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howto:hambasics:sections:test [2021/01/02 17:32] – [Complex Numbers] va7fihowto:hambasics:sections:test [2021/02/13 19:14] (current) – [Cartesian vs Polar] va7fi
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-~~NOTOC~~ 
 ====== Optional Math of Waves ====== ====== Optional Math of Waves ======
-This is a brief survey of the math required to analyze waves at the first or second year university level.+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 ===== ===== Real numbers =====
-If \$ (5) \times (5) = 25 \$ and \$ (-5) \times (-5) = 25 \$, what number (and it has to be the same number) can you put in the boxes so that:+If \$ (5)^2 = 25 \$ and \$ (-5)^2 = 25 \$, what number can you put in the box so that:
  
-\$$ \Box \times \Box = -25 \$$  +\$$ \Box ^2 = -25 \$$ 
- +
-It turns out that nowhere on the //real// number line is there a number such that when you multiply it by itself you get a negative number since two positive numbers give a positive number, and two negative numbers also give a positive number. +
- +
-But could we //invent// one?+
  
 +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 ===== ===== Complex Numbers =====
-Let's create an imaginary number called \$ i \$ such that:+Let's create an //imaginary// number called \$ i \$ such that:
  
 +<WRAP center box 30em>
 \$$ i = \sqrt{-1} \qquad \Rightarrow \qquad i^2 = -1 \$$ \$$ i = \sqrt{-1} \qquad \Rightarrow \qquad i^2 = -1 \$$
 +</WRAP>
  
-Even if \$ i \not\in \mathbb{R} \$, we can non-the-less perform interesting mathematical operations with it:+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:   * We can add it to a real number and create a //complex// number:
  
 \$$(1 + i) \$$ \$$(1 + i) \$$
  
-  * We can multiply complex number together:+  * We can multiply complex numbers together:
  
-\$$ \begin{align*} +\begin{align*} 
-(1+i)^2 &= 1 + 2i + i^2 \\+(1+i)^2  
 +&= (1+i)\cdot(1+i) \\ 
 +&= 1 + 2i + i^2 \\
 &= 1 + 2i - 1 \\ &= 1 + 2i - 1 \\
 &= 2i &= 2i
-\end{align*} \$$+\end{align*}
  
   * We can find roots:   * We can find roots:
  
-\$$ \begin{equation*} +\begin{equation*} 
-z^4 = 16 \Rightarrow z^2 = \left\{ +z^4 = 16 \Rightarrow z^2 = \left\{ \begin{array}{rl} 4 \Rightarrow z &= \pm 2 \\ 
-\begin{array}{rl} 4 \Rightarrow z &= \pm 2 \\ +-4 \Rightarrow z &= \pm 2i \end{array} \right. 
--4 \Rightarrow z &= \pm 2i \end{array} \right\} +\end{equation*}
-\end{equation*} \$$+
  
-\$$\begin{equation*} +==== A Little Philosophy ==== 
-z^8 256 \Rightarrow z^4 \left\{ \begin{array}{rl} 16 \Rightarrow z^2 &\left\{ \begin{array}{rl} 4 \Rightarrow z &\pm 2 \\ +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//?((Whether math is discovered or invented is a famous philosophical problem.  If you think it's invented: does that mean that 2 + 2 didn't equal 4 until someone invented that?  If you think it's discovered, what about computer programs?  At the root, all computing is basically just math.))
--4 \Rightarrow z &\pm 2i \end{array} \right. \\ +
--16 \Rightarrow z^2 &\left\{ \begin{array}{rl} 4i \Rightarrow z &\pm \sqrt{2}(1+i) \\ +
--4i \Rightarrow z &= \pm \sqrt{2}(1-i\end{array} \right. \\ +
-\end{array} \right. +
-\end{equation*} \$$+
  
-=== A Little Philosophy === +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?
-If these weird numbers follow all of the algebra rules without inconsistencies, does it mean the //exist// as much as the real numbers?  Aren't complex numbers a mere //creation// by mathematicians?  Is mathematics //discovered// or //created//?  In a certain way, negative numbers are just as weird as complex numbers: after all, we know what 5 cars look like, but −5 cars?+
  
-===== Plan complexe ===== +===== The Complex Plane ===== 
-Pour représenter un nombre complexe graphiquementPar exemple: \$ z^8 = 256 \$+In the same way that we can represent real numbers by a point on the real number line...((Number line picture from [[wp>Number_line |WikipediaNumber line]])):
  
