多元统计分析——泰勒展开式

目录

一、理解泰勒公式的由来及意义——一元函数的展开式

二、二元函数的展开式

三、多元函数的展开式

四、泰勒展开式的矩阵形式表达

五、哈密顿算子


一、理解泰勒公式的由来及意义——一元函数的展开式

问题:一个简单的三角函数y=sin(x),现在要求当x=1时的函数值。如果不借助计算机,要怎么求这个值呢?

泰勒的思路是:用多项式函数去近似拟合三角函数。

在回归分析中,我们以多项式函数拟合数据集,多项式的“项”越多,对数据集的拟合程度越好,如下图。

 于是这个问题就转换为求解一个多项式函数(“项”的个数越多拟合越好,可以无穷大),让这个多项式函数无限地和三角函数或者其他我们需要的函数等价。

推导过程如下:

我们定义f(x)=sin(x),我们塑造一个多项式函数:p(x)=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^{2}+...+a_{n}(x-x_{0})^{n}+o(x-x_{0})^{n+1},其中o(x-x_{0})^{n+1}为误差项,是p(x)f(x)的差值。

f(x)=sin(x)=p(x),则f(x)=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^{2}+...+a_{n}(x-x_{0})^{n}+o(x-x_{0})^{n+1}

 我们假设f(x)x=x_{0}点左右邻域内,各阶导数都存在(必要条件),则:

f(x_{0})=a_{0}

{f}'(x_{0})=a_{1}

{f}''(x_{0})=2!a_{2}

{f}'''(x_{0})=3!a_{3}

....

f^{(n)}(x_{0})=n!a_{n}

进而得:

a_{0}=f(x_{0})

a_{1}={f}'(x_{0})

a_{2}=\frac{​{f}''(x_{0})}{2!}

a_{3}=\frac{​{f}'''(x_{0})}{3!}

....

a_{n}=\frac{f^{(n)}(x_{0})}{n!}

将系数代入f(x)=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^{2}+...+a_{n}(x-x_{0})^{n}+o(x-x_{0})^{n+1}

\begin{align}f(x) & =f(x_{0})+{f}'(x_{0})(x-x_{0})\\&+\frac{1}{2!}{f}''(x_{0})(x-x_{0})^{2}+\frac{1}{3!}{f}'''(x_{0})(x-x_{0})^{3}...\\&+\frac{1}{n!}f^{(n)}(x_{0})(x-x_{0})^{n}+o(x-x_{0})^{n+1} \end{align}

化简得:

f(x)=\sum ^{\infty }_{n=0}\frac{1}{n!}f^{(n)}(x_{0})(x-x_{0})^{n}+o(x-x_{0})^{n+1},误差项o(x-x_{0})^{n+1}可去掉,得f(x)=\sum ^{\infty }_{n=0}\frac{1}{n!}f^{(n)}(x_{0})(x-x_{0})^{n}这就是泰勒公式,其中f^{(n)}(x_{0})代表f(x)n阶导数

案例1f(x)=sin(x) 

①先求其的n阶导数f^{(n)}(x_{0})

f(x_{0})=sinx_{0}=sin(x_{0}+\frac{\pi }{2}\cdot 0)

{f}'(x_{0})=cosx_{0}=sin(x_{0}+\frac{\pi }{2}\cdot 1)

{f}''(x_{0})=-sinx_{0}=sin(x_{0}+\frac{\pi }{2}\cdot 2)

{f}'''(x_{0})=-cosx_{0}=sin(x_{0}+\frac{\pi }{2}\cdot 3)

{f}^{4}(x_{0})=sinx_{0}=sin(x_{0}+\frac{\pi }{2}\cdot 4)

....

f^{(n)}(x_{0})=sin(x_{0}+\frac{\pi }{2}\cdot n)

已知:sin(0)=0,sin(\frac{\pi }{2})=1,sin(\pi)=0,sin(\frac{3\pi }{2})=-1,sin(2\pi)=0,.....

②我们将n阶导数f^{(n)}(x_{0})代入方程f(x)=\sum ^{\infty }_{n=0}\frac{1}{n!}f^{(n)}(x_{0})(x-x_{0})^{n},得

\begin{align}f(x)=p(x)&=\sum ^{\infty }_{n=0}\frac{1}{n!}sin(x_{0}+\frac{\pi }{2}\cdot n)(x-x_{0})^{n}\\&=\frac{1}{0!}sin(x_{0}+\frac{\pi }{2}\cdot 0)(x-x_{0})^{0}+\frac{1}{1!}sin(x_{0}+\frac{\pi }{2}\cdot 1)(x-x_{0})^{1}+\frac{1}{2!}sin(x_{0}+\frac{\pi }{2}\cdot 2)(x-x_{0})^{2} \\&+\frac{1}{3!}sin(x_{0}+\frac{\pi }{2}\cdot 3)(x-x_{0})^{3}+\frac{1}{4!}sin(x_{0}+\frac{\pi }{2}\cdot 4)(x-x_{0})^{4} +...+\frac{1}{n!}sin(x_{0}+\frac{\pi }{2}\cdot n)(x-x_{0})^{n} \end{align}

