Math 181H: PRACTICE Hour Exam 3 Spring 1992, S. Smith (based on Hour Exam 2 of quarter course Math 132, May 1990)

Problem 1: Find the sum (if convergent) of \frac{9}{100} + \frac{9}{100^{2}} + \frac{9}{100^{3}} + . . ..

(geometric series) \frac{9}{100} ∑_{n=0}^{∞}(\frac{1}{100})^{n} = \frac{9}{100} \frac{1}{1− \frac{1}{100}} = \frac{9}{100} \frac{100}{99} = \frac{1}{11}.

Problem 2: Test convergence of ∑_{k=0}^{∞} \frac{k^{2}}{k^{3}+3}.

Integral test: ∫_{1}^{∞} \frac{x^{2}}{x^{3}+3} = \frac{1}{3} ln(x^{3} + 3)|_{1}^{∞} = ∞, so diverges.

Problem 3: Test convergence of ∑_{n=1}^{∞} \frac{5}{n(n+1)}.

Comparison test: below 5 ∑_{n=1}^{∞} \frac{1}{n^{2}} < ∞, so converges.

Problem 4: Test convergence of ∑_{n=0}^{∞} \frac{n^{2}}{2^{n}}.

Root test: lim_{n}(\frac{n^{2}}{2^{n}})^{\frac{1}{n}} = \frac{1^{2}}{2} < 1, so converges. (Also ratio test).

Problem 5: Find the interval of convergence of ∑_{n=1}^{∞}(−1)^{n} \frac{(x+1)^{n}}{n(n+1)}.

(ratio:) lim_{n} \frac{|x+1|^{n+1}}{(n+1)(n+2)} \frac{n(n+1)}{|x+1|^{n}} = lim_{n} \frac{1}{1+ \frac{1}{n}} \frac{1+ \frac{1}{n}}{1+ \frac{2}{n}} |x + 1| = |x + 1|; so radius 1 about x = −1.

At x = −2, get ∑_{n=1}^{∞} \frac{(−1)^{n}(−1)^{n}}{n(n+1)} , converges by comparison < ∑ \frac{1}{n^{2}} . At x = 0, get ∑_{n=1}^{∞} \frac{(−1)^{n}1^{n}}{n(n+1)} , coverges by alternating test (decr, → 0). So the interval of convergence is [−2, 0].

Problem 6: Find the n-th Taylor polynomial of f(x) = cos x at a = \frac{π}{4} for n = 3. Compute f(x) = cos x,f′(x) = − sin x,f′′(x) = − cos x,f′′′(x) = sin x; at x = a = \frac{π}{4}, get values \frac{\sqrt{2}}{2}, −\frac{\sqrt{2}}{2}, −\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}. So \frac{\sqrt{2}}{2}[1 − (x − \frac{π}{4}) − \frac{1}{2!}(x − \frac{π}{4})^{2} + \frac{1}{3!}(x − \frac{π}{4})^{3}].

Problem 7: Express ∫_{0}^{1} \frac{sin x}{\sqrt{x}}dx as the sum of an infinite series, and compute the integral to three decimal places.

Usual sin x = x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!} − …, so \frac{sin x}{\sqrt{x}} = x^{\frac{1}{2}} − \frac{x^{\frac{5}{2}}}{3!} + \frac{x^{\frac{9}{2}}}{5!} − . . ..

Then ∫_{0}^{1} \frac{sin x}{\sqrt{x}}dx = [\frac{2x^{\frac{3}{2}}}{3} − \frac{2x^{\frac{7}{2}}}{7.3!} + \frac{2x^{\frac{11}{2}}}{11.5!} − …]_{0}^{1}. So series is 2 ∑_{n=0}^{∞} \frac{(−1)^{n}}{(4n+3)(2n+1)!}, approx 2(\frac{1}{3} − \frac{1}{7.3!} + \frac{1}{11.5!}), since |\frac{−2}{15.7!}| < \frac{1}{2000}.

Problem 8: Compute a power series at a = 0 for \frac{1}{(1−x^{2})}.

Shortcut: Use geometric series f(x) = \frac{1}{1−x} = 1 + x + x^{2} + … so \frac{1}{1−x^{2}} = f(x^{2}) = 1 + x^{2} + x^{4} + … = ∑_{n=0}^{∞} x^{2n}.

Problem 9: Find the Taylor Series of f(x) = cos 2x about x = \frac{π}{4}.

Sequence of derivs: cos 2x, −2 sin 2x, −4 cos 2x, 8 sin 2x, 16 cos 2x, −32 sin 2x, …; values at x = a = \frac{π}{4} are: 0, −2, 0, 8, 0, −32, . . .. So series is ∑_{n=0}^{∞} \frac{(−1)^{n+1}2^{2n+1}}{(2n+1)!} (x − \frac{π}{4})^{2n+1}.