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# 数学代写|傅里叶分析代写Fourier Analysis代考|Fourier Series on [0, L]

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## 数学代写|傅里叶分析代写Fourier Analysis代考|Fourier Series on [0, L]

IN MANY APPLICATIONS WE ARE INTERESTED in determining Fourier series representations of functions defined on intervals other than $[0,2 \pi]$. In this section, we will determine the form of the series expansion and the Fourier coefficients in these cases.

The most general type of interval is given as $[a, b]$. However, this often is too general.
More common intervals are of the form $[-\pi, \pi],[0, L]$, or $[-L / 2, L / 2]$. The simplest generalization is to the interval $[0, L]$. Such intervals arise often in applications. For example, for the problem of a one-dimensional string of length $L$, we set up the axes with the left end at $x=0$ and the right end at $x=L$. Similarly for the temperature distribution along a one-dimensional rod of length $L$ we set the interval to $x \in[0,2 \pi]$. Such problems naturally lead to the study of Fourier series on intervals of length $L$. We will see later that symmetric intervals, $[-a, a]$, are also useful.

Given an interval $[0, L]$, we could apply a transformation to an interval of length $2 \pi$ by simply rescaling the interval. Then we could apply this transformation to the Fourier series representation to obtain an equivalent one useful for functions defined on $[0, L]$.

We define $x \in[0,2 \pi]$ and $t \in[0, L]$. A linear transformation relating these intervals is simply $\mathrm{x}=2 \pi \mathrm{tL}$ as shown in Figure 2.7. So, $t=0$ maps to $x=0$ and $t=L$ maps to $x=2 \pi$. Furthermore, this transformation maps $f(x)$ to a new function $g(t)=f(x(t))$, which is defined on $[0, L]$. We will determine the Fourier series representation of this function using the representation for $f(x)$ from the previous section.

## 数学代写|傅里叶分析代写Fourier Analysis代考|Parseval’s Identity

ANOTHER APPROACH TO THE SUMMATION OF FOURIER SERIES is through the use of Parseval’s identity. It is named after Marc-Antoine Parseval (1755 – 1836) and a more general form and other derivations are presented in other sections of the text. As will be noted, this can be viewed as a generalization of the Pythagorean theorem in more general spaces as discussed in the next chapter. We will consider here the version of the Parseval identity as given for real-valued functions, $f(x), x \in[0, L]$.

Theorem 2.2. Let $f(x)$ have a uniformly convergent Fourier series representation for $x \in[0$, L]. Then,
$$2 \mathrm{~L} \int 0 \mathrm{~L}[\mathrm{f}(\mathrm{x}) 2 \mathrm{dx}]=\mathrm{a} 022+\sum \mathrm{n}=1 \infty(\mathrm{an} 2+\mathrm{bn} 2) .(2.56)$$
Proof. Let’s assume that $f(x)$ has the uniformly convergent Fourier series representation
$$f(x)=a 02+\sum n=1 \infty[\operatorname{ancos} 2 n \pi x L+b n \sin 2 n \pi x L] \cdot(2.57)$$
Treating this sum as a binomial, the square is given by ${ }^4$
$$[\mathrm{f}(\mathrm{x})] 2=\left[\mathrm{a} 02+\sum \mathrm{n}=1 \infty[\operatorname{ancos} 2 \mathrm{n} \pi \mathrm{xL}+\mathrm{bn} \sin 2 \mathrm{n} \pi \mathrm{xL}]\right] 2=\mathrm{a} 024+2 \mathrm{a} 02 \Sigma \mathrm{n}=1 \infty[a n \cos 2 \mathrm{n} \pi \mathrm{xL}+\mathrm{bn} \sin 2 \mathrm{n} \pi \mathrm{xL}]+\left(\sum \mathrm{n}=1 \infty[\operatorname{ancos} 2 \mathrm{n} \pi \mathrm{xL}+\mathrm{bn} \sin 2 \mathrm{n} \pi \mathrm{xL}]\right) 2 .(2.58)$$

