14 Exponential Function, Arc Measure and Radians, Defining Pi (Existence and Properties), Tangent and Cotangent, Limit of Functions

Lecture from: 10.04.2024 | Video: Video ETHZ

Clicker: $\cos (i)$

Since $e\approx 2.718$, $e^2+1\approx 7.389+1=8.389$, and $2e\approx 5.436$. Thus, $\cos (i)>1$. This shows that for complex arguments, the cosine function can take values greater than 1 in absolute value, unlike the real cosine function which is bounded by $[-1,1]$.

Corollary: Double-Angle Formulas

For $z\in \mathbb{C}$, the following hold:

$\cos (2z)={\cos }^2(z)-{\sin }^2(z)$ $\sin (2z)=2\sin (z)\cos (z)$

Intuition: These are direct consequences of the addition theorems for sine and cosine, which we derived in the previous lecture. Simply set $w=z$ in the formulas for $\cos (z+w)$ and $\sin (z+w)$.

Short Digression: The Exponential Function on $i\mathbb{R}$ and Arc Measure

For $x\in \mathbb{R}$, we know that $\cos (x)$ and $\sin (x)$ are real numbers. Euler’s formula states: $e^{ix}=\cos (x)+i\sin (x)$ The complex conjugate is: $\overline{e^{ix}}=\cos (x)-i\sin (x)=e^{-ix}$ From this, we can find the magnitude: $|e^{ix}|=\sqrt{(\cos (x){)}^2+(\sin (x){)}^2}=\sqrt{1}=1$ This means that for any real $x$, the complex number $e^{ix}$, or equivalently the point $(\cos (x),\sin (x))$ in the complex plane, lies on the unit circle $S^1=\{z\in \mathbb{C}\mid |z|=1\}$. The set $S^1$ can also be represented as $\{(a,b)\in {\mathbb{R}}^2\mid ||(a,b)||=1\}$, which is the circle with center $0$ and radius $1$, known as the “Einheitskreis” (unit circle).

The real number $x$ is a measure of the angle between the positive real axis and the vector representing $e^{ix}$. This angle is measured in radians (Bogenmass). It is precisely the length of the arc of the unit circle from the point $1$ (corresponding to $e^{i\cdot 0}$) to the point $e^{ix}$.

The function $\text{cis}:\mathbb{R}\to S^1$, defined by $x\mapsto e^{ix}$, “wraps” the real number line around the unit circle $S^1$. This wrapping is length-preserving (in terms of arc length on the circle) and happens infinitely often.

The Number $\pi$

Theorem: Existence of a Root for Sine

The sine function has at least one root (Nullstelle) in the interval $(0,\infty )$. (We know $\sin (0)=0$).

We define $\pi$ as the smallest positive root of the sine function: $\pi :=\inf \{x>0\mid \sin (x)=0\}$

Then the following hold:

1. $\sin (\pi )=0$, and $\pi \in (2,4)$ (Approximately $\pi \approx \mathrm{3.14...}$)
2. For all $x\in (0,\pi )$, $\sin (x)>0$.
3. $e^{i\pi }=-1$

Proof Strategy for Properties of $\pi$

Before diving into the full proof, let’s make some observations that will be useful. For $x\ge 0$, the series for $\sin (x)$ is: $\sin (x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\dots$ This is an alternating series.

Recall: Leibniz Criterion for Alternating Series If $(a_n{)}_{n\ge 1}$ is a monotonically decreasing sequence with $a_n\ge 0$ for all $n\ge 1$ and ${\lim }_{n\to \infty }{a}_n=0$, then the alternating series $S={\sum }_{k=1}^{\infty }(-1{)}^{k+1}{a}_k$ converges. Furthermore, we have the error bound, or more relevantly here, bounds on the sum: $a_1-a_2\le S\le a_1$.

Let’s examine if the terms of the alternating series for $\sin (x)$ are monotonically decreasing for certain $x$. The terms (ignoring the sign for a moment) are of the form $\frac{x^{2n-1}}{(2n-1)!}$. We need to check when $\frac{x^{2n+1}}{(2n+1)!}\le \frac{x^{2n-1}}{(2n-1)!}$.

This inequality holds if $x^2\le (2n)(2n+1)$.

• For the first two terms $a_1=x$ and $a_2=\frac{x^3}{3!}$: We need $\frac{x^3}{3!}\le x⟹x^2\le 6$.
• For $a_2=\frac{x^3}{3!}$ and $a_3=\frac{x^5}{5!}$: We need $\frac{x^5}{5!}\le \frac{x^3}{3!}⟹x^2\le 4\cdot 5=20$.
• In general, for the terms $\frac{x^{2n-1}}{(2n-1)!}$ and $\frac{x^{2n+1}}{(2n+1)!}$: We need $x^2\le 2n(2n+1)$.

