Question

In calculus you learned that \(e^{u}, \cos u,\) and \(\sin u\) can be represented by the infinite series $$ \begin{array}{c} e^{u}=\sum_{n=0}^{\infty} \frac{u^{n}}{n !}=1+\frac{u}{1 !}+\frac{u^{2}}{2 !}+\frac{u^{3}}{3 !}+\cdots+\frac{u^{n}}{n !}+\cdots \\ \cos u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n}}{(2 n) !}=1-\frac{u^{2}}{2 !}+\frac{u^{4}}{4 !}+\cdots+(-1)^{n} \frac{u^{2 n}}{(2 n) !}+\cdots, \end{array} $$ and $$ \sin u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}=u-\frac{u^{3}}{3 !}+\frac{u^{5}}{5 !}+\cdots+(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}+\cdots $$ for all real values of \(u\). Even though you have previously considered (A) only for real values of \(u\), we can set \(u=i \theta,\) where \(\theta\) is real, to obtain $$ e^{i \theta}=\sum_{n=0}^{\infty} \frac{(i \theta)^{n}}{n !} $$ Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in (D) converges for all real \(\theta\). (a) Recalling that \(i^{2}=-1,\) write enough terms of the sequence \(\left\\{i^{n}\right\\}\) to convince yourself that the sequence is repetitive: $$ 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, \cdots $$ Use this to group the terms in (D) as $$ \begin{aligned} e^{i \theta} &=\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4}+\cdots\right)+i\left(\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}+\cdots\right) \\ &=\sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n}}{(2 n) !}+i \sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n+1}}{(2 n+1) !} \end{aligned} $$ In calculus you learned that \(e^{u}, \cos u,\) and \(\sin u\) can be represented by the infinite series $$ \begin{array}{c} e^{u}=\sum_{n=0}^{\infty} \frac{u^{n}}{n !}=1+\frac{u}{1 !}+\frac{u^{2}}{2 !}+\frac{u^{3}}{3 !}+\cdots+\frac{u^{n}}{n !}+\cdots \\ \cos u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n}}{(2 n) !}=1-\frac{u^{2}}{2 !}+\frac{u^{4}}{4 !}+\cdots+(-1)^{n} \frac{u^{2 n}}{(2 n) !}+\cdots, \end{array} $$ and $$ \sin u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}=u-\frac{u^{3}}{3 !}+\frac{u^{5}}{5 !}+\cdots+(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}+\cdots $$ for all real values of \(u\). Even though you have previously considered (A) only for real values of \(u\), we can set \(u=i \theta,\) where \(\theta\) is real, to obtain $$ e^{i \theta}=\sum_{n=0}^{\infty} \frac{(i \theta)^{n}}{n !} $$ Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in (D) converges for all real \(\theta\). (a) Recalling that \(i^{2}=-1,\) write enough terms of the sequence \(\left\\{i^{n}\right\\}\) to convince yourself that the sequence is repetitive: $$ 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, \cdots $$ Use this to group the terms in (D) as $$ \begin{aligned} e^{i \theta} &=\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4}+\cdots\right)+i\left(\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}+\cdots\right) \\ &=\sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n}}{(2 n) !}+i \sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n+1}}{(2 n+1) !} \end{aligned} $$ (c) If \(\alpha\) and \(\beta\) are real numbers, define $$ e^{\alpha+i \beta}=e^{\alpha} e^{i \beta}=e^{\alpha}(\cos \beta+i \sin \beta) $$ Show that if \(z_{1}=\alpha_{1}+i \beta_{1}\) and \(z_{2}=\alpha_{2}+i \beta_{2}\) then $$ e^{z_{1}+z_{2}}=e^{z_{1}} e^{z_{2}} $$ (d) Let \(a, b,\) and \(c\) be real numbers, with \(a \neq 0 .\) Let \(z=u+i v\) where \(u\) and \(v\) are real-valued functions of \(x .\) Then we say that \(z\) is a solution of $$ a y^{\prime \prime}+b y^{\prime}+c y=0 $$ if \(u\) and \(v\) are both solutions of (G). Use Theorem \(5.2 .1(\mathbf{c})\) to verify that if the characteristic equation of (G) has complex conjugate roots \(\lambda \pm i \omega\) then \(z_{1}=e^{(\lambda+i \omega) x}\) and \(z_{2}=e^{(\lambda-i \omega) x}\) are both solutions of (G).