Next, we integrate $[f(x)]^2$ over the interval $[0, L]$. The first three expressions are easily integrated, leaving

Care should be taken when squaring the infinite series, since the result is a double sum. Therefore, we write
$\left(\sum \mathrm{n}=1 \infty[a n \cos 2 \mathrm{n} \pi \mathrm{xL}+\mathrm{bn} \sin 2 \mathrm{n} \pi \mathrm{xL}]\right) 2=\sum \mathrm{n}=1 \infty \sum \mathrm{m}=1 \infty a n a m \cos 2 \mathrm{n} \pi \mathrm{xL} \cos 2 \mathrm{~m} \pi \mathrm{xL}+\sum \mathrm{n}=1 \infty \sum \mathrm{m}=1 \infty a n b m \cos 2 \mathrm{n} \pi \mathrm{xL} \sin 2 \mathrm{~m} \pi \mathrm{xL}+\sum \mathrm{n}=1 \infty \sum \mathrm{m}=1 \infty \infty b n a m s \sin 2 \mathrm{n} \pi \mathrm{xL} \cos 2 \mathrm{~m}$
Integrating the products of these trigonometric functions, we make use of their orthogonality properties. This then gives,
$$2 \mathrm{~L} \int 0 \mathrm{~L}[\mathrm{f}(\mathrm{x})] 2 \mathrm{dx}=\mathrm{a} 022+\sum \mathrm{n}=1 \infty \sum \mathrm{m}=1 \infty\left[\text { anam+bnbm] } \mathrm{n} m=\mathrm{a} 022+\sum \mathrm{n}=1 \infty(\operatorname{an} 2+\mathrm{bn} 2) \cdot(2.60)\right.$$

## 数学代写|傅里叶分析代写Fourier Analysis代考|Parseval’s Identity

1836）的名字命名，更一般的形式和其他派生形式在文本的其他部分中介绍。正如将要指出的那样，

$$2 \mathrm{~L} \int 0 \mathrm{~L}[\mathrm{f}(\mathrm{x}) 2 \mathrm{dx}]=\mathrm{a} 022+\sum \mathrm{n}=1 \infty(\mathrm{an} 2+\mathrm{bn} 2) .(2.56)$$

$$f(x)=a 02+\sum n=1 \infty[\operatorname{ancos} 2 n \pi x L+b n \sin 2 n \pi x L] \cdot(2.57)$$

$$[\mathrm{f}(\mathrm{x})] 2=\left[\mathrm{a} 02+\sum \mathrm{n}=1 \infty[\operatorname{ancos} 2 \mathrm{n} \pi \mathrm{xL}+\mathrm{bn} \sin 2 \mathrm{n} \pi \mathrm{xL}]\right] 2=\mathrm{a} 024+2 \mathrm{a} 02 \Sigma \mathrm{n}=1 \infty[\mathrm{an} \cos 2 \mathrm{n} \pi \mathrm{xL}+\mathrm{bn} \sin 2 \mathrm{n} \pi 2$$

$\left(\sum \mathrm{n}=1 \infty[a n \cos 2 \mathrm{n} \pi \mathrm{xL}+\mathrm{bn} \sin 2 \mathrm{n} \pi \mathrm{xL}]\right) 2=\sum \mathrm{n}=1 \infty \sum \mathrm{m}=1 \infty \operatorname{anam} \cos 2 \mathrm{n} \pi \mathrm{xL} \cos 2 \mathrm{~m} \pi \mathrm{xL}+\sum \mathrm{n}=1 \infty \sum$ 整合这些三角函数的乘积，我们利用了它们的正交性。这然后给出，
$$2 \mathrm{~L} \int 0 \mathrm{~L}[\mathrm{f}(\mathrm{x})] 2 \mathrm{dx}=\mathrm{a} 022+\sum \mathrm{n}=1 \infty \sum \mathrm{m}=1 \infty[\text { anam }+\mathrm{bnbm}] \mathrm{n} m=\mathrm{a} 022+\sum \mathrm{n}=1 \infty(\mathrm{an} 2+\mathrm{bn} 2) \cdot(2.1$$

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