This means that for $x\in [0,\sqrt{6}]$, the terms of the sine series are monotonically decreasing (after the first term).

Corollary: Bounds for $\sin (x)$

Given that the conditions for the Leibniz Criterion are met for the sine series when $0\le x\le \sqrt{6}$, we can apply its error bounds. Specifically, if $S=\sin (x)$, $b_1=x$, and $b_2=\frac{x^3}{3!}$, then the sum $S$ satisfies $b_1-b_2\le S\le b_1$. Therefore, for $x\in [0,\sqrt{6}]$: $x-\frac{x^3}{3!}\le \sin (x)\le x$

Proof of Theorem (Properties of $\pi$)

The function $\sin :\mathbb{R}\to \mathbb{R}$, defined by its power series, is continuous on $\mathbb{R}$. This continuity is fundamental to the argument.

Existence of $\pi$

First, we establish the interval where $\pi$ lies. From the corollary, for $0<x\le \sqrt{6}$ (noting $\sqrt{6}\approx 2.449$), we have $\sin (x)\ge x-\frac{x^3}{3!}=x(1-\frac{x^2}{6})$. Since $x>0$ and $(1-\frac{x^2}{6})>0$ for $x<\sqrt{6}$, it follows that $\sin (x)>0$ for $x\in (0,\sqrt{6})$. For example, $\sin (2)>0$ since $2<\sqrt{6}$.

It has been shown (or can be shown by evaluating more terms of the series, as shown above) that $\sin (4)<0$.

Since $\sin (x)$ is continuous, $\sin (2)>0$, and $\sin (4)<0$, the Intermediate Value Theorem guarantees that $\sin (x)$ must have at least one root between $x=2$ and $x=4$.

We define $\pi :=\inf \{x>0\mid \sin (x)=0\}$. The existence of a root in $(2,4)$ ensures this set is non-empty and that $\pi \le 4$. Since $\sin (x)>0$ on $(0,\sqrt{6}]$, $\pi$ must be greater than or equal to $\sqrt{6}$, which is greater than 2. Thus, $\pi$ is well-defined and lies in the interval $(2,4)$. The continuity of $\sin (x)$ also ensures that $\sin (\pi )=0$.

$\sin (x)>0$ on $(0,\pi )$

Next, because $\pi$ is defined as the smallest positive root of $\sin (x)$, there can be no other roots in the interval $(0,\pi )$. Since $\sin (x)$ is continuous on this interval and we know $\sin (x)>0$ for values of $x$ close to zero (e.g., for $x\in (0,\sqrt{6})$), it must follow that $\sin (x)$ remains positive for all $x\in (0,\pi )$.

Finally, we prove the identity $e^{i\pi }=-1$. To do this, we first determine $\cos (\pi )$. We use the double angle formula $\sin (2z)=2\sin (z)\cos (z)$. Setting $z=\pi /2$, we get $\sin (\pi )=2\sin (\pi /2)\cos (\pi /2)$. Since $\sin (\pi )=0$, this implies $2\sin (\pi /2)\cos (\pi /2)=0$.

Given that $\pi \in (2,4)$, we have $\pi /2\in (1,2)$. Since $(1,2)\subset (0,\pi )$, and we’ve established that $\sin (x)>0$ for $x\in (0,\pi )$, it follows that $\sin (\pi /2)>0$. Because $\sin (\pi /2)\ne 0$, it must be that $\cos (\pi /2)=0$.

$e^{i\pi }=-1$

Now, using the fundamental identity ${\cos }^2(\pi /2)+{\sin }^2(\pi /2)=1$, and knowing $\cos (\pi /2)=0$, we have ${\sin }^2(\pi /2)=1$. Since $\sin (\pi /2)>0$, we conclude that $\sin (\pi /2)=1$.

With $\cos (\pi /2)=0$ and $\sin (\pi /2)=1$, we can evaluate $e^{i\pi /2}$: $e^{i\pi /2}=\cos (\pi /2)+i\sin (\pi /2)=0+i\cdot 1=i$. Therefore, $e^{i\pi }=e^{i(\pi /2)\cdot 2}=(e^{i\pi /2}{)}^2=i^2=-1$.