-FIXME: geogebra+{{ numberline.png }}
  
- Les points noires: \$ z = a ib \$ sont des solutions de l'équation \$ z^n = w \$. On remarque: +... 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:
-  * ... +
-  * ... +
-  * ... +
  
-=== Exercices === +<WRAP center round info 80%> 
-  * Graphiquement, quelles seraient les solutions de \$ z^3 = 8 \$ ? +You can move the point around to look at other complex numbers on the plane. 
-  * Comment pourraient-on calculer ces nombres algébriquement? +</WRAP>
-  * Algébriquement, calculer les racines de \$ z^2 = 2i \$ et vérifier vos réponses avec le graphique ci-dessous.+
  
-FIXME: Geogebra+{{ggb>/howto/hambasics/sections/polar.ggb}}
  
-===== Forme Polaire ===== +To convert between the Cartesian \$(a,b) \$ and the Polar \$ (r \angle \theta) \$ representations, only simple trigonometry and Pythagoras is needed.
-Comme pour les vecteursles nombres complexes peuvent être exprimés en coordonnées cartésiennes ou polaires.+
  
-FIXME: Geogebra +^ \$$ (a, b) \rightarrow (r\angle \theta\$$ ^ \$$ (r\angle \theta\rightarrow (a, b)\$$ |
- +
- +
-Pour convertir d'une représentation à l'autre, on utilise Pythagore et un peu de trigonométrie: +
-^ \$$ a + ib \rightarrow r\angle \theta \$$ ^ \$$ r\angle \theta \rightarrow a+ib \$$ |+
 | \$$ r^2 = a^2 + b^2 \$$  |  \$$ a = r\cos\theta \$$  | | \$$ r^2 = a^2 + b^2 \$$  |  \$$ a = r\cos\theta \$$  |
 | \$$ \tan \theta = \dfrac{b}{a} \$$ | \$$ b = r\sin\theta \$$ | | \$$ \tan \theta = \dfrac{b}{a} \$$ | \$$ b = r\sin\theta \$$ |
  
-=== Exercices === +Note that very often, we use radians instead of degrees for the angle.  There are a total 360° or 2π radians in a circle.  While most people are used to degrees, a radian is actually much easier to picture: 
-  * Convertir \$ 6 \angle \frac{\pi}{3} \$ en coordonnées cartésiennes+  * Imagine a circle
-  * Convertir \$ 8 - 2i \$ en coordonnées polaires+  * Now imagine the length from the centre to the circle (along the "radius"). 
 +  * Take that length and lay it down on the perimeter of the circle. 
 +  * The angle that this length covers is 1 radian (because of the length of the radius on the circle). 
 +  * That's why a circle has 2π radians (because the circumference is 2πr) 
  
 +==== 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 \$.
  
-===== Formule de Euler =====  +{{ggb>/howto/hambasics/sections/complexroots.ggb 850,500}} 
-La formule de Euler expose une relation très profonde entre les fonctions trigonométriques et les fonctions exponentiels ((Note, obtenir "la plus belle équation du monde"on met \$ \theta = \pi \$ dans la formule de Euler...)):+ 
 +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 graphically, you might be surprised by what you find. 
 + 
 +<hidden> 
 +{{ complexroots.png?600 }} 
 +So \$z 2\$ was to be expected since \$ 2^3 8 \$ but it looks like there are two more solutions. \\ \\ 
 + 
 +To find them, we first notice that the three solutions are spread out evenly around the circle, that is they are 120° apart. 
 +So in polar coordinates, the three solutions are \$z = 2\angle 0^\circ, \quad 2\angle 120^\circ,\quad 2\angle 240^\circ \$. \\ \\ 
 + 
 +We can now convert them to Cartesian: 
 +  * \$z = 2\angle 0^\circ = \big(2\cos(0^\circ), 2\sin(0^\circ)\big) = (2, 0) = 2\$ 
 +  * \$z = 2\angle 120^\circ = \big(2\cos(120^\circ), 2\sin(120^\circ)\big) = (-1, \sqrt{3}) = -1 + i\sqrt{3}\$ 
 +  * \$z = 2\angle 240^\circ = \big(2\cos(240^\circ), 2\sin(240^\circ)\big) = (-1, -\sqrt{3}) = -1 - i\sqrt{3}\$ 
 + 
 +Let's check that the second solution works: 
 +\begin{align*} 
 +(-1 + i\sqrt{3})^3 &= (-1 + i\sqrt{3})\cdot(-1 + i\sqrt{3})(-1 + i\sqrt{3}) \\ 
 +&= (-1 + i\sqrt{3})\cdot\big(1 -2i\sqrt{3} +(i^2)(3)\big) \\ 
 +&= (-1 + i\sqrt{3})\cdot\big(1 -2i\sqrt{3} +(-1)(3)\big) \\ 
 +&= (-1 + i\sqrt{3})\cdot(-2 -2i\sqrt{3}) \\ 
 +&= 2 + 2i\sqrt{3} - 2i\sqrt{3} -(2)i^2(3) \\ 
 +&= 2 -(2)(-1)(3) \\ 
 +&= 2 + 6 \\ 
 +&= 8  
 +\end{align*} 
 + 
 +</hidden> 
 +\\ 
 + 
 +===== The Euler Identity ===== 
 +The Euler identity exposes a deep relationship between trigonometric and exponential functions((Note, to obtain the [[wp> Euler's_identity#Mathematical_beauty |most beautiful equation in the world]]set \$ \theta = \pi \$ in the Euler identity: \\ 
 +\$$ 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.)): 
 + 
 +<WRAP center box 30em>
 \$$ e^{i \theta} = \cos \theta + i \sin \theta \$$ \$$ e^{i \theta} = \cos \theta + i \sin \theta \$$
  