x_{0}=0,得:

\begin{align}f(x)=sinx=p(x) &=0+x+0-\frac{1}{3!}x^{3}+0 +\frac{1}{5!}x^{5}+0-\frac{1}{7!}x^{7}...+\frac{1}{n!}sin(x_{0}+\frac{\pi }{2}\cdot n)(x-x_{0})^{n} \\&=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+...+(-1)^{n}\frac{x^{2n+1}}{(2n+1)!} \end{align}

案例2f(x)=e^{x} 

①我们知道{(e^{x})}'=e^{x}

\begin{align}f(x)=p(x)&=\sum ^{\infty }_{n=0}\frac{1}{n!}e^{x_{0}}(x-x_{0})^{n}\\&=\frac{1}{0!}e^{x_{0}}(x-x_{0})^{0}+\frac{1}{1!}e^{x_{0}}(x-x_{0})^{1}+\frac{1}{2!}e^{x_{0}}(x-x_{0})^{2} \\&+\frac{1}{3!}e^{x_{0}}(x-x_{0})^{3}+\frac{1}{4!}e^{x_{0}}(x-x_{0})^{4} +...+\frac{1}{n!}e^{x_{0}}(x-x_{0})^{n} \end{align}

x_{0}=0,得:

f(x)=e^{x}=p(x) =1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+...+\frac{x^{n}}{n!}

二、二元函数的展开式

由上面我们可知,一元函数f(x)在点x_{0}处的泰勒展开式为:

\begin{align}f(x) & =f(x_{0})+{f}'(x_{0})(x-x_{0})\\&+\frac{1}{2!}{f}''(x_{0})(x-x_{0})^{2}+\frac{1}{3!}{f}'''(x_{0})(x-x_{0})^{3}...\\&+\frac{1}{n!}f^{(n)}(x_{0})(x-x_{0})^{n}+o(x-x_{0})^{n+1} \end{align}

推广到二元函数,二元函数f(x,y)在点(x_{0},y_{0})处的泰勒展开式为:

\begin{align}f(x,y) & =f(x_{0},y_{0})+f_{x}'(x_{0},y_{0})(x-x_{0})+f_{y}'(x_{0},y_{0})(y-y_{0})\\&+\frac{1}{2!}\left \{ f_{xx}''(x_{0},y_{0})(x-x_{0})^{2} +2 f_{xy}''(x_{0},y_{0})(x-x_{0})(y-y_{0}) + f_{yy}''(x_{0},y_{0})(y-y_{0})^{2} \right \}\\&+\frac{1}{3!}\left \{ f_{xxx}'''(x_{0},y_{0})(x-x_{0})^{3} +3 f_{xxy}'''(x_{0},y_{0})(x-x_{0})^{2}(y-y_{0}) +3 f_{xyy}'''(x_{0},y_{0})(x-x_{0})(y-y_{0})^{2} + f_{yyy}''(x_{0},y_{0})(y-y_{0})^{3} \right \}\\&...\\&+o^{n} \end{align}

其中,o^{n}为误差项。

不同的教材对泰勒展开式有不同的写法,但是其原理其实是一样的。

\begin{align}f(x,y) & =f(x_{0},y_{0})+\frac{\partial f }{\partial x}(x-x_{0})+\frac{\partial f }{\partial y}(y-y_{0})\\&+\frac{1}{2!}\left \{ \frac{\partial^{2} f }{\partial x^{2}}(x-x_{0})^{2} +2 \frac{\partial^{2} f }{\partial x\partial y}(x-x_{0})(y-y_{0}) + \frac{\partial^{2} f }{\partial y^{2}}(y-y_{0})^{2} \right \}\\&+\frac{1}{3!}\left \{ \frac{\partial^{3} f }{\partial x^{3}}(x-x_{0})^{3} +3 \frac{\partial^{3} f }{\partial x^{2}\partial y}(x-x_{0})^{2}(y-y_{0}) +3 \frac{\partial^{3} f }{\partial x \partial y^{2}}(x-x_{0})(y-y_{0})^{2} +\frac{\partial^{3} f }{\partial y^{3}}(y-y_{0})^{3} \right \}\\&...\\&+o^{n} \end{align}

其中,\frac{\partial ^{2}f }{\partial x^{2}}=\frac{\partial f }{\partial x}\cdot \frac{\partial f }{\partial x}\frac{\partial ^{2}f }{\partial x \partial y}=\frac{\partial f }{\partial x}\cdot \frac{\partial f }{\partial y}......

