12.5 Bolzano theorem

Tutorial 8.pdf

1. Use Bolzano’s theorem to prove that the following equations have at least one real root:

   1. \(0 = x^3 + 5x^2 - 7x - 2\)

      \(f(0) = -2 < 0\)

      \(f(1) = 1 + 5 - 7 - 2 < 0\)

      \(f(-1) = -1 + 5 - 7 - 2 > 0\)

      \(f: [-1, 0] \to \mathbb{R}, f\) is continuous.

      Bolzano's Theorem\(\implies \exists c \in (-1, 0) \text{ s.t. } f(c) = 0\) 2. \(x^3 = 20 + \sqrt{x}\)

      \(f(x) = x^3 - 20 - \sqrt{x}\)

      \(\text{Dom}_f = [0, +\infty)\)

      \(g(x) = \sqrt{x}, \text{ continuous in } [0, +\infty)\)

      By algebra of continuous functions, \(f(x)\) is continuous.

      \(f(0) = -20 < 0\)

      \(f(3) = 3^3 - 20 - \sqrt{3} > 0\)

      By theorem, \(\exists c \in (0, 3) \text{ s.t. } f(c) = 0\)​ 3. \(e^x + x + 2 = 0\)

2. Prove that if \(f : \mathbb{R} \to \mathbb{R}\) is a polynomial function of odd degree, then it necessarily has at least one real root. In particular, any real polynomial of degree 3 has (at least) one real root.

   Proof

   Let \(P(x) = a_{2n+1}x^{2n+1} + a_{2n}x^{2n} + \dots + a_1x + a_0\), where \(a_{2n+1} \neq 0\).
   ​\(\lim_{x\to\infty}P(x)=\lim_{x\to\infty}\left(a_{2n+1}x^{2n+1}+a_{2n}x^{2n}+\ldots+a_1x+a_0\right)=\lim_{x\to\infty}x^{2n+1}\left(a_{2n+1}+a_{2n}\frac{1}{x}+\ldots+a_1\frac{1}{x^{2n}}+a_0\frac{1}{x^{2n+1}}\right)\)

   We are going to prove formally that (in a moment)
   ​\(\lim_{x \to +\infty} P(x) = +\infty\) and \(\lim_{x \to -\infty} P(x) = -\infty\)

   Then \(\exists\,x_1>0\,\text{such that}\,P(x_1)>0\,\text{and}\,\exists\,x_2<0\,\text{such that}\,P(x_2)<0\)

   \(P: [x_2, x_1] \to \mathbb{R} \, \text{continuous}\)

   Thus by theorem \(\,\exists\,c\in(x_2,x_1)\,\text{such that}\,P(c)=0\)

   \(P(x)\) has a real root

   ---

   Proposition:
   (1) If \(\lim_{x \to +\infty} f(x) = +\infty\) and \(\lim_{x \to +\infty} g(x) = L\), \(L > 0\), then \(\lim_{x\to+\infty}f(x)g(x)=\begin{cases}+\infty & \text{if }L>0\\ -\infty & \text{if }L<0\end{cases}\)​

   (2) If \(\lim_{x \to -\infty} f(x) = -\infty\) and \(\lim_{x \to -\infty} g(x) = L\), \(L \neq 0\), then \(\lim_{x\to-\infty}f(x)g(x)=\begin{cases}+\infty & \text{if }L<0\\ -\infty & \text{if }L>0\end{cases}\)​

   Let's prove (1) and suppose \(L > 0\)

   \(\lim_{x \to +\infty} f(x) = +\infty\), \(\forall M > 0\), \(\exists N\) such that \(\forall x > N\), we have \(f(x) > M\)

   \(\lim_{x \to +\infty} g(x) = L\), \(\forall \epsilon > 0\), \(\exists \delta > 0\) such that \(\forall x > \delta\), \(\left| g(x) - L \right| < \epsilon\)

   We want to show that \(\lim_{x \to +\infty} f(x)g(x) = +\infty\), so for any \(M > 0\), we must find \(N'\) such that \(f(x)g(x)>M,\forall x>N'\)

   Choosing \(\epsilon\) such that \(L - \epsilon > 0\), then \(T = L - \epsilon\), \(\forall x > \delta\)

   \(f(x)g(x) > M g(x) > MT\), \(\forall x > \delta\) and \(x > N\)

   Pick \(N' = \max(N, \delta)\), then \(f(x)g(x)>MT\), \(\forall x > N'\)

   The definition follows

3. Show that if \(f: [a, b] \to \mathbb{R}\) is continuous and strictly increasing or strictly decreasing, then \(f\) is bijective in \([a, b]\)

   Proof: Let's prove first that \(f\) is injective.

   Suppose for \(x, y \in [a, b]\), we have \(f(x) = f(y)\).

   Possibilities: \(\begin{cases}x<y\Rightarrow f\left(x\right)<f\left(y\right)\\x=y\\x>y\Rightarrow f\left(x\right)>f\left(y\right)\end{cases}\)​\(\Rightarrow x=y\Rightarrow f\) is injective.

   Let's prove that \(f\) is surjective

   Since \(f(a) < f(b)\), then \(f: [a, b] \to [f(a), f(b)]\)

   Let \(y \in [f(a), f(b)]\). By the intermediate value theorem, \(\exists x \in [a, b]\) such that \(f(x) = y\).