-Vérifions cette identité mystérieuse de deux façons.+</WRAP>
  
-=== La dérivée === +Let's use two different ways to verify that this mysterious identity is true.
-Si on sépare cette identité en deux fonctions et qu'on prend la dérivées de ces deux fonctions, on remarque que: +
-\$ \begin{align*} +
-\text{Si} && f(\theta) &= e^{i \theta} &\text{ et }&& g(\theta) &= \cos \theta + i \sin \theta \\ +
-\Rightarrow && f'(\theta) &= i e^{i \theta} &\text{ et }&& g'(\theta) &= -\sin \theta + i \cos \theta \\ +
-\Rightarrow && f'(\theta) &= i \cdot f(\theta) &\text{ et }&& g'(\theta) &= i \cdot g(\theta) +
-\end{align*} \$+
  
-On sait qu'il n'y a qu'une fonction \$ h(x) \$ qui satisfasse l'équation différentiel \$ h'(x) = ah(x) \$, et qu'elle est: \$ h(x) = e^{ax} \$ Ce que Euler à découvert est que quand \$ a = i \$ il y a une deuxième fonction qui satisfasse la même équation différentiel! Ces deux fonctions doivent donc être la même.+==== 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&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*}
  
-=== Taylor === +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.
-Une autre méthode de vérifier la formule d'Euler est d'utiliser les séries de Taylor:+
  
-\$ \begin{align*} +==== Taylor ==== 
-e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \\+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 \\ \sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \\
 \cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots
-\end{align*} \$+\end{align*}
  
-Si on remplace \$ x \$ par \$ i \theta \$ et qu'on additionne les séries de \$ \cos \theta \$ et \$ i \sin \theta \$ on obtiendra la série de \$ e^{i \theta} \$ +  - 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} \$ 
  
-=== Exercices === 
-  * Utiliser la formule de Euler pour obtenir ces deux résultats qui seront très pratique: 
-\$$ \cos \theta = \dfrac{e^{i \theta} + e^{-i \theta}}{2} \qquad \text { et } \qquad \sin \theta = \dfrac{e^{i \theta} - e^{-i \theta}}{2i} \$$ 
  
-  * Utiliser la formule de Euler pour démonter d'un seul coup que+===== 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) \$.  We saw that to convert between the Cartesian \$(a,b) \$ and the Polar \$ (r \angle \theta) \$ representations, only simple trigonometry and Pythagoras is needed
 + 
 +\$$ (a,b) \rightarrow (r\angle \theta) \$$ ^ \$$ (r\angle \theta) \rightarrow (a,b)\$$ | 
 +| \$$ r^2 = a^2 + b^2 \$$  |  \$$ a = r\cos\theta \$$  | 
 +| \$$ \tan \theta = \dfrac{b}{a} \$$ | \$$ b = r\sin\theta \$$ | 
 + 
 +This means that: 
 \begin{align*} \begin{align*}
-\cos (\theta + \phi) &= \cos \theta \cos \phi + \sin \theta \sin \phi \qquad \text{et}\\ +z &= a + ib \\ 
-\sin (\theta + \phi) &\sin \theta \cos \phi + \cos \theta \sin \phi+  &r\cos\theta + i r\sin\theta \\ 
 +  &= r\big\cos\theta + \sin\theta \big) \\ 
 +  &= r e^{i\theta}
 \end{align*} \end{align*}
-\$ 
  
