# 数学代写|偏微分方程代写partial difference equations代考|МАТH2415

## 数学代写|偏微分方程代写partial difference equations代考|Orthogonality

Recall that two vectors $\vec{a}$ and $\vec{b}$ in $\mathcal{R}^n$ are called orthogonal vectors if
$$\vec{a} \cdot \vec{b}=\sum_{i=1}^n a_i b_i=0 .$$
We would like to extend this definition to functions. Let $f(x)$ and $g(x)$ be two functions defined on the interval $[\alpha, \beta]$. If we sample the two functions at the same points $x_i, i=$ $1,2, \cdots, n$ then the vectors $\vec{F}$ and $\vec{G}$, having components $f\left(x_i\right)$ and $g\left(x_i\right)$ correspondingly, are orthogonal if
$$\sum_{i=1}^n f\left(x_i\right) g\left(x_i\right)=0 .$$
If we let $n$ to increase to infinity then we get an infinite sum which is proportional to
$$\int_\alpha^\beta f(x) g(x) d x .$$
Therefore, we define orthogonality as follows:
Definition 14. Two functions $f(x)$ and $g(x)$ are called orthogonal on the interval $(\alpha, \beta)$ with respect to the weight function $w(x)>0$ if
$$\int_\alpha^\beta w(x) f(x) g(x) d x=0 .$$
Example 1
The functions $\sin x$ and $\cos x$ are orthogonal on $[-\pi, \pi]$ with respect to $w(x)=1$,
$$\int_{-\pi}^\pi \sin x \cos x d x=\frac{1}{2} \int_{-\pi}^\pi \sin 2 x d x=-\left.\frac{1}{4} \cos 2 x\right|{-\pi} ^\pi=-\frac{1}{4}+\frac{1}{4}=0 .$$ Definition 15. A set of functions $\left{\phi_n(x)\right}$ is called orthogonal system with respect to $w(x)$ on $[\alpha, \beta]$ if $$\int\alpha^\beta \phi_n(x) \phi_m(x) w(x) d x=0, \quad \text { for } m \neq n .$$

## 数学代写|偏微分方程代写partial difference equations代考|Computation of Coefficients

Suppose that $f(x)$ can be expanded in Fourier series
$$f(x) \sim \frac{a_0}{2}+\sum_{k=1}^{\infty}\left(a_k \cos \frac{k \pi}{L} x+b_k \sin \frac{k \pi}{L} x\right) .$$
The infinite series may or may not converge. Even if the series converges, it may not give the value of $f(x)$ at some points. The question of convergence will be left for later. In this section we just give the formulae used to compute the coefficients $a_k, b_k$.
$$\begin{gathered} a_0=\frac{1}{L} \int_{-L}^L f(x) d x, \ a_k=\frac{1}{L} \int_{-L}^L f(x) \cos \frac{k \pi}{L} x d x \quad \text { for } k=1,2, \ldots \end{gathered}$$

$$b_k=\frac{1}{L} \int_{-L}^L f(x) \sin \frac{k \pi}{L} x d x \quad \text { for } k=1,2, \ldots$$
Notice that for $k=0$ (5.3.3) gives the same value as $a_0$ in (5.3.2). This is the case only if one takes $\frac{a_0}{2}$ as the first term in (5.3.1), otherwise the constant term is
$$\frac{1}{2 L} \int_{-L}^L f(x) d x .$$
The factor $L$ in (5.3.3)-(5.3.4) is exactly the square of the norm of the functions $\sin \frac{k \pi}{L} x$ and $\cos \frac{k \pi}{L} x$. In general, one should write the coefficients as follows:
\begin{aligned} &a_k=\frac{\int_{-L}^L f(x) \cos \frac{k \pi}{L} x d x}{\int_{-L}^L \cos ^2 \frac{k \pi}{L} x d x}, \ &b_k=\frac{\int_{-L}^L f(x) \sin \frac{k \pi}{L} x d x}{\int_{-L}^L \sin ^2 \frac{k \pi}{L} x d x}, \quad \text { for } k=1,2, \ldots \end{aligned} \quad \text { for } k=1,2, \ldots
These two formulae will be very helpful when we discuss generalized Fourier series.

# 偏微分方程代考

## 数学代写|偏微分方程代写partial difference equations代考|Orthogonality

$$\vec{a} \cdot \vec{b}=\sum_{i=1}^n a_i b_i=0 .$$

$$\sum_{i=1}^n f\left(x_i\right) g\left(x_i\right)=0 .$$

$$\int_\alpha^\beta f(x) g(x) d x .$$

$$\int_\alpha^\beta w(x) f(x) g(x) d x=0 .$$

$$\int_{-\pi}^\pi \sin x \cos x d x=\frac{1}{2} \int_{-\pi}^\pi \sin 2 x d x=-\frac{1}{4} \cos 2 x \mid-\pi^\pi=-\frac{1}{4}+\frac{1}{4}=0 .$$
$$\int \alpha^\beta \phi_n(x) \phi_m(x) w(x) d x=0, \quad \text { for } m \neq n \text {. }$$

## 数学代写|偏微分方程代写partial difference equations代考|Computation of Coefficients

$$f(x) \sim \frac{a_0}{2}+\sum_{k=1}^{\infty}\left(a_k \cos \frac{k \pi}{L} x+b_k \sin \frac{k \pi}{L} x\right) .$$

$$\begin{gathered} a_0=\frac{1}{L} \int_{-L}^L f(x) d x, a_k=\frac{1}{L} \int_{-L}^L f(x) \cos \frac{k \pi}{L} x d x \quad \text { for } k=1,2, \ldots \ b_k=\frac{1}{L} \int_{-L}^L f(x) \sin \frac{k \pi}{L} x d x \quad \text { for } k=1,2, \ldots \end{gathered}$$

$$\frac{1}{2 L} \int_{-L}^L f(x) d x .$$

$$a_k=\frac{\int_{-L}^L f(x) \cos \frac{k \pi}{L} x d x}{\int_{-L}^L \cos ^2 \frac{k \pi}{L} x d x}, \quad b_k=\frac{\int_{-L}^L f(x) \sin \frac{k \pi}{L} x d x}{\int_{-L}^L \sin ^2 \frac{k \pi}{L} x d x}, \quad \text { for } k=1,2, \ldots \quad \text { for } k=1,2, \ldots$$

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