三、多元函数的展开式

多元函数f(x^{1},x^{2},...,x^{n})在点(x^{1}_{0},x^{2}_{0},...,x^{n}_{0})处的泰勒展开式为:

\begin{align}f(x^{1},x^{2},...,x^{n}) & =f(x^{1}_{0},x^{2}_{0},...,x^{n}_{0})+\sum ^{n}_{i=1} f_{x^{i}}'(x^{1}_{0},x^{2}_{0},...,x^{n}_{0})(x^{i}-x^{i}_{0}) \\&+\frac{1}{2!}\sum ^{n}_{i,j=1} f_{x^{i}x^{j}}''(x^{1}_{0},x^{2}_{0},...,x^{n}_{0})(x^{i}-x_{0}^{i})(x^{j}-x_{0}^{j}) \\&+\frac{1}{3!}\sum ^{n}_{i,j,t=1} f_{x^{i}x^{j}x^{t}}'''(x^{1}_{0},x^{2}_{0},...,x^{n}_{0})(x^{i}-x_{0}^{i})(x^{j}-x_{0}^{j})(x^{t}-x_{0}^{t})\\&...\\&+o^{n} \end{align}

注意:多元函数f(x^{1},x^{2},...,x^{n})中的x^{n}表示的是第n个变量,并不是变量的n次方。

同样,展开式也可以这样表示:

\begin{align}f(x^{1},x^{2},...,x^{n}) & =f(x^{1}_{0},x^{2}_{0},...,x^{n}_{0})+\sum ^{n}_{i=1} \frac{\partial f }{\partial x^{i}}(x^{i}-x^{i}_{0}) \\&+\frac{1}{2!}\sum ^{n}_{i,j=1} \frac{\partial^{2} f }{\partial x^{i}\partial x^{j}}(x^{i}-x_{0}^{i})(x^{j}-x_{0}^{j}) \\&+\frac{1}{3!}\sum ^{n}_{i,j,t=1} \frac{\partial^{3} f }{\partial x^{i}\partial x^{j} \partial x^{t}} (x^{i}-x_{0}^{i})(x^{j}-x_{0}^{j})(x^{t}-x_{0}^{t})\\&...\\&+o^{n} \end{align}

其中,\frac{\partial^{2} f }{\partial x^{i}\partial x^{j}}即等同于上式中的f_{x^{i}x^{j}}''(x^{1}_{0},x^{2}_{0},...,x^{n}_{0}),以此类推。

四、泰勒展开式的矩阵形式表达

对以上的泰勒展开式我们取前两行,用矩阵的形式表示出来:

f(X)=f(X_{0})+[\nabla f(X_{0})]^{T}(X-X_{0})+\frac{1}{2!}[X-X_{0}]^{T}H(X_{0})[X-X_{0}]+o^{n}

其中,X=(x^{1},x^{2},...,x^{n})X_{0}=(x^{1}_{0},x^{2}_{0},...,x^{n}_{0})

H(X_{0})=\begin{bmatrix} \frac{\partial^{2}f}{\partial x_{1}^{2}} & \frac{\partial^{2}f}{\partial x_{1}\partial x_{2}} & ... & \frac{\partial^{2}f}{\partial x_{1}\partial x_{n}} \\ \frac{\partial^{2}f}{\partial x_{2}\partial x_{1}} & \frac{\partial^{2}f}{\partial x_{2}^{2}} & ... & \frac{\partial^{2}f}{\partial x_{2}\partial x_{n}}\\ & & & \\ . & . & . & . \\ \frac{\partial^{2}f}{\partial x_{n}\partial x_{1}} & \frac{\partial^{2}f}{\partial x_{n}\partial x_{2}} & ... & \frac{\partial^{2}f}{\partial x_{n}^{2}} \end{bmatrix}

注意:\nabla通常读作nabla,称为哈密顿算子。

五、哈密顿算子

在数学的世界中,有一个被称为向量分析的领域,其中有一个经常用到的符号\nabla\nabla称为哈密顿算子,其定义如下所示。

\nabla f=(\frac{\partial f}{\partial x_{1}},\frac{\partial f}{\partial x_{2}},...,\frac{\partial f}{\partial x_{n}})

则上式中的\nabla f(X_{0})即为X=X_{0}时的各变量的偏导组成的向量。

 

 

 

 

 

 

 

 

 

 

 

 


本文来自互联网用户投稿,文章观点仅代表作者本人,不代表本站立场,不承担相关法律责任。如若转载,请注明出处。 如若内容造成侵权/违法违规/事实不符,请点击【内容举报】进行投诉反馈!

相关文章

立即
投稿

微信公众账号

微信扫一扫加关注

返回
顶部