 +This offers another interpretation of the Euler identity as the algebraic conversion between Cartesian and Polar coordinates:
  
-  * Utiliser les deux résultats de la formule de Euler pour démontrer que:+^            Cartesian      ^  Polar  | 
 +^ Graphical |\$$(a, b) \$$    |\$$ (r\angle \theta) \$$  | 
 +^ Algebraic |\$$z = a + ib\$$ |\$$ z = re^{i\theta} \$$ | 
 + 
 +This now allows us to simplify a lot of difficult mathematics.  For example let's look at the root problem \$z^3 = 8 \$ again.  Since the number 8 on the complex plane is the point \$(8,0)\$, in polar coordinates, it can be any of the following: \$8\angle 0, 8\angle 2\pi, 8\angle 4\pi, \cdots \$  This is because we can go around the circle as many times as we want and return to the same point.  Since we expect three roots, let's use the first three polar representations of 8: 
 + 
 +\$$ 
 +z^3 = 8 = \left\{ 
 +\begin{array}{c} 
 +8e^{0 i} \\ 
 +8e^{2\pi i} \\ 
 +8e^{4\pi i}\\ 
 +\vdots  
 +\end{array} 
 +\right. 
 +\$$ 
 + 
 +\$$ 
 +\Rightarrow z = \left\{ 
 +\begin{array}{lcl} 
 +\left(8e^{0 i}\right)^{\frac{1}{3}} &=& 8^{\frac{1}{3}} e^{\frac{0}{3}} = 2\\ 
 +\left(8e^{2\pi i}\right)^{\frac{1}{3}} &=& 8^{\frac{1}{3}} e^{\frac{2\pi}{3}i} = 2 \left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right) = 2 \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = -1 + i\sqrt{3} \\ 
 +\left(8e^{4\pi i}\right)^{\frac{1}{3}} &=& 8^{\frac{1}{3}} e^{\frac{4\pi}{3}i} = 2 \left(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}\right) = 2 \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = -1 - i\sqrt{3} \\ 
 +&\vdots&  
 +\end{array} 
 +\right. 
 +\$$ 
 + 
 +If we had used more than 3 numbers, the roots would have started repeating.  If we had used less than 3, we would have missed some answers in the same way that there are two answers to \$z^2 = 9 \$ (namely \$z = \pm 3 \$). 
 + 
 +===== Cartesian vs Polar ===== 
 +Which is the best representation: Cartesian, \$z = a + i b \$, or Polar, \$z = re^{i\theta}\$.  As you might expect, it depends on what you're trying to do...  For example, let's take: 
 + 
 +\$$z_1 = 1 + i = \sqrt{2}e^\left(i\frac{\pi}{4}\right) \quad \text{and} \quad z_2 = -1 + i = \sqrt{2}e^\left(i\frac{3\pi}{4}\right) \$$ 
 + 
 +Imagine having to add, subtract, multiply, or divide these together.  Or raise them to a power, or take a root of them.  Which of the two representations do you think would be easiest to use for each operation? 
 + 
 +<hidden> 
 +|< 100% - 50% 50% >| 
 +^Operation                Cartesian  ^  Polar  ^ 
 +^Addition        Easiest: \$$ z_1 + z_2 = (1 + i) + (-1 + i) = 2i \$$  |  I don't know how to do it.  \$$ z_1 \pm z_2 = \sqrt{2}e^\left(i\frac{\pi}{4}\right) \pm \sqrt{2}e^\left(i\frac{3\pi}{4}\right) = ??  \$$  | 
