前几天博主刚考完试，只能说感觉大学生活就这样过去了，还没有学到多少东西。这次博客也没想好去写什么，不过倒是修改了一些bug（文章日期竟然是根据美国时间发布的，所以我的日期相应会有所延后），以及添加新内容：tags。之后会增加关于搜索，和分类的内容方便查找.
傅里叶变换的简介
摘自维基百科：
傅里叶变换（法语：Transformation de Fourier，英语：Fourier transform，缩写：FT）是一种线性变换，通常定义为一种积分变换。其基本思想是一个函数可以用（可数或不可数，可数的情况对应于傅里叶级数）无穷多个周期函数的线性组合来逼近，(博主：就像是泰勒展开一样)从而这些组合系数在保有原函数的几乎全部信息的同时，还直接地反映了该函数的“频域特征”。
傅里叶级数（英语：Fourier series）是把类似波的函数表示成简单谐波的方式。更正式地说，对于满足狄利克雷定理的周期函数，其傅里叶级数是由一组正弦与余弦函数的加权和表示的方法。具体一些知识可以去看3b1b的视频，或许能获得众多灵感。
一些前提知识
Definition.
If f, g are real values functions that are Riemann integrable on [a,b], then we define the inner product $$ (f,g):=\int_a^b f(x) \cdot g(x) , dx $$ then for a collection of functions
$$ \varphi_0, \varphi_1, \varphi_2, \ldots $$
S is orthogonal system on [a,b] if $ (\varphi_m, \varphi_n)=0 $ for $m \neq n $
Definition. A function is periodic with period T if $$ f(x+T)=f(x) $$ and for the point which is discontinuous, we check x= $\xi$, i.e. $$ f(\xi) = \frac{1}{2} \bigl[f(\xi^+) + f(\xi^-)\bigr] $$
As we have known about the oscillaation sin($\omega x$), $\omega$ =$\frac{2\pi}{t}$, if we add another oscillation of frequency $2 \omega$, $$S_2(x)=A_1sin(\omega x) +A_2 sin(2 \omega x)$$ If we add more and more, we could get a trigonometric polynomial: $$S_n(x)=\frac{1}{2}a_0+\sum_{k=1}^n [a_kcos(k \omega x )+ b_k sin(k \omega x)]$$
$a_n$ = $\frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx)\ dx$
$b_n$ = $\frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx)\ dx$
Definition. Euler’s relation cos($\theta$)+isin($\theta$)=$e^{i\theta}$
and
$$\int_{-\pi}^{\pi} e^{inx} , dx = \begin{cases} 2\pi & \text{if } n = 0 \ 0 & \text{if } n \ne 0 \end{cases}$$
Definition.
Dirichlet Conditions:
Not all ( u(t) ) can be expressed as a Fourier series.
Peter Dirichlet derived a set of sufficient conditions.
The function ( u(t) ) must satisfy:
- It is periodic and single-valued.
- $ \int_{0}^{T} |u(t)|,dt < \infty $
- A finite number of maxima and minima within one period.
- A finite number of finite discontinuities per period.
关于傅里叶级数
But how do we found $a_n’s \text{ and } b_n’s$ ? The answer is that we could use a clever trick relies on taking averages: let <x(t)> equals the average of x(t) over any integer number of periods: $$ <x(t)>=\frac{1}{T}\int_{t=0}^T x(t)dt$$ and also $$<sin(2n\pi Ft)>=0$$ $$<cos(2n\pi Ft)>=\begin{cases} 0 & \text {if } n \neq 0 \ 1 & \text{if } x = 0 \end{cases}$$
Thus, if we multiply both sides by $\cos(2\pi n F t)$ and integrate over one period,
we can find the coefficients of the Fourier series:
$$ a_n = 2<u(t)cos(2n\pi Ft)>=\frac{2}{T} \int_0^T u(t) \cos(2\pi n F t) , dt $$
$$ b_n = 2<u(t)sin(2n\pi Ft)>=\frac{2}{T} \int_0^T u(t) \sin(2\pi n F t) , dt $$
一些相关定理
$$ c_n(x) = \frac{1}{2} + \sum_{k=1}^n \cos(kx) = \frac{\sin\left(\frac{n+1}{2} x\right)}{2 \sin\left(\frac{x}{2}\right)} $$
Riemann-Lebesgue Lamma $$ I_\lambda = \int_a^b g(x) \sin(\lambda x) , dx $$
$$ \int_0^\infty \frac{\sin z}{z} , dz = \frac{\pi}{2} $$
Parseval’s Theorem $$ \frac{1}{\pi} \int_{-\pi}^{\pi} f^2(x) , dx = \frac{a_0^2}{2} + \sum_{n=1}^\infty \left( a_n^2 + b_n^2 \right). $$ 未完待续。。。。