Calculus II, Spring 2016

Midterm 1

Mar 28, 2016 Show all details.

1. (16 pts) Are the integrals ∫_{0}^{1} \frac{1}{\sqrt{x + x^{3}}} dx and ∫_{1}^{∞} tan(\frac{1}{x})dx convergent? Explain. Answer. ∵ lim_{x→0} \frac{\frac{1}{\sqrt{x+x^{3}}}}{\frac{1}{\sqrt{x}}} = 1 (3 pts) and ∫_{0}^{1} \frac{1}{\sqrt{x}} dx converges. (3 pts) ∴ Converges. (2 pts)

∵ lim_{x→∞} \frac{tan(1/x)}{1/x} = 1 (3 pts) and ∫_{1}^{∞} \frac{1}{x} dx diverges. (3 pts) ∴ Diverges. (2 pts)

2. (6 pts) Give formal definition of lim_{n→∞} a_{n} = L.

Answer.

∀ ϵ > 0, ∃ N ∈ ℕ such that |a_{n} − L| < ϵ ∀ n ≥ N. (6 pts)

3. (8 pts) Evaluate lim_{n→∞} (\frac{n − 1}{n + 1})^{n}.

Answer.

lim_{n→∞} (\frac{n − 1}{n + 1})^{n} = lim_{n→∞} (1 − \frac{2}{n + 1})^{n+1} \frac{1}{1 − \frac{2}{n+1}} (6 pts) = e^{−2}. (2 pts)

4. (6+10 pts) Show that ∑_{k=1}^{∞} k^{−2} converges and evaluate^{∗} lim_{n→∞} \frac{log ( ∑_{k=n}^{∞} k^{−2})}{log n}. Give details. (If the limit is p, this means that ∑_{k=n}^{∞} k^{−2} is approximately n^{p}. Find p and prove it.) Hint: recall the proof of one of the convergence tests. Answer.

∵ f(k) = k^{−2} ≥ 0 ↘ ctu (2 pts) and ∫_{1}^{∞} \frac{1}{x^{2}} dx converges. (2 pts) ∴ Converges. (2 pts) ∵

\frac{1}{n} = ∫_{n}^{∞} \frac{1}{x^{2}} dx (2 pts) ≤ ∑_{k=n}^{∞} \frac{1}{k^{2}} ≤ \frac{1}{n^{2}} + ∫_{n+1}^{∞} \frac{1}{x^{2}} dx = \frac{1}{n^{2}} + \frac{1}{n + 1} ≤ \frac{2}{n + 1} (2 pts)

⇒ −1 (2 pts) ≤ \frac{log ( ∑_{k=n}^{∞} k^{−2})}{log n} ≤ \frac{log 2}{log n} − \frac{log(n + 1)}{log n} (2 pts) → −1.

∴ p = −1. (2 pts)

5. (8 pts) State (need not prove) Taylor’s Theorem (or Taylor’s formula). Assume the function f has derivatives of all orders on ℝ. Write down n terms approximation and the formula of the remainder term. Answer. If f has derivatives of all orders in an open interval I containing a, then ∀ n ∈ ℕ, ∀ x ∈ I, ∃ c between a and x such that

f(x) = f(a)+f′(a)(x−a)+…+ \frac{f^{(n)}(a)}{n!}(x−a)^{n} (4 pts) + \frac{f^{(n+1)}(c)}{(n + 1)!}(x−a)^{n+1}. (4 pts)

6. (6+12 pts)

(a) Show that the series 1 − \frac{1}{2·1!} + \frac{1}{4·2!} − ⋯ + (−1)^{n} \frac{1}{2^{n}·n!} + ⋯ converges absolutely.

(b) Find the sum of the series in (a). \underline{Prove your answer} (that is, the equality holds).

Answer.

