KindiakMath

Exponential Fourier Series

The function f is defined by f(t) = e^t for -1 < t \leq 1 and extended periodically by f(t) = f(t+2).

Problem 1. For any nonnegative integer n, evaluate the integrals

\displaystyle \int f(t) \cos (n \pi t)\, \mathrm dt,\quad \int f(t) \sin(n \pi t)\, \mathrm dt.

(Click for Solution)

Solution. Integrating by parts twice,

\begin{aligned} \int e^{t} \cos (n \pi t)\, \mathrm dt &= \underbrace{e^t}_{\mathrm I} \cdot \underbrace{\cos (n \pi t)}_{\mathrm S} - \int \underbrace{e^t}_{\mathrm I} \cdot \underbrace{(-n\pi \cdot \sin(n \pi t))}_{\mathrm D}\, \mathrm dt \\ &= e^t \cos(n\pi t) + n \pi \int e^t \sin(n \pi t)\, \mathrm dt \\ &= e^t \cos(n \pi t) + n\pi \left( \underbrace{e^t}_{\mathrm I} \cdot \underbrace{\sin (n \pi t)}_{\mathrm S} - \int \underbrace{e^t}_{\mathrm I} \cdot \underbrace{(n\pi \cdot \cos(n \pi t))}_{\mathrm D}\, \mathrm dt \right) \\ &= e^t ( \cos(n \pi t) + n\pi \sin(n \pi t)) - n^2 \pi^2 \int e^t \cos(n\pi t)\, \mathrm dt. \end{aligned}

By careful algebruh,

\displaystyle \int e^{t} \cos (n \pi t)\, \mathrm dt = \frac{e^t ( \cos(n \pi t) + n\pi \sin(n \pi t))}{1 + n^2 \pi^2} + C_1.

In a similar though tedious manner,

\displaystyle \int e^{t} \sin (n \pi t)\, \mathrm dt = \frac{e^t ( \sin(n \pi t) - n\pi \cos(n \pi t))}{1 + n^2 \pi^2} + C_2.

Remark 1. For an alternate solution, recall Euler’s formula which states that

e^{i n \pi t} = \cos(n \pi t) + i \sin(n \pi t).

Multiplying by f(t) then integrating on the left-hand side yields

\begin{aligned} \int f(t) e^{i n \pi t} \, \mathrm dt &= \int e^t \cdot e^{i n \pi t}\, \mathrm dt \\ &= \int e^{(1 + i n \pi )t}\, \mathrm dt \\ &= \frac {e^{(1 + i n \pi )t}}{1 + i n \pi} + C \\ &= e^t \cdot \frac{1 - i n \pi}{1 + n^2 \pi^2} (\cos (n \pi t) + i \sin(n \pi t)) + C \\ &= e^t \cdot \frac{(\cos (n \pi t) + n \pi \sin(n \pi t)) + i (\sin(n \pi t) - n \pi \cos(n \pi t))}{1 + n^2 \pi^2} + C \\ &= e^t \cdot \frac{\cos (n \pi t) + n \pi \sin(n \pi t)}{1 + n^2 \pi^2} + i \cdot e^t \cdot \frac{\sin(n \pi t) - n \pi \cos(n \pi t)}{1 + n^2 \pi^2} + C . \end{aligned}

The real and imaginary parts then agree with our solutions in Problem 1.

Problem 2. Denoting T = 2, the Fourier series of f is defined by

\displaystyle f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos\left( \frac{2n\pi t}{T} \right) + b_n \sin\left( \frac{2n\pi t}{T} \right) \right),

where the Fourier coefficients a_n, b_n are defined by

\displaystyle a_n = \frac 2{T} \int_{-T/2}^{T/2} f(t) \cos\left( \frac{2n\pi t}{T} \right)\, \mathrm dt,\quad b_n = \frac 2{T} \int_{-T/2}^{T/2} f(t) \sin\left( \frac{2n\pi t}{T} \right)\, \mathrm dt.

Use Problem 1 to evaluate the Fourier series of f.

(Click for Solution)

Solution. By Problem 1,

\begin{aligned} a_n &= \frac 2{2} \int_{-2/2}^{2/2} f(t) \cos \left( \frac{2n\pi t}{2} \right)\, \mathrm dt \\ &= \int_{-1}^1 e^t \cos (n \pi t)\, \mathrm dt \\ &= \left[ \frac{e^t ( \cos(n \pi t) + n\pi \sin(n \pi t))}{1 + n^2 \pi^2} \right]_{-1}^1 \\ &= \frac{(-1)^n ( e - e^{-1} )}{1 + n^2 \pi^2} . \end{aligned}

Similarly,

\begin{aligned} b_n &= \frac 2{2} \int_{-2/2}^{2/2} f(t) \sin \left( \frac{2n\pi t}{2} \right)\, \mathrm dt \\ &= \int_{-1}^1 e^t \sin (n \pi t)\, \mathrm dt \\ &= \left[ \frac{e^t ( \sin(n \pi t) - n\pi \cos(n \pi t))}{1 + n^2 \pi^2} \right]_{-1}^1 \\ &= -n \pi \cdot \frac {(-1)^n (e - e^{-1} )}{1 + n^2 \pi^2} = -n\pi a_n. \end{aligned}

By the definition of the Fourier series,

\displaystyle \begin{aligned} f(t) &= \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos\left( \frac{2n\pi t}{T} \right) + b_n \sin\left( \frac{2n\pi t}{T} \right) \right) \\ &= \frac{a_0}{2} + \sum_{n=1}^\infty a_n (( \cos (n \pi t) - \pi n \sin (n \pi t) ) \\ &= \frac{e - e^{-1}}{2} + \sum_{n=1}^\infty \frac{(-1)^n ( e - e^{-1} )}{1 + n^2 \pi^2} (( \cos (n \pi t) - \pi n \sin (n \pi t) ). \end{aligned}

Remark 2. If instead we used the complex-exponential form of the Fourier series

\displaystyle f(t) = \sum_{n = -\infty}^\infty c_n e^{i \cdot 2\pi n t/T},\quad c_n := \frac 1{T} \int_{-T/2}^{T/2} f(t) \cdot e^{i \cdot 2\pi n t/T}\, \mathrm dt,

then complex-valued integration yields

\displaystyle c_n = \frac 12 \left[ \frac{e^{(1 + i \cdot \pi n) t}}{1 + i \cdot \pi n} \right]_{-1}^1 = \frac {(-1)^n (e - e^{-1})(1 - (\pi n) \cdot i)}{2(1 + n^2 \pi^2)},

recovering the results in Problem 2 via the relationships

a_n = c_n + \bar{c}_n,\quad b_n = -i \cdot (c_n - \bar{c}_n) =-n\pi a_n.

—Joel Kindiak, 14 May 25, 1911H

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