### Fourier Representation and Exponential Form#### a) Fourier Series RepresentationAfter writing the Fourier representation:\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos \frac{n \pi x}{3} + b_n \sin \frac{n \pi x}{3} \right) \]where\[a_n = \frac{1}{3} \int_{-3}^{3} f(x) \cos \frac{n \pi x}{3} \, dx, \quad n = 0, 1, 2, \ldots \]\[b_n = \frac{1}{3} \int_{-3}^{3} f(x) \sin \frac{n \pi x}{3} \, dx, \quad n = 1, 2, \ldots \]Using the exponential forms:\[ \cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} \]of the cosine and sine functions to express the representation in exponential form:\[f(x) = \sum_{-\infty}^{\infty} A_n \exp \left( i \frac{n \pi x}{3} \right)\]where\[A_0 = \frac{a_0}{2}, \quad A_n = \frac{a_n - ib_n}{2}, \quad A_{-n} = \frac{a_n + ib_n}{2} \quad (n = 1, 2, \ldots)\]#### b) Single Formula for $A_n$Then, use the expression of $a_n$ and $b_n$ to obtain a single formula:\[A_n = \frac{1}{6} \int_{-3}^{3} f(x) \exp \left( -i \frac{n \pi x}{3} \right) \, dx \quad (n = 0, \pm1, \pm2, \ldots)\]

Answered: a f(z) = % + Em (an cos nga + bn sin…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

### Fourier Representation and Exponential Form

#### a) Fourier Series Representation

After writing the Fourier representation:

\[
f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos \frac{n \pi x}{3} + b_n \sin \frac{n \pi x}{3} \right)
\]

where

\[
a_n = \frac{1}{3} \int_{-3}^{3} f(x) \cos \frac{n \pi x}{3} \, dx, \quad n = 0, 1, 2, \ldots
\]

\[
b_n = \frac{1}{3} \int_{-3}^{3} f(x) \sin \frac{n \pi x}{3} \, dx, \quad n = 1, 2, \ldots
\]

Using the exponential forms:

\[
\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}
\]

of the cosine and sine functions to express the representation in exponential form:

\[
f(x) = \sum_{-\infty}^{\infty} A_n \exp \left( i \frac{n \pi x}{3} \right)
\]

where

\[
A_0 = \frac{a_0}{2}, \quad A_n = \frac{a_n - ib_n}{2}, \quad A_{-n} = \frac{a_n + ib_n}{2} \quad (n = 1, 2, \ldots)
\]

#### b) Single Formula for $A_n$

Then, use the expression of $a_n$ and $b_n$ to obtain a single formula:

\[
A_n = \frac{1}{6} \int_{-3}^{3} f(x) \exp \left( -i \frac{n \pi x}{3} \right) \, dx \quad (n = 0, \pm1, \pm2, \ldots)
\]

Expert Solution

Step 1

For (a),

We are given the following Fourier representation.

$f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} \cos (\frac{n πx}{3}) + b_{n} \sin (\frac{n πx}{3}))$ ------------(1)

with

$a_{n} = \frac{1}{3} \int_{- 3}^{3} f (x) \cos (\frac{n πx}{3}) d x, n = 0, 1, 2, . . . b_{n} = \frac{1}{3} \int_{- 3}^{3} f (x) \sin (\frac{n πx}{3}) d x, n = 1, 2, . . .$ . ------------(2)

Now, we will us the exponential forms $\cos θ = \frac{e^{i θ} + e^{- i θ}}{2}, \sin θ = \frac{e^{i θ} - e^{- i θ}}{2 i}$ of the cosine and sine functions to put that representation in exponential form:

$f (x) = \sum_{n = - \infty}^{\infty} A_{n} e x p (i \frac{n πx}{3})$ -------------(3)

where

$A_{0} = \frac{a_{0}}{2}, A_{n} = \frac{a_{n} - i b_{n}}{2}, A_{- n} = \frac{a_{n} + i b_{n}}{2} (n = 1, 2, . . .)$ ---------(4)

From equation (1),

$f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} (\frac{\exp (i \frac{nπx}{3}) + \exp (- i \frac{nπx}{3})}{2}) + b_{n} (\frac{\exp (i \frac{nπx}{3}) - \exp (- i \frac{nπx}{3})}{2 i})) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (\exp (i \frac{nπx}{3}) (\frac{a_{n}}{2} + \frac{b_{n}}{2 i}) + \exp (- i \frac{nπx}{3}) (\frac{a_{n}}{2} - \frac{b_{n}}{2 i})) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (\exp (i \frac{nπx}{3}) (\frac{a_{n}}{2} - \frac{i b_{n}}{2}) + \exp (- i \frac{nπx}{3}) (\frac{a_{n}}{2} + \frac{i b_{n}}{2})) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (\exp (i \frac{nπx}{3}) (\frac{a_{n} - i b_{n}}{2}) + \exp (- i \frac{nπx}{3}) (\frac{a_{n} + i b_{n}}{2}))$

Let us use equation (4),

$= A_{0} + \sum_{n = 1}^{\infty} (\exp (i \frac{nπx}{3}) (A_{n}) + \exp (- i \frac{nπx}{3}) (A_{- n})) = A_{0} + \sum_{n = 1}^{\infty} (A_{n} \exp (i \frac{nπx}{3}) + A_{- n} \exp (- i \frac{nπx}{3})) = A_{0} + \sum_{n = 1}^{\infty} (A_{n} \exp (i \frac{nπx}{3}) + A_{- n} \exp (i \frac{(- n) πx}{3})) = \sum_{n = - \infty}^{\infty} A_{n} \exp (i \frac{nπx}{3})$