Corollary: Further Properties related to $\pi$

1. $e^{i\pi }=-1$, $e^{2i\pi }=(e^{i\pi }{)}^2=(-1{)}^2=1$.
2. For $x\in \mathbb{R}$:
   1. $\sin (x+\pi /2)=\sin (x)\cos (\pi /2)+\cos (x)\sin (\pi /2)=\sin (x)\cdot 0+\cos (x)\cdot 1=\cos (x)$
   2. $\cos (x+\pi /2)=\cos (x)\cos (\pi /2)-\sin (x)\sin (\pi /2)=\cos (x)\cdot 0-\sin (x)\cdot 1=-\sin (x)$
3. For $x\in \mathbb{R}$:
   1. $\sin (x+\pi )=\sin (x)\cos (\pi )+\cos (x)\sin (\pi )=\sin (x)(-1)+\cos (x)(0)=-\sin (x)$
   2. $\cos (x+\pi )=\cos (x)\cos (\pi )-\sin (x)\sin (\pi )=\cos (x)(-1)-\sin (x)(0)=-\cos (x)$
4. And from these,
   1. $\sin (x+2\pi )=\sin ((x+\pi )+\pi )=-\sin (x+\pi )=-(-\sin (x))=\sin (x)$
   2. $\cos (x+2\pi )=\cos ((x+\pi )+\pi )=-\cos (x+\pi )=-(-\cos (x))=\cos (x)$

This shows that $\sin$ and $\cos$ are $2\pi$-periodic.

Behavior of Sine and Cosine on $[0,2\pi ]$

Based on the properties above, we can deduce the behavior of $\sin (x)$ and $\cos (x)$ on the interval $[0,2\pi ]$:

• $\sin (x)=0⟺x\in \{0,\pi ,2\pi \}$
• $\sin (x)>0⟺x\in (0,\pi )$
• $\sin (x)<0⟺x\in (\pi ,2\pi )$
• $\cos (x)=0⟺x\in \{\pi /2,3\pi /2\}$
• $\cos (x)>0⟺x\in [0,\pi /2)\cup (3\pi /2,2\pi ]$
• $\cos (x)<0⟺x\in (\pi /2,3\pi /2)$

The behavior on the entire real line $\mathbb{R}$ is then obtained by shifting by multiples of $2\pi$.

Proof of $\sin (x+\pi /2)=\cos (x)$ (one of the identities from Corollary 2)

Using the exponential form for $\sin (z)$: $\sin (x+\pi /2)=\frac{e^{i(x+\pi /2)}-e^{-i(x+\pi /2)}}{2i}=\frac{e^{ix}{e}^{i\pi /2}-e^{-ix}{e}^{-i\pi /2}}{2i}$ Since $e^{i\pi /2}=i$ and $e^{-i\pi /2}=-i$: $=\frac{e^{ix}(i)-e^{-ix}(-i)}{2i}=\frac{i(e^{ix}+e^{-ix})}{2i}=\frac{e^{ix}+e^{-ix}}{2}=\cos (x)$

Definition: Tangent and Cotangent Functions

• For $x\notin \{\frac{\pi }{2}+k\pi \mid k\in \mathbb{Z}\}$ (i.e., where $\cos (x)\ne 0$), we define the tangent function: $\tan (x)=\frac{\sin (x)}{\cos (x)}$

• For $x\notin \{k\pi \mid k\in \mathbb{Z}\}$ (i.e., where $\sin (x)\ne 0$), we define the cotangent function: $\cot (x)=\frac{\cos (x)}{\sin (x)}$

Limits of Functions

We now turn our attention to the concept of limits for functions $f:D\to \mathbb{R}$. We want to investigate the behavior of $f(x)$ as $x$ approaches some point $x_0$.

As “limit points” $x_0$, we must also allow certain points that are outside of $D$.

Definition: Accumulation Point (Häufungspunkt)

Let $D\subseteq \mathbb{R}$. A point $x_0$ is an accumulation point (or limit point) of $D$ if $x_0\in \mathbb{R}$, or $x_0=\infty$, or $x_0=-\infty$, with the property that there exists a sequence $(a_n{)}_{n\in \mathbb{N}}$ in $D\setminus \{x_0\}$ such that ${\lim }_{n\to \infty }{a}_n=x_0$.

Intuition: An accumulation point $x_0$ of a set $D$ is a point that can be “approached” by points in $D$ other than $x_0$ itself. This means every neighborhood of $x_0$ contains infinitely many points of $D$. We also extend this concept to include $\infty$ and $-\infty$ if $D$ is unbounded.

Continue here: 15 Limit of Functions and Rules, One-Sided Limits