Short answer

Question: Prove that \(e^{z_1+z_2} = e^{z_1} e^{z_2}\), where \(z_1\) and \(z_2\) are complex numbers. Answer: Given two complex numbers \(z_1 = \alpha_1 + i\beta_1\) and \(z_2 = \alpha_2 + i\beta_2\), we will prove that \(e^{z_1+z_2} = e^{z_1} e^{z_2}\). We have: \(e^{z_1+z_2} = e^{(\alpha_1+\alpha_2) + i(\beta_1+\beta_2)} = e^{\alpha_1+\alpha_2}(\cos (\beta_1+\beta_2) +i\sin (\beta_1+\beta_2)) = e^{\alpha_1} e^{\alpha_2} (\cos \beta_1\cos \beta_2-\sin\beta_1\sin\beta_2 + i(\cos\beta_1\sin\beta_2+\sin\beta_1\cos\beta_2)) = (e^{\alpha_1}(\cos\beta_1 + i\sin\beta_1))(e^{\alpha_2}(\cos\beta_2+ i\sin\beta_2)) = e^{z_1} e^{z_2}\)

Step-by-step solution

(a) Sequence of \(i^n\) and Grouping Terms

First, we write down the sequence of \(i^n\) for a few values of n: $$ 1, i, -1, -i, 1, i, -1, -i, 1, i, -1, -i, 1, i, -1, -i, \cdots $$ We observe that the given sequence repeats after every four terms. Now we will write the expression for \(e^{i\theta}\) and group the terms: $$ \begin{aligned} e^{i \theta} &=\sum_{n=0}^{\infty} \frac{(i \theta)^{n}}{n !} \\\ &=1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} +\cdots \\\ &=\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4}+\cdots\right)+i\left(\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}+\cdots\right) \\\ &=\sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n}}{(2 n) !}+i \sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n+1}}{(2 n+1) !} \end{aligned} $$

(b) Define exponential of complex numbers

We define exponential function of complex numbers in the form of \(e^{\alpha + i\beta}\) as: $$ e^{\alpha+i \beta}=e^{\alpha} e^{i \beta}=e^{\alpha}(\cos \beta+i \sin \beta) $$

(c) Prove the Exponential Property

We have two complex numbers \(z_1 = \alpha_1 + i\beta_1\) and \(z_2 = \alpha_2 + i\beta_2\). We need to prove that \(e^{z_1+z_2} = e^{z_1} e^{z_2}\): $$ \begin{aligned} e^{z_1+z_2} &= e^{(\alpha_1+\alpha_2) + i(\beta_1+\beta_2)} \\ &= e^{\alpha_1+\alpha_2}(\cos (\beta_1+\beta_2) +i\sin (\beta_1+\beta_2)) \\ &= e^{\alpha_1} e^{\alpha_2} (\cos \beta_1\cos \beta_2-\sin\beta_1\sin\beta_2 + i(\cos\beta_1\sin\beta_2+\sin\beta_1\cos\beta_2)) \\ &= (e^{\alpha_1}(\cos\beta_1 + i\sin\beta_1))(e^{\alpha_2}(\cos\beta_2+ i\sin\beta_2)) \\ &= e^{z_1} e^{z_2} \end{aligned} $$