 +^Subtraction    |  Easiest: \$$ z_1 - z_2 = (1 + i) - (-1 + i) = 2 \$$   |  :::  | 
 +^Multiplication |  Moderate: \$$ z_1 \cdot z_2 = (1 + i) \cdot (-1 + i) \$$ \$$ = (1)(-1) + (1)(i) + (i)(-1) + (i)(i) \$$  \$$ = -1 + i - i -1 = -2 \$$  |  Easiest:  \$$ z_1 \cdot z_2 = \sqrt{2}e^\left(i\frac{\pi}{4}\right) \cdot \sqrt{2}e^\left(i\frac{3\pi}{4}\right)  \$$  \$$ = \sqrt{2}\cdot \sqrt{2} e^\left(i\frac{\pi}{4} + i\frac{3\pi}{4}\right) \$$  \$$ = 2 e^{i\pi} = -2 \$$  | 
 +^Division |  Tedious: \$$ \frac{z_1}{z_2} = \frac{(1 + i)}{(-1 + i)} = \frac{(1 + i)(-1 - i)}{(-1 + i)(-1 - i)} \$$  \$$ = \cdots = \frac{-2i}{2} = -i  \$$  |  Easiest:  \$$ \frac{z_1}{z_2} = \frac{\sqrt{2}e^\left(i\frac{\pi}{4}\right)}{\sqrt{2}e^\left(i\frac{3\pi}{4}\right)}  = e^\left(i\frac{\pi}{4} - i\frac{3\pi}{4}\right) \$$  \$$ = e^\left(-i\frac{\pi}{2}\right) = -i \$$  | 
 +^Exponentiation |  The bigger the exponent, the more tedious: \$$z_1^{100} = (1 + i)^{100} = \cdots \$$  |  Easy no matter how big the exponent:  \$$z_1^{100} =  \left(\sqrt{2}e^\left(i\frac{\pi}{4}\right)\right)^{100} \$$  \$$ =  2^{50}e^{25\pi i} = 2^{50}e^{12(2\pi i)} e^{\pi i} \$$ \$$ = 2^{50}(1)(-1) = -2^{50}\$$ 
 +^Roots          |  I don't know if it's even possible: \$$ \sqrt[3]{z_2} = \sqrt[3]{-1 + i}\$$ \$$ (-1 + i)^\left(\frac{1}{3}\right) = ??  \$$  |  As easy as exponentiation:  \$$ \sqrt[3]{z_2} = \left(\sqrt{2}e^\left(i\frac{3\pi}{4}\right) \right)^\left(\frac{1}{3} \right) \$$ \$$ = \sqrt[6]{2}e^\left(i\frac{\pi}{4}\right) = \frac{1 + i}{\sqrt[3]{2}}  \$$  | 
 +</hidden> 
 +\\ 
 + 
 +The lesson here is that since the polar representation uses exponents, and exponents turn multiplication into addition((\$ x^a \cdot x^b = x^{a+b} \$)), the polar representation is easiest for multiplication, division, exponentiation, and roots.  It's essentially why the [[howto/hambasics/sections/mathbasics#the_decibel |dB scale]] is so useful.  But addition and subtraction is intrinsically easier in Cartesian coordinates. 
 +===== Important Algebraic Results ===== 
 +  Use the Euler identity to get the following two useful results: 
 + 
 +<WRAP center box 30em> 
 +\$$ \cos \theta = \dfrac{e^{i \theta} + e^{-i \theta}}{2} \qquad \text { & } \qquad \sin \theta = \dfrac{e^{i \theta} - e^{-i \theta}}{2i} \$$ 
 +</WRAP> 
 + 
 +<hidden> 
 +  * Modify the Euler identity to see what happens when the angle is \$ (-\theta) \$ 
 + 
 +\begin{align*} 
 +&& e^{i \theta}  &= \cos \theta + i \sin \theta \\ 
 +\Rightarrow && e^{i (-\theta)} &= \cos (-\theta) + i \sin (-\theta) \\ 
 +\Rightarrow && e^{-i \theta} &= \cos \theta - i \sin \theta 
 +\end{align*} 
 + 
 +  * Add the original Euler identity to the new one for \$ (-\theta) \$ and simplify: 
 +\begin{align*} 
 +&& e^{i \theta}  &= \cos \theta + i \sin \theta \\ 
 ++ && \underline{e^{-i \theta}} &= \underline{\cos \theta - i \sin \theta} \\ 
