Math 213 Advanced Calculus Midterm Exam II Solution Set

November 23, 2000 Ali Sinan Sertöz

1) Construct a function f : [1, ∞) → ℝ which is unbounded on [1, ∞) but is improperly integrable there in the sense that lim_{R→∞} ∫_{1}^{R} f(x)dx exists and is finite.

Several such constructions are possible. One such function is given by

f(x) = \{ \begin{matrix}2^{n} & if n < x ≤ n + \frac{1}{2^{2n}} \\ 0 & otherwise. \end{matrix}

Clearly f is unbounded. Notice however that ∫_{n}^{n+1} f(x)dx = 1/2^{n}, so ∫_{1}^{∞} f(x)dx = lim_{N→∞} ∑_{n=1}^{N} ∫_{n}^{n+1} f(x)dx = lim_{n→∞} ∑_{n=1}^{N} \frac{1}{2^{n}} = 1.

2) Assume that f : [3, 5] → ℝ is one-to-one. Assume further that f′ exists and is integrable on [3, 5], and that f(3) = 7, f(5) = 8. Calculate ∫_{3}^{5} f(x)dx + ∫_{7}^{8} f^{−1}(x)dx.

Putting x = f(t) we find that ∫_{7}^{8} f^{−1}(x)dx = ∫_{3}^{5} f^{−1}(f(t))f′(t)dt = ∫_{3}^{5} tf′(t)dt. Then ∫_{3}^{5} f(x)dx + ∫_{7}^{8} f^{−1}(x)dx = ∫_{3}^{5} f(x)dx + ∫_{3}^{5} xf′(x)dx = ∫_{3}^{5}(xf(x))′dx = 5f(5) − 3f(3) = 19.

3) Let {a_{n}} be a sequence of real numbers such that for some real number p > 1 we have; lim_{n→∞} n^{p}a_{n} = A, with A ∈ ℝ. Show that ∑_{n=1}^{∞} a_{n} converges.

Let ϵ > 0 be chosen arbitrarily. Then there is an N such that for all n ≥ N we have

A − \frac{ϵ}{2} < n^{p}a_{n} < A + \frac{ϵ}{2}.

−\frac{ϵ}{2} \frac{1}{n^{p}} < a_{n} − \frac{A}{n^{p}} < \frac{ϵ}{2} \frac{1}{n^{p}}  (1)

Let L_{K} = ∑_{n=N}^{K} \frac{1}{n^{p}}, and L = lim_{K→∞} L_{K}. Now sum up all sides of equation 1 from n = N to K to obtain

−\frac{ϵ}{2}L_{K} < ∑_{n=N}^{K} a_{n} − AL_{K} < \frac{ϵ}{2}L_{K}

and since 0 < L_{K} < L we have

−\frac{ϵ}{2}L < ∑_{n=N}^{K} a_{n} − AL_{K} < \frac{ϵ}{2}L

−\frac{ϵ}{2} < \frac{1}{L} ∑_{n=N}^{K} a_{n} − \frac{1}{L}AL_{K} < −\frac{ϵ}{2}

|\frac{1}{L} ∑_{n=N}^{K} a_{n} − \frac{1}{L}AL_{K}| < \frac{ϵ}{2}.  (2)

Since lim_{K→∞} \frac{A}{L}L_{K} = A, there is an index K_{0} such that for all K ≥ K_{0} we have

|\frac{A}{L}L_{K} − A| < \frac{ϵ}{2}.  (3)

Now combining equations 2 and 3 we have for all K ≥ K_{0}

|\frac{1}{L} ∑_{n=N}^{K} a_{n} − A| ≤ |\frac{1}{L} ∑_{n=N}^{K} a_{n} − \frac{A}{L}L_{K}| + |\frac{A}{L}L_{K} − A| < \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ

which is equivalent to saying that ∑_{n=N}^{∞} a_{n} = AL, and which in turn implies that the series ∑_{n=1}^{∞} a_{n} converges.

4) Show that cos(1) is irrational.

Assume cos(1) = \frac{m}{n} for some integers m and n > 0. Using the series expansion of cosine we have

\frac{m}{n} = 1 − \frac{1}{2!} + ⋯ + (−1)^{k} \frac{1}{(2k)!} + ⋯

Assuming 2k ≤ n < 2k + 1, multiply both sides by (−1)^{k+1}n! to obtain

(−1)^{k+1}(n − 1)!m = [(−1)^{k+1}n!(1 − \frac{1}{2!} + ⋯ + (−1)^{k} \frac{1}{(2k)!})]

+\frac{n!}{(2k + 2)!} − \frac{n!}{(2k + 4)!} + ⋯

Letting the expression inside the square brackets on the right hand side to be A and setting B = (−1)^{k+1}(n − 1)!m − A, we see that B is an integer which satisfies

0 < \frac{n!}{(2k + 2)!} − \frac{n!}{(2k + 4)!} < B < \frac{n!}{(2k + 2)!} < 1

which is a contradiction. Hence cos(1) is irrational.

5) Consider the set A = ∪_{n=1}^{∞}{(x, y) ∈ ℝ^{2}|x^{2} + y^{2} ≥ 1 + 1/n}. Describe the interior, the closure and the boundary of A. Is A closed, open or neither?

A = {(x, y) ∈ ℝ^{2}| x^{2} + y^{2} > 1 }. Therefore A is open and the interior of A is equal to A. The closure of A is {(x, y) ∈ ℝ^{2}|x^{2} + y^{2} ≥ 1}, and the boundary of A is the unit circle x^{2} + y^{2} = 1.