Dual basis and Taylor series

In this essay, I will introduce the connection between the Dual basis and the Taylor series(For real analytic function)

Review the definition of a real analytic function,

it is a kind of function on $\mathbb{R}$$\mathbb R$, which can be written as convergence power series at an open set $D\subseteq \mathbb{R}$$D\subseteq \mathbb R$

And the Dual basis is elements of dual space for a vector space $V$$V$

Give a basis for $V$$V$ like $v_1,v_2,v_3,...v_n$$v_1,v_2,v_3,... v_n$

we can define dual basis like $\begin{array}{c}{\psi }_i(v_j)=\{\begin{array}{l}1,i=j\\ 0,i\ne j\end{array}\end{array}$$\psi_i(v_j)=\begin{cases}1,i=j\\0,i\ne j\end{cases}$

witch gives the coordinate of the vector under the basis vector.

For example, $v\in V,v=c_1{v}_1+c_2{v}_2+c_3{v}_3+...c_n{v}_n$$v\in V, v=c_1v_1+c_2v_2+c_3v_3+...c_nv_n$

${\psi }_i(v)=\psi (c_1{v}_1+c_2{v}_2+c_3{v}_3+...c_n{v}_n)=c_i{\psi }_i(v_1)+c_2\psi (v_2)+....c_i\psi (v_i)+c_n{\psi }_i(v_n)=c_i$$\psi_i(v)=\psi(c_1v_1+c_2v_2+c_3v_3+...c_nv_n)=c_i\psi_i(v_1)+c_2\psi(v_2)+....c_i\psi(v_i)+c_n\psi_i(v_n)=c_i$

And actually, it is how we get the Matrix representation for Linear Map.

$a_{ij}={\psi }_{i}T(v_j)$$a_{ij}=\psi_i T(v_j)$, and you can also consider the connection between dual basis, determinate and inverse matrix, and coordinate change...

Now we can consider a linear space $\mathcal{P}$$\mathcal P$

the basis is $1,x,x^2,x^3,...x^n,...$$1,x,x^2,x^3,...x^n,...$

real analytic function space $\mathcal{A}$$\mathcal A$ is a subspace of $\mathcal{P}$$\mathcal P$

and the dual basis is ${\displaystyle {\left(\frac{d}{dx}\right)}^0,{\left(\frac{d}{dx}\right)}^1,\frac{1}{2!}{\left(\frac{d}{dx}\right)}^2,\frac{1}{3!}{\left(\frac{d}{dx}\right)}^3...}$$\displaystyle\left(\frac{d}{dx}\right)^0,\left(\frac{d}{dx}\right)^1,\frac{1}{2!}\left(\frac{d}{dx}\right)^2,\frac{1}{3!}\left(\frac{d}{dx}\right)^3...$(derivative at 0)

So consider a function $f\in \mathcal{A}$$f\in \mathcal A$

$f(x)=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+...$$f(x)=c_0+c_1x+c_2x^2+c_3x^3+...$

Using the dual basis, we can get $c_i={\displaystyle \frac{f^n(0)}{n!}}$$c_i=\displaystyle \frac{f^{n}(0)}{n!}$

You can generalize the idea by considering $(x-a{)}^i,i\in \mathbb{N}$$(x-a)^i,i\in\mathbb N$

You can think of The Fourier series in the same way. The dual basis is given by inner product and orthogonal basis.

But using the beautiful approach, we can not get the general Taylor series.

For a function $f(x):(a,b)\to \mathbb{R}$$f(x):(a,b)\to\mathbb R$, $f(x)$$f(x)$ can be derivative $n+1$$n+1$ times, we can use integral by part

$f(x){\displaystyle =f(x_0)+f(x)-f(x_0)=f(x_0)+{\int }_{x_0}^x{f}^{\prime }(t)d(t-x)}$$f(x)\displaystyle=f(x_0)+f(x)-f(x_0)=f(x_0)+\int_{x_0}^{x} f'(t)d(t-x)$

$={\displaystyle f(x_0)-{\int }_{x_0}^x{f}^{\prime }(t)d(x-t)=f(x_0)-f^{\prime }(t)(x-t){|}_{x_0}^x-\frac{1}{2}{\int }_{x_0}^{x}f(x)d(x-t)}$$=\displaystyle f(x_0)-\int_{x_0}^{x} f'(t)d(x-t)=f(x_0)-f'(t)(x-t)|_{x_0}^x-\frac{1}{2}\int_{x_0}^xf(x)d(x-t)$

...

Finally, you can get $f(x)={\displaystyle \sum _{i=0}^n{f}^{(i)}(x_0)\frac{x^n}{n!}+R_n(x),R_n(x)={\displaystyle \frac{1}{n!}{\int }_{x_0}^x{f}^{(n+1)}(t)(x-t{)}^{n}dx}}$$f(x)=\displaystyle\sum_{i=0}^n f^{(i)}(x_0)\frac{x^n}{n!}+R_n(x),R_n(x)=\displaystyle\frac{1}{n!}\int_{x_0}^x f^{(n+1)}(t)(x-t)^ndx$