(15.1, 15.2, 15.4) Determine which of the following series converge. Justify your answers.
\(\displaystyle\sum { \left[ \sin \left( \frac {n \pi}{6} \right)\right]^n }\)
This series diverges, as the terms do not converge to zero (\(a_{6k-3} = (-1)^{k+1}\) for all \(k \in \mathbb N\); this subsequence does not converge)
\(\displaystyle\sum { \frac{(-1)^n}{n} }\)
This series converges, by the alternating series test.
\(\displaystyle\sum_{n=2}^\infty { \frac {1}{\sqrt n \log n} }\)
This series diverges, for example by comparison with the series \(\sum {1/n}\)
\(\displaystyle\sum_{n=2}^\infty { \frac {\log n}{n^2} }\)
This series converges, for example by comparison with the series \(\sum {1/n^{3/2}}\)
(17.4) Prove that the function \(\sqrt x\) is continuous on its natural domain \([0, \infty)\). Hint: Think about Example 5 in §8.
Proof. In Example 5, §8 it is proved that for any convergent sequence \((x_n)\) of nonnegative real numbers, \(\lim_{n\to\infty}{\sqrt x_n} = \sqrt{ \lim_{n\to\infty} x_n}\).
This is the definition of continuity of the function \(\sqrt x\). \(\Box\)
(17.5)
Prove that \(x^m\) is a continuous function for any \(m \in \mathbb N\).
Use induction, and that \(fg\) is continuous on a set \(S\) if both \(f\) and \(g\) are.
Prove that every polynomial function \(\sum_{i=0}^n{a_i x^i}\) is continuous.
Use induction, and that \(kf\) and \(f+g\) are continuous on a set \(S\) if both \(f\) and \(g\) are, for any real number \(k\).
(17.13)
Let \(f(x) = 1\) for rational numbers \(x\) and \(f(x) = 0\) for irrational numbers. Show that \(f\) is discontinous at every \(x\) in \(\mathbb R\).
Proof. Consider \(q \in \mathbb Q\), and let \(\alpha\) be an irrational number. Then the sequence \((q + \alpha/n)\) is a sequence of irrational numbers converging to \(q\). But the sequence \((f(q + \alpha/n))\) converges to 0 even though \(f(q) = 1\). Hence \(f\) is not continuous at \(q\).
Now, consider \(\alpha\) any irrational number. By density of the rationals, there exists a rational number \(q_n\) for which \(\alpha - 1/n < q_n < \alpha\). Observe that \(q_n\) converges to \(\alpha\), and that \((f(q_n))\) converges to 1 while \(f(\alpha) = 0\). Hence \(f\) is not continuous at \(\alpha\). \(\Box\)
Let \(h(x) = x\) for rational numbers \(x\) and \(h(x) = 0\) for irrational numbers. Show that \(h\) is continuous at \(0\) and at no other point.
Proof. Much of the reasoning in (a) holds, with minor changes. The meaningful thing to see here is why \(h\) is continuous at \(0\).
Let \((x_n)\) be any sequence of real numbers converging to 0. If only finitely many of \(x_n\) are irrational, it is easy to see that \((h(x_n))\) converges to 0, as all \(h(x_n)\) are identically 0 for large \(n\).
Suppose there are an infinite number of rational \(x_n\) and let these define the sequence \((x_{n_k})\). Observe both that \(\lim h(x_{n_k}) = \lim x_{n_k} = 0\) and \(\limsup h(x_n) \le \limsup h(x_{n_k}) = 0\) and that \(\liminf h(x_n) \ge \liminf h(x_{n_k}) = 0\). So \(\lim h(x_n) = 0 = h(0)\), and we conclude that \(h\) is continuous at 0. \(\Box\)
Let \(E \subseteq \mathbb R\) be a set of real numbers. An element \(s_0 \in E\) is interior to \(E\) if for some \(r > 0\) we have
\[\{ s \in \mathbb R : |s - s_0| < r\} = (s_0 - r, s_0 + r) \subset E.\]The set \(E\) is open in \(\mathbb R\) if every point in \(E\) is interior to \(E\).
Let \(f: \mathbb R \to \mathbb R\) be a function, and for any subset \(V \subset \mathbb R\), let \(f^{-1}(V) \subset \mathbb R\) be the set;
\[f^{-1}(V) = \{ x \in \mathbb R : f(x) \in V \}.\]Show that \(f\) is continuous if and only if, for every open set \(U \subset \mathbb R\), \(f^{-1}(U)\) is also an open set.
Proof. (\(\Rightarrow\)) Suppose \(f\) is continuous and let \(U\) be any open set in \(\mathbb R\) (viewed as the range of \(f\)!) Let \(x \in f^{-1}(U)\). Notice that as \(f(x) \in U\), and \(U\) is open, there is some \(\epsilon > 0\) for which, for any \(y \in \mathbb R\), \(|y - f(x)| < \epsilon\) implies that \(y \in U\).
Now, as \(f\) is continuous, there exists a number \(\delta > 0\) so that whenever \(t \in \mathbb R\) and \(|t - x| < \delta\), \(|f(t) - f(x)| < \epsilon\); in other words, \(f(t) \in U\)! So \(t \in f^{-1}(U)\).
We have just shown that for all \(x \in f^{-1}(U)\) there exists \(\delta > 0\) so that \(|t - x| < \delta\) implies that \(t \in f^{-1}(U)\). That is, we have just shown that \(f^{-1}(U)\) is open, as desired.
(\(\Leftarrow\)) Suppose that for all open sets \(U \subseteq \mathbb R\), we know that \(f^{-1}(U)\) is open.
Observe that any open interval \((a,b)\) is an open set. (Let \(x \in (a,b)\) and observe that \(r = \min{|x-a|, |x-b|}) > 0\) satisfies the correct condition)
Let \(\epsilon > 0\), let \(x \in \mathbb R\) (the domain of \(f\)!) and consider the open interval \(I = (f(x) - \epsilon, f(x) + \epsilon) = \{y \in \mathbb R : |f(x) - y| < \epsilon\}\) which is an open set in the range of \(f\).
We have that \(f^{-1}(I)\) is itself open by assumption. In other words, there exists \(\delta > 0\) so that for any \(x \in f^{-1}(I)\), if \(t \in \mathbb R\) and \(|t - x| < \delta\), \(t \in f^{-1}(I)\). So in fact, \(f(t) \in I\), meaning that \(|f(t) - f(x)| < \epsilon\).
Hence \(f\) is continuous by the \(\epsilon,\delta\) definition of continuity. \(\Box\)