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Table of Contents
Optional Math of Waves
This is a brief survey of the math required to analyze waves at the first or second year university level. If you did well in grade 12 high school math, you'll probably be able to follow this and learn some new and really cool math.
Real numbers
If \$ (5)^2 = 25 \$ and \$ (-5)^2 = 25 \$, what number can you put in the box so that:
\$$ \Box ^2 = -25 \$$
It turns out that there is no real number such that when you multiply it by itself you get a negative number, but could we invent an imaginary one?
Complex Numbers
Let's create an imaginary number called \$ i \$ such that:
\$$ i = \sqrt{-1} \qquad \Rightarrow \qquad i^2 = -1 \$$
Even though \$ i \$ is nowhere on the real line (in math, we say that: \$ i \not\in \mathbb{R} \$), we can none-the-less perform interesting mathematical operations with it:
- We can add it to a real number and create a complex number:
\$$(1 + i) \$$
- We can multiply complex numbers together:
\begin{align*} (1+i)^2 &= 1 + 2i + i^2 \\ &= 1 + 2i - 1 \\ &= 2i \end{align*}
- We can find roots:
\begin{equation*} z^4 = 16 \Rightarrow z^2 = \left\{ \begin{array}{rl} 4 \Rightarrow z &= \pm 2 \\ -4 \Rightarrow z &= \pm 2i \end{array} \right. \end{equation*}
A Little Philosophy
If these weird numbers follow all of the algebra rules without inconsistencies, does it mean they exist as much as the real numbers? Aren't complex numbers a mere creation by mad mathematicians? How about mathematics itself: is it discovered or invented?1)
In a certain way, negative numbers are just as weird as complex numbers: after all, we know what 5 cars look like, but what does −5 cars mean? And yet, in certain context (like temperature), we have no problem using negative numbers. Could it be that there are contexts where complex numbers make sense?
The Complex Plane
In the same way that we can represent real numbers by a point on the real number line...2):
... we can also represent a complex number graphically on a complex plane by using the horizontal axis for the real part and the vertical axis for the imaginary part. For example, \$ (1 + i) \$ would be represented as a point 45° up the horizontal axis and \$ \sqrt{2} \$ away from the origin:
Download polar.ggb
You can move the point around to see what the polar representation \$ (r \angle \theta) \$ is.
To convert between the Cartesian \$(a,b) \$ and the Polar \$ (r \angle \theta) \$ representations, only simple trigonometry and Pythagoras is needed.
\$$ a + ib \rightarrow r\angle \theta \$$ | \$$ r\angle \theta \rightarrow a+ib \$$ |
---|---|
\$$ r^2 = a^2 + b^2 \$$ | \$$ a = r\cos\theta \$$ |
\$$ \tan \theta = \dfrac{b}{a} \$$ | \$$ b = r\sin\theta \$$ |
Roots
The complex plane has many useful applications, but one of them allows us to visualize roots of the form \$ z^n = w \$. For example, if we set \$ w = 9 \$ and \$ n = 2 \$ on the graph below, we'll see that the roots of \$z^2 = 9 \$ are \$ z = \pm 3 \$.
Download complexroots.ggb
Without using the graph above, what do you expect the solution(s) to \$ z^3 = 8 \$ will be? That is, what number(s), when multiplied by itself three times gives 8?
Now move \$ w = 8 \$ and \$n = 3 \$ to have a look at the solutions.
The Euler Identity
The Euler identity exposes a deep relationship between trigonometric and exponential functions3):
\$$ e^{i \theta} = \cos \theta + i \sin \theta \$$
Let's use two different ways to verify that this mysterious identity works.
The Derivatives
If we separate this identity into two functions and take their derivatives, we notice that: \begin{align*} && f(\theta) &= e^{i \theta} &\text{ & }&& g(\theta) &= \cos \theta + i \sin \theta \\ \Rightarrow && f'(\theta) &= i e^{i \theta} &\text{ & }&& g'(\theta) &= -\sin \theta + i \cos \theta \\ \Rightarrow && f'(\theta) &= i \cdot f(\theta) &\text{ & }&& g'(\theta) &= i \cdot g(\theta) \end{align*}
We know that there's only one functions \$ h(x) \$ that satisfies the differential equation \$ h'(x) = ah(x) \$, and it is \$ h(x) = A e^{ax} \$ What Euler discovered is that when \$ a = i \$ , there's a second function that also satisfies the same differential equation! These two functions must therefore be one and the same.