(a)

lim_{n→∞} \frac{\frac{1}{2^{n+1}(n+1)!}}{\frac{1}{2^{n}n!}} → 0. (6 pts)

(b)

e^{−1/2}. (2 pts) |R_{n}(−\frac{1}{2})| ≤ \frac{e^{−c}}{(n + 1)!} \frac{1}{2^{n+1}} → 0 where c ∈ (0, \frac{1}{2}) . (10 pts)

7. (8 pts) Find ∑_{n=1}^{∞} nx^{n} and ∑_{n=1}^{∞} n^{2}x^{n} on |x| < 1 using computational rules of power series.

Need NOT prove your equality holds as in previous question. Answer.

1 + x + x^{2} + … = \frac{1}{1 − x} (2 pts)

⇒ x + 2x^{2} + 3x^{3} + … = x (\frac{1}{1 − x})′ = \frac{x}{(1 − x)^{2}} (3 pts)
⇒ x + 2^{2}x^{2} + 3^{2}x^{3} + … = x (\frac{x}{(1 − x)^{2}} )′ = \frac{x(1 + x)}{(1 − x)^{3}} (3 pts).

8. (6 pts) Find a power series that converges on [1, 5) and diverges elsewhere. Explain.

Answer.

∑_{n} \frac{1}{n} (\frac{x − 3}{2})^{n} . (2 pts)

lim_{n→∞} \frac{n}{n + 1} \frac{1}{2} = \frac{1}{2} ⇒ R = 2. (2 pts)

For x = 1,

∑_{n} \frac{(−1)^{n}}{n} converges. (1 pt)

For x = 5,

∑_{n} \frac{1}{n} diverges. (1 pt)

9. (12 pts) Give an approximation of ∫0^{\frac{1}{2}} sin(x^{2})dx to within 10^{−8}. Give the formula of the approximation, but need not find the numerical value. Explain why the error is less than 10^{−8}. Answer.

| ∫0^{\frac{1}{2}} (−1)^{n} \frac{x^{2(2n+1)}}{(2n + 1)!} dx| (4 pts) ≤ \frac{1}{(4n + 3) · 2^{4n+3} · (2n + 1)!} < 10^{−8} (4 pts).

n ≥ 3 will do (2 pts). The approximation is

∑_{k=0}^{2}(−1)^{k} \frac{1}{(4k + 3) · 2^{4k+3} · (2k + 1)!} (2 pts).

(error ≈ 4 × 10^{−10}) Giving optimal n, (n = 3): (extra 2 pts).

10. (8+10+10 pts) True or False? Prove it if true, give a counter example if false.

(a) If ∑ a_{n} converges, then ∑ na_{n} converges.

(b) ^{∗} If ∑ a_{n}x^{n} converges on |x| < 1, then ∑ \sqrt{n}a_{n}x^{n} also converges on |x| < 1.

(c) If g(x) = f(0) + ∑_{n=1}^{∞} \frac{f^{(n)}(0)}{n!}x^{n} converges on |x| < 1, then f(x) = g(x) on |x| < 1.

Answer.

(a) False (2 pts). ∑_{n} \frac{(−1)^{n}}{n} converges (3 pts), but ∑_{n}(−1)^{n} diverges (3 pts).

(b) True (2 pts). Given |x| < 1. Take y with |x| < |y| < 1. Then |\sqrt{n}a_{n}x^{n}| ≤ |a_{n}y^{n}| for n large enough since \sqrt{n}|x/y|^{n} → 0 as n → ∞ (6 pts). Furthermore, ∑_{n} |a_{n}y^{n}| converges (2 pts).

(c) False (2 pts). Take f(x) = e^{−1/x^{2}} for x ≠ 0, and f(x) = 0 for x = 0 (3 pts).

Then f^{(n)}(0) = 0, ∀ n ≥ 0 (2 pts). Therefore f(0) + ∑_{n=1}^{∞} \frac{f^{(n)}(0)}{n!}x^{n} = 0 ≠ f(x),

if x ≠ 0 (3 pts).