L :











Lx = 2(x − 2) − 2λ(x − 1) − 2µx = 0 ⇒ (1 − λ − µ)x = 2 − λ Ly = 2y − 2λy − 2µy = 0 ⇒ (1 − λ − µ)y = 0

Lz = 2(z − 2) − 2µz = 0 ⇒ z(1 − µ) = 2

Lλ = −((x − 1)²+ y²− 1) = 0 ⇒ (x − 1)²+ y²= 1 Lµ = −(x²+ y²+ z²− 4) = 0 ⇒ x²+ y²+ z²= 4.

(3pts)

(A) If y = 0, then x = 0 or x = 2, and it implies (x, y, z) = (0, 0, 2) and (0, 0, −2). (Remark that (2, 0, 0) does not satisfies Lz= 0.) The distance will be 2 and 2√

5, respectively. (2pts) (B) If λ + µ = 1, then λ = 2, µ = −1, and it gives z = 1 and then x = 3

2 and y = ±

√3

2 . The distance will be√

2. (2pts)

Hence the nearest points are (3 2, ±

√3

2 , 1). (1pt) The farthest points is (0.0, −2). (1pt)

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8. (15%) Let f (x) = ˆ x

−1

√ 1

t²+ 2t + 2dt.

(a) Find the Taylor series for f (x) centered at a = −1. (Hint: Complete the square first.) (b) Find f⁽⁹⁾(−1) and f⁽¹⁰⁾(−1).

(c) Write down the 3rd-degree Taylor polynomial T3(x) for f (x) centered at a = −1, and calculate T3

−1 2

. Estimate the error

f

−1 2

− T³

−1 2

by some estimation theorem.

Solution:

(a) Note that x²+ 2x + 1 = (x + 1)²+ 1.

f^′(x) = 1

√x²+ 2x + 2 = 1 + (x + 1)²⁻¹2

=

∞

X

n=0

−¹2

n

(x + 1)²n

(x + 1)² < 1

=

∞

X

n=0

−¹2

n

(x + 1)²ⁿ |x + 1| < 1

f (x) = c0+

∞

X

n=0

−¹₂ n

(x + 1)²ⁿ⁺¹

2n + 1 |x + 1| < 1

f (−1) = c0+ 0 = ˆ ⁻1

−1

√ 1

t²+ 2t + 2 dt = 0 Thus c0= 0 and

f (x) =

∞

X

n=0

−¹2

n

(x + 1)²ⁿ⁺¹

2n + 1 |x + 1| < 1

It is also OK to write−¹2

n

= (−1)ⁿ(2n!) 4ⁿ(n!)²

Although not grading, one can find the interval of convergence is [−2, 0]. The proof is at the last part.

(b)

f (x) =

∞

X

k=0

f^(k)(−1)

k! (x + 1)^k=

∞

X

n=0

−2¹

n

1

2n + 1(x + 1)²ⁿ⁺¹ By comparing the coefficient, we have k = 9 = 2n + 1, n = 4 and then

f⁽⁹⁾(−1) = 9! ·−¹2

4

1

2 · 4 + 1 = 8! ·

−1

2 ·⁻₂³·⁻₂⁵· ⁻₂⁷

4 · 3 · 2 · 1 = (1 · 3 · 5 · 7)²= 11025 and k = 10 = 2n + 1, n /∈ N thus f⁽¹⁰⁾(−1) = 0

(c) Take the terms until power 3,

T3(x) = 0 +−¹2

0

(x + 1)¹

1 + 0 +−¹2

1

(x + 1)³ 3

= (x + 1) −1

6(x + 1)³ T3

−1 2

= 1 2

−1 6

1 2

3

= 23 48 Let R(x) = f (x) − T³(x). We want to find an estimate to

R

−1 2

.

• (Method 1) Use Taylor’s Inequality.

If f

(4)(x)

≤ M for |x + 1| ≤ 1

2, then |R3(x)| ≤ M

4!|x + 1|⁴. Thus

R

−1 2

=

R3

−1 2

≤M

4!

−1

2 + 1

4

= M

384.

Page 9 of 12

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Now we are going to try to find some M satisfies f

(4)(x)

≤ M for ALL |x + 1| < 1

2. (Not only x = −1 or −1

2. There is a remark later.) f⁽²⁾(x) = −1

2 1 + (x + 1)²⁻³₂

· 2(x + 1) f⁽³⁾(x) = 3

4 1 + (x + 1)²⁻⁵₂

· 4(x + 1)²+−1

2 1 + (x + 1)²⁻³₂

· 2 f⁽⁴⁾(x) = −15

8 1 + (x + 1)²⁻⁷₂

· 8(x + 1)³+3

4 1 + (x + 1)²⁻⁵₂

· (8 + 4)(x + 1)

= 1 + (x + 1)²⁻⁷₂

−15(x + 1)³+ 9(x + 1) 1 + (x + 1)²

=3(x + 1) 3 − 2(x + 1)² q

(1 + (x + 1)²)⁷

Note that 0 ≤ |x + 1| ≤ 1 2 and

f

(4)(x) ≤

3 ·¹₂· (3 − 2 · 0) q

(1 + 0²)⁷

=9 2

Take M = 9 2 and

R

−1 2

≤ 3

256

• (Method 2) Use Alternating Series Estimation Theorem. Let

bn= (−1)ⁿ−¹2

n

1

2n + 1

−1 2+ 1

2n+1

= cn

1 (2n + 1)

1 2²ⁿ⁺¹

Where cn= (−1)ⁿ−¹2

n

. Then c0= 1 and cn= 2n − 1 2n cn−1. Note that both bn and cn are always positive.