Taylor
Another method to verify the Euler identity is to use Taylor series:
\begin{align*} e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots \\ \sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \\ \cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \end{align*}
- Replace \$ x \$ with \$ i \theta \$ in the Taylor series of \$ e^x \$
- Replace \$ x \$ with \$ \theta \$ in the Taylor series for \$ \sin x \$ and \$ \cos x \$
- Add the Taylor series of \$ \cos \theta \$ and \$ i \sin \theta \$ together and you'll get the Taylor series for \$ e^{i \theta} \$
Euler Identity and Polar-Cartesian Representations
In the previous section, we saw that a complex number \$z = a + ib \$ could be represented as a point \$(a, b)\$ on the complex plane, which could also be viewed in polar coordinates as \$r\angle \theta \$
Important Algebraic Results
- Use the Euler identity to get the following two useful results:
\$$ \cos \theta = \dfrac{e^{i \theta} + e^{-i \theta}}{2} \qquad \text { & } \qquad \sin \theta = \dfrac{e^{i \theta} - e^{-i \theta}}{2i} \$$
- Use the Euler identity to show that:
\$$ \cos (\theta + \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi\\ \sin (\theta + \phi) = \cos \theta \sin \phi + \sin \theta \cos \phi \$$
- Use the two earlier results to show that:
\$$ \sin (\theta + \Delta \theta) + \sin (\theta - \Delta \theta) = 2 \cos \Delta \theta \sin \theta \$$
Équations différentiels
Dans la physique des ondes, on aura bientôt à trouver la solutions d'une équation différentiel de ce type: \$$ a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \$$
Dans nos applications, les paramètres \$ a, b, c \$ serons tous des quantités réels et positive. Même sans avoir étudier le sujet d'équation différentiels en profondeurs, on peut imaginer qu'une solution possible serait \$ x(t)= e^{rt} \$ puisque la dérivé d'une fonction exponentiel est elle même une fonction exponentiel, ce qui est encourageant... L'étape suivante est donc d'essayer cette fonction dans l'équation différentiel, pour trouver les valeurs de \$ r \$ qui fonctionnent. Donc:
\$ \begin{align*} & x (t) = e^{rt} \\ \Rightarrow \qquad & \dot{x}(t) = r e^{rt} \\ \Rightarrow \qquad & \ddot{x}(t) = r^2 e^{rt} \end{align*} \$
et notre équation différentiel devient:
\$ \begin{align*} & a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \\ \Rightarrow \qquad & a (r^2 e^{rt}) + b (r e^{rt}) + c (e^{rt}) = 0 \\ \Rightarrow \qquad & e^{rt} (a r^2 + b r + c ) = 0 \\ \Rightarrow \qquad & a r^2 + b r + c = 0 \end{align*} \$
\$ \Rightarrow \qquad r = - \dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} \$
Puisque \$ r \$ contient une racine carré, il peut être réel, ou complexe. Pour simplifier la notation, disons que: \$ \alpha = \dfrac{b}{2a} \qquad \text{et} \qquad \beta = \dfrac{\sqrt{|{b^2 - 4ac}|}}{2a} \$
Nous pourrons donc dire que: \$ \begin{equation*} r = \left\{ \begin{array}{rl} -\alpha \pm \beta & \text{si } b^2 - 4ac > 0,\\ -\alpha \pm i \beta & \text{si } b^2 - 4ac < 0, \end{array} \right. \end{equation*} \$
Examinons les deux cas en plus de détails.