(d) Verify solutions of 2nd order ODE

Given a 2nd order homogeneous differential equation: $$ a y^{\prime \prime}+b y^{\prime}+c y=0 $$ We will verify if \(z_1=e^{(\lambda+i \omega) x}\) and \(z_2=e^{(\lambda-i \omega) x}\) are solutions of the ODE when the characteristic equation has complex conjugate roots \(\lambda \pm i \omega\). From Theorem \(5.2 .1(\mathbf{c})\), given \(u+v\) and \(u-v\) as solutions of the ODE, we have the following equation for the derivative of the product of the functions: $$ (u+v)'(u-v)' = u'^2 - v'^2 $$ For \(z_1\) and \(z_2\), we have: $$ \begin{aligned} z_1' &= (\lambda + i\omega)e^{(\lambda+i\omega)x} \\ z_2' &= (\lambda - i\omega)e^{(\lambda-i\omega)x} \end{aligned} $$ We will compute the products \((z_1'+z_2')(z_1'-z_2')\) and \(z_1'^2 - z_2'^2\): $$ \begin{aligned} (u+v)'(u-v)' &= (\lambda^2 + \omega^2)e^{2\lambda x} \\ u'^2 - v'^2 &= (\lambda^2 + \omega^2)e^{2\lambda x} \end{aligned} $$ Since we have \((z_1'+z_2')(z_1'-z_2') = z_1'^2 - z_2'^2\), \(z_1=e^{(\lambda+i\omega)x}\) and \(z_2=e^{(\lambda-i\omega)x}\) are both verified solutions of the given 2nd order homogeneous differential equation with complex conjugate roots.

Key concepts

Infinite Series

The concept of an infinite series is central to many areas of mathematics, including the field of differential equations. An infinite series is essentially the sum of the terms of an infinite sequence. Common functions can be expressed as infinite series expansions, providing a powerful tool for both analysis and computation.

For example, the exponential function, sine, and cosine, fundamental in solving differential equations, have their own series representations. These representations are particularly useful when dealing with complex numbers, as they allow for the manipulation of terms in a structured manner.

In the context of differential equations, infinite series often arise when seeking solutions to equations that cannot be solved explicitly. They can be used to approximate solutions or even to express exact solutions in series form, under proper mathematical conditions. It is crucial, therefore, to understand the convergence of these series, since not all series will converge to a finite value for every input. The concept of convergence is closely tied to the validity and application of series solutions in various problems, including those involving differential equations.

Complex Numbers

When delving into complex numbers, we enter a realm that broadens the scope of traditional real-number calculus. A complex number is composed of a real part and an imaginary part, typically denoted as \( a + bi \), where \( i \) is the imaginary unit, satisfying \( i^2 = -1 \).

Imaginary and complex numbers are not just mathematical abstractions; they are essential in various scientific fields, including engineering, physics, and signal processing. In the problem at hand, the use of complex numbers allows for the extension of the exponential function to complex arguments, leading to Euler's formula, which links exponential functions to trigonometric functions through the relation \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \).

The manipulation of complex numbers is rooted in their algebraic properties, which in turn makes it possible to solve a wider range of problems, such as the behavior of alternating currents in electrical engineering or the analysis of oscillations in mechanical systems. Mastering complex numbers and their properties, including the exponential form, is thus indispensable for understanding and solving a broad spectrum of problems in science and engineering.

Second-Order Linear Homogeneous Differential Equations

The second-order linear homogeneous differential equations are a cornerstone in the study of differential equations. These equations are of the form \( ay'' + by' + cy = 0 \) where \( y \) is the unknown function of the variable \( x \), and \( a \), \( b \), and \( c \) are constant coefficients. The term 'homogeneous' means that there is no independent function of \( x \) on the right-hand side of the equation.

These equations are particularly important in physics and engineering, describing systems ranging from simple harmonic oscillators to electrical circuits. The general solution to these equations is a combination of exponential functions, which, in the case of complex roots to the characteristic equation, involve complex numbers. These solutions directly result in oscillatory motion, synchronizing beautifully with the concepts of infinite series and complex numbers.

The understanding of how complex roots translate to oscillatory solutions via the exponential function is crucial. Moreover, these principles illustrate the deep interconnectivity between various mathematical concepts and their practical applications, providing a powerful toolkit for tackling a wide range of theoretical and practical problems. Recognizing and applying the connection between the solutions of differential equations and exponential functions involving complex numbers is fundamental to solving second-order linear homogeneous differential equations.