 +\Rightarrow && e^{i \theta} + e^{-i \theta} &= 2\cos \theta \\ 
 +\Rightarrow && \cos \theta &= \frac{e^{i \theta} + e^{-i \theta}}{2} \\ 
 +\end{align*} 
 + 
 +  * To get the second equation, take the original Euler identity and subtract the new one for \$ (-\theta) \$ from it and simplify: 
 +\begin{align*} 
 +&& e^{i \theta}  &= \cos \theta + i \sin \theta \\ 
 +- && \underline{e^{-i \theta}} &= \underline{\cos \theta - i \sin \theta} \\ 
 +\Rightarrow && e^{i \theta} - e^{-i \theta} &= 2i\sin \theta \\ 
 +\Rightarrow && \sin \theta &= \frac{e^{i \theta} - e^{-i \theta}}{2i} \\ 
 +\end{align*} 
 +</hidden> 
 +\\ 
 + 
 +  * Use the Euler identity to show that: 
 + 
 +<WRAP center box 30em> 
 +\$$ 
 +\cos (\theta + \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi\\ 
 +\sin (\theta + \phi) = \cos \theta \sin \phi + \sin \theta \cos \phi 
 +\$$ 
 +</WRAP> 
 + 
 +<hidden> 
 +  * Multiply the Euler identities for \$\theta\$ and \$\phi\$ and simplify. 
 + 
 +\begin{align*} 
 +&& e^{i \theta}  &= \cos \theta + i \sin \theta \\ 
 +\times && \underline{e^{i \phi}}  &= \underline{\cos \phi + i \sin \phi} \\ 
 +\Rightarrow && e^{i \theta} e^{i \phi}  &= \big(\cos \theta + i \sin \theta\big)\cdot \big(\cos \phi + i \sin \phi \big) \\ 
 +\Rightarrow && e^{i\theta + i\phi}  &= (\cos \theta) (\cos \phi) + (\cos \theta) (i\sin \phi) + (i \sin \theta)(\cos \phi) +  (i \sin \theta)(i \sin \phi)\\ 
 +\Rightarrow && e^{i (\theta + \phi)}  &= \cos \theta \cos \phi + i\cos \theta \sin \phi + i \sin \theta \cos \phi +  i^2 \sin \theta \sin \phi\\ 
 +\Rightarrow && e^{i (\theta + \phi)}  &= \cos \theta \cos \phi + i (\cos \theta \sin \phi + \sin \theta \cos \phi) - \sin \theta \sin \phi\\ 
 +\Rightarrow && e^{i (\theta + \phi)}  &= \color{green}{(\cos \theta \cos \phi - \sin \theta \sin \phi)} + i \color{blue}{(\cos \theta \sin \phi + \sin \theta \cos \phi)} 
 +\end{align*} 
 + 
 +But the Euler identity for angle \$ (\theta + \phi) \$ is: 
 +\$$ 
 +e^{i (\theta + \phi)} = \color{green}{\cos(\theta + \phi)} + i\color{blue}{\sin(\theta + \phi)} 
 +\$$ 
 + 
 + 
 +So comparing the <fc #008017>real</fc> and <fc #0000FF>imaginary</fc> parts, we can conclude that: 
 +\$$ 
 +\color{green}{\cos(\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi} \\ 
 +\color{blue}{\sin(\theta + \phi) = \cos \theta \sin \phi + \sin \theta \cos \phi} 
 +\$$ 
 +</hidden> 
 +\\ 
 + 
 + 
 +  * Use the two earlier results to show that: 
 + 
 +<WRAP center box 30em>
 \$$ \sin (\theta + \Delta \theta) + \sin (\theta - \Delta \theta) = 2 \cos \Delta \theta \sin \theta \$$ \$$ \sin (\theta + \Delta \theta) + \sin (\theta - \Delta \theta) = 2 \cos \Delta \theta \sin \theta \$$
-(ce qui sera très important quand nous ferons de l'interférence d'onde...)+</WRAP>
  