   Thus \(f\) is surjective.

   Hence, \(f\) is bijective.

4. Prove that the image of exp function is \((0, \infty)\), then prove \(\exp: \mathbb{R} \to (0, +\infty)\) is bijective

   Observations \(\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e\)

   \(\exp(x) = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = \sum_{n=0}^{\infty} \frac{x^n}{n!}\)

   1. \(\exp(x) > 0, \, \forall x \in \mathbb{R}\)
   2. \(\exp(x + y) = \exp(x) \cdot \exp(y)\) (Cauchy's product of series)

   c. \(\exp(x)\) is continuous in \(\mathbb{R}\)

   ---

   Let's show that

   1. \(\exp(x)\) is strictly increasing in \(\mathbb{R}\)

      Suppose \(x, y \in \mathbb{R}\) such that \(x < y\) Then \(0 < y - x \implies \exp(y - x) > 1\)

      \(\exp(y) = \exp(x) \cdot \exp(y - x) \implies \exp(y) > \exp(x)\)​\(\implies \exp(x)\) is strictly increasing 2. \(\text{Im} \, \exp = (0, +\infty)\)

      Since \(\exp(x)\) is strictly increasing and \(e^{x}>2^{x},\lim_{x\to\infty}2^{x}=\infty\), then \(\lim_{x\to\infty}e^{x}=\infty\)

      Then \(\lim_{-x\to\infty}2^{x}=\lim_{x\to-\infty}\frac{1}{2^{-x}}=0\)​

      Thus the \(\text{Im} \, \exp = (0, +\infty)\) 3. Injectivity

      Take \(e^{x_1}=e^{x_2}\), then \(e^{x_1-x_2}=1\), then \(x_1-x_2=0\)​ 4. Surjectivity

      Take any \(y\in (0,+\infty)\), then \(\exists\) not defined??

   ‍

   1. Suppose \(x, y \in \mathbb{R}\) such that \(x < y\) Then \(0 < y - x \implies \exp(y - x) > 1\)

      \(\exp(y) = \exp(x) \cdot \exp(y - x) \implies \exp(y) > \exp(x)\)​\(\implies \exp(x)\) is strictly increasing 3. \(\exp(0) = 1 \iff 1 \in \text{Im} \, \exp\) \(\exp(1) = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e > 1\)

      We are going to prove that \((0, +\infty) \subseteq \text{Im} \, \exp\) \(\text{Im} \, \exp \subseteq (0, +\infty)\) (Exercise)

      For \(n \in \mathbb{N}\): \(\exp(n)=\exp(1+1+\dots+1)\,(\text{n times})=\exp(1)\cdot\exp(1)\cdot\dots\cdot\exp(1)=e^{n}>1,\forall n\)

      For \(y > 1\), \(\exists n \in \mathbb{N}\) such that \(e^n > y\)

      So \(1<y<e^{n}\)

      \(\exp: [0, n] \to [1, e^n]\) continuous function

      By the intermediate value theorem, \(\exists x \in [0, n]\) such that \(\exp(x) = y\)

      \(\implies y \in \text{Im} \, \exp\)

      \(\implies\left\lbrack1,+\infty)\subseteq\text{Im}\,\exp\right.\)

      ---

      Let's consider \(0<y<1\), then \(\frac1y>1\)

      using the previous cases

      \(1 = \exp(0) = \exp(x' + (-x')) = \exp(x') \cdot \exp(-x')\) \(1 = \frac{1}{y} \cdot \exp(-x')\) \(y = \exp(-x')\)

      \(\implies y \in \text{Im} \, \exp\)

      \(\implies (0, 1] \subseteq \text{Im} \, \exp\)

      \(\implies (0, +\infty) \subseteq \text{Im} \, \exp\)

      \(\exp: \mathbb{R} \to (0, +\infty)\) is an injective, surjective function and continuous function \(\implies\) the inverse \(\exp^{-1}\) exists and it is continuous.

\(\exp^{-1}:(0,+\infty)\to\mathbb{R}---\exp^{-1}(y)=x\;\iff\;\exp(x)=y\)

We will denote \(\exp^{-1}(y) = \ln(y) = \log_e(y),\) where \(\ln\) is the Napierian logarithm, the logarithm of base \(e\)

Properties of logarithm:

1. \(\ln(ab) = \ln(a) + \ln(b)\)

2. \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)

3. \(\ln(a^b) = b \ln(a)\)

Arbitrary exponential functions: for \(a > 0, \ x \in \mathbb{R}\) we have \(a^x := \exp(\ln(a) \cdot x) = \text{composition of continuous functions.}\)

Properties:

1. \(n \in \mathbb{N}, \ a^n = a \cdot a \cdots a \ (n \text{-times})\)

2. \(a^{b+c} = a^b \cdot a^c\)

3. \((a^b)^c = a^{bc}\)

4. \(a^{\frac{1}{n}} = \sqrt[n]{a}, \ n \in \mathbb{N}\)

5. \(a^{\frac{m}{n}} = \sqrt[n]{a^m}, \ m \in \mathbb{Z}, \ n \in \mathbb{N}\)