# 数学代写|偏微分方程代写partial difference equations代考|Math442

## 数学代写|偏微分方程代写partial difference equations代考|Relationship to Least Squares

In this section we show that the Fourier series expansion of $f(x)$ gives the best approximation of $f(x)$ in the sense of least squares. That is, if one minimizes the squares of differences between $f(x)$ and the $n^{\text {th }}$ partial sum of the series
$$\frac{a_0}{2}+\sum_{k=1}^{\infty}\left(a_k \cos \frac{k \pi}{L} x+b_k \sin \frac{k \pi}{L} x\right)$$
then the coefficients $a_0, a_k$ and $b_k$ are exactly the Fourier coefficients given by (5.3.6)-(5.3.7).
Let $I\left(a_0, a_1, \cdots, a_n, b_1, b_2, \cdots, b_n\right)$ be defined as the “sum” of the squares of the differences, i.e.
$$I=\int_{-L}^L\left[f(x)-s_n(x)\right]^2 d x$$
where $s_n(x)$ is the $n^{\text {th }}$ partial sum
$$s_n(x)=\frac{a_0}{2}+\sum_{k=1}^n\left(a_k \cos \frac{k \pi}{L} x+b_k \sin \frac{k \pi}{L} x\right) .$$
In order to minimize the integral $I$, we have to set to zero each of the first partial derivatives,
$$\begin{array}{lrl} & \frac{\partial I}{\partial a_0}=0, \ \frac{\partial I}{\partial a_j}=0, & j=1,2, \cdots, n \ \frac{\partial I}{\partial b_j}=0, & j=1,2, \cdots, n . \end{array}$$
Differentiating the integral we find
\begin{aligned} \frac{\partial I}{\partial a_0} &=\int_{-L}^L 2\left[f(x)-s_n(x)\right] \frac{\partial}{\partial a_0}\left[f(x)-s_n(x)\right] d x \ &=-2 \int_{-L}^L\left[f(x)-s_n(x)\right] \frac{1}{2} d x \ &=-\int_{-L}^L\left[f(x)-\frac{a_0}{2}-\sum_{k=1}^n\left(a_k \cos \frac{k \pi}{L} x+b_k \sin \frac{k \pi}{L} x\right)\right] d x \end{aligned}

## 数学代写|偏微分方程代写partial difference equations代考|Term by Term Differentiation

In order to check that the solution obtained by the method of separation of variables satisfies the PDE, one must be able to differentiate the infinite series.

1. A Fourier series that is continuous can be differentiated term by term if $f^{\prime}(x)$ is piecewise smooth. The result of the differentiation is the Fourier series of $f^{\prime}(x)$.
2. A Fourier cosine series that is continuous can be differentiated term by term if $f^{\prime}(x)$ is piecewise smooth. The result of the differentiation is the Fourier sine series of $f^{\prime}(x)$.
3. A Fourier sine series that is continuous can be differentiated term by term if $f^{\prime}(x)$ is piecewise smooth and $f(0)=f(L)=0$. The result of the differentiation is the Fourier cosine series of $f^{\prime}(x)$.
Note that if $f(x)$ does not vanish at $x=0$ and $x=L$ then the result of differentiation is given by the following formula:
$$f^{\prime}(x) \sim \frac{1}{L}[f(L)-f(0)]+\sum_{n=1}^{\infty}\left{\frac{n \pi}{L} b_n+\frac{2}{L}\left[(-1)^n f(L)-f(0)\right]\right} \cos \frac{n \pi}{L} x .$$
Note that if $f(L)=f(0)=0$ the above equation reduces to term by term differentiation.
Example 12
Given the Fourier sine series of $f(x)=x$,
$$x \sim 2 \sum_{n=1}^{\infty} \frac{L}{n \pi}(-1)^{n+1} \sin \frac{n \pi}{L} x .$$
Since $f(L)=L \neq 0$, we get upon differntiation using (5.7.1)
$$1 \sim \frac{1}{L}[L-0]+\sum_{n=1}^{\infty}{\frac{n \pi}{L} \underbrace{2 \frac{L}{n \pi}(-1)^{n+1}}_{b_n}+\frac{2}{L}(-1)^n L} \cos \frac{n \pi}{L} x$$
The term in braces is equal
$$2(-1)^{n+1}+2(-1)^n=0 .$$
Therefore the infinite series vanishes and one gets
$$1 \sim 1 \text {, }$$
that is, the Fourier cosine series of the constant function 1 is 1 .

# 偏微分方程代考

## 数学代写|偏微分方程代写partial difference equations代考|Relationship to Least Squares

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

$$I=\int_{-L}^L\left[f(x)-s_n(x)\right]^2 d x$$

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

$$\frac{\partial I}{\partial a_0}=0, \frac{\partial I}{\partial a_j}=0, \quad j=1,2, \cdots, n \frac{\partial I}{\partial b_j}=0, \quad j=1,2, \cdots, n .$$

$$\frac{\partial I}{\partial a_0}=\int_{-L}^L 2\left[f(x)-s_n(x)\right] \frac{\partial}{\partial a_0}\left[f(x)-s_n(x)\right] d x \quad=-2 \int_{-L}^L\left[f(x)-s_n(x)\right] \frac{1}{2} d x=-\int_{-L}^L[f($$

## 数学代写|偏微分方程代写partial difference equations代考|Term by Term Differentiation

1. 一个连续的傅里叶级数可以逐项微分如果 $f^{\prime}(x)$ 是分段光滑的。微分的结果是傅里叶级数 $f^{\prime}(x)$.
2. 连续的傅立叶余弦级数可以逐项微分如果 $f^{\prime}(x)$ 是分段光滑的。微分的结果是傅立叶正弦级数 $f^{\prime}(x)$.
3. 连续的傅立叶正弦级数可以逐项微分如果 $f^{\prime}(x)$ 是分段光滑的并且 $f(0)=f(L)=0$. 微分的结果是傅 立叶余弦级数 $f^{\prime}(x)$.
请注意，如果 $f(x)$ 不会消失在 $x=0$ 和 $x=L$ 则微分结果由下式给出:
请注意，如果 $f(L)=f(0)=0$ 上述等式简化为逐项微分。
例 12
给定的傅立叶正弦级数 $f(x)=x$ ，
$$x \sim 2 \sum_{n=1}^{\infty} \frac{L}{n \pi}(-1)^{n+1} \sin \frac{n \pi}{L} x .$$
自从 $f(L)=L \neq 0$, 我们使用 (5.7.1) 进行微分
$$1 \sim \frac{1}{L}[L-0]+\sum_{n=1}^{\infty} \frac{n \pi}{L} \underbrace{2 \frac{L}{n \pi}(-1)^{n+1}}_{b_n}+\frac{2}{L}(-1)^n L \cos \frac{n \pi}{L} x$$
大括号中的项是相等的
$$2(-1)^{n+1}+2(-1)^n=0$$
因此无限级数消失，一个人得到
$$1 \sim 1$$
也就是说，常数函数 1 的傅立叶余弦级数为 1 。

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