 +<hidden>
 +  * Use the last result for \$ \sin(\theta + \phi) \$ but replace \$ \phi \$ with \$ \Delta \theta \$
  
-===== Équations différentiels =====+\$$ 
 +\sin (\theta + \Delta \theta)  \cos \theta \sin \Delta \theta + \sin \theta \cos \Delta \theta 
 +\$$
  
-Dans la physique des ondeson aura bientôt à trouver la solutions d'une équation différentiel de ce type: +Similarily
-\$$ a \ddot{x}(t+ b \dot{x}(t) + c x(t) = \$$ +\begin{align*} 
 +&& \sin (\theta - \Delta \theta&=  \cos \theta \sin(-\Delta \theta) + \sin \theta \cos (-\Delta \theta\\ 
 +\Rightarrow && \sin (\theta - \Delta \theta) & -\cos \theta \sin\Delta \theta + \sin \theta \cos \Delta \theta \\ 
 +\end{align*}
  
-Dans nos applicationsles 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:+Adding both of these togetherwe get: 
 +\begin{align*} 
 +\sin (\theta + \Delta \theta) + \sin (\theta - \Delta \theta& \cos \theta \sin \Delta \theta + \sin \theta \cos \Delta \theta  \\ 
 +    & \quad - \cos \theta \sin\Delta \theta + \sin \theta \cos \Delta \theta \\ 
 +&= 2\sin \theta \cos \Delta \theta  \\ 
 +&= 2\cos \Delta \theta \sin \theta  \\ 
 +\end{align*} 
 +</hidden> 
 +\\
  
 +This last result is the basis behind why [[howto/hambasics/sections/wavemodulationmath |modulating the amplitude]] of a carrier produces side bands.
  
-\$ \begin{align*}+ 
 +===== Differential Equations ===== 
 + 
 +In the physics of wave, we often have to find solutions to the following type of differential equations: 
 +<WRAP center box 30em> 
 +\$$ a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \$$ 
 +</WRAP> 
 + 
 +<WRAP center round info 80%> 
 +  * A differential equation is an equation that relates a function to its derivatives in some ways and the question is: given some information about the system, what's the function (or family of functions) that satisfy the differential equation. 
 + 
 +  * In physics we often use a dot above the function to indicate a derivative with respect to time, where as in math, we'll often use an apostrophe.  Physicists don't like the apostrophe too much because they sometimes use it to denote a different coordinate system.  So don't let the notation confuse you: 
 +\$$ \dot{x}(t) = x'(t) = \frac{dx}{dt} \quad \text{and} \quad \ddot{x}(t) = x''(t) = \frac{d^2x}{dt^2} \$$ 
 +</WRAP> 
 + 
 + 
 +In our applications, the parameters  \$ a, b, c \$ are all real and positive quantities.  Even without having studied different equations in any depth, we can imagine that a possible solution to the above differential equation would be: \$ x(t)= e^{rt} \$  since the derivative of an exponential function is itself an exponential function, which is encouraging. 
 + 
 +The next step is to try this ``test function'' in the differential equation and see if we can find the values of \$ r \$ that make it work.  First we'll need derivatives of the test function: 
 + 
 +\begin{align*}
 & x (t) = e^{rt} \\  & x (t) = e^{rt} \\ 
 \Rightarrow \qquad & \dot{x}(t) = r e^{rt} \\  \Rightarrow \qquad & \dot{x}(t) = r e^{rt} \\ 
 \Rightarrow \qquad & \ddot{x}(t) = r^2 e^{rt}  \Rightarrow \qquad & \ddot{x}(t) = r^2 e^{rt} 
-\end{align*} \$+\end{align*}
  
- et notre équation différentiel devient+When we put these into the differential equation, we get:
  
-\$ \begin{align*}+\begin{align*}
 & a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \\  & 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 & 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 & e^{rt} (a r^2 + b r + c ) = 0 \\ 
-\Rightarrow \qquad & 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} 
 +\end{align*}
  
-\$ \Rightarrow \qquad r = - \dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} \$+So what does that result mean?  Remember, what we're looking for is the function \$x(t)\$ that satisfies the differential equation \$ a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \$
  
-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.+What we've go so far says that our test function \$x(t) = e^{rt}\$ will satisfy the differential equation if \$r\$ is given by above equation.  There is still a lot to unpack however.  For example, since \$r\$ contains a square root, it could be real or complex depending on the values of \$a, b,\$ and \$c\$.  And as we saw above, if \$r\$ is real, then \$x(t)\$ will be a real exponential function.  But if \$r\$ is complex, then we can expect \$x(t)\$ to be some sort of sinusoidal function (recall the Euler Identity).
  