Now (i) cn≤ cⁿ⁻¹, thus {cn} is decreasing, so does {bn} (ii) cn≤ 1, thus 0 ≤ lim

n→∞bn ≤ lim

n→∞

1 (2n + 1)

1 2²ⁿ⁺¹ = 0 Thus

∞

X

n=0

(−1)ⁿbn is an alternating series. Therefore

R

−1 2

= |R1| ≤ b2= (−1)²−¹2

2

· 1 5· 1

2⁵ = 3 1280 This method is simpler and 5 times accurate then the previous one.

Grading

(a) Total 5 pts.

(1 pt) Find f^′(x).

(2 pts) Find the Taylor series of f^′(x).

(1 pt) Write out the Taylor series of f (x).

(1 pt) For integrate coefficient.

Calculating f (x) = sinh⁻¹(x + 1) does not count. Those ONLY calculating the integral without a series get (only) 2 pts.

Radius of convergence does not count, but costs 1 pt if answered incorrect.

Interval of convergence does not count, even answered incorrect.

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(1 pt) Find T3

−1 2

(1 pt) State out which estimation theorem is used.

(1 pt) State or use the theorem correctly.

(1 pt) Find an suitable upper bound for |f(−1

2) − T3(−1 2)|

For Method 1, writing M without a way how to find it costs 1 pt.

For Method 2, not fully checking the criterion of alternating costs 1 pt.

Accurate answer is not available. Use a rational number to estimate.

Remarks

• Cn^mis not good when m is not a nonnegative integer. Use m n

instead.

• The interval of convergence in (a) can be found as following:

Let (−1)ⁿcn=−¹2

n

, cn = (−1)ⁿ−¹2

n

, then 1. {cn} is nonnegative.

2. c0= 1 and cn

cn−1

=2n − 1 2n . 3. We now prove cn≤ 1

√n + 1.

Clearly case n = 0 holds. Now n ≥ 1, if cn−1≤ 1

√n, then

cn= 2n − 1

2n cn−1≤2n − 1 2n

√1

n = p(2n − 1)²(n + 1) p(2n)²(n)

√ 1

n + 1≤ 1

√n + 1

Since (2n − 1)²(n + 1) = 4n³− 3n + 1 ≤ 4n³= (2n²)(n).

Therefore, both |x + 1| = 1 or −1 cases, we have

∞

X

n=0

−¹2

n

1

2n + 1 ≤

∞

X

n=0

√ 1 n + 1

1 2n + 1 ≤

∞

X

n=0

1 2n√

n Which is absolute convergent by p-series test.

Then the interval of convergence is [−2, 0].

The reason choosing 1

√n + 1 will be clear if one knows Stirling’s Formula, which gives an approximation of the factorial n!.

• In Method 1 in (c), Taylor’s inequality needs |f⁽⁴⁾(x)| ≤ M for all |x + 1| ≤ 1

2. Actually, if |f⁽⁴⁾(−1

2 )| is the global maximum, then all will be fine. Unfortunately, that is not the case.

Figure 3: Both x = −1/2 (orange) and x = −3/2 (green) not global extreme.

• In (c), When estimating error of f(x), use ”≤” but not ”≈”. The latter one is really dangerous, since the error of f (x) may be even smaller then the error of approximating, leading to an inaccurate result.

Page 11 of 12

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9. (12%) Consider the power series f (x) =

∞

X

n=2

1

n(n − 1)3ⁿ(x − 2)ⁿ. (a) Find the interval of convergence for f (x).

(b) Write down the power series representation for d

dxf (x) and find its sum in the interior of the interval of convergence.

Solution:

(a) By ratio test lim

n→∞|an+1

an | =|x − 2|

3 < 1 if −1 < x < 5 (4 pts) At x = 5,

∞

X

n=2

1

n(n − 1) converge. (1 pts) At x = −1,

∞

X

n=2

(−1)ⁿ

n(n − 1) converge. (1 pts) Thus interval of convergence is −1 ≤ x ≤ 5

(b) d

dx(f (x)) = 1 3

∞

X

n=2

1

(n − 1)(x − 2

3 )ⁿ⁻¹(2 pts) Now, compute g(x) =

∞

X

n=2

1

(n − 1)(x)ⁿ⁻¹ then d

dxf (x) =1

3g(x − 2 3 ) First, 1

1 − x = 1 + x + x²+ · · ·

⇒ − ln(1 − x) = x +x² 2 +x³

3 + · · · RHS is g(x) Thus d

dx(f (x)) = −1

3ln(5 − x

3 ) (4 pts)