Cas 1:
Quand \$ b^2 - 4ac > 0 \$ , \$ r \$ est réel et la solutions général sera:
\$ \begin{align*} x(t) &= A_1 e^{r_1 t} + A_2 e^{r_2 t} \\ &= A_1 e^{( -\alpha + \beta) t} + A_2 e^{( -\alpha - \beta) t} \\ &= A_1 e^{-\alpha t} e^{\beta t} + A_2 e^{-\alpha t} e^{-\beta t} \end{align*} \$
\$$ x(t) = e^{-\alpha t} ( A_1 e^{\beta t} + A_2 e^{-\beta t} ) \$$
C'est normal d'avoir deux constantes d'intégrations puisque notre équation différentiel a une dérivé du second degré. Pour trouver ces constantes, ça nous prendrait des valeurs initiales.
Cas 2:
Quand \$ b^2 - 4ac < 0 \$ , \$ r \$ est complexe et nous utiliserons la formule de Euler pour simplifier notre solutions.
\$ \begin{align*} x(t) &= A_1 e^{r_1 t} + A_2 e^{r_2 t} \\ &= A_1 e^{( -\alpha + i \beta) t} + A_2 e^{( -\alpha - i \beta) t} \\ &= A_1 e^{-\alpha t} e^{i \beta t} + A_2 e^{-\alpha t} e^{-i \beta t} \\ &= e^{-\alpha t} ( A_1 e^{i \beta t} + A_2 e^{-i \beta t} ) \\ &= e^{-\alpha t} \Big( A_1 \big(cos( \beta t) + i \sin( \beta t) \big) + A_2 \big(cos( -\beta t) + i \sin( -\beta t) \big) \Big) \\ &= e^{-\alpha t} \Big( A_1 \big(cos(\beta t) + i \sin(\beta t) \big) + A_2 \big(\cos(\beta t) - i \sin(\beta t)\big)\Big) \\ &= e^{-\alpha t} \Big( (A_1 + A_2) \cos(\beta t) + i (A_1 - A_2) \sin(\beta t) \Big) \\ &= e^{-\alpha t} \Big( a_1 \cos(\beta t) + a_2 \sin(\beta t) \Big) \\ &= e^{-\alpha t} \Big( A \sin \phi \cos(\beta t) + A \cos \phi \sin(\beta t) \Big) \end{align*} \$
Dans les deux dernière ligne, nous avons ré-écrit les constantes d'intégration:
\$ \begin{align*} a_1 &= A_1 + A_2 & , a_2 &= i(A_1 - A_2) \\ a_1 &= A \sin \phi & , a_2 &= A \cos \phi \end{align*} \$
Pour finalement pouvoir utiliser une identité trigonométrique:
\$$ x(t) = A e^{-\alpha t} \sin(\beta t + \phi) \$$
Exemple
Nous avons donc deux types de solutions complètement différents qui dépendent de trois paramètres \$ a, b, c \$. Pour voir comment ces paramètres affectent le graphique, imaginons qu'une de nos conditions initiales est \$ \phi = \frac{\pi}{2} \$ . Ça veut dire que:
\$ \begin{align*} & a_1 = A \sin \pi/2 = A & & a_2 = A \cos \pi/2 = 0 \\ \Rightarrow \qquad & A_1 + A_2 = A & & A_1 - A_2 = 0 \\ \Rightarrow \qquad & A_1 = A/2 & & A_2 = A/2 \end{align*} \$
Dans ce cas particulier, nous avons donc:
\$ \begin{equation*} x(t) = \left\{ \begin{array}{rl} A e^{-\alpha t} \dfrac{e^{\beta t} + e^{-\beta t}}{2} & \text{si } b^2 - 4ac > 0 ,\\ A e^{-\alpha t} \cos(\beta t) & \text{si } b^2 - 4ac < 0 ,\\ \end{array} \right. \end{equation*} \$
Geogebra
\$$ e^{i \pi} = \cos (\pi) + i \sin (\pi) = -1 + 0i = -1\$$ \$$ \Rightarrow e^{i \pi} = -1\$$ which is amazingly beautiful because it relates \$ e = 2.71828... \$, \$i = \sqrt{-1} \$, \$\pi = 3.14159... \$, and \$-1\$ in the most surprising and elegant way.