-=== Cas 1=== +To simplify the notation, let's define \$\alpha\$ and \$\beta\$ as
-Quand \$ b^2 - 4ac > 0 \$ , \r \$ est réel et la solutions général sera: +\$$ \alpha = \dfrac{b}{2a} \qquad \text{and} \qquad \beta = \dfrac{\sqrt{|{b^2 - 4ac}|}}{2a} \$$
  
-\$ \begin{align*}+Notice how the absolute value under the square root ensures that \$\beta\$ is always real. 
 + 
 +\$r\$ then:  
 + 
 +\$$ r = \left\{ \begin{array}{ll} -\alpha \pm \beta & \text{if } b^2 - 4ac > 0,\\  
 + -\alpha \pm i \beta & \text{if } b^2 - 4ac < 0, \end{array} \right. \$$ 
 + 
 +Let's examine both of these cases in more detail. 
 + 
 +==== Case 1: Over Damped Oscillation ==== 
 +When \$ b^2 - 4ac > 0 \$ , \$ r \$ is real and the general solution is:  
 + 
 +\begin{align*}
 x(t) &= A_1 e^{r_1 t} + A_2 e^{r_2 t} \\  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 + \beta) t} + A_2 e^{( -\alpha - \beta) t} \\ 
 &= A_1 e^{-\alpha t} e^{\beta t} + A_2 e^{-\alpha t} e^{-\beta t}  &= A_1 e^{-\alpha t} e^{\beta t} + A_2 e^{-\alpha t} e^{-\beta t} 
-\end{align*} \$+\end{align*}
  
 \$$ x(t) = e^{-\alpha t} ( A_1 e^{\beta t} + A_2 e^{-\beta t} ) \$$ \$$ 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 une dérivé du second degréPour trouver ces constantesça nous prendrait des valeurs initiales.+It'normal to have two constants of integration since our differential equation has a second degree derivative in it To find these constantswe'd need to know more about the system's initial conditions.
  
-=== Cas 2: === +==== Case 2: Under Damped ==== 
-Quand \$ b^2 - 4ac < 0 \$ , \$ r \$ est complexe et nous utiliserons la formule de Euler pour simplifier notre solutions.+When \$ b^2 - 4ac < 0 \$ , \$ r \$ is complex and we'll be using the Euler identity to simplify our solutions
    
-\$ \begin{align*}+\begin{align*}
 x(t) &= A_1 e^{r_1 t} + A_2 e^{r_2 t} \\  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 + i \beta) t} + A_2 e^{( -\alpha - i \beta) t} \\ 
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 &= 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 + 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_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)  +&= e^{-\alpha t} \Big( A \sin \phi \cos(\beta t) + A \cos \phi \sin(\beta t) \Big)  \\ 
-\end{align*} \$+&= Ae^{-\alpha t} \Big(\sin \phi \cos(\beta t) + \cos \phi \sin(\beta t) \Big)  \\ 
 +\end{align*}
  
-Dans les deux dernière lignenous avons ré-écrit les constantes d'intégration:+In the last three lineswe've redefined the constants of integration a few times so that:
  
-\$ \begin{align*}+\begin{align*}
 a_1 &= A_1 + A_2 & , a_2 &= i(A_1 - A_2) \\  a_1 &= A_1 + A_2 & , a_2 &= i(A_1 - A_2) \\ 
 a_1 &= A \sin \phi & , a_2 &= A \cos \phi  a_1 &= A \sin \phi & , a_2 &= A \cos \phi 
-\end{align*} \$ +\end{align*}
- +
-Pour finalement pouvoir utiliser une identité trigonométrique:+
  
 +And we finally use one of the trig identities we proved earlier to write the solution as:
 \$$ x(t) = A e^{-\alpha t} \sin(\beta t + \phi) \$$ \$$ x(t) = A e^{-\alpha t} \sin(\beta t + \phi) \$$
  
-=== Exemple ===+==== Case 3: Critically Damped ==== 
 +When \$ b^2 - 4ac = 0 \$ , \$ r = -\frac{b}{2a} \$ is real and negative but our test solution is under determined.  We'll instead propose a solution of the following type and test that it works: 
 +\begin{align*} 
 + && x(t) &= e^{rt}(A + Bt) \\ 
 +\Rightarrow && \dot x(t) &= re^{rt}(A + Bt) + Be^{rt} \\ 
 +            &&           &= e^{rt}\big(r(A + Bt) + B)\big) \\ 
 +            &&           &= e^{rt}(rA + B + Brt) \\ 
 +\Rightarrow &&\ddot x(t) &= re^{rt}(rA + B + Brt) + e^{rt}Br \\ 
 +            &&           &= e^{rt}\big(r(rA + B + Brt) + Br\big) \\ 
 +            &&           &= e^{rt}(Ar^2 + 2Br + Br^2t) \\ 
 +\end{align*} 
 + 
 + 
 + 
 + 
 +==== 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: 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:
  
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howto/hambasics/sections/test.1609637529.txt.gz · Last modified: 2021/01/02 17:32 by va7fi